mathematical analysis ii integral calculus

MATHEMATICAL ANALYSIS II

INTEGRAL CALCULUS

SEVER ANGEL POPESCU

To my family...........................................................................................

To those who really try to put their lives in the Light ofTruth.

............................................... ...............................................Dedicated to the memory of my late Professor and

Mentor, Dr. Doc. Nicolae Popescu.

Contents

Preface 7

Chapter 1. Indefinite integrals (Primitives, Antiderivatives) 11. Definitions, some properties and basic formulas 12. Some results on polynomials 103. Primitives of rational functions 214. Primitives of irrational and trigonometric functions 275. Problems and exercises 47

Chapter 2. Definite integrals 491. The mass of a linear bar 492. Darboux sums and their applications 523. Lebesgue criterion and its applications 594. Mean theorem. Newton-Leibniz formula 635. The measure of a figure in Rn 686. Areas of plane figures bounded by graphics of functions 757. The volume of a rotational solid 858. The length of a curve in R3 879. Approximate computation of definite integrals. 9010. Problems and exercises 101

Chapter 3. Improper (generalized) integrals 1051. More on limits of functions of one variable 1052. Improper integrals of the first type 1123. Improper integrals of the second type 1224. Problems and exercises 135

Chapter 4. Integrals with parameters 1371. Proper integrals with parameters 1372. Improper integrals with parameters 1473. Euler’s functions gamma and beta 1674. Problems and exercises 179

Chapter 5. Line integrals 1811. The mass of a wire. Line integrals of the first type. 1812. Line integrals of the second type. 194

3

4 CONTENTS

3. Independence on path. Conservative fields 2014. Computing plane areas with line integrals 2115. Supplementary remarks on line integrals 2166. Problems and exercises 221

Chapter 6. Double integrals 2251. Double integrals on rectangles. 2252. Double integrals on an arbitrary bounded domain 2333. Green formula. Applications. 2444. Change of variables in double integrals. Polar coordinates. 2515. Problems and exercises 263

Chapter 7. Triple integrals 2651. What is a triple integral on a parallelepiped? 2652. Triple integrals on a general domain. 2713. Iterative formulas for a general space domain 2784. Change of variables in a triple integral 2865. Problems and exercises 303

Chapter 8. Surface integrals 3051. Deformation of a 2D-domain into a space surface 3052. Surface integral of the first type. The mass of a 3D-lamina. 3133. Flux of a vector field through an oriented surface. 3194. Problems and exercises 335

Chapter 9. Gauss and Stokes formulas. Applications 3391. Gauss formula (divergence theorem) 3392. A mathematical model of the heat flow in a solid 3483. Stokes Theorem 3494. Problems and exercises 357

Chapter 10. Some remarks on complex functions integration 3591. Complex functions integration 3592. Applications of residues formula 3773. Problems and exercises 385

Appendix A. Exam samples 3871. June, 2010 3872. June, 2010 3873. June, 2010 3874. June, 2010 3885. September, 2010 3886. September, 2010 388

Appendix B. Basic antiderivatives 389

CONTENTS 5

Appendix. Index 391

Appendix. Bibliography 397

Preface

Here is the second volume "Mathematical Analysis II. Integral Cal-culus" of my course of Advanced Calculus for Engineers and beginningMathematicians. The first volume "Mathematical Analysis I. Differ-ential Calculus" appeared in 2009 (see [Po]). I invite the reader tocarefully meditate on the main ideas I discussed in the Preface of thisfirst volume. The matter is about the way I think Mathematical Analy-sis (or Advanced Calculus) have to be taught to an engineering student.First of all I insist on a strong and correct intuitive basis of any math-ematical notion and statement. Then I proceed step by step to a rig-orously mathematical model of this notion or statement. My aim inthis course (primarily intended for engineering students) is to create anintuitive thinking to the reader who must feel the simplicity and thenaturalness of the introduced notion or concept. Abstract Mathemat-ics need not destroy the intuition of the future engineer. Mathematicalcourses must make this intuition stronger, more mature and very solid.This is why this course appeared to be "a spoken course". I wanted itto be so! In general, I like a lot "live" Mathematics. A "dead" abstractnotion, without at least one small motivation, generally will remaindead and nonuseful for an engineering student.

I also blame the opposite side, that mathematical teaching whichconsists only on a sequence of examples, motivations, etc., without ageneral definition, without a mathematical correct reasoning, based on"stories", etc. A creative middle way teaching is to be preferred. All ofthese belong to something which is beyond Mathematics itself, namelyto the "Art of Teaching".

For instance, many times in this volume, the idea of the convergenceof a system A of numbers to a number I, was substituted with the ideaof a "well approximation" of I with numbers of A. Surely, I could usefrom the very beginning the definition "with ε”and everything couldappeared formally rigorous for an "extremist" mathematician, but thiscourse is not intended for such a people! If I had done so, I wouldhave killed the intuitive power of the idea of approximation, extremelyuseful for a future engineer and not only for him!

7

8 PREFACE

In fact, in all my courses of Mathematical Analysis, I carried on thebasic idea of Prof. Dr. Gavril Paltineanu to make appropriate mathe-matical courses for engineers and not for mathematicians. I think thateven a student in abstract Mathematics could find a good opportunityto read such an "applicable" course.

This book contains ten chapters and an appendix with exam sam-ples and a table with some basic formulas for primitives (antideriva-tives). Since I successively introduced in the first volume ([Po]) basicelements of complex functions theory, I continued here with some ba-sic remarks on integrals of functions with a complex variable, residuestheory and some of their applications.

I started with a serious review of the calculus of primitives, usuallyassumed to be known from high school by the Romanian student.

I go on with definite or Riemann integrals, improper integrals, in-tegrals with parameters, line integrals, double and triple integrals, sur-face integrals, Gauss and Stokes formulas and I finish with complexfunctions integrals.

Any chapter contains many motivation-examples, worked exercisesand proposed exercises. I insisted on using instead of an abstract math-ematical notion more intuitive objects taken from Physics, Mechanics,Strains of Materials, etc. Don’t say that the word "deformation" is notbetter for an engineering student then the corresponding mathemati-cal notion of "vector transformation", "parametric path", "parametricsheet", etc.

Many people contributed directly or indirectly to this volume.The late Professor Dr. Doc. Nicolae Popescu, my Professor and

my Master, was teaching me to do "live" Mathematics.My colleague and my friend Prof. Dr. Gavriil Paltineanu challenged

me to investigate the interesting way to teach Analysis for engineers.He gave me many deep advises on this fascinating domain which isMathematical Analysis. He also carried this difficult task to read thishuge volume and to push me in making many corrections.

Prof. Dr. Octav Olteanu and Prof. Dr. Nicolae Danet read care-fully the entire manuscript and made some very useful suggestions.

I am also grateful to my younger colleagues Dr. Emil Popescu, Dr.Viorel Petrehus, Dr. Marilena Jianu and Dr. Valentin Popescu forhelping me a lot with the proofreading.

At last, but not the least, I will always express my thanks to twopersons: to my wife Angela for encouraging me to "go on" with thisdifficult task and to my 11

2years old grand-daughter Monica Izabela

who succeeded to refresh a lot my tired mind before chapter 4. This

PREFACE 9

is why the reader must be grateful to her for the easy (but not short)way I succeeded to present the improper integrals with parameters.

Finally, I mention the special help that trees and birds of my closepark gave me during my silent walk around the lake in those hotevenings of August 2010. All of these are of a great importance whenone is thinking of mathematical affairs.

Prof. Dr. Sever Angel PopescuTechnical University of Civil Engineering BucharestDepartment of Mathematics and Computer ScienceB=dul Lacul Tei 124, Sector 2, Bucharest, [email protected], 2011

CHAPTER 1

Indefinite integrals (Primitives, Antiderivatives)

1. Definitions, some properties and basic formulas

We shall see later that in many problems coming from Geometry,Statistics, Mechanics, Physics, Chemistry, Engineering, Biology, Econ-omy, etc. we have to compute the area of a plane figure bounded by thelines x = a, x = b, y = 0 and the graphic of a nonnegative continuousfunction y = f(x), where x ∈ [a, b], a < b (see Fig.1).

F 1.

Such a figure is usually called a curvilinear trapezoid. The greatEnglish scientist Sir Isaac Newton (1642-1727) discovered the mathe-matical relation between the function y = f(x) and this area.

T 1. (I. Newton) Let f : [a, b] → R+ be a continuousfunction defined on a real closed interval [a, b] with nonnegative realvalues. Let x be a number in (a, b] and let F (x) be the area of thecurvilinear trapezoid [AxMD] (the area under the graphic of f "up tothe point x" (see Fig.2). Then this area function is differentiable andits derivative F ′(x) at the point x is exactly f(x), the value of the initialfunction f at the same point x.

1

2 1. INDEFINITE INTEGRALS (PRIMITIVES, ANTIDERIVATIVES)

F 2.

P. (intuitively proof; you will see later a mathematically rig-orous proof) Let us fix a point x0 in (a, b]. We are going to prove thatthe derivative F ′(x0) exists and it is equal to f(x0). By definition

(1.1) F ′(x0) = limx→x0

F (x) − F (x0)

x− x0,

if this limit exists. Let us take for instance x ∈ (a, b], x > x0 (ifx0 = b, then take x < x0 and do a completely similar reasoning!). ButF (x)−F (x0) is exactly the area of the curvilinear trapezoid [M0x0xM ](see also Fig.2). Since f is continuous, this area can be well approxi-mated by the area of the rectangle [M0x0xM

′]. Indeed, in the case ofour figure (Fig.2), the least value of f on the interval [x0, x] is real-ized at the point x0 and the greatest is realized at the point x (usethe boundedness Weierstrass theorem ([Po], Theorem 32), for a moregeneral situation). So, the area of the curvilinear trapezoid [M0x0xM ]is a number between the area of the rectangle [M0x0xM

′] and the areaof the rectangle [M ′

0x0xM ]. When M becomes closer and closer toM0,these last two areas tends to give the same number, so the area of thecurvilinear trapezoid can be well approximated by the area of the rec-tangle [M0x0xM

′]. Thus, using now the corresponding mathematicalexpressions, we get:

(x− x0)f(x0) ≤ F (x) − F (x0) ≤ (x− x0)f(x),

1. DEFINITIONS, SOME PROPERTIES AND BASIC FORMULAS 3

or

(1.2) f(x0) ≤F (x) − F (x0)

(x− x0)≤ f(x).

Since f is continuous, taking limits when x → x0 in (1.2), we get thatF ′(x0) exists and that it is equal to f(x0).

The area function constructed above is called the Newton area func-tion of f.

C 1. Let f : [a, c) → R be a continuous function whichdoes not change its sign on the interval [a, c) (here c may be even ∞!).Then, there is a differentiable function F defined on this interval (a, c)such that F ′(x) = f(x) for any x ∈ (a, c).

P. It is sufficient to take f to be nonnegative (otherwise re-place f by −f). Then we define like above F (x) = the area of thecurvilinear trapezoid [AxMD] (see Fig.2).

E 1. Try to substitute the above interval [a, c) for an openfinite or infinite interval (a, c). Moreover, try to substitute this lastinterval for any open subset of R.

All of the discussion above is a real motivation for the followinggeneral definition.

D 1. Let A be an open subset of the real line R. Let f :A→ R be a function defined on A with values in R. Any differentiablefunction F : A → R such that its derivative F ′(x) is equal to f(x) forany x in A (simply if F ′ = f) is called a primitive of f on A. It is alsocalled an antiderivative or an integral of f on A.

In the following all functions are defined only on subsets A of R,which have no isolated points. We shall simply note A for a subsetsatisfying this last property. We introduced this restriction becauseit is a nonsense to speak about the limit of a function at an isolatedpoint. In particular, about its derivative!

If we remember the definition of the notion of a differential dF (a)of a function F at a point a of A, namely dF (a) which is the linearmapping defined on R with values in the same R, such that dF (a)(h) =F ′(a)h for any h in R, or dF (a) = F ′(a)dx, where dx is the differentialof the variable x (in our case the identity mapping-see Analysis I, [Po]),we can easily prove the following very useful result.

T 2. Let f : A→ R be a function defined on A (as above).A function F : A → R is a primitive of f on A if and only if it is


differentiable on A and dF = fdx, i.e. its differential dF (a) at anypoint a of A is equal to f(a)dx.

A primitive F of f is denoted by the notation∫f(x)dx and it is

also called an integral of f with respect to the variable x. Moreover, anexpression of the form fdx is called a differential form of order one inone variable x. A primitive of such a differential form is a function U :A→ R such that U is differentiable on A and dU = fdx, equivalentlyU ′(x) = f(x) for any x in A. Now, the notation

∫f(x)dx means such

a primitive U for the differential form fdx. It is useful for us in thefollowing to speak the primitives of differential forms language, insteadof using the language of primitives for usual functions.

To decide if a differential form has a primitive or not is in general avery difficult problem. We shall prove later a more general result thanthat one of exercise 1. Namely, one can rigorously prove (see Chapter2) that any continuous function f on an interval I has a primitive Fon I. This means that for a continuous function f on an interval I, thedifferential form fdx has a primitive F, i.e. dF = fdx.

For instance, a primitive of f(x) = cosx is F (x) = sinx becaused(sin x) = cosxdx. We see that a function of the form sin x+C, whereC is an arbitrary constant (does not depend on x. It may depend onany other variable distinct of x), is also a primitive of cosx. Indeed,

d(sin x+ C) = d(sin x) + dC = cosxdx+ 0 = cosxdx.

So, if a function f has a primitive F on an open subset A of R, then ithas an infinite number of primitives, namely, any function of the form:F + C, where C is a constant w.r.t. x (prove this as in the case off(x) = cosx).

T 3. (a structure theorem for all primitives) Let f : I → Rbe a function defined on an interval I and let F be a fixed primitiveof it (on I). Then any other primitive G of f on the interval I is ofthe form G = F + C, where C is a constant number (which dependsonly on G). This is why the set of all primitives of f is denoted by∫f(x)dx+ C. Here

∫f(x)dx means any fixed primitive of f on I.

P. Since F ′(x) = f(x) = G′(x) for any x in I, one can seethat the difference function H = G−F has its derivative equal to zeroat any point of I. We are going to prove that in this case H must be aconstant, i.e. it cannot depend on the variable x. This means that forany points a and b of I we must have that H(a) = H(b). Indeed, let ussuppose that a < b and let us apply Lagrange theorem on the closedinterval [a, b] (which is included in I, why?):

H(b) −H(a) = H ′(c)(b− a),


for a point c in (a, b). Since H ′(x) = 0 for any x of I, one gets thatH(b) = H(a). Let us denote by C this common value of H(x). Hence,G− F = C, or G = F + C. This means that for this fixed number C,we have that G(x) = F (x) + C for any x ∈ I.

R 1. If the set A is not an interval, then the above result isnot always true. For instance, let A = (0, 1)∪ (2, 3) and f(x) = 2x forany x of A. It is easy to see that the following two functions

F (x) =

x2, for x ∈ (0, 1)

x2 + 1, for x ∈ (2, 3),

G(x) =

x2, for x ∈ (0, 1)x2, for x ∈ (2, 3)

are primitives of f. It is clear enough that there is no constant numberC such that G(x) = F (x) + C for any x in A. Indeed, for x ∈ (0, 1),C = 0 and for x ∈ (2, 3), C = −1, so we cannot have the same constantC on the entire set A.

E 1. Let us find a primitive for the following continuousfunction

f(x) =

x, if x ∈ [−1, 1]x3, if x ∈ (1, 2)

x2 + 4, if x ∈ [2, 3),

f : [−1, 3) → R. Since(x2

2

)′= x,

(x4

4

)′= x3 and

(x3

3+ 4x

)′= x2+4,

any primitive of f is of the following form:

F (x) =

x2

2+ C1, if x ∈ [−1, 1]

x4

4+ C2, if x ∈ (1, 2)

x3

3+ 4x+ C3, if x ∈ [2, 3)

,

Let us force now F to be continuous at x = 1 (F is differentiable, thusit must be continuous):

1

2+ C1 =

1

4+ C2,

so we get that C2 = 14

+ C1. The continuity of F at x = 2 implies that

24

4+

1

4+ C1 =

23

3+ 8 + C3,

thus C3 = −7712

+ C1. Hence, all the primitives of f are:

F (x) =

x2

2, if x ∈ [−1, 1]

x4

4+ 1

4, if x ∈ (1, 2)

x3

3+ 4x− 77

12, if x ∈ [2, 3)

+ C1 ,


where C1 is an arbitrary constant (with respect to x) number. It is easyto see that F (x) is differentiable and F ′(x) = f(x) for any x ∈ [−1, 3).

E 2. Here is an example of a discontinuous function whichhas no primitive at all. Let

f(x) =

1, for x ∈ [0, 1]0, for x ∈ (1, 2]

.

It is easy to see that a primitive of f must be of the form:

F (x) =

x+ C1, for x ∈ [0, 1]C2, for x ∈ (1, 2]

.

The continuity of F at x = 1 implies that C2 = 1 + C1, thus,

F (x) =

x+ C1, for x ∈ [0, 1]1 + C1, for x ∈ (1, 2]

.

But this last function is not differentiable at x = 1 for no value of C1

(why?). Hence, F cannot be a primitive of f on [0, 2]!

But, do not you hurry to conclude that all the discontinuous func-tions have no primitive. Here is a counterexample!

E 3. The function

f(x) =

2x sin 1

x− cos 1

x, for x ∈ (0, 1]

0, for x = 0,

f : [0, 1] → R. Since limx→0

2x sin 1x

= 0 (prove it!) and since limx→0

cos 1x

does not exist (xn = 12nπ

and x′n = 2(4n+1)π

are convergent to 0, but

cos 1xn

= 1 → 1 and cos 1x′n

= 0 → 0), we see that function f has no

limit at x = 0. In particular, it is not continuous at x = 0. At the sametime, it is easy to prove that the function

F (x) =

x2 sin 1

x, for x ∈ (0, 1]

0, for x = 0

is a primitive for f on [0, 1].

In general, to decide if a given noncontinuous function has a prim-itive or not is a difficult task.

The main problem which is to be considered in the following isthe one of the effective construction of primitives for some classes ofcontinuous functions.

Here is a table with the primitives of the basic elementary functions,mostly used in Mathematics and in its applications. We write downonly one primitive (the immediate one, obtained by a direct process of


anti-differentiation). The others can be obtained by adding a constantnumber to this last one.

(1.3)

∫xαdx =

xα+1

α + 1, if α = −1, (x > 0, if α is not an integer)

(1.4)

∫1

xdx = lnx, if x ∈ I,

where I is any interval contained in (0, ∞).

(1.5)

∫1

xdx = ln(−x), if x ∈ J,

where J is any interval contained in (−∞, 0).

(1.6)

∫exdx = ex

(1.7)

∫axdx =

ax

ln a, if a > 0 and a = 1.

(1.8a)

∫1

x2 + 1dx = arctan x

(1.9)

∫1

x2 + a2dx =

1

aarctan

x

a, if a = 0.

(1.10)

∫sin xdx = − cosx

(1.11)

∫cosxdx = sin x

(1.12)

∫1√

1 − x2dx = arcsin x, if x ∈ (−1, 1).

(1.13)

∫1√

a2 − x2dx = arcsin

x

a, if x ∈ (−a, a).

(1.14)

∫1√

x2 + αdx = ln(x+

√x2 + α),

if α = 0, x2 + α > 0 and x+√x2 + α > 0.

(1.15)

∫1

cos2 xdx = tan x,


if x belongs to any interval which does not contain any number of theform nπ + π

2, n ∈ Z.

(1.16)

∫1

sin2 xdx = − cotx,

if x belongs to any interval which does not contain any number of theform nπ, n ∈ Z.

(1.17)

∫sinhxdx = cosh x,

∫coshxdx = sinh x,

where sinh x = ex−e−x

2and cosh x = ex+e−x

2.

All of the above formulas can be directly verified by differentiat-ing the function from the right side of each equality. Let’s verify forinstance formula (1.14):(

ln(x+√x2 + α)

)′=

1

x+√x2 + α)

(1 +

2x

2√x2 + α

)=

1√x2 + α

.

T 4. (linearity) Let f, g : I → R be two functions definedon the same interval I and let α, β be two real numbers. Let

∫f(x)dx

and∫g(x)dx be two primitives of f and g respectively (on I). Then

α∫f(x)dx+ β

∫g(x)dx is a primitive of αf + βg on I.

P. Since(∫

f(x)dx)′

= f and(∫

g(x)dx)′

= g (use only thedefinition!), we get that

(α

∫f(x)dx+ β

∫g(x)dx

)′=

= α

(∫f(x)dx

)′+ β

(∫g(x)dx

)′= αf + βg

and the proof is complete (why?). Here we used the linearity of thedifferential operator h→ h′.

This last result says that the "mapping" f →∫f(x)dx is linear.

Here is a concrete example of the way we can work with this lastproperty and with the above basic formulas.

E 4. Let us compute∫ (

x5 − 3x+ 2 3√x− 7

5√x3

)dx. The

linearity of the integral operator∫(see theorem 4) and formula (1.3)

imply that∫ (

x5 − 3x+ 2 3√x− 7

5√x3

)dx =

∫x5dx− 3

∫xdx+


+2

∫x13dx− 7

∫x−

35dx =

x6

6− 3

x2

2+ 2

x43

43

− 7x25

25

=

=x6

6− 3

2x2 +

3

2x43 − 35

2x25 .

Let J be another interval and let u : J → I, x = u(t), be a functionof class C1(J). Then dx = u′(t)dt (see Analysis I, [Po]). Let f : I → Rbe a continuous function of the variable x ∈ I. Let F (x) be a primitiveof f on I. Then F (u(t)) is a primitive of the function f(u(t))u′(t) of t,i.e.

(1.18)

∫f(u(t))d(u(t)) =

(∫f(x)dx

)(u(t)).

Indeed,

[F (u(t))]′ = F ′(u(t))u′(t) = f(u(t))u′(t).

Formula (1.18) is called the change of variable formula for integralcomputation. This formula says that if in an integral

∫h(x)dx one can

put in evidence an expression u = u(x) such that h(x)dx = f(u(x))duand if one can compute

∫f(u)du then, put instead of u, u(x) in this

last primitive and we obtain a primitive∫h(x)dx for the differential

form hdx.

E 5. Let us compute∫

13x+2

dx, where 3x + 2 < 0. If wedenote u = 3x+ 2 then, du = 3dx (why?). Thus, our integral becomes13

∫1

3x+2d(3x+2). Since u = 3x+2 < 0, a primitive for

∫1udu is ln(−u)

(see formula (1.5)). So, a primitive of 13x+2

dx is 13

ln(−3x− 2).

Sometimes it is easier to directly compute dx as a function of t anddt. For instance, let us compute

∫tan3 xdx. Let us change the variable:

t = tan x, x ∈ (−π2, π

2) (why this restriction?). Since

dt =1

cos2 xdx =

(1 + tan2 x

)dx = (1 + t2)dx,

we get that∫

tan3 xdx =

∫t3

1 + t2dt =

∫t3 + t− t

t2 + 1dt =

∫tdt−

∫tdt

t2 + 1=

=t2

2− 1

2

∫d(t2 + 1)

t2 + 1=t2

2− 1

2ln(t2 + 1) =

=1

2

[tan2 x− ln(tan2 x+ 1)

].


Let f, g : I → R be two functions of class C1(I), where I is a realinterval. We know the "product formula" :

(fg)′ = f ′g + fg′.

Applying the integral operator to both sides, we get:

(1.19)

∫f(x)g′(x)dx = fg −

∫f ′(x)g(x)dx.

or, in language of differentials,

(1.20)

∫fdg = fg −

∫gdf

These last two formulas are called the integral by parts formulas.When do we use them? If one has to compute the integral

∫h(x)dx

and if we can write h(x) = f(x)g′(x) and if the integral∫f ′(x)g(x)dx

can be easier computed, then formula (1.19) works. For instance, if wedifferentiate the function f(x) = ln x, we get 1

xwhich is an "easier"

function from the point of view of integral calculus. Practically, let ususe this philosophy to compute the integral: In =

∫xn ln xdx, where n

is a natural number. For n = 0 we get

I0 =

∫ln xdx =

∫(lnx)(x)′dx.

In formula (1.19) one can put f(x) = lnx and g(x) = x. Thus,

I0 = x ln x−∫

1

x· xdx = x ln x− x.

Let us use the same "trick" for computing In, n > 0.

In =

∫xn ln xdx =

∫(ln x)

(xn+1

n + 1

)′dx =

xn+1

n+ 1ln x−

∫1

x

xn+1

n + 1dx =

=xn+1

n + 1ln x− 1

n + 1

∫xndx =

xn+1

n+ 1lnx− xn+1

(n+ 1)2.

2. Some results on polynomials

We begin with some facts on polynomial functions.A polynomial function ( or simply a polynomial) is a function P of

one variable x, defined on R by the following formula

P (x) = a0 + a1x+ a2x2 + ...+ anx

n, x ∈ R,where a0, ..., an are fixed real numbers. These numbers are said to bethe coefficients of P. We shortly write that P (x), or P ∈ R[x], whereR[x] is the set (a ring!) of all polynomials with real coefficients. For

2. SOME RESULTS ON POLYNOMIALS 11

instance, P (x) = 2x − 1 is a polynomial. Here n = 1, a0 = −1 anda1 = 2. If an = 0 (except the case P = 0− this means that ALL itscoefficients are zero, we always assume that the dominant coefficient anis not zero) we say that the degree of P is n and write this as degP = n.If P (x) = a0 = 0, a nonzero constant, then degP = 0. If P (x) = 0for any x (this is equivalent to saying that all its coefficients are zero),then the degree of P is by definition −∞. This value was chosen topreserve the formula degPQ = degP + degQ for P = 0 and Q = 0.

If an = 1 we say that the polynomial P is monic. For instance,P (x) = x− 1 is monic, but Q(x) = −x− 1 is not monic. If P has thedegree greater or equal to 1 and if it cannot be written as a productof two polynomials of degrees greater or equal to 1, it is said to beirreducible. For instance, P (x) = x2 + 2, P (x) = 2x−1 are irreducible,but Q(x) = x2 − 1 or Q(x) = x3 are reducible, i.e. they can bedecomposed into at least two factors of degrees greater than zero.

The following result is fundamental in Algebra.

T 5. (Euclid’s division algorithm) Let P and Q = 0 be twopolynomials. Then there are another two polynomials C and R suchthat P = CQ + R and degR < degQ. Here the polynomials C and Rare uniquely defined by P and Q.

P. An "abstract" general proof can be found in [La1]. But,...nothing is "abstract" here! We simply use the "division algorithm" forpolynomials, which is very known even from elementary school Algebra.If degP < degQ, then take C = 0 and R = P. Assume that n =degP ≥ m = degQ. We use mathematical induction with respect ton. If n = 0, then m = 0 (Q = 0) and P = k, Q = h, where k and h aretwo real numbers and h = 0 (why?). Since

k =k

hh+ 0,

one can take C = khand R = 0 (degR = −∞ < degQ = 0). Let

now n be grater than 0 and let us assume that we have just proved thetheorem for any k = 0, 1, ..., n− 1. Let us prove it for k = n. SupposeP (x) = anx

n+an−1xn−1+...+a0 andQ(x) = bmx

m+bm−1xm−1+...+b0,

where an = 0 and bm = 0. Then, it is easy to see that

(2.1) P (x) = Q · anbmxn−m + P1(x),

where

P1(x) =

(an−1 −

anbm−1

bm

)xn−1 + ...+

(an−m − anb0

bm

)xn−m+


+an−m−1xn−m−1 + ...+ a0

is a polynomial of degree at most n− 1. Let us use now the inductionhypothesis and write

P1(x) = C1Q +R,

where C1 and R are polynomials, degR < degQ. Let us come back toformula (2.1) with this expression of P1(x) and find

P (x) =

(anbmxn−m + C1

)Q+R.

If we put now C = anbmxn−m+C1, we just obtained the statement of the

theorem for n. Hence the proof of the theorem is complete.The uniqueness can be derived as follows. Let P = DQ+ S, where

deg S < degQ. Then, R−S = (D−C)Q. If R = S, then the inequality,deg(R−S) = deg(D−C)Q ≥ degQ, give rise to a contradiction (why?).Thus, R = S and so (D − C)Q = 0 implies D = C (Q = 0).

C 2. (the P -expansion of Q) Let P be a nonconstantpolynomial and let Q be another arbitrary polynomial. Then Q can beuniquely written as

(2.2) Q = A0 + A1P + ... + AnPn,

where A0, A1, ..., An are polynomials of degrees at most degP − 1. Wesay that "we write Q in base P". In particular, for any nonzero naturalnumber m, one get:

(2.3)Q

Pm=

A0

Pm+

A1

Pm−1+ ...+

AnPm−n

,

if m > n and

(2.4)Q

Pm=

A0

Pm+

A1

Pm−1+ ...+

Am−1

P+ S,

where S is a polynomial, if n ≥ m.

P. We apply again mathematical induction on the degree ofQ. If Q is a nonzero constant (degree zero) polynomial, then Q = Q(A0 = Q). Assume that degQ > 0 and that the statement is true forany polynomial Q1 of degree < degQ. Let us apply Euclid’s divisionalgorithm for Q and P :

(2.5) Q = CP + A0,

where degA0 < degP. Since degC is less than degQ (degP > 0),using the induction hypothesis, we get C = A1 + A2P + ...+ AnP

n−1,where degAj < degP, j = 1, 2, ..., n. Coming back to formula (2.5)with this expression of C, we obtain exactly formula (2.2).


Let us apply this theory to the polynomials Q(x) = x3 + 2 andP (x) = x− 1. Since

x3 + 2 = C(x)(x− 1) +R,

putting x = 1 we get R = 3. Thus,

x3 − 1 = C(x)(x− 1).

Hence, C(x) = x2 + x+ 1. Now,

x2 + x+ 1 = C1(x)(x− 1) +R1.

Making x = 1, we get R1 = 3. So,

x2 + x+ 1 − 3 = x2 + x− 2 = C1(x)(x− 1).

Thus,

C1(x) = x+ 2 = 1 · (x− 1) + 3.

Coming back, step by step, we obtain

x3 + 2 = 3 + (x− 1)[3 + 1 · (x− 1)(x− 1) + 3] =

= 3 + 3(x− 1) + 3(x− 1)2 + (x− 1)3.

Let C = a+ bi : a, b ∈ R, i =√−1, be the field of complex num-

bers. A fundamental result in Algebra (C. F. Gauss) [La1] says thatthe roots of any polynomial P ∈ R[x] are complex numbers, namelyelements of the form a + bi, with a, b real numbers (see also theorem100 in this book).

T 6. (Bezout’s theorem) Let P = P (x) be a nonconstantpolynomial and let α ∈ C be a root of P (P (α) = 0). Then P (x) =A(x − α)Q(x), where A is a constant number and Q(x) is a monicpolynomial. Moreover, if α1, ..., αs are all the distinct roots of P, then

(2.6) P (x) = A(x− α1)n1(x− α2)

n2 ...(x− αs)ns,

where A is the dominant coefficient of P (A = an) and n1, ..., ns arethe algebraic multiplicities of the roots α1, ..., αs. All of these numbersare uniquely determined only by P.

P. Let P (x) = a0 +a1x+a2x2 + ...+anx

n, be our polynomialand let A = an. Then

(2.7) P (x) = AT (x)

where T (x) is the monic polynomial xn + an−1an

xn−1 + ... + a0an. Let us

use Euclid’s division algorithm for dividing T (x) by x− α :

(2.8) T (x) = (x− α)Q(x) +R,


where Q(x) is a polynomial and R is a constant number. Since T (α) =0 (why?), making x = α in (2.8) one gets R = 0. Thus, T (x) =(x− α)Q(x) and, coming back to formula (2.7), we finally obtain that

P (x) = A(x− α)Q(x).

Now we apply the same procedure to the monic polynomial Q(x) andanother root β of it:

P (x) = A(x− α)(x− β)Q1(x),

where Q1 is a polynomial with degP > degQ > degQ1. We continuein this way and finally obtain

P (x) = A(x− α)(x− β)...(x− ω),

where α, β, ..., ω are all the n roots of P. Let us put together the equalroots and so we get the required expression (2.6).

A bijection σ : C→ C which is a field morphism, i.e.

(2.9) σ(z + w) = σ(z) + σ(w), σ(zw) = σ(z)σ(w)

is called an automorphism of C. An R-automorphism µ of C is anautomorphism of C which does not change the elements of R, i.e. µ(a+0i) = a, if a ∈ R. For instance, it is easy to see (prove it!) thatthe identity e : C → C, e(z) = z and the conjugation e : C → C,e(a + bi) = a− bi are R-automorphisms of C. We also denote e(z) byz, the conjugate of z. For instance, 3 − 4i = 3 + 4i and 3.5 = 3.5.

L 1. Let

P (x) = a0 + a1x+ a2x2 + ...+ anx

n

be a polynomial with real coefficients and let α ∈ C be a root of it, i.e.P (α) = a0 + a1α + a2α

2 + ...+ anαn = 0.

Let σ be a R-automorphism of C. Then, the complex number σ(α) isalso a root of the polynomial P.

P. Let us apply σ to the equality

a0 + a1α+ a2α2 + ...+ anα

n = 0.

Taking into account the algebraic properties (2.9) of σ and the factthat it preserves the real coefficients a0, ..., an, we get

a0 + a1σ(α) + a2σ(α)2 + ...+ anσ(α)n = 0,

i.e. the complex number σ(α) verifies the equality: P (x) = 0, thusσ(α) is another root of P.


This lemma says that any R-automorphism σ of C permutes theroots between them. For instance, since i is a root of the equationX2 + 1 = 0, σ(i) = i or σ(i) = −i. If σ(i) = i, σ(a + bi) = a + bi, i.e.σ = e, the identity. If σ(i) = −i, then σ(a+ bi) = a−bi, i.e. σ = e, theconjugation automorphism of C. These considerations lead us directlyto the following basic result.

T 7. Any monic irreducible polynomial P with real coef-ficients is either of the form P (x) = x − a, or of the form P (x) =x2 + bx+ c, where a, b, c are real numbers.

P. If our polynomial P is of degree 1 everything is clear. IfdegP > 1, let α be one of its root in C. Use lemma 1 and find thatits conjugate α is also a root of P (x). If α = α, i.e. if α is a realnumber, then P (x) is divisible by the polynomial Q(x) = x − α (seetheorem 6). Since degP > 1 and since Q(x) is a factor of P (x), weobtain a contradiction (P was considered to be irreducible). Hence,the root α is not real. Let us denote b = −(α + α) and c = αα. Sinceα+ α = α + α and αα = αα we see that b, c ∈ R. The polynomialH(x) = (x−α)(x−α) = x2+bx+c is a divisor of P. Indeed, let us divideP by H. Thus, P = HU + V, where deg V < 2. Moreover, V (α) = 0and V (α) = 0, i.e. V is a polynomial of degree at most 1 which has twodistinct roots! The only possibility is that V be identically with zero,so P = HU. Since P is irreducible, U must be a constant polynomialand, since P and H are both monic, this last constant polynomial mustbe 1. Thus,

P = H = x2 + bx+ c.

The following theorem (see [La1] for instance) is a basic result inAlgebra.

T 8. (factorial theorem). Let P be a nonconstant polyno-mial with real coefficients. Then P can be uniquely written as

(2.10) P = AP n11 P n22 ...P nkk ,

where A is a constant number and P1, P2, ..., Pk are distinct monicirreducible polynomials of degrees 1 or 2. "Uniquely" here means thatif

P = BQm11 Qm2

2 ...Qmhh ,

is a decomposition of P of the same type (B constant and Q1, Q2, ..., Qhare monic and irreducible polynomials), then B = A, k = h and foreach Q

mj

j , there is a Pi such that Qj = Pi and mj = ni.


P. In formula (2.6) some α′s could be real numbers and somecould be complex numbers. If for instance α1 is a real number, thenP1 = (x − α1)

n1 . If αi is a complex number (nonreal), then there isexactly an αj between the roots α1, ..., αs of P such that αj = αi (seelemma 1) and ni = nj (why?). Thus, the polynomial (x− αi)(x− αj)is a monic irreducible polynomial with real coefficients (why?). Let usdenote it by P2. Hence, the contribution of αi and αj to P is exactlyP n22 , where n2 is the common value of ni and nj . We continue thisreasoning on the roots of P up to obtaining the formula (2.10). SinceA = B = an, the dominant coefficient of P , since any root αh of P isa root of a Pt and of a Qv at the same time and since Pt and Qv aremonic irreducible polynomials, we see that Pt = Qv. Indeed, if αh is areal number, then Pt(x) = x− αh = Qv(x). If αh is a complex nonrealnumber, since Pt and Qt are monic irreducible polynomials with realcoefficients, we see that both of them must be equal to (x− αh)(x−αh)(why?). So, we have the required uniquenees.

D 2. (the greatest common divisor) Let P and Q be twononzero polynomials. The monic polynomial D of maximum degreesuch that D is a divisor of P and of Q is said to be the greatest commondivisor of P and Q. We denote D by (P,Q).

For instance, if P = x4(x + 1)2(x2 + x + 1) and Q = x2(x + 1)5,then D = (P,Q) = x2(x+ 1)2. In general, if

P = AP n11 P n22 ...P nkk ,

andQ = BQm1

1 Qm22 ...Qmt

t ,

are the decompositions of P and respectively of Q, of type (2.10), suchthat P1 = Q1, P2 = Q2, ..., Pr = Qr and all the others are distinct oneto each other, then (why?)

D = Pmin(n1,m1)1 P

min(n2,m2)2 ...Pmin(nr ,mr)

r .

Moreover, the roots of D are exactly the common roots of P and Qwith the least multiplicities (look at the above example!). If D is 1,we say that P and Q are coprime. For instance, P = 2x2 + 2 andQ = x3 + 1 are coprime, but P = x+ 1 and Q = x3 + 1 are not coprime(why?).

T 9. D is the greatest common divisor of P and Q (P andQ are not zero) if and only if there exists two polynomials U0 and V0

such that

(2.11) D = PU0 +QV0


and D is monic with the least degree such that D can be written asin (2.11). In particular, if P and Q are coprime, then there are twopolynomials U0 and V0 such that

(2.12) 1 = PU0 +QV0

P. Let us define the following set of polynomials(2.13)S = H = PU +QV : U and V are arbitrary polynomials in R[x].It is easy to see that the sum between two polynomials of S is also apolynomial in S.

If we multiply a polynomial of S by an arbitrary polynomial of R[x],we also get a polynomial of S (prove these two last statements). LetD be a (monic) nonzero polynomial of S of the least degree and letM be another polynomial of S. Let us apply the Euclid’s algorithm fordividing M by D.We get

M = CD +R,

where degR < degD. Since M,D ∈ S, then R = M − CD ∈ S. Thus,R = 0 and so M = CD. Therefore D is a divisor of any polynomial ofS. In particular, D divides P and Q (P,Q ∈ S!) Since D ∈ S, thereare two polynomials U0 and V0 such that D = PU0 +QV0.

=⇒) Let now G be the greatest common divisor of P and Q. SinceG divides P and Q, G also divides D. But D also divides P and Q,thus the degree of D is at most equal to degG. Since G divides D, wehave that degG = degD. Since D and G are both monic, we musthave that D = G. Thus, for the greatest common divisor we have therelation (2.11).

⇐=) Let D0 be a monic nonzero polynomial of minimal degree suchthat

D0 = PU1 +QV1 ∈ S.

Exactly like above we prove that D0 is the greatest common divisor ofP and Q.

The last statement is obvious.

Practically, we can find the greatest common divisor and its ex-pression (2.11) by transferring the problem from polynomials P andQ to another pair of polynomials Q and R, where degR < degQ (ifdegP ≥ degQ). Then to another pair R and R1, where degR1 <degR, etc. This idea directly comes from the Euclid’s division algo-rithm. Indeed, assume that degP ≥ degQ (if degQ > degP, change Pwith Q, etc.) and divide P by Q. There are polynomials C and R suchthat P = CQ+R and degR < degQ. It is easy to see that the greatest


common divisor of P and Q is equal to the greatest common divisor ofQ and R (prove slowly!). Now divide Q by R and get Q = C1R + R1,with degQ > degR > degR1. The greatest common divisor of Q andR is equal to the greatest common divisor of R and R1, and so on... upto a Rj = 0. Then Rj−1 is exactly the greatest common divisor of Pand Q eventually multiplied by a constant (in our above definition thegreatest common divisor is a monic polynomial!) (prove all of these!).

Let for instance P (x) = x4 − x and Q(x) = x2 − 1. It is not neces-sarily to use Euclid’s division algorithm, but to successively diminishthe degrees of P or of Q. Let us write:

x4 − x = x2(x2 − 1) + x2 − x = x2(x2 − 1)+

+(x2 − 1) − (x− 1) = (x2 + 1)(x2 − 1) − (x− 1).

Since (P,Q) = (x2 − 1, x− 1) we go on with division:

x2 − 1 = (x+ 1)(x− 1),

the greatest common divisor of P and Q is x− 1 and

x− 1 = (x4 − x)(−1) + (x2 − 1)(x2 + 1).

Thus, the polynomials U0 and V0 are

U0(x) = (−1), and V0(x) = x2 + 1

and they can be chosen such that degU0 < degQ and deg V0 < P.A representation ofD as in (2.11) with degU0 < degQ and deg V0 <

P is called a canonical representation.

T 10. In a canonical representation for D = 1, U0 and V0

are uniquely determined.

P. Indeed, if

1 = PU0 + QV0 = PU ′0 +QV ′0 ,

then

(2.14) P (U0 − U ′0) = Q(V ′0 − V0)

Since P and Q have no nontrivial factors (of degree greater than 1) incommon, we see that V ′

0 − V0 is divisible by P and U0 −U ′0 is divisibleby Q. Since deg(U0 − U ′0) < degQ and deg(V ′0 − V0) < degP, we mustconclude that V ′0 = V0 and U0 = U ′0.

Let P and Q be two polynomials such that P = 0. The fractionQP

is called a rational fraction. A rational fraction APm , where P is an

irreducible polynomial and degA < degP is called a simple rationalfraction. An arbitrary polynomial is also called a simple fraction (withdenominator 1). If P is an irreducible polynomial, n is a nonzero


natural number and Q is an arbitrary polynomial, then the fractionQPn is called a simple block (of fractions). Formulas (2.3) and (2.4) of

corollary 2 says that any simple block QPm can be uniquely decomposed

into simple fractions:

(2.15)Q

Pm=

A0

Pm+

A1

Pm−1+ ...+

Am−1

P+ S,

where A0, A1, ..., Am−1, S are polynomials (some of them may be zero!)and degAj < degP for any j = 0, 1, ....m− 1.

T 11. (decomposition into partial fractions) Let Q and Pbe arbitrary polynomials with P = 0 and let

(2.16) P = AP n11 P n22 ...P nkk ,

be a decomposition of P into a product of irreducible factors and aconstant A. Then the fraction Q

Pcan be uniquely (up to the order of the

terms!) written as a sum of blocks:

(2.17)Q

P=

1

A

[Q1

P n11

+Q2

P n22

+ ...+QkP nkk

],

where Q1, Q2, ..., Qk are (uniquely determined) polynomials. Since eachblock Qi

Pnii

can be uniquely represented as a sum of simple fractions:

(2.18)QiP nii

=A

(i)0

P nii+

A(i)1

P ni−1i

+ ...+A

(i)ni−1

Pi+ Si,

we get the final decomposition of the fraction QPas a sum of simple

fractions:

Q

P=S

A+

1

A

k∑

i=1

[A

(i)0

P nii+

A(i)1

P ni−1i

+ ... +A

(i)ni−1

Pi

],

where A is a constant (the dominant coefficient of P ), S is a polynomial

(S =∑ki=1 Si) and A

(i)j , i = 1, 2, ..., k and j = 0, 1, ..., ni − 1 are

polynomials with degA(i)j < degPi for each i = 1, 2, ..., k and j =

0, 1, ..., ni−1. This decomposition is unique up to the order of terms in

sum. Moreover, the degrees of the polynomials A(i)j are 0 or 1 (because

the degrees of Pi are 1 or 2).

P. We have only to prove formula (2.17). Let us use mathe-matical induction on the number k of the irreducible factors P n11 , P n22 ,... , P nkk . If k = 1 we have nothing to prove. Assume that k > 1 anddenote P n22 P n33 ...P nkk by L. Suppose that the statement of the theoremis true for any nonzero natural number less than k and we shall prove it


for k. Since P n11 , P n22 , ... , P nkk are monic irreducible and distinct one toeach other, P n11 and L have no common roots. Indeed, if α is a root ofP n11 and of L, then there is a j = 1 such that α is a root of P1 and of Pj;if α is real, P1 and Pj are both equal to x−α, a contradiction! (why?).If α is not real, let α be the conjugate of α. Then P1 and Pj are bothequal to (x− α)(x− α), again a contradiction! (why?). Thus P n11 andL are coprime and we can apply theorem 9, formula (2.12). So thereare two uniquely defined polynomials V and U with deg V < degL anddegU < degP n11 such that

P n11 V + LU = 1

(see theorem 10). Let us multiply both sides by QP

= Q

APn11 L

. We get:

(2.19)Q

P=

1

A

[QV

L+QU

P n11

].

Let us put Q1 instead of QU and let us see that the number of irre-ducible factors of L is k − 1. Applying the induction hypothesis weobtain the representation (L is monic, so A = 1 in this case!):

(2.20)QV

L=

Q2

P n22

+ ...+QkP nkk

,

where Q2, Q3, ..., Qk are uniquely defined polynomials. If we come backto formula (2.19) with this expression of QV

L, we get exactly formula

(2.17). The other statement is a direct computational consequence ofthis last formula.

E 6. Let us find the decomposition into simple fractionsof the rational function Q(x)

P (x)= x+1

x4+x2. Since P (x) = x2(x2 + 1), we

have two blocks of simple fractions, one corresponding to the irreduciblepolynomial P1(x) = x and the other corresponding to the irreduciblepolynomial P2(x) = x2 + 1. Hence,

(2.21)x+ 1

x4 + x2=A

x2+B

x+Cx+D

x2 + 1.

Thus,

x+ 1

x4 + x2=A(x2 + 1) +Bx(x2 + 1) + (Cx+D)x2

x4 + x2,

orx+ 1 = (B + C)x3 + (A+D)x2 +Bx+ A.

We identify the coefficients of both sides and obtain:

1 = A,

1 = B,

3. PRIMITIVES OF RATIONAL FUNCTIONS 21

0 = A+D,

0 = B + C,

so A = 1, B = 1, C = −1 and D = −1. Therefore our decompositionis

(2.22)x+ 1

x4 + x2=

1

x2+

1

x+

−x− 1

x2 + 1.

3. Primitives of rational functions

We shall use now the basic properties of the integrals (primitives)and the above sketchy theory of rational functions in order to computetheir primitives.

Let us compute for instance a primitive for the rational functionwhich appeared in example 6, namely

∫x+1x4+x2

dx. We use the linearity

of∫in formula (2.22):

∫x+ 1

x4 + x2dx =

∫1

x2dx+

∫1

xdx+

∫ −x− 1

x2 + 1dx =

=

∫x−2dx+

∫x−1dx− 1

2

∫2x

x2 + 1dx−

∫1

x2 + 1dx =

=x−3

−3+ ln |x| − 1

2ln(x2 + 1) − arctanx,

because∫2x

x2 + 1dx =

∫d(x2 + 1)

x2 + 1=

∫du

u= ln |u| = ln(x2 + 1).

Since any simple fraction has one of the following forms:

(3.1)A

x− α, where A,α ∈ R,

(3.2)A

(x− α)n, where A,α ∈ R, n ∈ N∗, n ≥ 2,

(3.3)Ax+B

x2 + bx+ c, where A, b, c ∈ R,∆ = b2 − 4c < 0,

(3.4)Ax+B

(x2 + bx+ c)n, where A, b, c ∈ R,∆ = b2 − 4c < 0, n ≥ 2,

we must show how to find a primitive for each of these cases.Case 1

(3.5)

∫A

x− αdx = A

∫1

x− αdx = A ln |x− α| .


Case 2 ∫A

(x− α)ndx = A

∫1

(x− α)ndx =(3.6)

A

∫(x− α)−ndx = A

(x− α)−n+1

−n+ 1, if n ≥ 2.

Case 3 Since∫Ax+B

x2 + bx+ cdx =

A

2

∫2x+ b

x2 + bx+ cdx+

+

(−Ab

2+B

)∫1

x2 + bx+ cdx

and since ∫2x+ b

x2 + bx+ cdx = ln(x2 + bx+ c),

one can reduce everything to the computation of a primitive of thefollowing type:

∫A

x2 + bx+ cdx = A

∫1

(x+ b2)2 + a2

dx,

where a =√c− b2

4. Thus,

∫A

x2 + bx+ cdx = A

∫d(x+ b

2)

(x+ b2)2 + a2

=(3.7)

= A1

aarctan

x+ b2

a.

Here we use the basic formula∫

1x2+a2

dx = 1a

arctan xa(see formula

(1.9)).Case 4 Using a similar reasoning as in Case 3 one can reduce the

computation of the primitive of the simple fraction of formula (3.4) tothe following primitive:

∫1

(x2 + bx+ c)ndx =

∫d(x+ b

2)[

(x+ b2)2 + a2

]n .

Denoting x + b2by a new variable u we must compute the primitive∫

du(u2+a2)n

, for n ≥ 2. For convenience we use the same variable x. Let

us denote In =∫

dx(x2+a2)n

for any n > 0 this time (a = 0). We know

that

I1 =1

aarctan

x

a.


Let n > 1. We shall construct now a recurrence formula for In :

In =1

a2

∫x2 + a2 − x2

(x2 + a2)ndx =

1

a2

∫1

(x2 + a2)n−1dx−

− 1

a2

∫x2

(x2 + a2)ndx =

1

a2In−1 −

1

2a2

∫x

(x2 + a2)nd(x2 + a2) =

=1

a2In−1 −

1

2a2

[∫xd

((x2 + a2)−n+1

−n+ 1

)].

Let us use now the formula of integrating by parts (see (1.20)) andcompute∫xd

((x2 + a2)−n+1

−n+ 1

)= x

(x2 + a2)−n+1

−n + 1− 1

−n+ 1

∫1

(x2 + a2)n−1dx,

Thus,

In =1

a2In−1 −

1

2a2

[x

(x2 + a2)−n+1

−n+ 1− 1

−n + 1In−1

],

or

(3.8) In =

[1

a2− 1

2a2(n− 1)

]In−1 +

x

2a2(n− 1)(x2 + a2)n−1.

For instance, let us compute I2 =∫

dx(x2+a2)2

.

(3.9) I2 =1

2a3tan−1 x

a+

x

2a2(x2 + a2).

Here tan−1 x is another notation for arctan x. Formula (3.8) can be usedto compute In "from up to down", i.e. the computation of In reducesto the computation of In−1. The computation of In−1 reduces to thecomputation of In−2, etc., up to the computation of I1 = 1

aarctan x

a.

In the following examples we use the ideas and experience just ex-posed in the above four cases.

E 7. What is a primitive of 34x+5

, where x runs over an

interval I which does not contain x = −54.

∫3

4x+ 5dx =

3

4

∫d(4x+ 5)

4x+ 5=

3

4

∫du

u=

3

4ln |u| =

3

4ln |4x+ 5| .

E 8. Let us compute a primitive of f(x) = 1(2x+5)5

, where

x belongs to an interval I which does not contain x = −52.

∫1

(2x+ 5)5dx =

1

2

∫(2x+ 5)−5d(2x+ 5) =

1

2

∫u−5du =


=1

2

u−5+1

−5 + 1=

(2x+ 5)−4

−8.

E 9. What is the set of all primitives of the function f(x) =3x+2x2+x+1

, x ∈ R?∫

3x+ 2

x2 + x+ 1dx = 3

∫x+ 2

3

x2 + x+ 1dx =

3

2

∫2x+ 4

3

x2 + x+ 1dx =

=3

2

∫2x+ 1 + 1

3

x2 + x+ 1dx =

3

2

∫(x2 + x+ 1)

′

x2 + x+ 1dx+

1

2

∫1

x2 + x+ 1dx =

=3

2

∫d (x2 + x+ 1)

x2 + x+ 1+

1

2

∫1(

x+ 12

)2+ 3

4

d

(x+

1

2

)=

=3

2ln(x2 + x+ 1) +

1

2

∫1

u2 +(√

32

)2du =3

2ln(x2 + x+ 1)+

+1

2

1√

32

arctanu√

32

=3

2ln(x2 + x+ 1) +

1√3

arctan2(x+ 1

2

)√

3=

3

2ln(x2 + x+ 1) +

1√3

arctan2x+ 1√

3.

So the set of all primitives of f(x) = 3x+2x2+x+1

is

3

2ln(x2 + x+ 1) +

1√3

arctan2x+ 1√

3+ C

,

where C is an arbitrary constant.

E 10. Let us find a primitive for f(x) = x+3(x2+4x+6)3

, x ∈ R.

I =

∫x+ 3

(x2 + 4x+ 6)3dx =

1

2

∫2x+ 4

(x2 + 4x+ 6)3dx+

(3.10) +

∫1

(x2 + 4x+ 6)3dx =

1

2

∫(x2 + 4x+ 6)

′

(x2 + 4x+ 6)3dx+

+

∫1

[(x+ 2)2 + 2]3d(x+ 2) =

1

2

∫u−3du+

∫dv

[v2 + (

√2)2

]3 ,

where u = x2 + 4x+ 6 and v = x+ 2. Let

I3 =

∫dv

[v2 + (

√2)2

]3


and let us use successively the recurrence formula (3.8) (put v insteadof x, 3 instead of n and

√2 instead of a):

I3 =

[1

2− 1

4(3 − 1)

]I2 +

v

4(3 − 1)(v2 + 2)2=

3

8I2 +

v

8(v2 + 2)2.

But, from formula (3.9) we get

I2 =1

4√

2arctan

v√2

+v

4(v2 + 2),

thus

I3 =3

32√

2arctan

v√2

+3v

32(v2 + 2)+

v

8(v2 + 2)2.

Let us come back with this value of I3 in (3.10) and find:

I =1

2

∫u−3du+

∫dv

[v2 + (

√2)2

]3 =

1

2

u−2

−2+

3

32√

2arctan

v√2

+3v

32(v2 + 2)+

v

8(v2 + 2)2.

Finally, we make u = x2 + 4x+ 6 and v = x+ 2 :

I = − 1

4 (x2 + 4x+ 6)2+

3

32√

2tan−1 x+ 2√

2+

+3(x+ 2)

32 [(x+ 2)2 + 2]+

x+ 2

8 [(x+ 2)2 + 2]2.

E 11. Compute I =∫

dx(x2+a2)(x2+b2)

for all values of a and

b.Case 1. a = 0 = b. Then

I =

∫x−4dx =

x−3

−3.

Case 2. a = 0, b = 0. Then we must decompose into simple frac-

tions the integrand function: 1x2(x2+b2)

. Let us denote x2 by u :

1

u(u+ b2)=A

u+

B

u+ b2,

or

1 = A(u+ b2) +Bu = (A+B)u+ Ab2.

So A = b−2 and B = −b−2. Coming back to x, we get

1

x2(x2 + b2)=b−2

x2− b−2

x2 + b2.


Thus,

(3.11)

∫1

x2(x2 + b2)dx = b−2

∫x−2dx− b−2

∫1

x2 + b2dx =

= b−2x−1

−1− b−2 1

barctan

x

b.

Case 3. a = 0, b = 0. Then I =∫

1x2(x2+a2)

dx. Use now formula

(3.11) with a instead of b and find

∫1

x2(x2 + a2)dx = a−2x

−1

−1− a−2 1

aarctan

x

a.

Case 4. a = 0, b = 0, a = b. Let us take the fraction 1(x2+a2)(x2+b2)

and

decompose it into simple fractions. Since this last expression is in facta rational function of x2, let us put u = x2 (it it easier to work withfactors of degree one at denominator-why?). So

1

(x2 + a2)(x2 + b2)=

A

u+ a2+

B

u+ b2.

Thus,

1 = A(u+ b2) +B(u+ a2) = (A+B)u+ Ab2 +Ba2.

Hence, A+ B = 0 and Ab2 + Ba2 = 1, or A = 1b2−a2 and B = − 1

b2−a2 .Finally,

∫1

(x2 + a2)(x2 + b2)dx =

=1

b2 − a2

[∫1

x2 + a2dx−

∫1

x2 + b2dx

]=

=1

b2 − a2

[1

aarctan

x

a− 1

barctan

x

b.

]

Case 5. a = b = 0. Then I =∫

1(x2+a2)2

= I2, with the notation of

formula (3.9). We use this last formula and find

I =1

2a3arctan

x

a+

x

2a2(x2 + a2).

4. PRIMITIVES OF IRRATIONAL AND TRIGONOMETRIC FUNCTIONS 27

4. Primitives of irrational and trigonometric functions

What is an irrational function? The big temptation were to say thatsuch a function is a function "which is not rational"! But, it is not so!For instance f(x) = exp(x) is not a rational function (why?-prove it!)but it is also not an irrational one. We say that it is a transcendentalfunction. "Easily" speaking, a function is irrational if it is "rational"and contains some radicals...The exact definition is the following. Afunction of two variables

P (x, y) =n∑

i=0

m∑

j=0

aijxiyj = a00 + a10x+ a01y + ... + anmx

nym,

where aij ∈ R, is called a polynomial of two variables. A ratio-nal function of two variables R(x, y) is a quotient of two polynomi-

als P (x, y), Q(x, y) of two variables: R(x, y) = P (x,y)Q(x,y)

. An algebraic

function of one variable is a root y = f(x) of a nonzero polynomialP (x, y) = a0(x) + a1(x)y + ...+ am(x)ym, where ai(x) are polynomialsof one variable with coefficients in R, i.e. P (x, f(x)) = 0 for any realnumber x in the definition domain of f. For instance, y =

√x2 + 1

is algebraic because it is a root of the equation 1 + x2 − y2 = 0. Anintegral of the form

∫R(x, y)dx, where R(x, y) is a rational function

and y is an algebraic function is called an abelian integral ("abelian"comes from the name of the great mathematician Niels Abel who sys-tematically studied such integrals). If the algebraic function y = f(x)can be expressed by radicals we say that R(x, y) is an irrational func-

tion. For instance, R(x,√x2 + 1) = x+

√x2+1

1+3√x2+1

is an irrational function.

Since primitives of some classes of irrational functions appears in manyapplications, we present here some modalities to compute them.

A. Primitives of the form∫R(x,

√ax+ b)dx, a = 0

We can reduce the computation of such a primitive to the compu-tation of a primitive of a rational function by the following change ofvariable: u2 = ax + b for ax + b ≥ 0, a = 0 (a, b are fixed constants).Our primitive becomes

F (u) =

∫R

(u2 − b

a, u

)2u

adu.

Then G(x) = F(√

ax+ b)is a primitive of R(x,

√ax+ b) (see formula

(1.18)).


E 12. Compute I =∫

11+√

2x+3dx. Write u2 = 2x + 3,

2udu = 2dx =⇒ dx = udu, so

I =

∫u

1 + udu =

∫u+ 1 − 1

u+ 1du =

∫1du−

∫1

1 + udu =

= u− ln |u+ 1| =√

2x+ 3 − ln(√

2x+ 3 + 1).

B. Primitives of the form∫R(x,

√ax2 + bx+ c)dx, a = 0,where

ax2 + bx+ c is not a perfect squareAssume that a > 0. Then

ax2 + bx+ c =

(√ax+

b

2√a

)2

+ c− b2

4a.

Let us make the following convention: if c− b2

4a> 0, we denote it by δ2

(any positive real number is a square!!); if c− b2

4a< 0, we denote it by

−δ2. Since ax2 + bx+ c is not a perfect square, c− b2

4acannot be zero.

Thus, if we denote√ax + b

2√a

= u, then du =√adx and our integral

becomes

(4.1)

∫R

(u− b

2√a√

a,√u2 ± δ2

)1√adu.

Assume now that a < 0. Then

ax2 + bx+ c = −(√

−ax− b

2√−a

)2

+ c− b2

4a.

Now, always c− b2

4ais positive because ax2 + bx+ c ≥ 0 (otherwise we

cannot define the square root of it!). So this time we put c− b2

4a= δ2.

Making the change of variable u =√−ax − b

2√−a , du =

√−adx, weget

∫R

(u+ b

2√−a√

−a ,√δ2 − u2

)1√−adu.

Hence, we have to study the following three types of such integrals:

B1 :∫R(x,

√x2 + δ2)dx

B2 :∫R(x,

√x2 − δ2)dx and

B3 :∫R(x,

√δ2 − x2)dx.

In order to eliminate the radical in B1 and in B2 one can use the(Euler) substitution

(4.2)√x2 ± δ2 = x+ t,


where t is a new variable. Squaring in (4.2) we get

x =±δ2 − t2

2t,√x2 ± δ2 =

±δ2 + t2

2tand dx =

∓δ2 − t2

2t2dt.

Thus our primitives B1 and B2 become∫R

(±δ2 − t2

2t,±δ2 + t2

2t

) ∓δ2 − t2

2t2dt

which is a primitive of a rational function in the new variable t. Com-pute it by the methods described in the previous section and finally

put t =√x2 ± δ2 − x (see formula (4.2)).

E 13. Let us compute I =∫ √

3x2 − 4x+ 1dx. First of allwe must reduce the computation of I to one of the form B1 or B2. Since

3x2 − 4x+ 1 =

(√3x− 2√

3

)2

− 1

3.

Thus for u =√

3x− 2√3, dx = 1√

3du and our primitive becomes

1√3

∫ √u2 − 1

3du.

Let us make now an Euler substitution

(4.3)

√u2 − 1

3= u+ t, u = −

13

+ t2

2t, du = −t

2 − 13

2t2dt,

so1√3

∫ √u2 − 1

3du = − 1

4√

3

∫t4 − 2

3t2 + 1

9

t3dt =

= − 1

4√

3

[t2

2− 2

3ln |t| − t−2

18

].

Coming back to the initial variable x we get (t =√u2 − 1

3− u, see

formula (4.3)):

I = − 1

4√

3[

(√u2 − 1

3− u

)2

2− 2

3ln

∣∣∣∣∣

√u2 − 1

3− u

∣∣∣∣∣−

−

(√u2 − 1

3− u

)−2

18].

To find the expression of the primitive I in x we must put in this lastexpression u =

√3x − 2√

3, etc. (go on with computations up to the

end).


In order to eliminate the radical in B3 we can make the substitution:

(4.4)√δ2 − x2 = xt+ δ.

Squaring and dividing by x we get:

−x = xt2 + 2tδ =⇒ x = − 2tδ

t2 + 1, dx =

2t2δ − 2δ

(t2 + 1)2dt,(4.5)

√δ2 − x2 =

−t2δ + δ

t2 + 1.(4.6)

So

(4.7)

∫R(x,

√δ2 − x2)dx =

∫R

(− 2tδ

t2 + 1,−t2δ + δ

t2 + 1

)2t2δ − 2δ

(t2 + 1)2dt,

which is a primitive of a rational function in the variable t. After finding

it we put instead of t its expression from (4.4), t =

√δ2−x2−δx

.

E 14. Let us use formula (4.7) to find the primitive ofI =

∫1

x+√

4−x2dx.

For this we change the variable x with a new one t (like in (4.4)):

(4.8)√

22 − x2 = xt+ 2.

So, using (4.5) we get∫

1

x+√

4 − x2dx =

∫1

− 4tt2+1

+ −2t2+2t2+1

4t2 − 4

(t2 + 1)2dt =

−2

∫t2 − 1

(t2 + 2t− 1)(t2 + 1)dt.

Since t2 + 2t − 1 = (t + 1 +√

2)(t + 1 −√

2), we have the followingdecomposition into simple fractions

t2 − 1

(t2 + 2t− 1)(t2 + 1)=

A

t + 1 +√

2+

B

t+ 1 −√

2+Ct +D

t2 + 1.

An ugly computation leads to

(4.9) A = − 1 +√

2

4 + 2√

2, B = − 1

4 + 2√

2, C = D =

1

2.

Thus

I = −2

∫t2 − 1

(t2 + 2t− 1)(t2 + 1)dt = −2A

∫1

t+ 1 +√

2dt−

−2B

∫1

t + 1 −√

2dt− C

∫d(t2 + 1)

t2 + 1− 2D

∫1

t2 + 1dt =


(4.10)

= −2A ln∣∣∣t + 1 +

√2∣∣∣−2B ln

∣∣∣t+ 1 −√

2∣∣∣−C ln(t2 +1)−2D arctan t,

where A,B,C,D have the numerical values from formula (4.9). From(4.8) we get

t =

√4 − x2 − 2

x.

With this last value of t we come in (4.10) and obtain

I = −2A ln

∣∣∣∣√

4 − x2 − 2

x+ 1 +

√2

∣∣∣∣−

−2B ln

∣∣∣∣√

4 − x2 − 2

x+ 1 −

√2

∣∣∣∣−

−C ln

([√4 − x2 − 2

x

]2

+ 1

)− 2D arctan

√4 − x2 − 2

x.

R 2. To compute the primitive B1 we also can use trigono-

metric substitutions. For instance, if in I =∫R(x,

√x2 + δ2)dx we

put x = δ tan t, we get

dx =δ

cos2 tdt, I =

∫R(δ tan t,

δ

|cos t|)δ1

cos2 tdt.

This last primitive is a rational function of sin x and cosx. We shellsee later how to integrate a rational function of sin x and cosx.

E 15. Let us make x = a tan t (a > 0) and compute

I =

∫ √x2 + a2dx = a2

∫1

cos3 tdt = a2

∫d(sin t)

cos4 t= a2

∫du

(1 − u2)2,

where u = sin t. Since1

(1 − u2)2=

A

(1 − u)2+

B

1 − u+

C

(1 + u)2+

D

1 + u,

we find A = B = C = D = 14. So

I =a2

4

[∫du

(1 − u)2+

∫du

1 − u+

∫du

(1 + u)2+

∫du

1 + u

]=

a2

4

[1

1 − u− ln(1 − u) − 1

1 + u+ ln(1 + u)

]=a2

4

(2u

1 − u2+ ln

1 + u

1 − u

).

Coming back to t we obtain

I =a2

4

(2 sin t

cos2 t+ ln

1 + sin t

1 − sin t

).


But t = arctan xa, thus I finally becomes:

I =a2

4

(2 sin

[arctan x

a

]

cos2[arctan x

a

] + ln1 + sin

[arctan x

a

]

1 − sin[arctan x

a

]).

R 3. To compute∫R(x,

√x2 − δ2)dx we can use sinhx and

cosh x. Recall that sinhx = expx−exp(−x)2

and coshx = expx+exp(−x)2

. It iseasy to see that

(4.11) cosh2 x− sinh2 x = 1.

Hence, if we put x = δ cosh t we get dx = δ sinh tdt and√x2 − δ2 =

δ |sinh t| . Thus our integral becomes a rational integral in sinh x andcosh x.

To compute∫R(x,

√δ2 − x2)dx we can use the trigonometric sub-

stitution x = δ sin t. Thus dx = δ cos tdt and√δ2 − x2 = δ |sin t| . So

our integral reduces to an integral of a rational function in sinx andcosx.

C. Primitives of the form∫xm(axn + b)pdx, where m,n, p are

nonzero rational numbers and a, b = 0Let m = m1

m2, n = n1

n2and p = p1

p2be three nonzero rational numbers

(fractions are simplified) withm1, n1, p1,m2, n2, p2 natural numbers andm2, n2, p2 > 0. The expression xm(axn + b)pdx is called a binomialdifferential form. If there is a differential function F (x) such thatdF (x) = xm(axn + b)pdx we say that the binomial differential is exactand the function F (x) is a primitive of it. A substitution or a change ofthe variable x with a new variable t is an equation of the type g(x) = t,where g is a diffeomorphism on a fixed interval J (g is of class C1 andhas an inverse x = g−1(t) also of class C1). A substitution can beinterpreted as a plane curve h(x, t) = g(x) − t = 0. The substitutionis called "rational" if one can find a rational parameterization of thecurve h(x, t) = 0, i.e. if one can find two rational functions R1(u),R2(u), where u ∈ L, a real interval, such that h(R1(u), R2(u)) = 0

for any u ∈ L. For instance, (3x+ 5)53 = t is a rational substitution

because for x = R1(u) = u3−53, and t = R2(u) = u5, rational functions

of u, one has: (3R1(u) + 5)53 = R2(u). At the end of 19-th century, the

great Russian mathematician P. L. Cebyshev proved a basic results onsuch binomial differential forms (on the existence of their primitives).


T 12. (Cebyshev) One can find an integral of a rationalfunction starting from

∫xm(axn + b)pdx, by making "rational" substi-

tutions, if and only if m,n and p are in one of the following threesituations:

Case 1. p is an integer (p2 = 1).Case 2. p is not an integer but m+1

nis an integer.

Case 3. p is not an integer, m+1n

is not an integer, but m+1n

+ p isan integer.

P. We prove only the fact that if we are in one of the threecases above, then the integral can be "rationalized", i.e. there is asequence of "rational substitutions" of variables such that our binomialdifferential form xm(axn+b)pdx becomes a rational differential form, i.e.a differential form of the type Q(u)du, where Q is a rational function ina new variable u. The reverse part of the theorem cannot be proved byelementary tools. It involves deep knowledge of Algebraic Geometry.

First of all let us make the natural substitution: xn = t (if n = 1; ifn = 1 we pass directly to the second step of the following reasoning).

Thus x = t1n and dx = 1

nt1n−1dt. Hence, our binomial integral becomes

(4.12)1

n

∫tq(at+ b)pdt,

the canonical form of the binomial integral. Here we denoted m+1n

− 1by q. In general, since q = m+1

n− 1 and p are rational numbers we

have in the expression tm+1n−1(at + b)p two radicals. If we had only

one radical, it could be easy to change the variable t with another oneu such that the new differential form becomes rational. For instance,in

∫t−4(3t + 4)−

13dt we have only one radical, so we "kill" it by the

obvious substitution 3t+ 4 = u3, or t = u3−43

and dt = u2du. Thus ourlast integral becomes

34

∫ (u3 − 4

)−4u−1u2du = 81

∫u

(u3 − 4)4du,

which is a primitive of a rational function. We decompose u(u3−4)4

into

simple fractions, etc.Let us come back to our general case of the primitive (from formula

(4.12)).Case 1. p is an integer.If q is also an integer, we have nothing to do, the integrand tq(at+b)p

being rational.


If q = m+1n

− 1 is not an integer, i.e. if m+1n

is not an integer, q =q1q2, simplified and q1, q2 integers with q2 = 1, then the obvious substitu-

tion t = uq2 makes the binomial differential form tq(at+ b)pdt rational.Indeed,

tq(at + b)pdt = uq1 (auq2 + b)p q2uq2−1du

is rational because q1, q2 and p are integers.Case 2. p is not an integer, but m+1

nis an integer. Let p = p1

p2,

simplified and p1, p2 integers, with p2 = 1. Then the substitutionat+ b = up2 give rise to a rational differential form. Indeed, this timeq = m+1

n− 1 is an integer, t = up2−b

aand dt =

p2

aup2−1du, so

tq(at+ b)pdt =

(up2 − b

a

)qup1

p2

aup2−1du

is rational because p1, p2 and q are integers.Case 3. p = p1

p2is not an integer and m+1

nis not an integer too.

In this case we make the following "trick" in the canonical differ-ential form:

(4.13) tq(at + b)pdt = tq+p(at+ b

t

)pdt.

We apply now the same idea like above.If q + p = m+1

n− 1 + p is an integer, i.e. if m+1

n+ p is an integer,

then the obvious substitution at+bt

= up2 , where p2 is the denominator

of p, leads to a rational differential form. Indeed, this time

t =b

up2 − a= b (up2 − a)−1 ,

dt = −bp2up2−1 (up2 − a)−2 du,

so

tq+p(at + b

t

)pdt = bq+p (up2 − a)−(q+p) up1 (−bp2)up2−1 (up2 − a)−2 du.

Since q + p, p1 and p2 are integers, then this last differential formis rational and the implication "⇐=" of the theorem is completelyproved.

R 4. The case 3 which naturally appeared during the proofof the above theorem 12 can also be manipulated in the following way.Instead of the "trick" used in formula (4.13) we can use the followingone:

(4.14) tq(at+ b)pdt =

(t

at + b

)q(at+ b)p+qdt.


Since q + p is an integer, and since q = q1q2is simplified with q2 = 1

we make the natural substitution tat+b

= uq2 in order to "kill" the onlyradical which appears in (4.14) (because q is not an integer!!). Now,

t = b (u−q2 − a)−1 and dt = bq2u−q2−1 (u−q2 − a)−2 du. Thus formula

(4.14) becomes(

t

at+ b

)q(at+b)p+qdt = uq1 (1 − auq2)−(q+p) bq2u

−q2−1(u−q2 − a

)−2du,

which is a rational differential form because q+p, q1 and q2 are integers.

We give now some examples in which we practically use the theoryexposed above.

E 16. Reduce to an integral of a rational function and thencompute the following integral:

∫3√x (2 + 3

√x)−2dx. Here m = 1

3, n =

12and p = −2. Thus we are in Case 1 of the Cebyshev theorem. Let us

make the canonical substitution

t = x12 or x = t2, dx = 2tdt.

So ∫3√x(2 + 3

√x)−2

dx = 2

∫t53 (2 + 3t)−2 dt.

Now we "kill" the radical t53 by making the new substitution t = u3.

Thus

2

∫t53 (2 + 3t)−2 dt = 6

∫u5(2 + 3u3)−3u2du

by parts=

−1

3

∫u5

[(2 + 3u3)−2

]′du =

−1

3

u5(2 + 3u3)−2 − 5

∫u4(2 + 3u3)−2du

=

−1

3

u5(2 + 3u3)−2 +

5

9

∫u2

[(2 + 3u3)−1

]′du

by parts

=

(4.15) −1

3

u5(2 + 3u3)−2 +

5

9

[u2(2 + 3u3)−1 − 2

∫u

2 + 3u3du

].

This last integral∫

u

2 + 3u3du =

1

3

∫u

23

+ u3du =

1

3

∫u

a3 + u3du,


where a = 3

√23, is an integral of a rational function. We can integrate it

by the usual methods of decomposing of the integrand ua3+u3

into simplefractions:

u

a3 + u3= − 1

3a

1

u+ a+

1

3a

u+ a

u2 − ua+ a2.

Hence,

1

3

∫u

a3 + u3du = − 1

9a

∫1

u+ adu+

1

18a

∫2u− a + 3a

u2 − ua+ a2du =

(4.16) − 1

9aln |u+ a| +

1

18aln

∣∣u2 − ua+ a2∣∣ +

1

6

∫1

u2 − ua+ a2du.

Let us evaluate this last integral:(4.17)∫

1

u2 − ua + a2du =

∫d(u− a

2)

(u− a

2

)2+

(a√

32

)2 =2

a√

3arctan

2u− a

a√

3.

Coming back with this result to formula (4.16) we get:

1

3

∫u

a3 + u3du = − 1

9aln |u+ a| +

1

18aln

∣∣u2 − ua+ a2∣∣+

+1

3a√

3arctan

2u− a

a√

3.

Now we go back to formula (4.15) and find:

2

∫t53 (2 + 3t)−2 dt = −1

3u5(2 + 3u3)−2+

+5

9[u2(2 + 3u3)−1 − 2

3− 1

9aln |u+ a| +

1

18aln

∣∣u2 − ua+ a2∣∣+

+1

3a√

3tan−1 2u− a

a√

3].

Now, instead of u let us put t13 and obtain:

2

∫t53 (2 + 3t)−2 dt = −1

3t 53 (2 + 3t)−2 +

5

9[t23 (2 + 3t)−1−

−2

3− 1

9aln

∣∣∣t 13 + a∣∣∣ +

1

18aln

∣∣∣t 23 − t13a + a2

∣∣∣ +1

3a√

3tan−1 2t

13 − a

a√

3].

But t = x12 , so our initial integral is:

∫3√x(2 + 3

√x)−2

dx = −1

3x 5

6 (2 + 3x12 )−2 +

5

9[x

13 (2 + 3x

12 )−1−


−2

3− 1

9aln

∣∣∣x 16 + a

∣∣∣+1

18aln

∣∣∣x 13 − x

16a + a2

∣∣∣+1

3a√

3tan−1 2x

16 − a

a√

3].

where a = 3

√23.

R 5. The last example 16 is a very particular example of amore general situation. Let x1, x2, ..., xn be n independent variables. Amonomial in these variables x1, x2, ..., xn is an expression of the form

ak1,k2,...,knxk11 x

k22 ...x

knn ,

where ak1,k2,...,kn is a real number and k1, k2, ..., kn are nonnegative inte-gers (they may be also zero). A finite sum of such expressions is calleda polynomial P = P (x1, x2, ..., xn) in the n variables x1, x2, ..., xn. Arational function

R = R(x1, x2, ..., xn)

in n variables is a quotient R = PQof two polynomials in n variables.

A rational expression of simple radicals is obtained from a rationalfunction

R(x1, x2, ..., xn) in n variables by substituting the variable xi with a

radical expression of the form(ax+bcx+d

)piqi , where x is our "old variable

of integration", a, b, c, d are real numbers with det

(a bc d

)= 0 and

pi, qi are nonzero natural numbers. Here i goes from 1 up to n. Such arational expression of simple radicals usually appears like:

(4.18) R

((ax+ b

cx+ d

)p1q1

,

(ax+ b

cx+ d

)p2q2

, ...,

(ax+ b

cx+ d

)pnqn

).

To integrate such a function of x we make the natural substitution

(4.19)ax+ b

cx+ d= tq,

where q is the least common multiple (lcm) of all the denominatorsq1, q2, ..., qn of the powers of

ax+bcx+d

. It is clear that the new obtaineddifferential form in t is a rational one. Indeed, starting from

∫R

((ax+ b

cx+ d

)p1q1

,

(ax+ b

cx+ d

) p2q2

, ...,

(ax+ b

cx+ d

)pnqn

)dx,

after substitution, we get:

q (ad− bc)

∫R (ts1, ts2 , ..., tsn)

tq−1

(ctq−1 − a)2dt,


because, from (4.19) x = tqd−ba−tqc and dx = q (ad− bc) tq−1

(ctq−1−a)2dt. Here

si = piqiq are natural numbers (why?). Thus, the obtained integral is an

integral of a rational function in the variable t.For instance, in our example 16 the common expression ax+b

cx+dun-

der radicals is simply x. Now, the rational function is R(x1, x2) =x1 (2 + 3x2)

−2 , p1q1

= 13, p2q2

= 12and q = lcm(3, 2) = 6. So the nat-

ural substitution is x = t6 and we get:∫

3√x(2 + 3

√x)−2

dx = 6

∫t2

(2 + 3t3

)−2t5dt,

etc.

E 17. Let us compute the integral I =∫x3 3√

1 + x2dx.Since m = 3, n = 2 and p = 1

3, p is not an integer and m+1

n= 2

is an integer, we are in Case 2. We can see that it is possible to putdirectly

(4.20) 1 + x2 = u3 , u =3√

1 + x2

(without making first of all x2 = t, why?). Thus,

2xdx = 3u2du =⇒ dx =3u2

2xdu.

Let us come back to our differential form x3 3√

1 + x2dx and performthese new substitutions:

x3 3√

1 + x2dx = x3u3u2

2xdu =

3

2x2u3du =

3

2(u3 − 1)u3du.

So∫

3

2(u3 − 1)u3du =

3

2

(u7

7− u4

4

)=

3

14

(1 + x2

) 73 − 3

8

(1 + x2

) 43 .

E 18. Let us compute now I =∫ √

1+x2

x4dx. This is a bino-

mial integral with m = −4, n = 2 and p = 12. Since p is not an integer,

m+1n

= −32is also not an integer, but m+1

n+p = −1 is an integer (Case

3), we can directly make the substitution 1+x2

x2= u2 (because t = x2 in

formula (4.13)). So

1 + x2 = u2x2, x2 =1

u2 − 1, 2xdx = 2x2udu+ 2u2xdx,

dx =x2u

x− u2xdu =

xu

1 − u2du.


Thus, our differential form becomes (we consider only the case x > 0and u ∈ (1,∞)):

√1 + x2

x4dx =

ux

x4

xu

1 − u2du =

u2

x2(1 − u2)du =

=u2

(1

u2−1

)(1 − u2)

du = −u2du.

Finally we get:∫ √

1 + x2

x4dx = −

∫u2du = −u

3

3= −1

3

(1 + x2

x2

) 32

.

The best idea is to make "formal" computations (not taking count ofthe definition domains of different expressions which appear during sub-stitutions!) and, in the end, to verify by a direct differentiation theobtained result. In our example,

[−1

3

(1 + x2

x2

) 32

]′=

√1 + x2

x4, x = 0.

Hence, in spite of the required limitations which appeared during a par-

ticular computation, a primitive of the differential form√

1+x2

x4dx, on

each interval which does not contain 0, is −13

(1+x2

x2

) 32

.

R 6. In general, a primitive of the form∫R(x,

√anxn + an−1xn−1 + ... + a0)dx,

where an = 0 and n > 2 is not an elementary function. By an ele-mentary function we understand a function which is a composition ofrational functions, trigonometric functions, exponential functions andlogarithm functions. If n = 3 our primitive is called an elliptic inte-gral. In general, if the integrand function is a rational expression whichcontains radicals of polynomials, then it is called an abelian integral (inmemory of the great Norway mathematician, Niels Abel). For instance,

∫x2

(2 + 5x3

) 35 dx =

1

15

∫ (2 + 5x3

) 35 d(2 + 5x3) =

=1

15

∫u35du =

1

15

u35+1

35

+ 1=

1

24

(2 + 5x3

) 85 ,

is an elementary function. But,∫ √

1 + x4dx is not an elementary

function. Indeed, let us make x4 = t. So, x = t14 , dx = 1

4t−

34dt and the


integral becomes:1

4

∫t−

34 (1 + t)

12 dt.

The last integral is a binomial integral with m = −34, n = 1 and p = 1

2.

Since p is not an integer, m+1n

= 14is also not an integer and m+1

n+ p

= 34is not an integer, we are in no one of the three cases of the Cebysev

theorem. So we cannot reduce this primitive to a rational primitive byrational substitutions. O. K., but the question is still alive! Is therean elementary differentiable function F (x) such that F ′(x) =

√1 + x4?

The answer is no, but we need a lot of higher Mathematics to prove it!

D. Primitives of the form∫R(cosx, sin x)dx, where R(x1, x2)

is a rational functionThe general method consists in the following change of variable.

Assume that x belongs to an interval J on which the function x→ tan x2

is invertible and its inverse is of class C1. Since

cos x =1 − tan2 x

2

1 + tan2 x2

, sinx =2 tan x

2

1 + tan2 x2

,

it is naturally to put t = tan x2, x = 2 arctan t, dx = 2

1+t2dt, so

∫R(cosx, sin x)dx =

∫R

(1 − t2

1 + t2,

2t

1 + t2

)2

1 + t2dt,

which is an integral of a rational function, etc.

E 19. Let us use this substitution to compute

I =

∫1

1 + sin x+ cosxdx.

I =

∫1

1 + 2t1+t2

+ 1−t21+t2

2

1 + t2dt =

∫1

t+ 1dt =

= ln |t+ 1| = ln∣∣∣tan

x

2+ 1

∣∣∣ .

R 7. Sometimes this substitution is not indicated because itleads to a very complicated computation.

If R(cosx, sin x) can be written as S(cosx) sin x or as T (sin x) cosx,where S(y) and T (y) are rational functions, then our integral becomeseither

(4.21)

∫R(cosx, sinx)dx =

∫S(cosx) sin xdx =

= −∫S(cosx)d(cosx) = −

∫S(u)du,


where u = cosx, or

(4.22)

∫R(cosx, sin x)dx =

∫T (sin x) cosxdx =

=

∫T (sin x)d(sin x) =

∫T (v)dv,

where v = sin x.

E 20. If we want to compute I =∫

sin4 x cos3 xdx by thegeneral substitution t = tan x

2, we get a very complicated integral of a

rational function in t (why is it so complicated?)

I =

∫ (2t

1 + t2

)4 (1 − t2

1 + t2

)32

1 + t2dt.

But, if we use the substitution described in (4.22), we get:

I =

∫sin4 x cos2 x cosxdx =

∫sin4 x(1 − sin2 x)d(sin x) =

=

∫v4(1 − v2)dv =

v5

5− v7

7=

sin5 x

5− sin7 x

7.

R 8. Suppose now that cosx and sin x appear to even powersin

R(cosx, sin x). Then always one can write R(cosx, sinx)dx as

S(tanx)1

cos2 xdx = S(tanx)d(tan x) = S(u)du,

where u = tan x and S is a rational function of u (why all of these?).The substitution u = tan x can also be done even in other cases. Forinstance,

∫sinxdx

sin3 x+cos3 xdx can be "rationalized" by puting tan x = u.

E 21. Let us compute I =∫

cos2 x1+sin2 x

dx by this last method

(the general substitution t = tan x2is not good at all!-why?). Since

1 + tan2 x = 1cos2 x

, and since u = tan x implies that x = tan−1 u anddx = 1

1+u2du, we get

∫cos2 x

1 + sin2 xdx =

∫1

1cos2 x

+ tan2 xdx =

∫1

(1 + 2u2) (1 + u2)du.

We must decompose into simple fractions the rational expression

1

(1 + 2u2) (1 + u2).


Do not you hurry to search for a decomposition of a general type:

1

(1 + 2u2) (1 + u2)=Au+B

1 + 2u2+Cu+D

1 + u2,

because we can think of this rational expression as a rational functionof t = u2, so we simply search for a decomposition of the following type:

1

(1 + 2u2) (1 + u2)=

A

1 + 2u2+

B

1 + u2.

So A = 2 and B = −1. Hence,∫

cos2 x

1 + sin2 xdx = 2

∫1

1 + 2u2du−

∫1

1 + u2du =

=√

2

∫1

1 +(√

2u)2d

(√2u

)−− arctan u =

=√

2 arctan(√

2u)− arctan u =

√2 arctan

(√2 tan x

)− x.

Sometimes it is useful to extend the computations with primitivesto functions of real variables but with values in the complex numberfield C. Let f(x) = f1(x) + if2(x), where i =

√−1 and f1, f2 : I → R,

are two continuous functions of real variable x. They are usually calledthe (real) components of f. So f : I → C is continuous (see Analysis I,[Po], sequences of complex numbers, etc.). Moreover, f is differentiableon I if and only if f1 and f2 are differentiable on I and then f ′ = f ′1+if ′2(see for instance Analysis I, [Po]). Thus

∫f(x)dx =

∫f1(x)dx+ i

∫f2(x)dx.

(why?). Let us compute for instance∫

exp(ix)dx =

∫cosxdx+ i

∫sin xdx = sinx− i cosx =

=1

i(cosx+ i sin x) =

1

iexp(ix).

We see from here that the usual rules for computing primitives of thereal valued functions extends naturally to complex valued functions ofreal variables. This is true because such extension works in the case ofthe differential calculus (see Analysis I, [Po]). So we can directly write

∫exp(ix)dx =

1

i

∫exp(ix)d(ix) =

1

i

∫exp udu =

1

iexpu =

=1

iexp(ix) = −i exp(ix).


(i is a constant complex number). These simple ideas are very helpfulin computing usual primitives of real valued functions.

E 22. Let a, b be two real numbers and letI(a, b) =

∫exp(ax) sin(bx)dx be a primitive of exp(ax) sin(bx). As-

sume that a, b = 0 (otherwise the integral is trivial!). We can computeI(a, b) by integrating by parts two times:

I(a, b) =

∫exp(ax)

[−1

bcos(bx)

]′dx =

by parts= −1

bexp(ax) cos(bx) +

a

b

∫exp(ax) cos(bx)dx =

−1

bexp(ax) cos(bx) +

a

b2

∫exp(ax) [sin(bx)]′ dx =

by parts= −1

bexp(ax) cos(bx) +

a

b2exp(ax) sin(bx) − aI(a, b) =

=a

bexp(ax)

[sin(bx)

b− cos(bx)

a

]− a2

b2I(a, b).

So (1 +

a2

b2

)I(a, b) =

a

bexp(ax)

[sin(bx)

b− cos(bx)

a

],

or

I(a, b) =ab

a2 + b2exp(ax)

[sin(bx)

b− cos(bx)

a

]=

=1

a2 + b2exp(ax) [a sin(bx) − b cos(bx)] .

Another smarter way to compute this integral is based on the aboveideas. Let J(a, b) =

∫exp(ax) cos(bx)dx and let the complex number

J(a, b) + iI(a, b) =

∫exp(ax) exp(ibx)dx =

∫exp [(a + bi)x] dx =

1

a+ biexp [(a + bi)x] =

a− bi

a2 + b2exp(ax) exp(ibx) =

exp(ax)

a2 + b2(a− ib) [cos(bx) + i sin(bx)] =

exp(ax)

a2 + b2[a cos(bx) + b sin(bx)] + i

exp(ax)

a2 + b2[a sin(bx) − b cos(bx)] .

Thus

J(a, b) =exp(ax)

a2 + b2[a cos(bx) + b sin(bx)]


and

I(a, b) =exp(ax)

a2 + b2[a sin(bx) − b cos(bx)] .

R 9. A way to compute primitives of the form∫R(cosmx, sinnx)dx,

where m,n are natural nonzero numbers is to use Euler’s formulas:

(4.23) cosmx =exp(imx) + exp(−imx)

2,

(4.24) sinnx =exp(inx) − exp(−inx)

2i

and to reduce the computation to a primitive of a complex valued func-tion (of a real variable).

E 23. Let us compute a primitive for the trigonometricdifferential form sin 3x cos4 xdx.

Since sin 3x = exp(3ix)−exp(−3ix)2i

and since

cos2 x =1 + cos 2x

2=

1

2+

exp(2ix) + exp(−2ix)

4=

=1

4[2 + exp(2ix) + exp(−2ix)] ,

cos4 x =1

16[6 + exp(4ix) + exp(−4ix) + 4 exp(2ix) + 4 exp(−2ix)] ,

one has ∫sin 3x cos4 xdx =

1

32i

∫[exp(3ix) − exp(−3ix)]×

× [6 + exp(4ix) + exp(−4ix) + 4 exp(2ix) + 4 exp(−2ix)] dx =

=1

32i

∫[6 exp(3ix) + exp(7ix) + exp(−ix) + 4 exp(5ix) + +4 exp(ix)−

−6 exp(−3ix)− exp(ix)− exp(−7ix)− 4 exp(−ix)− 4 exp(−5ix)]dx =

=1

16

∫[6 sin(3x) + sin(7x) + 3 sin x+ 4 sin 5x]dx =

−1

8cos 3x− 1

112cos 7x− 3

16cosx− 1

20cos 5x.


E 2. Let In =∫

sinn xdx, where n > 2, and n = 2k is even(k > 1). If n is odd it is easier to write

sinn xdx = − sinn−1 x d(cosx) = −(1 − u2)n−12 du,

etc. Let us compute In by firstly finding a recurrence formula:

I2k =

∫sin2k xdx =

∫ (sin2k−2 x

)(1 − cos2 x)dx =

I2k−2 −∫

sin2k−2 x cosxd(sin x)by parts

=

= I2k−2 − [sin2k−1 x cosx−

−∫

sinx

(2k − 2) sin2k−3 x cos2 x− sin2k−1 xdx] =

I2k−2 − sin2k−1 x cosx+ (2k − 2)

∫sin2k−2 x(1 − sin2 x)dx− I2k =

= (2k − 1)I2k−2 − (2k − 1)I2k − sin2k−1 x cosx,

or

(4.25) I2k =2k − 1

2kI2k−2 −

1

2ksin2k−1 x cosx, k ≥ 1,

because it is easy to see that this last formula works even for k = 1.

E 24. For instance, since

I2 =

∫sin2 xdx =

1

2

∫(1 − cos 2x)dx =

1

2

(x− sin 2x

2

)=

1

2(x− sin x cosx) ,

one has from (4.25) that

(4.26)

∫sin4 xdx =

3

4

[1

2(x− sinx cosx)

]− 1

4sin3 x cosx.

Another way to compute such an integral is to express sinn x as apolynomial of different powers of exp(ix) or exp(−ix) :

sinn x =

[exp(ix) − exp(−ix)

2i

]n=

=1

2nin[exp(inx) −

(n

1

)exp i(n− 2)x+

+

(n

2

)exp i(n− 4)x− ...].


So ∫sinn xdx =

=1

2nin[

∫exp(inx)dx−

(n

1

)∫exp i(n− 2)xdx+

+

(n

2

)∫exp i(n− 4)xdx− ...].

For instance,

∫sin4 xdx =

1

16

[∫exp(4ix)dx− 4

∫exp(2ix)dx

]+

+6

16

[∫dx− 4

∫exp(−2ix)dx+

∫exp(−4ix)dx

]=

1

16

[exp(4ix)

4i − 4exp(2ix)

2i︸︷︷︸+6x− 4

exp(−2ix)

−2i︸︷︷︸+

exp(−4ix)−4i

]=

1

16

[1

2sin 4x− 4 sin 2x+ 6x

].

An elementary trigonometric computation tells us that this expressionis exactly that one from (4.26). Moreover, it is more beautiful, becauseit has no powers of trigonometric functions! These last ones are "notdesirable" during the integration computation.

R 10. It is very useful to know that the following primitivesare not elementary functions:

∫expxxdx;

∫sinxxdx;

∫cos xxdx;

∫sinhxxdx;∫

coshxxdx;

∫sin x2dx;

∫cos x2dx;

∫exp(−x2)dx;

∫1

lnxdx;

∫dϕ√

1−k2 sin2 ϕ;

∫ √1 − k2 sin2 ϕdϕ;

∫dϕ

(1+h sin2 ϕ)√

1−k2 sin2 ϕ, where k, h ∈ (−1, 1) \

0 (elliptic integrals). Proofs for such statements belong to very highMathematics. It implies deep knowledge of Algebraic Geometry and Al-gebraic Functions Theory. We note that not all the above statementsare "independent" one to each other". For instance, if

∫expxxdx is not

an elementary function, then∫

1lnxdx is also not an elementary func-

tion. Indeed, let F (x) =∫

1lnxdx, a primitive of the differential form

1lnxdx. Let us make the substitution x = exp(t) in this last differential

form:1

ln xdx =

exp t

tdt.

So a primitive of exp ttdt is F (exp(t)) on any interval I which does

not contain 0 (prove it!). But, if F were elementary, then F (exp(t)

5. PROBLEMS AND EXERCISES 47

would also be elementary (as a composition between two elementaryfunctions), a contradiction (explain everything slowly!). Try to findother pairs of "dependent" primitives in the above row of primitives!

Remark 10 is very useful in practice. For instance, if in the state-ment of an exercise there is a mistake: instead of

∫1

x lnxdx, x > 0, one

omitted "an x” and it appears∫

1lnxdx, you may spend a lot of time "to

compute" this last primitive!! All this effort is for nothing! Because"to compute" usually means to work only with elementary functions!Indeed, the primitive

∫1

lnxdx is not an elementary function (see remark

10), while∫1

x ln xdx =

∫1

ln xd(lnx) =

∫1

udu = ln |u| = ln |ln x| ,

is an elementary one!

E 25. We know that a canonical parametrization of the

ellipse x2

a2+ y2

b2= 1, a, b > 0, and a < b, is the following

x = x(t) = a cos t, y = y(t) = b sin t, t ∈ [0, 2π).

To compute the length of an arc of this ellipse, one reaches the followingprimitive (see Chapter 2, Section 8):∫ √

a2 sin2 t + b2 cos2 tdt =

∫ √a2 sin2 t + b2(1 − sin2 t)dt =

=

∫ √b2 − (b2 − a2) sin2 tdt = b

∫ √1 − k2 sin2 tdt,

where k =√b2−a2b

. This last primitive is not an elementary function(see remark 10). It is called an "elliptic integral" because of this aboveproblem which generated such a primitive.

5. Problems and exercises

1. Prove the following formulas and indicate the maximum defi-nition domains of their existence (recall that by

∫f(x)dx we mean a

primitive of the differential form f(x)dx):a)

∫1

x2−a2dx = 12a

ln∣∣x−ax+a

∣∣ , a = 0.

b)∫

1x2+a2

dx = 1a

arctan xa, a = 0.

c)∫

1√x2+a

dx = ln(x+√x2 + a), a = 0.

d)∫

1√a2−x2dx = arcsin x

a, a = 0.

e)∫ √

a2 − x2dx = x2

√a2 − x2 + a2

2arcsin x

a, a = 0.

f)∫ √

x2 ± a2dx = x2

√x2 ± a2 ± a2

2ln

(x+

√x2 ± a2

), a = 0.

g)∫

1sin2 x

dx = − cot x;∫

1cos2 x

dx = tanx.


h)∫

tan xdx = − ln |cosx| .i)

∫cotxdx = ln |sin x| .

j)∫

1sinx

dx = ln∣∣tan x

2

∣∣ .k)

∫1

cosxdx = ln

∣∣tan(π4− x

2

)∣∣l)

∫1

cosh2 xdx = sinhx

coshx(= tanhx).

m)∫

1sinh2 x

dx = −coshxsinhx

(= − cothx).

n)∫|x| dx = sign(x)x

2

2, where sign(x) =

1, if x > 00, if x = 0−1, if x < 0

.

2. Use basic formulas, basic properties and basic methods to com-pute (where your result works on?):

a)∫x4+ 3√x+3x2−1√

xdx; b)

∫x3+x2+2x+1

x−1dx; c)

∫n√

3 − 2xdx, n = 2, 3, ...;

d)∫

x+1√x2+1

dx; e)∫

1√7−5x2

dx; f)∫

32x−13x

dx; g)∫

xcos2(5x)

dx;

h)∫

5−3x√4−3x2

dx; i)∫

1

x(4−ln2 x)dx; j)

∫1

1+cos2 xdx; k)

∫ exp(2x)exp(4x)+5

dx;

l)∫

1√x(1+ 3√x)

dx; m)∫

1A2 sin2 x+B2 cos2 x

dx, where AB = 0;

n)∫

1x2√x2+a2

dx, a = 0; o)∫

1

(a2−x2) 32dx; p)

∫1

1+ 3√1+xdx; q)

∫1

x√

1+x2dx;

r)B =∫x2 exp(x) sin xdx; Hint: consider A =

∫x2 exp(x) cosxdx

and compute by parts A+ iB.s)

∫exp(ax) sin(bx)dx;

3. Prove that∫P (x) exp(ax)dx = exp(ax)

[Pa− P ′

a2+ P ′′

a3− ...

],

where a = 0 and P is a polynomial. Use this formula to compute∫x6 exp(5x)dx. Find similar formulas for computing

∫P (x) sin(ax)dx

and∫P (x) sin(ax)dx. Use them to find

∫(x6 + x5 + 1) sin 7xdx.

4. Justify the formula:∫u(x) [v(x)](n+1) dx = u(x) [v(x)](n) − u′(x) [v(x)](n−1) + ...+

+(−1)n+1

∫[u(x)](n+1) v(x)dx.

Use it to compute∫

(x5 + x+ 1) exp(2x)dx.

5. Are the primitives∫

ln2 x√x5dx, x > 0 and

∫ x2 exp(x)(x+2)2

dx elementary

functions? If yes, compute them!6. Use the recurrence formula for In =

∫dx

(x2+a2)nto compute∫

dx(x2+2x+5)3

.

CHAPTER 2

Definite integrals

1. The mass of a linear bar

By a linear bar we mean a closed interval [a, b] and a nonnegativecontinuous function f : [a, b] → R+. We call this last function f thedensity function of the bar. Let us divide the interval [a, b] into n subin-tervals [xi−1, xi], where a = x0 < x1 < x2 < ... < xn = b, i = 1, 2, ..., n.This last sequence xii of real numbers is called a division ∆ of theinterval [a, b]. The greatest length of all of the segments [xi−1, xi],i = 1, 2, ..., n is said to be the norm ‖∆‖ of the division ∆. Let usdenote by

(1.1) ωf [xi−1, xi] = supx′,x′′∈[xi−1,xi]

|f(x′) − f(x′′)| ,

the variation of f over the interval [xi−1, xi], i.e. the length of the inter-val f([xi−1, xi]) if f is continuous. Recall that a function f : [a, b] → Ris uniformly continuous if for any small real number ε > 0 there is a realnumber δε > 0 such that whenever x′, x′′ are in [a, b] and |x′ − x′′| < δε,one has that |f(x′) − f(x′′)| < ε. It is clear enough that a uniformlycontinuous function is also a continuous one (why?). Since our con-tinuous function f is also uniformly continuous on [a, b] (see [Po], Th.59), for any small number ε > 0, one can find a δε > 0, such that forany division

∆ : a = x0 < x1 < x2 < ... < xn = b

with ‖∆‖ < δε, the variation ωf [xi−1, xi] is less then ε. This means thatone can well approximate the density function f on the subinterval[xi−1, xi] with any fixed value f(ξi) of f at a fixed point ξi ∈ [xi−1, xi].The set of the fixed points ξi, ξi ∈ [xi−1, xi], i = 1, 2, ..., n is calleda set of marking points for the division ∆. Thus, the mass of the barcan be well approximated by the sum:

(1.2) S(f ; ∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1).

Such a sum S(f ; ∆; ξi) is called the Riemann sum associated tothe function f : [a, b] → R, the division ∆ and to the set of marking

49

50 2. DEFINITE INTEGRALS

points ξi.We also can interpret the Riemann sum S(f ; ∆; ξi as thesum of the hatched areas of rectangles in Fig.1.

F 1.

Thus, these sums can well approximate the area of the trapezoidaABb (see Fig.1) If the set of all these Riemann sums (when ∆ and ξivary such that ‖∆‖ → 0) have one and only one limit point I(f), we saythat this number I(f) is exactly the mass of the bar (think of this intu-itively!). Here it was an example which justify the following definition.We recall that a limit point for a nonempty subset A of R is a pointa of R such that for any small real number ε > 0 the ε-neighborhood(a − ε, a + ε) contains an infinite number of elements which are in A.For instance, 0 is a limit point of the set A = 1, 1/2, 1/3, ...1/n, ....

D 3. We say that a function f : [a, b] → R is (Riemann)integrable on [a, b] if there exist a real number I(f), called the definite

integral of f from a to b and denoted by∫ baf(x)dx, such that for any

ε > 0 there is a δε > 0 with the following property: if ∆ is a divisionof [a, b] with ‖∆‖ < δε, then |S(f ; ∆; ξi) − I(f)| < ε for any set ofmarking points ξi of the division ∆.

The above defined number I(f) =∫ baf(x)dx can be well approxi-

mated with Riemann sums of type (1.2). It is easy to see that such anumber I(f) is unique (do it!). The above definition says that if welook at a Riemann sum as a function of the division ∆ and of ξi, aset of marking points of ∆, this last defined function "has a limit" as‖∆‖ → 0 if and only if f is Riemann integrable on [a, b].

1. THE MASS OF A LINEAR BAR 51

E 26. Let k be a constant number and let f : [a, b] → Rbe the constant function: f(x) = k for any x ∈ [a, b]. For a division∆ : a = x0 < x1 < x2 < ... < xn = b and for a set of marking pointsξi of it, the corresponding Riemann sum is

S(f ; ∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1) = kn∑

i=1

(xi − xi−1) = k(b− a),

thus all the Riemann sums are equal to the constant number k(b − a),i.e. f is integrable on [a, b] and its integral I(f) = k(b− a).

Even for not complicated functions, to prove the integrability andto compute directly I(f), usually is not an easy task.

E 27. For instance, let f(x) = x, x ∈ [0, 1]. Then

S(f ; ∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1) =n∑

i=1

ξi(xi − xi−1).

Since ξi ∈ [xi−1, xi], i = 1, 2, ..., n, we can writen∑

i=1

ξi(xi − xi−1) =n∑

i=1

(ξi −

xi + xi−1

2

)(xi − xi−1)+

+n∑

i=1

xi + xi−1

2(xi − xi−1)

Sincen∑

i=1

xi + xi−1

2(xi − xi−1) =

1

2

and since for any fixed ε > 0 and for any division ∆ with ‖∆‖ < ε∣∣∣∣∣n∑

i=1

(ξi −

xi + xi−1

2

)(xi − xi−1)

∣∣∣∣∣ < ε,

one sees that ∣∣∣∣∣n∑

i=1

ξi(xi − xi−1) −1

2

∣∣∣∣∣ < ε

for any division ∆ with ‖∆‖ < ε. This means (see the definition above)

that∫ 1

0xdx = 1

2.

The main problems we are concerned with are: 1) When a functionf : [a, b] → R is Riemann integrable? and 2) If f is integrable, how

do we compute its definite integral∫ baf(x)dx? Let us begin with the

following remark.


T 13. Let f : [a, b] → R be a Riemann integrable function.Then it is bounded, i.e. ‖f‖ = sup

x∈[a,b]|f(x)| is a finite number.

P. Let I(f) =∫ baf(x)dx and let ε > 0 be a small number,

ε = 1/10 for instance. Let ∆ : a = x0 < x1 < x2 < ... < xn = b be adivision of [a, b] such that

(1.3) |I(f) − S(f ; ∆; ξi)| < 1/10,

for any set of marking points ξi of ∆. Suppose that f is unboundedto ∞. Then f is unbounded at least on a subinterval [xj−1, xj]. Let

us change the marking point ξj with ξ(1)j , ξ

(2)j , ..., ξ

(k)j , ... such that

f(ξ(k)j ) → ∞, when k → ∞. But

S(f ; ∆; ξi) =

[n∑

i=1,i=jf(ξi)(xi − xi−1)

]+ f(ξ

(k)j )(xj − xj−1) → ∞,

which contradicts the inequality (1.3). Thus f must be bounded.

2. Darboux sums and their applications

D 4. Let f : [a, b] → R be a bounded function and let∆ : a = x0 < x1 < x2 < ... < xn = b be a division of [a, b]. Letm = inf

x∈[a,b]f(x), the least value of f on [a, b], M = sup

x∈[a,b]f(x), the

greatest value of f on [a, b], mi = infx∈[xi−1,xi]

f(x), the least value of f on

[xi−1, xi] and Mi = supx∈[xi−1,xi]

f(x), the greatest value of f on [xi−1, xi],

i = 1, 2, ..., n. The following sums

(2.1) s∆(f) =n∑

i=1

mi(xi − xi−1), S∆(f) =n∑

i=1

Mi(xi − xi−1)

are called the inferior and respectively the superior Darboux sums as-sociated to function f and to division ∆.

Now, s∆(f) and S∆(f) can be interpreted as the sum of the hatchedareas of rectangles in Fig.2 and respectively Fig.3.

It is clear enough that

(2.2) m(b− a) ≤ s∆ ≤ S(f ; ∆; ξi) ≤ S∆ ≤M(b− a),

for any set of marking points ξi, ξi ∈ [xi−1, xi], i = 1, 2, ..., n (see alsoFig.4).

Since the set or real numbers s∆, when ∆ runs on the set ofall divisions of [a, b], is upper bounded by M(b − a) (see (2.2)), there

2. DARBOUX SUMS AND THEIR APPLICATIONS 53

F 2.

F 3.

is the least upper bound I∗(f) of s∆. This last number is called theinferior Darboux integral of f. Since the set or real numbers S∆, when∆ runs on the set of all divisions of [a, b], is lower bounded by m(b−a)(see (2.2)), there is the greatest lower bound I∗(f) of S∆. This lastnumber is called the superior Darboux integral of f. There are examplesfor which these two Darboux integrals are not equal. For instance, letf : [a, b] → R be the mapping which associates to a rational numberof [a, b] the value 0 and to an irrational number of [a, b] the value 1.Then, I∗(f) = 0 and I∗(f) = 1. To make things clearer, let us consider


F 4.

two divisions ∆ and ∆′ of [a, b] such that ∆′ contains the points of ∆and maybe some additional other points. We write this as: ∆ ≺ ∆′.Adding successively to ∆ only one point c at a time, we can easilyprove (do it using a picture!-see Fig.5) that:

(2.3) s∆ ≤ s∆′ ≤ S∆′ ≤ S∆.

F 5

Taking now two arbitrary divisions ∆′ and ∆′′ of [a, b], we canconstruct the "least" division ∆′′′ = ∆′ ∪ ∆′′ which contains ∆′ and∆′′ at the same time: ∆′ ≺ ∆′′′ and ∆′′ ≺ ∆′′′. Thus (2.3) becomes:

s∆′ ≤ s∆′′′ ≤ S∆′′′ ≤ S∆′′,


so s∆′ ≤ S∆′′ for any two arbitrary divisions ∆′ and ∆′′ of [a, b].Thus the inequality I∗(f) ≤ I∗(f) is clear! Moreover, since mi =

infx∈[xi−1,xi]

f(x), keeping the division ∆ fixed, one can find a sequence

ξ(n)i of numbers in [xi−1,xi] such that f(ξ(n)i ) → mi as n→ ∞. Thus,

as close as we want to s∆ one can find Riemann sums of the formS(f ; ∆; ξ(n)i ) for a fixed ∆. The same is also true for S∆. Let us assume

now that f : [a, b] → R is Riemann integrable with I(f) =∫ baf(x)dx

and let ε > 0 be a small real number. Let δε > 0 be small enough suchthat if ‖∆‖ < δε then

(2.4) |I(f) − S(f ; ∆; ξi)| < ε

4for any marking points ξi of the division ∆. Take now two sets ofmarking points ξ′i and ξ′′i with the property that:

(2.5) |S∆ − S(f ; ∆; ξ′i)| < ε

4and |S(f ; ∆; ξ′′i ) − s∆| <

ε

4.

So, one can write:

S∆ − s∆ ≤ |S∆ − S(f ; ∆; ξ′i)| + |S(f ; ∆; ξ′i) − I(f)|++ |I(f) − S(f ; ∆; ξ′′i )| + |S(f ; ∆; ξ′′i ) − s∆| <

ε

4+ε

4+ε

4+ε

4= ε.

Thus S∆ − s∆ < ε for any division ∆ with ‖∆‖ < δε. We just provedan implication of the following basic result.

T 14. (Darboux Criterion) f : [a, b] → R is Riemann in-tegrable if and only if I∗(f) = I∗(f), i.e. if and only if for any smallreal number ε > 0, there is another small real number δε > 0 such thatif ∆ is any division with ‖∆‖ < δε, one has that S∆ − s∆ < ε.

P. It remains to prove the converse. We denote the commonvalue I∗(f) = I∗(f) by I(f) and let us prove that I(f) is the uniquelimit point of all the Riemann sums S(f ; ∆; ξi), when ‖∆‖ → 0. Letagain ε > 0 be a small real number and let δε > 0 be the correspondingsmall real number such that if ∆ is any division with ‖∆‖ < δε, onehas that S∆ − s∆ < ε. Since s∆ ≤ S(f ; ∆; ξi) ≤ S∆ (see 2.2) andsince s∆ ≤ I∗(f) = I∗(f) = I(f) ≤ S∆ one obtains that

|S(f ; ∆; ξi) − I(f)| ≤ S∆ − s∆ < ε

for any division ∆ with ‖∆‖ < δε and for any set of marking points(ξi) of ∆.

R 11. (Riemann Criterion) It is not difficult to prove thatf : [a, b] → R is Riemann integrable if and only if there is a realnumber I(f) such that for any sequence ∆n of divisions of [a, b] with


‖∆n‖ → 0 as n → ∞ and for any set of marking points ξ(n)i of ∆n,

n = 1, 2, ..., one has that S(f ; ∆n; ξ(n)i ) → I(f), when n→ ∞. Thus,if we know that f is integrable on [a, b], we can use some special typesof divisions, for instance the equidistant divisions, i.e. those divisions∆n : a = x0 < x1 < x2 < ... < xn = b for which xi − xi−1 = b−a

nfor

any i = 1, 2, ..., n. Here we see that xi = a + i b−anso, ∆n depends only

of n. If n < m, then ∆n ≺ ∆m and ‖∆n‖ = b−an

→ 0, when n → ∞.

Thus, one can find I(f) as the limit of a sequence S(f ; ∆n; ξ(n)i )for some particular values of the marking points. For instance, one can

take ξ(n)i = 1

2(xi−1 + xi) = a + 2i−1

2b−anand the approximation:

∫ b

a

f(x)dx ≈ R(f ;n)def= S(f ; ∆n; ξ(n)i ) =(2.6)

=b− a

n

n∑

i=1

f

(a +

2i− 1

2

b− a

n

)(2.7)

for n large enough, is called "the rectangle method" (explain why?, bydrawing all...).

Darboux Criterion is analogous to Cauchy Criterion for numeri-cal sequences. Since we usually cannot guess in advance the value of∫ baf(x)dx, or we cannot exactly compute it, at least we need to know

if it exists. To prove its existence Darboux Criterion is needed. Thenwe can approximate it by different methods. One cannot approximatesomething on which we are not sure that it exists! Let us now apply itin some particular but basic situations.

T 15. Let f : [a, b] → R be a monotonous (increasing ordecreasing) function. Then f is Riemann integrable on [a, b].

P. Suppose that f is a decreasing function and that it is not

a constant function (we saw that∫ bakdx = k(b − a) in example 26).

Let us use Darboux Criterion (see theorem 14). Since Mi = f(xi−1)and mi = f(xi) (f is decreasing) one has that

S∆ − s∆ =n∑

i=1

[f(xi−1) − f(xi)] (xi − xi−1) ≤(2.8)

‖∆‖n∑

i=1

[f(xi−1) − f(xi)] = ‖∆‖ (f(a) − f(b)) .(2.9)

For a small real number ε > 0 it is sufficient to take δε = εf(a)−f(b) .

Here f(a) = f(b) because f was supposed not to be a constant function.


Indeed, if ‖∆‖ < δε = εf(a)−f(b) , then in (2.8) we get: S∆ − s∆ <

εf(a)−f(b) (f(a) − f(b)) = ε and Darboux Criterion works.

T 16. Let f : [a, b] → R be a continuous function. Then fis Riemann integrable.

P. We again apply Darboux Criterion. Let ε > 0 be a smallreal number and let δε > 0 be another small real number (which de-pends on ε) such that if |x′ − x′′| < δε, one has that |f(x′) − f(x′′)| <εb−a (f is uniformly continuous, see [Po], Th. 59). Let us take a di-vision ∆ : a = x0 < x1 < x2 < ... < xn = b with ‖∆‖ < δε. ThenMi−mi = f(x′i)− f(x′′i ) <

εb−a for x′i, x

′′i ∈ [xi−1, xi], i = 1, 2, ..., n (see

[Po], Th. 32). Hence,

S∆ − s∆ =n∑

i=1

(Mi −mi)(xi − xi−1) <

<ε

b− a(x1 − a+ x2 − x1 + x3 − x2 + ...+ b) = ε.

i.e. f is Riemann integrable on [a, b].

R 12. a) Let f : [a, b] → R be such that f is continuous on[a, b) and f has the limit y0 = lim

xրb,x =bf(x) at b, y0 = f(b), i.e. f is not

continuous at b. Let f : [a, b] → R,

f(x) =

f(x), x ∈ [a, b)y0, x = b

.

It is easy to see that the Riemann sums of f and of f respectively arethe same with the only exception when ξn = b. In this last case, thecontributions of this marking point to both sums are f(b)(b−xn−1) andy0(b − xn−1) respectively. But, when ‖∆‖ → 0, b − xn−1 → 0, thus

both integrals exists or not simultaneously and∫ baf(x)dx =

∫ baf(x)dx.

The same is true if f is not continuous at a, but it has finite side limiton the right at a. Moreover, if c ∈ (a, b) and if f is continuous on[a, c) ∪ (c, b], but it is not continuous at c, but having finite side limitsat c then, using extensions like above for f |[a,c) (f restricted to theinterval [a, c)) and for f |(c,b] respectively, we obtain that both integrals∫ caf(x)dx and

∫ bcf(x)dx exist. Moreover, it is not difficult to see that∫ b

af(x)dx exists and

(2.10)

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.


In particular, when f is continuous on [a, b], then this last relationbecomes:

(2.11)

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

Combining all these above observations, we can recursively provethat if c1, c2, ..., cN , a ≤ c1 < c2 < ... < cN ≤ b are the unique N pointsof discontinuity for f at which side limits exist and are finite, then onehas:

(2.12)

∫ b

a

f(x)dx =

∫ c1

a

f(x)dx+

∫ c2

c1

f(x)dx+ ...+

∫ b

cN

f(x)dx.

Here we clearly put∫ aaf(x)dx = 0 if such situation appears. The func-

tion f is the extension of f at the ends of the corresponding intervalsby the values of the side limits of f . Moreover, it is easy to prove fromhere that two continuous (but a finite number of points eventually, atwhich they have finite side limits) functions f and g, which differs atmost on a finite set of points in [a, b], are or not simultaneously inte-grable and

(2.13)

∫ b

a

f(x)dx =

∫ b

a

g(x)dx.

For instance,

g(x) =

x, x = 1

20, x = 1

2

, g : [0, 1] → R,

is integrable and ∫ 1

0

g(x)dx =

∫ 1

0

xdx =1

2,

as we just proved in example 27. The morale is that if the function isbounded, we do not care with it at a finite number of points from [a, b],when we are interested in the integration process of this function. Forinstance, if

f(x) =

1, x ∈ [0, 1)2 x ∈ [1, 2]

.

Since∫ 2

0f(x)dx =

∫ 1

0dx+

∫ 2

12dx = 1 + 2 = 3, (see example 26) we do

not care of the value of the function in its discontinuity point x = 1.

We can give a more general frame for all of the questions discussedabove, by introducing a new basic notion.

3. LEBESGUE CRITERION AND ITS APPLICATIONS 59

3. Lebesgue criterion and its applications

D 5. We say that a subset A of R has Lebesgue measure(L-measure) zero (we write L(A) = 0) if it can be covered by a finite ora countable set of intervals In, n = 1, 2, ..., with sum of their lengths∑∞n=1 l(In) as small as we want. This means that for any small real

number ε > 0, there is a set (finite or countable) of intervals In suchthat A ⊂ ∪∞n=1In and

∑∞n=1 l(In) < ε.

E 28. For instance, if A = a, a point, has the L-measurezero. Indeed, let ε > 0 be a small real number and let the intervalI = (a − ε/4, a + ε/4). Since a ∈ I and l(I) = ε/2 < ε, we getthat the Lebesgue measure of A is zero, i.e. L(A) = 0. It is easyto see that if B,C are two sets with B ⊂ C and L(C) = 0, thenL(B) = 0. Moreover, if A1, A2, ..., At have Lebesgue measures zero,then L (∪ti=1Ai) = 0 (prove it!). What happens if instead of a finite setof L-measure zero sets we have a countable set of L-measure zero sets?Is their union again a L-measure zero set?

D 6. A mathematical object (a function for instance) hasa property "P" almost everywhere (a.e.) on a subset D of R if thesubset A ⊂ D of all the points at which this property "P" fails hasLebesgue measure L(A) = 0.

For instance, if D is a domain in R, a function f : D → R is a.e.continuous on D if the set A of discontinuity points of f has Lebesguemeasure zero. For instance, if A is a finite set of points.

A famous criterion was proved by a French mathematician, HenriLebesgue (1875-1941).

T 17. (Lebesgue criterion) Let f : [a, b] → R be a function.Then f is Riemann integrable if and only if it is bounded and the subsetA of [a, b] of all the points at which f is discontinuous has the Lebesguemeasure zero, i.e. if and only if it is bounded and a.e. continuous on[a, b].

The proof of this result is very technical and it cannot be givenhere. One can find it for instance in [Nik], §12.10. We shall apply thisbasic theorem in order to prove some other properties of the definiteintegral.

T 18. Let f : [a, b] → R be a bounded piecewise continuousfunction (it is continuous but a finite number of points at which it hasfinite sided limits). Then it is integrable.


P. Using theorem 17 it is enough to see that a set of a finitenumber of points has Lebesgue measure zero. But this last comes easilyfrom example 28.

T 19. Let f : [a, b] → R be a Riemann integrable functionand let g : [a, b] → R be a bounded function which is equal to f almost

everywhere. Then g is also integrable and∫ baf(x)dx =

∫ bag(x)dx. In

particular, if f is equal to zero almost everywhere, then f is integrable

and∫ baf(x)dx = 0.

P. Let A be the subset of [a, b] on which f is not continuous,let B be the subset of [a, b] on which g is not continuous and let C bethe subset of those x ∈ [a, b] at which f(x) = g(x). We can see thatthe discontinuity points of g are contained in A ∪ C and the Lebesguemeasure of this last union is zero (see example 28). So g is integrable(see theorem 17).To prove the second statement, it is enough to see

that both real numbers I =∫ baf(x)dx and J =

∫ bag(x)dx can be ap-

proximated by the same type of Riemann sums. In any ε-neighborhoodof I we can find a number of the form (a Riemann sum):

S(f ; ∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1),

where we can choose ξi arbitrary in [xi−1, xi]. Since this last intervalcannot have the Lebesgue measure equal to zero (why?), it cannot becontained in C, so there is at least a ξi ∈ [xi−1, xi] such that ξi /∈C. Thus, f(ξi) = g(ξi) and S(f ; ∆; ξi) = S(g; ∆; ξi). If ‖∆‖ issufficiently small, S(g; ∆; ξi) belongs to the ε-neighborhood of J. Sothe intersection (I−ε, I+ε)∩ (J−ε, J+ε) is not empty for any ε > 0.Hence I = J.

T 20. Let f : [a, b] → R be a Riemann integrable functionand let |f | : [a, b] → R be the function defined by |f | (x) = |f(x)| (theabsolute value of f). Then |f | is Riemann integrable and

∣∣∣∫ baf(x)dx

∣∣∣ ≤∫ ba|f(x)| dx.P. Since f is Riemann integrable, then f is bounded (see

theorem 13). So there isM > 0 such that |f(x)| ≤M for any x ∈ [a, b].But this also means that |f | is bounded. Let x0 be a continuity pointof f. Then x0 is a continuity point for |f | . Indeed, if xn → x0, thenf(xn) → f(x0) and, since ||f(xn)| − |f(x0)|| ≤ |f(xn) − f(x0)| (proveit!), one has that |f(xn)| → |f(x0)| , i.e. x0 is also a continuity point for|f | . Hence, if z is a discontinuity point of |f |, it cannot be a continuity

3. LEBESGUE CRITERION AND ITS APPLICATIONS 61

point for f. Finally we get that the set A of the discontinuity points of|f | is contained in the set B of the discontinuity points of f. Since f isRiemann integrable, the L-measure of B is zero. Thus the L-measureof A is also zero (see example 28).

Let now ∆n be a sequence of divisions of [a, b], ∆n : a < x(n)1 <

x(n)2 < ... < x

(n)kn

= b, such that ‖∆n‖ → 0. Then S(f ; ∆n; ξ(n)i ) →I(f) and S(|f | ; ∆n; ξ(n)i ) → I(|f |). But

∣∣∣S(f ; ∆n; ξ(n)i )∣∣∣ =

∣∣∣∣∣n∑

i=1

f(ξ(n)i )(x

(n)i − x

(n)i−1)

∣∣∣∣∣ ≤

≤n∑

i=1

∣∣∣f(ξ(n)i )

∣∣∣ (x(n)i − x

(n)i−1) = S(|f | ; ∆n; ξ(n)i ).

So |I(f)| ≤ I(|f |), i.e.∣∣∣∫ baf(x)dx

∣∣∣ ≤∫ ba|f(x)| dx.

In the following we put together some basic properties of the Rie-mann integrable functions space Int[a, b], defined on a fixed interval[a, b].

T 21. Let Int[a, b] be the set of all Riemann integrablefunctions on [a, b].

a) Then Int[a, b] is a vector subspace of the vector space of all func-tions defined on [a, b]. Moreover, if f, g ∈ Int[a, b] and α, β ∈ R, then

(3.1)

∫ b

a

[αf(x) + βg(x)]dx = α

∫ b

a

f(x)dx+ β

∫ b

a

g(x)dx.

b) If f, g ∈ Int[a, b] and f(x) ≤ g(x) for any x ∈ [a, b], then∫ baf(x)dx ≤

∫ bag(x)dx. In particular, if f(x) ≥ 0, then

∫ baf(x)dx ≥ 0.

In general,∫ baf(x)dx may be zero but f(x) = 0 for at least one point

x.c) If c ∈ (a, b) and if f ∈ Int[a, c] and if f ∈ Int[c, b], then f ∈

Int[a, b] and

(3.2)

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx

If [a, b]∩ [c, d] = ∅, we define∫[a,b]∪[c,d] f(x)dx =

∫ baf(x)dx+

∫ dcf(x)dx.

We also put∫ aaf(x)dx = 0 and

∫ abf(x)dx = −

∫ baf(x)dx. With this

notation, let l, m, n be three arbitrary real numbers in the interval [a, b],then for f : [a, b] → R, integrable, one has:

(3.3)

∫ n

l

f(x)dx =

∫ m

l

f(x)dx+

∫ n

m

f(x)dx


d) Let g : [a, b] → [0,∞) be a continuous function such that∫ bag(x)dx = 0. Then g(x) = 0 for any x in [a, b].

P. a) Let z be a continuity point for f and for g, where f, g ∈Int[a, b] and let α, β be two arbitrary real numbers. Then we knowthat z is also a continuity point for αf + βg, where (αf + βg)(x) =αf(x)+βg(x) for any x ∈ [a, b]. Thus, the discontinuity set of points forαf+βg is a subset of the union of the set of discontinuity points of f andthe set of discontinuity points of g. Since these last two sets have theL-measure zero, we see that the L-measure of the set of discontinuitypoints of αf + βg is also zero. Now, since f and g are bounded, it iseasy to see that αf + βg is also bounded. Let us use now theorem 17and find that αf + βg is integrable. Formula (3.1) is true for Riemannsums, so it is also true for their limits.

b) Let ∆n, ∆n : a < x(n)1 < x

(n)2 < ... < x

(n)kn

= b, be a sequence of

divisions with ‖∆n‖ → 0 and let ξ(n)i ∈ [x

(n)i−1, x

(n)i ], i = 1, 2, ..., kn, be a

corresponding sequence of marking points of it. Since f(x) ≤ g(x) onehas that

S(f ; ∆n; ξ(n)i ) =n∑

i=1

f(ξ(n)i )(x

(n)i − x

(n)i−1) ≤

≤n∑

i=1

g(ξ(n)i )(x

(n)i − x

(n)i−1) = S(g; ∆n; ξ(n)i ).

Taking limits here when n→ ∞, we get that∫ baf(x)dx ≤

∫ bag(x)dx.

c) Since the discontinuity set of f on [a, b] is the union of the dis-continuity sets of f on [a, c] and of f on [c, b] respectively, we seethat the L-measure of the first one is zero, i.e. f is integrable on[a, b], if it is integrable on [a, c] and on [c, b] respectively. Let ∆′

n,∆′n : a < x

′(n)1 < x

′(n)2 < ... < x

′(n)k′n

= c, be a sequence of divisions

of [a, c], with ‖∆′n‖ → 0 and let ξ

′(n)i ∈ [x

′(n)i−1, x

′(n)i ], i = 1, 2, ..., k′n,

be a corresponding sequence of marking points of it. Let also ∆′′n,

∆′′n : a < x

′′(n)1 < x

′′(n)2 < ... < x

′′(n)k′′n

= c, be a sequence of divisions

of [c, b], with ‖∆′′n‖ → 0 and let ξ

′′(n)i ∈ [x

′′(n)i−1 , x

′′(n)i ], i = 1, 2, ..., kn,

be a corresponding sequence of marking points of it. We see that

∆n = ∆′n ∪ ∆′′

n is a division of [a, b] with ξ(n)i = ξ′(n)i ∪ ξ′′(n)i as acorresponding set of marking points. Since ‖∆n‖ → 0, and because

S(f ; ∆n; ξ(n)i ) = S(f ; ∆′n; ξ′(n)i ) + S(f ; ∆′′

n; ξ′′(n)i ),

one has that∫ baf(x)dx =

∫ caf(x)dx +

∫ bcf(x)dx. Let us use now this

last equality and the definition∫ abf(x)dx = −

∫ baf(x)dx for a < b, in

4. MEAN THEOREM. NEWTON-LEIBNIZ FORMULA 63

order to prove (3.3). Assume for instance that m < n < l. Then, from(3.2) we get:

∫ l

m

f(x)dx =

∫ n

m

f(x)dx+

∫ l

n

f(x)dx.

So ∫ n

l

f(x)dx = −∫ l

n

f(x)dx =

∫ n

m

f(x)dx−∫ l

m

f(x)dx,

or∫ nlf(x)dx =

∫ mlf(x)dx+

∫ nmf(x)dx.

d) Assume that there is a c ∈ [a, b] such that g(c) > 0 (if g(x) < 0,the reasoning is completely analogous). Since g is continuous, there isan entire interval [c − ε, c + ε] ⊂ [a, b], ε > 0, such that g(x) > 0 forany x ∈ [c− ε, c+ ε] (see [Po], theorem 34). Moreover, there is a pointx0 ∈ [c − ε, c + ε] with g(x0) = inf

x∈[c−ε,c+ε]g(x) (see [Po], theorem 32).

Thus, g(x0) > 0. But∫ b

a

g(x)dx =

∫ c−ε

a

g(x)dx+

∫ c+ε

c−εg(x)dx+

∫ b

c+ε

g(x)dx ≥ 2ε·g(x0) > 0,

which contradicts the hypothesis. Hence, g(x) = 0 for any x ∈ [a, b].

R 13. We say that a number r ∈ R can be well approximatedwith elements from a given subset A (the "approximation set") of R iffor any small real number ε > 0 one can find an element aε of Asuch that |r − aε| < ε. For instance, any real number r can be wellapproximated with rational numbers (here A is Q, the set or rationalnumbers). Let now f : [a, b] → R be a Riemann integrable function.

Then I(f) =∫ baf(x)dx can be well approximated by an element of the

set of all Riemann sums of the form S(f ; ∆; ξi) (see definition 3).

4. Mean theorem. Newton-Leibniz formula

In order to estimate a definite integral∫ baf(x)dx, sometimes we

need "mean formulas". They are also useful to prove some basic resultswhich appear in many branches of pure or applied mathematics.

T 22. (Mean formulas) Let f : [a, b] → R be a continu-ous function and let g : [a, b] → [0,∞) be a nonnegative and nonzerocontinuous function. Then there is a real number ξ ∈ [a, b] such that

(4.1)

∫ b

a

f(x)g(x) dx = f(ξ)

∫ b

a

g(x) dx.


In particular, if g(x) = 1 for any x ∈ [a, b], we get the classical "meanformula":

(4.2)

∫ b

a

f(x) dx = f(ξ)(b− a).

This last formula says that the trapezoidal area [AabB] (see Fig.6)is equal to the area of the rectangle with the base [a, b] and the heightequal to the ordinate of f at a point ξ ∈ [a, b].

F 6

P. Since f is continuous on [a, b], it is bounded and its leastand upper bounds are realized, i.e. there are x1 and x2 in [a, b] suchthat m = f(x1) = inf

x∈[a,b]f(x) and M = f(x2) = sup

x∈[a,b]f(x) (see [Po],

Th. 32). Since m ≤ f(x) ≤M and g(x) ≥ 0, one has that

mg(x) ≤ f(x)g(x) ≤Mg(x),

for any x ∈ [a, b]. Let us use now theorem 21 and find:

m

∫ b

a

g(x) dx ≤∫ b

a

f(x)g(x) dx ≤M

∫ b

a

g(x) dx.

Since g(x) is continuous and not identical to zero, one has that∫ b

a

g(x) dx > 0

(see theorem 21), so we can divide the last inequalities by∫ bag(x) dx :

m ≤∫ baf(x)g(x) dx∫ bag(x) dx

≤M.


We use now Darboux theorem (see [Po], Th. 33) for the function f

(f([a, b]) = [m,M ]) and find a ξ ∈ [a, b] such that∫ ba f(x)g(x) dx∫ b

a g(x) dx= f(ξ),

i.e. formula (4.1).

E 29. Let us use the classical mean formula in order toprove that I =

∫ 100

10e−x

2dx < 1

293, so I is a very small number. Indeed,

from formula (4.2) we find that I = 90e−ξ2, where ξ ∈ [10, 100]. But

the biggest value of e−x2is e−100 < 2−100, so I < 90

2100< 27

2100= 1

293.

Let us now reformulate in a more general case the basic theorem ofNewton (see theorem 1).

T 23. (Newton-Leibniz formula) Let f : [a, b] → R be acontinuous function. Then F (x) =

∫ xaf(t) dt is a primitive for f, i.e.

F is differentiable and F ′(x) = f(x) for any x ∈ [a, b]. Moreover, ifG(x) is any primitive of f on [a, b], then

(4.3)

∫ b

a

f(x) dx = G(b) −G(a).

P. Take x0 ∈ [a, b] and let us study the following limit:

(4.4) limx→x0

F (x) − F (x0)

x− x0= limx→x0

∫ xaf(t) dt−

∫ x0af(t) dt

x− x0.

But, using theorem 21, c), we get∫ x

a

f(t) dt−∫ x0

a

f(t) dt =

∫ a

x0

f(t) dt +

∫ x

a

f(t) dt =

∫ x

x0

f(t) dt,

so, applying the classical mean formula (4.2) in (4.4), we get

(4.5) limx→x0

F (x) − F (x0)

x− x0

= limx→x0

f(ξx)(x− x0)

x− x0

= limx→x0

f(ξx).

Since ξx is between x and x0, then ξx → x0, whenever x → x0. Sincef is continuous, the last limit in (4.5) is f(x0). Thus, we just provedthat "the area function" F of Newton is a primitive of f on [a, b].

Let now G be another primitive. Theorem 3 says that G(x) =F (x) + C, where C is a real constant. Thus,

G(b) −G(a) = F (b) − F (a) =

∫ b

a

f(x) dx.

Newton-Leibniz formula (4.3) is true even for noncontinuous func-tions.


T 24. Let f : [a, b] → R be an integrable function and let

F be a primitive of f on [a, b]. Then∫ baf(x) dx = F (b) − F (a).

P. For any division ∆ = a = x0, x1, ..., xn = b of the inter-val [a, b] one has:

n∑

i=1

[F (xi) − F (xi−1)] = F (b) − F (a).

But, using the Lagrange formula (see [Po], Ch.4, Cor.5) on the interval[xi−1, xi], we find ci ∈ [xi−1, xi] such that F (xi)− F (xi−1) = f(ci)(xi −xi−1). Thus,

n∑

i=1

f(ci)(xi − xi−1) = F (b) − F (a).

If we take divisions ∆ with ‖∆‖ → 0, we finally find that∫ b

a

f(x) dx = F (b) − F (a).

E 30. Let us compute∫ 1

0x2dx. To use Newton-Leibniz for-

mula we need a primitive function for f(x) = x2. Since∫x2dx = x3

3,

one has that∫ 1

0x2dx = x3

3

∣∣∣1

0= 1

3− 0

3= 1

3.

E 31. Let us compute limx→0

∫ x0 e

−t2dt

x. Since we have the non-

determinate case 00, we can apply l’Hôspital rule. But the function

G(x) =∫ x0e−t

2dt is a primitive of g(x) = e−x

2(see theorem 23). So

G′(x) = g(x) and our limit is equal to limx→0

e−x2

= 1.

E 32. A point is moving on the segment [0, π2] in a linear

field of forces F (x) = x sin x. Find the work of F on [0, π2].

The work

W =

∫ π2

0

x sin xdx = −∫ π

2

0

x(cosx)′dxby parts

=

− x cosx|π20 +

∫ π2

0

cosxdx = 0 + sin x |π20 = 1.

E 33. On the segment [0, 2π] we have a density functionf(x) = 1

1+cos2 x. Let us find the mass of the wire ([0, 2π], f(x)) . If we


change the variable x with a new variable t by the formula tan x = t,we get :

mass =

∫ 2π

0

1

1 + cos2 xdx =

∫ 0

0

... = 0 !,

which is not true! Where did we make a mistake? Since the mappingx→ tan x = t is not injective on [0, 2π], it is not a "change of variable"on this last interval. The good interval for this change of variable is(−π

2, π

2

). So let us make a translation x → x − π

2= z of the interval

[0, 2π] :

(4.6)

∫ 2π

0

1

1 + cos2 xdx =

∫ 3π2

−π2

1

1 + sin2 zdz,

because cosx = cos(z + π2) = − cos(π − z − π

2) = − sin z. The last

integral in (4.6) can be written∫ 3π

2

−π2

1

1 + sin2 zdz =

∫ π2

−π2

1

1 + sin2 zdz +

∫ 3π2

π2

1

1 + sin2 zdz.

If in the last integral of the last sum of integrals we make the translation(change of variable) z → z − π = u, we get

∫ 3π2

π2

1

1 + sin2 zdz =

∫ π2

−π2

1

1 + sin2 udu,

because sin z = sin(π + u) = sin(π − π − u) = − sin u. Thus∫ 3π

2

−π2

1

1 + sin2 zdz = 2

∫ π2

−π2

1

1 + sin2 zdz = 4

∫ π2

0

1

1 + sin2 zdz.

In this last integral we can make the change of variable tan z = t andobtain:

4

∫ π2

0

1

1 + sin2 zdz = 4

∫ ∞

0

1

1 + 2t2dt =

4√2

∫ ∞

0

1

1 +(√

2t)2d

(√2t)

=4√2

arctan√

2t

∣∣∣∣∞

0

=√

2π,

because sin2 z = t2

1+t2, z = arctan t and dz = 1

1+t2dt for z ∈ [0, π

2). Here

∫ ∞

0

1

1 + 2t2dt

def= lim

A→∞

∫ A

0

1

1 + 2t2dt,

i.e. it is a first example of an improper integral of the first type (seethe next chapter).


5. The measure of a figure in Rn

Recall that the n-dimensional real vector space Rn is a "geometri-cal" space with the Euclidean distance

d(x,y) =√

(y1 − x1)2 + (y2 − x2)2 + ... + (yn − xn)2,

where x = (x1, x2, ..., xn) and y = (y1, y2, ..., yn) are two arbitraryelements of Rn. An open ball with centre a ∈ Rn and radius r > 0, inRn, is a subset B(a, r) of Rn, of the following form:

B(a, r) = x ∈ Rn : d(x,a) < rA point a of a subset A in Rn is said to be an interior point of A(relative to Rn) if there is a real number r > 0 such that the entireopen ball B(a, r) is contained in A. If all the point of A are interiorpoints we say that A is an open subset of Rn. The boundary ∂A of asubset A of Rn is the union of all the points b in Rn such that for anyr > 0 the ball B(b, r) contains at least one point from A and at leastone point from the outside of A at the same time. A∪ ∂A is called the(topological) closure of A in Rn. If A = A∪∂A, i.e. if the boundary ∂Aof A is contained in A, we say that A is closed. We say that a subset Ais bounded in Rn if there is a sufficiently large open ball B(0, R) whichcontains A. Here 0 = (0, 0, ..., 0) ∈ Rn. Let a and b be two points inRn and t a variable point (number) in the real interval [0, 1].The closedsubset of Rn

[a,b] = x ∈ Rn : x = (1 − t)a+ tb, for a t ∈ [0, 1]is called an (oriented) segment in Rn with extremities points a and b. Asubset P = ∪i=ni=1 [ai−1, ai] of Rn is said to be a polygonal line generatedby the "vertices" a0, a1, ...,an. We also say that the points a0 and anare connected by the polygonal line P. A subset A of Rn is said to beconnected if any two points of it can be connected by a polygonal lineP. A figure B in Rn is a bounded subset of Rn. A domain in Rn is anopen and connected subset of Rn. Since all these above notions justappeared and were discussed in [Po], we only ask the reader to try tofind examples and couterexamples of all of these notions, by drawingeverything, in Rn for n = 1, 2, and 3.

D 7. A subset A in Rn of the form

A = [a1, b1] × [a2, b2] × ...× [an, bn]

is said to be a parallelepiped in Rn. A finite union E = ∪ni=1Ai ofparallelepipeds Ai, such that for any i = j the intersection subset Ai ∩Aj has no interior point in Rn, is called an elementary figure. By

5. THE MEASURE OF A FIGURE IN Rn 69

definition, the measure (or the volume) of the above parallelepiped A isthe number

m(A) = (b1 − a1)(b2 − a2)...(bn − an).

The measure m(E) of the above elementary figure E is by definition

m(E) =n∑

i=1

m(Ai).

We say that a subset B of Rn is Jordan measurable (or has a volume)in Rn and its measure (or its volume) is the nonnegative real numberm(B) if for any ε > 0 there are two elementary figures Eεint and E

εext

with the following properties:1) Eεint ⊆ B ⊆ Eεext2) m(Eεint) ≤ m(B) ≤ m(Eεext) and3) m(Eεext) −m(Eεint) < ε.These three properties say that B is measurable and its measure

(volume) is the nonnegative real number m(B) if and only if B canbe well approximated (or covered) "from interior (inside)" and "from

exterior (outside)" by two sequences E(m)int m, E

(m)ext m of elementary

figures in Rn :

E(1)int ⊆ E

(2)int ⊆ ... ⊆ E

(p)int ⊆ ... ⊆ B ⊆(5.1)

... ⊆ E(q)ext ⊆ ... ⊆ E

(2)ext ⊆ E

(1)ext,(5.2)

such that

(5.3) m(B) = limm→∞

m(E(m)ext ) = lim

m→∞m(E

(m)int ).

For instance, a parallelepiped in R is a closed interval [a, b]. Twosuch intervals [a, b] and [c, d] have no interior points in common if andonly if their intersection is empty or it is simply a point (Why cannotit contain two distinct points?). The measure (the length in this case)of an interval [a, b] is exactly its length b − a. The measure of a finitesubset of points in R is equal to zero. But what about an infinite subsetof points in R ? For instance, let us prove that the set C = 1

nn has

measure zero in R. Its enough to construct two sequences like in (5.1).

We take for all E(p)int, p = 1, 2, ... the empty set ∅. For E(q)

ext we take theunion

E(q)ext = ∪2q

n=1

[1

n− 1

2(n+2)q,

1

n+

1

2(n+2)q

]∪[0,

1

2q + 1

]

It is easy to see that this union is a disjoint one, i.e. the intersectionbetween any two distinct segments of this union is empty. Moreover,

the length of E(q)ext is equal to

∑2q

n=11

2(n+2)q−1 + 1

2q+1. Since 1

2(n+2)q−1 ≤ 1

23q


one has that m(E(q)ext) ≤ 1

23q−q + 1

2q+1→ 0, when q → ∞. Hence the

measure of C exists and it is equal to zero. An easy example of anonmeasurable subset of R is the set F = [0, 1] ∩Q. Since F containsno interior points w.r.t. R, it cannot contain a nontrivial interval. So,if F were measurable, its measure would be zero. But this is not true

because we cannot construct a sequence (E(q)ext)q, where each E

(q)ext is a

finite union of intervals, two of which having in common at most one

point, F ⊂ E(q)ext for each q and m(E

(q)ext) → 0, where q → 0. Indeed,

in these conditions [0, 1] ⊂ E(q)ext for all q = 1, 2, ..., thus m(E

(q)ext) ≥ 1

and so the sequence (E(q)ext)q cannot tend to 0. In the same manner one

can prove that if a subset A is measurable in R and has measure zero,then it cannot have interior points. Conversely, if it is measurable andhas no interior points, then its measure must be zero (Why?). Themeasure of a measurable subset A of R is simply called its length,l(A). What is the connection between the notion of a subset of R ofLebesgues measure zero (see Definition 5) and the notion of a subsetof R of measure zero? It is clear that a subset of R of measure zero(it is enough to be measurable!) must be bounded. It is also clearthat a subset of R of measure zero has also the Lebesgue measure zero(look carefully to both definitions!). The set N∗ = 1, 2, ..., n, ... canbe covered by the disjoint union

∪∞n=1[n− 1

2ns+1, n +

1

2ns+1]

for any s ≥ 1. Since the sum of the lengths of all intervals of this unionis

∞∑

n=1

1

2ns=

1

2s· 1

1 − 12s

=1

2s − 1

and since 12s−1

→ 0, when s → ∞, we see that the Lebesgue measureof N∗ is zero and it is not bounded, i.e. it cannot be measurable, inparticular it cannot have the measure zero.

Another interesting example of a zero Jordan measure set of R isthe famous Cantor set. Take the interval A0 = [0, 1], divide it into 3equal subintervals, get out the middle open subinterval (1/3, 2/3) andobtain

A1 = [0, 1/3] ∪ [2/3, 1] ⊂ A0.

Take now each of the both disjoint intervals of A1 and proceed in thesame way. We obtain

A2 = [0, 1/32] ∪ [2/32] ∪ [2/3, 7/32] ∪ [8/32, 1] ⊂ A1 ⊂ A0.


We continue in this way and get a tower of subsets

A0 ⊃ A1 ⊃ ...An ⊃ An+1 ⊃ ...

The intersection C = ∩∞n=1An is an infinite subset of [0, 1]. This subsetis called a Cantor subset. Since m(An) =

(23

)n → 0, when n → ∞, itsmeasure is zero. In particular its Lebesgue measure is also zero.

A parallelepiped in R2 is simply a rectangular surface (let us abbre-viate with "rectangle") with its sides parallel to the coordinates axes(see Fig.7). Its measure is equal to its area. Using the area of sucha rectangle we can deduce formulas for the areas of a parallelogram,a triangle or even for a disc (as the limit of the areas of some regularpolygons inscribed in the circumference of this disc). Look at Fig.7.

F 7

Indeed, the area of the parallelogram [EFGH] is equal to the areaof the rectangle [HH ′JG] = l(EF ) · l(HH ′). The area of the triangle[KLM ] is a half from the area of the parallelogram [KLNM ]. To findthe area of a disc with centre at W and of radius R, we approximatethe disc with the surface bounded by a regular polygon with n sidesinscribed in the boundary circle of this disc. Now, the angle α, inradians (see Fig.7), is equal to 2π

2n= π

n. So the area of the polygon is

equal to n · 2R sin πn·R cos π

n· 1

2. Hence, the area of the disc is

limn→∞

R2 cosπ

n· sin π

nπn

· π = πR2.

The length of the boundary circle of the above disc is

limn→∞

n · 2R sinπ

n= 2πR lim

n→∞

sin πn

πn

= 2πR.


E 3. Starting with the formula for the area of a rectangleand of a triangle, find formulas for the side areas of a cylinder, ofa cone and of a frustum of a cone (πR2h, πR

√R2 + h2 and π(R +

r)√

(R− r)2 + h2), where h is the height of the corresponding solid)

The area of a region bounded by a closed polygonal line can becomputed as a sum of areas of triangles (see Fig.8).

F 8

The intersection of two rectangles which belong to an elementaryfigure has no interior points w.r.t. R2 if and only if either it is empty,it is a point or a segment of a line parallel to one of the axes. For aplane elementary figure see Fig.9.

In Fig.10 we see how to cover "from interior" with an elementaryfigure a general plane figure A.

In Fig.11 we see how to cover "from exterior" with an elementaryfigure another general figure A.

In Fig.12 we see the simultaneous process of approximation "frominterior" and "from exterior" with elementary figures Ei and Eo respec-tively, a general plane figure B.

The measure of a measurable figure A in R2 (a plane figure!) issimply called its area σ(A). The areas of a finite union of points or ofa finite union of segments of lines are zero (Why?). The same is truefor a finite union of curves of class C1 (piecewise smooth curves!).

R 14. To define the area σ(A) of a measurable subset A ofR2 we used the above finite unions of rectangles as elementary figures.Since any rectangle is obvious a union of two triangles (as surfaces!)with their intersections a segment of zero areas and since any triangle


F 9

F 10

can be well covered by rectangles, we can use in definition 7 trianglesinstead of rectangles, namely an elementary figure can be a finite unionof triangles. The same is true if instead of rectangles we use discs, orregular polygons, etc.

A parallelepiped in R3 is an usual parallelepiped [a, b]× [c, d]× [e, f ]which has its side faces parallel to the coordinate planes. The intersec-tion between two such parallelepipeds contains no interior points w.r.t.R3 if and only if it is either empty, a point, a segment of a line or a rec-tangular surface parallel to one of the coordinates planes. The measure


F 11

F 12

vol(A) of a measurable subset A of R3 is simply called its volume. Thevolume of a finite set of points, a piecewise smooth curve or a piece-wise smooth surfaces (define them by analogy with piecewise smoothcurves) is zero. In this case we also can substitute parallelepipeds withtetrahedrons, balls, regular polyhedrons, etc.

6. AREAS OF PLANE FIGURES BOUNDED BY GRAPHICS OF FUNCTIONS75

E 4. Starting with the formula of volume of a parallelepiped,find formulas to compute the volume of an arbitrary oblique paral-lelepiped, of a tetrahedron, a cylinder, a cone and a frustum of a cone(the area of the basis×h, 1

3×the area of the basis×h, πR2(area of the

basis)×h, 13× πR2(area of the basis)×h and πh

3(R2 + r2 +Rr)), where

h is the height of the corresponding solid).

It is not so difficult to see that if A,B are two measurable figure (Cmeasurable ⇒ C bounded) in Rn, then A ∪ B, A ∩ B, A \ B are alsomeasurable and

(5.4) m(A ∪B) = m(A) +m(B) −m(A ∩B).

We also see that if A ⊆ B and both are measurable, then m(A) ≤m(B). If a subset A of R is measurable and its measure is zero, then itsLebesgue measure is also zero. Conversely, it is not true (see A = N).

The following result is fundamental in what follows.

T 25. A figure A of Rn is measurable if and only if itsboundary is measurable and its measure is equal to zero. In particular,if one can find a piecewise smooth parameterization for the boundary∂A of a figure A of Rn, then A itself is measurable in Rn.

For the case n = 2 one can find a proof in [Nik] or in [Pal]. Ingeneral, the proof follows a similar way. For an intuitive argument, seealso Fig.12.

6. Areas of plane figures bounded by graphics of functions

Let f : [a, b] → R+ be a bounded function defined on the interval[a, b] with nonnegative values. Let Df be the plane figure bounded bythe lines x = a, x = b, the segment [a, b] and the graphic of f (see thehatched figure [ABCD] in Fig.1, Ch.1).

T 26. With the above notation f is Riemann integrable on[a, b] if and only if the plane figure Df has a measure (an area) in R2

and then

(6.1) m(Df) =

∫ b

a

f(x)dx.

P. Assume that f is integrable on [a, b]. Then, for any division

∆ : a = x0 < x1 < ... < xn = b,

the union of corresponding rectangles which give rise to the inferiorDarboux sum

s∆(f) =n∑

i=1

mi(xi − xi−1)


is an elementary figure which covers Df "from interior". Its area iss∆(f) (see Fig.8). The elementary figure generated by the union of allrectangles which appear in the superior Darboux sum

S∆(f) =n∑

i=1

Mi(xi − xi−1)

coversDf "from exterior". The area of this last figure is equal to S∆(f).The integrability Darboux criterion (see theorem 14) says that Df ismeasurable and its area is the common limit of s∆(f) and S∆(f)when ‖∆‖ → 0, i.e. m(Df) =

∫ baf(x)dx.

Assume now that Df is measurable and its area is m(Df ). Let ususe now theorem 25 and, by looking at Fig.12, we see that for anyε > 0, starting with the elementary figures Eεint and E

εext which appear

in definition 7, we can find a division ∆0 of the interval [a, b] such that

S∆0(f) − s∆0(f) < ε. For any other division ∆ ≻ ∆0, let Es∆(f)int and

ES∆(f)ext be the two elementary figures generated by the rectangles which

give rise to the corresponding Darboux sums from the superior index.Then

(6.2) Eεint ⊂ Es∆(f)int ⊂ Df ⊂ E

S∆(f)ext ⊂ Eεext

and

(6.3) m(ES∆(f)ext )(= S∆(f)) −m(E

s∆(f)int )(= s∆(f)) < ε.

Taking now an arbitrary division ∆′ of [a, b] and denoting by ∆ =∆′ ∪ ∆0 we see that ‖∆‖ ≤ ‖∆0‖ and ∆ ≻ ∆0. So S∆(f) − s∆(f) < ε.By looking at (6.2) and (6.3) we see that s∆(f) and S∆(f) have acommon limit when ‖∆‖ → 0. And this last limit is m(Df). Hence f

is integrable on [a, b] and∫ baf(x)dx = m(Df ).

R 15. For instance, if f(x) ≥ 0 is bounded and continuouson [a, b] except maybe a finite number of points, then Df is measurableand

∫ b

a

f(x)dx = m(Df).

E 34. Let f(x) = sinx, x ∈ [0, 2π]. We are interested inthe computation of the hatched area from Fig.13.

Applying the remark 15 we see that the area of the "+" part is equalto

∫ π0

sin xdx = − cosx |π0= 2. Now, if we take a division

∆ : π = x0 < x1 < ... < xn = 2π


F 13

and a set of marking points ξii=1,n, ξi ∈ [xi−1, xi], then the corre-sponding Riemann’s sum

Sf(∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1)

is negative, because f(ξi) = sin ξi ≤ 0. Thus the area (which is a non-negative number) is the limit of the following sums

−Sf(∆; ξi) = −n∑

i=1

f(ξi)(xi − xi−1) =

n∑

i=1

|f(ξi)| (xi − xi−1) = S|f |(∆; ξi).

Hence its value is −∫ 2π

πsinx dx = cosx |2ππ = 2. Thus, the hatched area

is equal to 4 and it is not equal to∫ 2π

0sinxdx = 0!.

In general, if f : [a, b] → R is continuous almost everywhere (thenoncontinuity set of points A has the Lebesgue measure 0), then thearea bounded by the graphic of f, [a, b] and the lines x = a, x = b isequal to the hatched area in Fig.14, i.e.

(6.4) area(Df ) =

∫ b

a

|f(x)| dx,

in this more general case.


F 14

E 35. Let us compute area(Df) for f : [−2, 1] → R, where

f(x) =

x+ 1, x ∈ [−2,−1),x3, x ∈ [−1, 1]

.

Since

|f(x)| =

−x− 1, x ∈ [−2,−1),−x3, x ∈ [−1, 0),x3, x ∈ [0, 1],

one has

area(Df ) =

∫ 1

−2

|f(x)| dx =

∫ −1

−2

(−x− 1)dx+

∫ 0

−1

(−x3)dx+

∫ 1

0

x3dx

= −(x2

2+ x

)∣∣∣∣−1

−2

− x4

4

∣∣∣∣0

−1

+x4

4

∣∣∣∣1

0

= 1.

Suppose that f, g : [a, b] → R are two functions continuous almosteverywhere and f(x) ≥ g(x) for any x in [a, b]. Then it is clear that thehatched area in Fig.15 is equal to

∫ b

a

f(x)dx−∫ b

a

g(x)dx =

∫ b

a

[f(x) − g(x)]dx.


F 15

But whenever on a subinterval [c, d] ⊂ [a, b], g(x) ≥ f(x), x ∈ [c, d],we compute the area "between g and f" by the formula

∫ d

c

[g(x) − f(x)]dx =

∫ d

c

|f(x) − g(x)| dx.

Thus, in general, the hatched area areaf,g between f and g in Fig.16can be computed by the formula

areaf,g =

∫ b

a

|f(x) − g(x)| dx.

Indeed,

areaf,g =

∫ c

a

[g(x) − f(x)]dx+

∫ d

c

[f(x) − g(x)]dx+

+

∫ e

d

[g(x) − f(x)]dx+

∫ b

e

[f(x) − g(x)]dx =

∫ b

a

|f(x) − g(x)| dx.

E 36. Let us compute the area between f(x) = sinx andg(x) = sin 2x, x ∈ [0, π]. The best way is to look at the graphics of fand g in Fig.17.

Since the intersection points of these graphics are x = 0, x = π3and

x = π, one can write:

areaf,g =

∫ π

0

|sin x− sin 2x| dx.


F 16

F 17

Since a continuous function keeps the same sign between two consec-utive zeros of it (Why?-see Darboux theorem in [Po]), the sign of thefunction h(x) = sin x − sin 2x is the sign of h(π

4) = 1√

2− 1 < 0 on

the interval [0, π3]. The sign of h(x) on the interval [π

3, π] is the sign of

h(3π4

) = 1√2

+ 1 > 0. Thus,

areaf,g =

∫ π3

0

(sin 2x− sinx)dx+

∫ π

π3

(sin x− sin 2x)dx =

− cos 2x

2

∣∣∣∣π3

0

+ cos x |π30 − cosx |ππ

3+

cos 2x

2

∣∣∣∣π

π3

=


1

4+

1

2+

1

2− 1 + 1 +

1

2+

1

2+

1

2=

11

4.

Assume now that we have a plane parametric curve

(Γ) :

x = x(t)y = y(t)

, x ∈ [a, b],

such that x(t) is an increasing function of class C1 and y(t) is a non-negative continuous function (see Fig.18).

F 18

Then the area of the domain D(Γ), bounded by (Γ), the segment[x(a), x(b)] and the lines x = x(a), x = x(b), can be approximated byRiemann’s sums of the form:

(6.5) S(Γ)(∆; (ξi)) =

n∑

i=1

y(ξi) (x(ti) − x(ti−1)) ,

where ∆ : a = t0 < t1 < ... < b = tn is a division of the interval [a, b]and (ξi), i = 1, 2, ..., n, ξi is an arbitrary set of fixed marking pointsfor ∆. One can use Lagrange’s formula for function x(t) restricted to[ti−1, ti] and find:

x(ti) − x(ti−1) = x′(ηi)(ti − ti−1)

for an ηi ∈ (ti−1, ti). When ‖∆‖ → 0, ηi and ξi become closer andcloser. Since x′(t) is continuous, x′(ηi) and x′(ξi) become closer andcloser. So the area above can be well approximated by sums of theform:

S∗(Γ)(∆; (ξi)) =n∑

i=1

y(ξi)x′(ξi) (ti − ti−1) .


But these last sums are Riemann’s sums for a new function f(t) =y(t)x′(t), t ∈ [a, b]. Hence the area above can be computed by theformula

(6.6) area(D(Γ)) =

∫ b

a

y(t)x′(t)dt.

If x(t) is increasing or decreasing almost everywhere (it is increasingor decreasing on each subinterval of a fixed division a = c0 < c1 < ... <ck = b), then obviously formula (6.6) must be substituted with a moregeneral formula

(6.7) area(D(Γ)) =

∫ b

a

y(t) |x′(t)| dt.

The same formula works if in addition to this last generalization, x(t)is a smooth piecewise function on [a, b] and (or) y(t) is piecewise con-tinuous on the same interval.

E 37. Let us compute the area bounded by the axis Ox andthe arc of the cycloid

(Γ) :

x = a(t− sin t)y = a(1 − cos t)

, t ∈ [0, 2π],

where a > 0 is a parameter. Since x′(t) = a(1 − cos t) ≥ 0, we canapply formula (6.6) and find

area = a2

∫ 2π

0

(1 − cos t)2dt = a2

∫ 2π

0

(1 − 2 cos t+ cos2 t)dt =

= 2πa2 + 0 +a2

2

∫ 2π

0

(1 + cos 2t)dt = 3πa2.

Sometimes our curves are represented in polar coordinatesM(ρ, θ),where ρ is the distance from O to M and θ is the angle in radians

between Ox-axis and−−→OM. So ρ ≥ 0 and θ can be in general any real

number. For any fixed θ in a fixed interval [α, β], we give ρ = ρ(θ), i.e.we prescribe the distance fromM to O on the positive direction of thehalf axis corresponding to that θ. Let us compute the area boundedby the graphic of a continuous (or piecewise continuous) mapping θ →ρ(θ), θ ∈ [α, β] and the rays θ = α, θ = β (see Fig.19).

Let us consider a division ∆ : α = θ0 < θ1 < ... < θn = β ofthe interval [α, β] and an arbitrary set ξi, ξi ∈ [θi−1, θi] of markingpoints for ∆. Let us approximate the small area bounded by the arc(θi−1θi) and the rays θ = θi−1 and θ = θi by the area of the sector


F 19

of the disc of radius ρ(ξi) and centre O, bounded by the arc (θi−1θi) .A very known elementary formula says that this last area is equal to12[ρ(ξi)]

2(θi − θi−1), where θi − θi−1 is measured in radians. This lastformula is in fact the area of a triangle of basis ρ(ξi)(θi−θi−1) (the reallength of the arc (θi−1θi) of that circle) and height ρ(ξi). So the totalarea can be well approximated by the following sum

Sρ(θ)(∆; ξi) =n∑

i=1

1

2ρ2(ξi)(θi − θi−1).

But this last sum is exactly a Riemann sum for the function θ → 12ρ2(θ),

θ ∈ [α, β]. Hence our area is equal to the corresponding Riemann inte-gral

(6.8) area =1

2

∫ β

α

ρ2(θ)dθ.

E 38. The cardioid is the curve ρ = ρ(θ) = a(1 + cos θ),θ ∈ [0, 2π) and a > 0 is a parameter (see Fig.20). Let us use formula(6.8) in order to compute the area bounded by this cardioid:

area =1

2

∫ 2π

0

a2(1 + cos θ)2dθ =a2

2

[2π +

∫ 2π

0

cos2 θdθ

]=

= πa2 +a2

4

∫ 2π

0

(1 + cos 2θ)dθ =3πa2

2.

E 39. Let us consider the snail ρ = ρ(θ) = 3θ, where θ ∈[0, 5π

2] (see Fig.21). Let us find the area of the hatched region of Fig.21.


F 20

Since the ray−−→OM, M(ρ(θ), θ) covers the double hatched region I two

times the area of the hatched region is:

area =1

2

∫ 5π2

0

9θ2dθ − 1

2

∫ π2

0

9θ2dθ =93π3

4.

F 21

E 40. The graphic of the curve ρ = ρ(θ) = a sin 3θ, whereθ ∈ [0, 2π] has the form of a clover with three leaflets (see Fig.22).Since sin 3θ must be greater or equal to zero, we have that

θ ∈ [0,π

3] ∪ [

2π

3, π] ∪ [

4π

3,5π

3].

7. THE VOLUME OF A ROTATIONAL SOLID 85

Compute the area of the hatched region in Fig.22. The symmetry ofthe figure with respect to the three distinct lines (see Fig.22) gives riseto

area = 3 · 1

2

∫ π3

0

a2 sin2 3θ dθ =πa2

4.

F 22

7. The volume of a rotational solid

Let f : [a, b] → R+ be a continuous function defined on the interval[a, b] with nonnegative real values. Let D be the solid obtained by therotation of the arc y = f(x), x ∈ [a, b], around Ox-axis. An arbitrarydivision ∆ : a = x0 < x1 < ... < xn = b gives rise to a division ofD into n solids D1,D2, ...,Dn. Di can be obtained by the rotation ofthe same arc y = f(x), but restricted to the interval [xi−1, xi]. We canwell approximate this arc y = f(x), x ∈ [xi−1, xi] with the segment[Mi−1Mi], where Mi−1(xi−1, f(xi−1)) and Mi(xi, f(xi)). So the solid Dican be approximated by a frustum of a cone with r = f(xi−1), R =f(xi) and h = xi − xi−1. (look at Fig.23 and make the same reasoningfor cylinders instead of frustums).

Using now a formula from the exercise 4 we find that

vol(Di) ≈π

3(xi − xi−1)

[f(xi−1)

2 + f(xi)2 + f(xi−1)f(xi)

].


F 23

Hence,

(7.1) vol(D) ≈ π

3

n∑

i=1

[f(xi−1)

2 + f(xi)2 + f(xi−1)f(xi)

](xi − xi−1).

Let us fix now a set of marking points (ξi)i, ξi ∈ [xi−1, xi], i = 1, 2, ..., n.Since f is continuous, if the norm ‖∆‖ of the division ∆ is smaller andsmaller, then f(xi) and f(xi−1) are closer and closer to f(ξi). Thusformula 7.1 becomes

(7.2) vol(D) ≈ π

n∑

i=1

f(ξi)2(xi − xi−1) = Sπf2(∆; (ξi)),

a Riemann sum for the function πf2, the division ∆ and the set ofmarking points (ξi). So

(7.3) vol(D) = π

∫ b

a

f2(x)dx.

Formula (7.2) says that we can well approximate the volume of the ro-tational solid D with the sum of volumes of the cylinders Ci, generatedby the rotation of the line y(x) = f(ξi) for any x ∈ [xi−1, xi], aroundOx-axis (see Fig.24).

E 41. Let us find the volume of a ball of radius R > 0, inR3.

Such a ball is the rotational solid generated by the rotation of thearc y =

√R2 − x2, x ∈ [−R,R] of the circle of radius R, x2 + y2 = R2,

8. THE LENGTH OF A CURVE IN R3 87

F 24

around Ox-axis. Thus,

vol = π

∫ R

−R(R2 − x2)dx = 2π

∫ R

0

(R2 − x2)dx =

= 2π

(R2x− x3

3

)∣∣∣∣R

0

=4πR3

3.

8. The length of a curve in R3

Let (Γ) be a curve in R3 represented by the parametric path

(8.1) (Γ) :

x = x(t)y = y(t)z = z(t)

, t ∈ [a, b].

Let ∆ : a = t0 < t1 < ... < tn = b be a division of the interval[a, b] and let Mi(x(ti), y(ti), z(ti)), i = 0, 1, ..., n, be the correspondingdivision on (Γ), i.e. the image through the vector mapping −→r (t) =(x(t), y(t), z(t)) , t ∈ [a, b], (the deformation mapping of the segment[a, b] onto the curve (Γ) as a subset of R3) of the division ∆. Let l(∆)be the length of the polygonal line [M0M1...Mn].

D 8. We say that the curve (Γ) is rectifiable if there existsa nonnegative real number l(Γ), such that for any ε > 0, there existsδε > 0 with the following property: if ∆ is a division of the interval[a, b], ‖∆‖ < δε, then |l(∆) − l(Γ)| < ε. This nonnegative real numberl(Γ) is called the length of (Γ) and it can be well approximated by lengthsof polygonal lines of the form [M0M1...Mn] (described above).


It is not difficult to prove that the property to be rectifiable doesnot depend on the parametric representation of (Γ). The same is truefor the length of (Γ). This number l(Γ) defined above is unique. Letus estimate now l(Γ).

Let us consider a parametric representation of a smooth curve (Γ)(see (8.1)), a division ∆ : a = t0 < t1 < ... < tn = b of the interval [a, b]and a set of marking points (ξi)i, ξi ∈ [ti−1, ti] for any i = 1, 2, ..., n.By definition 8 we can approximate the length l(Γ) with

(8.2) S =n∑

i=1

∥∥∥−−−−−→Mi−1Mi

∥∥∥ =

=n∑

i=1

√[x(ti) − x(ti−1)]2 + [y(ti) − y(ti−1)]2 + [z(ti) − z(ti−1)]2.

(see Fig. 25).

F 25

Since x(t), y(t) and z(t) are of class C1, applying Lagrange formulato them on each interval [ti−1, ti], we get

x(ti) − x(ti−1) = x′(ai)(ti − ti−1) ≈ x′(ξi)(ti − ti−1),

y(ti) − y(ti−1) = y′(bi)(ti − ti−1) ≈ y′(ξi)(ti − ti−1),

z(ti) − z(ti−1) = z′(ci)(ti − ti−1) ≈ z′(ξi)(ti − ti−1),

8. THE LENGTH OF A CURVE IN R3 89

where ai, bi, ci ∈ [ti−1, ti] for each i = 1, 2, ..., n. Thus, S can be wellapproximated (when ‖∆‖ → 0) with

S1 =n∑

i=1

√x′(ξi)

2 + y′(ξi)2 + z′(ξi)

2(ti − ti−1).

Indeed, since the function

H(u, v, w) =√x′(u)2 + y′(v)2 + z′(w)2,

u, v, w ∈ [a, b]×[a, b]×[a, b] is uniformly continuous, then for any ε > 0,there exists δε > 0 such that

|H(u′, v′, w′) −H(u′′, v′′, w′′)| < ε

b− a

if

|u′ − u′′| , |v′ − v′′| , |w′ − w′′| < δε

. Thus, if ‖∆‖ < δε one has that

|S − S1| ≤n∑

i=1

|H(ai, bi, ci) −H(ξi, ξi, ξi)| (ti − ti−1) ≤

≤ ε

b− a· (b− a) = ε.

But this sum S1 is a Riemann sum for the function

f(t) =√x′(t)2 + y′(t)2 + z′(t)2.

So our curve (Γ) is rectifiable and its length can be computed by usingthe formula

(8.3) l(Γ) =

∫ b

a

√x′(t)2 + y′(t)2 + z′(t)2dt.

The expression ds =√x′(t)2 + y′(t)2 + z′(t)2dt is called the element

of length on the arc Γ, i.e. it is "the limit" of the length of the arc

Mi−1Mi on Γ, when the distance between Mi−1Mi is small enough.Let us compute the length of the astroide

x = a cos3 ty = a sin3 t

, t ∈ [0, 2π], a > 0

This curve is a plane curve, so z(t) = 0 in the above formula. We seethat the Cartesian form of the parametric equation of the astroide isx23 +y

23 = a

23 . Thus the curve is symmetric relative to Ox and Oy axes.

Hence its length is


l = 4

∫ π2

0

√9a2 cos4 t sin2 t+ 9a2 sin4 t cos2 tdt = 12a

∫ π2

0

√cos2 t sin2 tdt =

= 12a

∫ π2

0

|cos t sin t| dt = 6a

∫ π2

0

sin 2tdt = −3a cos 2t|π20 = 6a.

E 42. Let M(x) be a moving point on the segment [a, b]

in a field of forces−→f parallel to the Ox-axis oriented like the direct

orientation of [a, b], i.e. "from a to b”. (see Fig.26)

F 26

Let the norm (modulus) of this field be a continuous function f(x),x ∈ [a, b]. If

∆ : a = x0 < x1 < x2 < ... < xn = b

is an arbitrary division of [a, b] and ξi is a set of marking pointsof it, then we can well approximate the work of

−→f along the oriented

segment [xi−1, xi] with f(ξi)(xi − xi−1). Thus, the work of−→f along

[a, b] can be well approximated with the Riemann sum S(f ; ∆; ξi) =∑ni=1 f(ξi)(xi − xi−1). If the function f is integrable, then the work

W[a,b]

−→f of

−→f along the segment [a, b] is just

∫ baf(x)dx.

9. Approximate computation of definite integrals.

Let f : [a, b] → R be an integrable function. The definition itselfof the notion of integrability (see Definition 3) is equivalent to the fact

9. APPROXIMATE COMPUTATION OF DEFINITE INTEGRALS. 91

that we can well approximate the number I =∫ baf(x)dx with Riemann

sums of the form

Sf (∆; ξi) =n∑

i=1

f(ξi)(xi − xi−1),

where ∆ : a = x0 < x1 < ... < xn = b is a division of [a, b] with‖∆‖ small enough and ξi is an arbitrary set of marking points, ξi ∈[xi−1, xi] for any i = 1, 2, ..., n. Let us take a particular type of divisions,namely equidistant divisions. Let n be large enough and let h = b−a

n,

"the increment" of the division

∆n = a = x0 < a+ h = x1 < ... < a+ ih = xi < ... < xn = a+ nh = b.

This division is an example of an equidistant division because xi −xi−1 = h, a constant number. Let us take ξi = xi−1+xi

2, the midpoint of

the interval [xi−1, xi]. Thus,

ξi =a + (i− 1)h + a+ ih

2= a +

2i− 1

2h.

So we can use in practice the following approximation:

(9.1)

∫ b

a

f(x)dx ≈ R(f ;n) =b− a

n

n∑

i=1

f

(a+

2i− 1

2h

).

This formula is called the rectangles formula. In Fig.27 the integral∫ baf(x)dx is just the area of the plane surface bounded by the graphic

of f(x), when x ∈ [a, b], the Ox-axis and the vertical lines x = a, y = b.The approximate R(f ;n) from formula (9.1) is exactly the hatched areaof Fig.27, i.e. the sum of the areas of all the rectangles Di, where Dihas as a basis the segment [xi−1, xi] and as height f(ξi), ξi being themidpoint of [xi−1, xi].

We can also be interested in the error err = |I − R(f ;n)| . If ourfunction is integrable, this error becomes smaller and smaller, whenn→ ∞.

E 43. Let us use the rectangles formula to evaluate thedefinite integral I =

∫ 1

0e−x

2dx. Higher Mathematics show that function

f(x) = e−x2has not any primitive which could expressed by elementary

functions. For an approximate computation, take n = 5; then ξ1 =x0+x1

2= 0.1, ξ2 = x1+x2

2= 0.3, ξ3 = x2+x3

2= 0.5, ξ4 = x3+x4

2= 0.7 and

ξ5 = 0.9. So,

I ≈ 1

5[exp (−0.01) + exp (−0.09) + exp (−0.25) +(9.2)

+ exp (−0.49) + exp (−0.81)].(9.3)


F 27

But what about the error we just have made in this last approxi-mation?

Let us evaluate the error in the case of rectangles formula (9.1),when f is a function of class C2 on [a, b].

For this, we fix an i ∈ 1, 2, ..., n and we try to estimate the errorerri made by substituting

∫ xixi−1

f(x)dx with the area f(xi−1+xi

2

)(xi −

xi−1) of the rectangle which has the basis the segment [xi−1, xi] andthe height f(ci), where ci = xi−1+xi

2is the midpoint of this last segment

(see Fig.28).

F 28


Let us denote byM2 = supx∈[a,b]

|f ′′(x)| and let us apply Taylor formula

of order 2 around the point ci, i.e.

(9.4) f(x) = f(ci) +f ′(ci)

1!(x− ci) +

f ′′(c)

2!(x− ci)

2,

where ci is a point in the interval [xi−1, xi]. Thus,

erri =

∣∣∣∣∫ xi

xi−1

f(x)dx− f

(xi−1 + xi

2

)(xi − xi−1)

∣∣∣∣ =

=

∣∣∣∣∫ xi

xi−1

[f(x) − f(ci)]dx

∣∣∣∣ ≤|f ′(ci)|

1!

∣∣∣∣∫ xi

xi−1

(x− ci)dx

∣∣∣∣+

+M2

2

∣∣∣∣∫ xi

xi−1

(x− ci)2dx

∣∣∣∣ = 0 +M2

2

[(xi − ci)

3

3− (xi−1 − ci)

3

3

]=

=M2

2

[(xi − xi−1+xi

2)3

3− (xi−1 − xi−1+xi

2)3

3

]= M2

h3

24=

M2

24n3(b− a)3.

The global error

err =

∣∣∣∣∣I −n∑

i=1

f(ci)(xi − xi−1)

∣∣∣∣∣ =

∣∣∣∣∣n∑

i=1

∫ xi

xi−1

(f(x) − f(ci)) dx

∣∣∣∣∣ ≤

≤n∑

i=1

erri ≤ n · M2

24n3(b− a)3 =

M2

24n2(b− a)3.

Thus

(9.5) err ≤ M2

24n2(b− a)3,

where M2 = supx∈[a,b]

|f ′′(x)| .

Let us come back to example 43 and evaluate the error we have

made by approximating I =∫ 1

0e−x

2dx with R

(e−x

2; 5

). Let us evalu-

ate M2 :

f ′(x) = −2xe−x2

; f ′′(x) = (−2 + 4x2)e−x2

.

Thus, |f ′′(x)| ≤ 2 when x ∈ [0, 1] and

err ≤ 2

24 · 25· 13 =

1

300<

1

100.

Hence, the sum from (9.2) approximates our integral with at least 2exact decimals.

The general idea above is to substitute the original function f :[a, b] → R with another one, called an interpolating function, usually


more elementary then the initial one. In our case of rectangles ap-

proximation, we substituted f with f such that f(x) = f(ci) for any

x ∈ [xi−1, xi), i = 1, 2, ..., n. Then we approximated I =∫ 1

0e−x

2dx with

∫ b

a

f(x)dx =n∑

i=1

∫ xi

xi−1

f(ci)dx =n∑

i=1

f(ci)(xi − xi−1) = R(f, n).

In (9.1) we interpolated the function f(x), x ∈ [xi−1, xi] with a polyno-mial of degree 0, which has the same value as our function at the pointci = xi−1+xi

2, the midpoint of [xi−1, xi]. If instead of a polynomial of de-

gree 0, we interpolate f(x), x ∈ [xi−1, xi] with the unique polynomial ofdegree 1 (a segment of a line), which has 2 points in common with thegraphic of f, namely the points Mi−1(xi−1, f(xi−1)) and Mi(xi, f(xi))(see Fig.29), we get

I =

∫ b

a

f(x)dx ≈∫ b

a

f(x)dx =

=n∑

i=1

∫ xi

xi−1

f(x)dx =n∑

i=1

area[xi−1xiMi−1Mi].

Since the figure [xi−1xiMi−1Mi] is a trapezoid, its area is equal to

F 29

b−a2n

[f(xi−1) + f(xi)] . Thus,

(9.6) I ≈n∑

i=1

b− a

2n[f(xi−1) + f(xi)] =


=b− a

2n

[f(a) + f(b) + 2

n∑

i=1

f(xi)

]def= T (f, n).

This last formula is called the trapezoids formula. One can also estimatethe error err = |I − T (f, n)| in this case:

(9.7) err ≤ M2

12n2(b− a)3,


|f ′′(x)| . For a proof of this last inequality see for

instance [GG], pag.139.Even our feeling says that trapezoids formula gives a better approx-

imation than the rectangles formula, the mathematical estimation oferrors (9.5) and (9.7) say contrary, namely, the estimation of error inthe case of the rectangles formula is two times smaller than the estima-tion in the case of the trapezoids formula. In practice, sometimes theerror in the case of the trapezoids approximation formula can be lessthan that one in which we use rectangle formula (for the same equidis-tant division). Sometimes it is greater! (by drawing, find out someexamples!). Since both R(f, n) and T (f, n) are convergent to I, whenn→ ∞, for n large enough one can use either rectangles or trapezoidsformula.

In order to obtain "better" approximation formulas we need to "in-terpolate" our integrand function f(x) by a polynomial P (x) of a fixeddegree n. This means to fix n + 1 distinct points x0 < x1 < ... < xnin [a, b] and to find a polynomial P (x) of degree n such that P (xi) =f(xi) for any i = 0, 1, ..., n. We shall see that this last polynomialis unique and it is called the interpolation polynomial of f at thenodes x0, x1, ..., xn. In practice, we usually do not know the ana-lytical expression of the function f(x). In fact we measure some valuesy0, y1, ..., yn of it at a given finite number of points x0 < x1 < ... < xn.Thus, f(xi) = yi for i = 0, 1, ..., n. Having such an interpolation poly-nomial we can force the approximation f(x) ≈ P (x), x ∈ [a, b]. One canprove that if the number of nodes is greater and greater and if they are"uniformly" distributed (for instance if we use equidistant divisions of[a, b]), then the error ‖f − P‖ = sup

x∈[a,b]|f(x) − P (x)| becomes smaller

and smaller (see [GG] for instance). Let us use this last approximationin order to find an approximation of the integral

∫ b

a

f(x)dx ≈∫ b

a

P (x)dx.


This last integral can be easily computed even the degree of P is verylarge (we can derive an elementary formula which is a function of thecoefficients of P and of a and b). Thus, the approximate computation

of∫ baf(x)dx reduces to an easy task.

Let us reformulate our general problem. Given n+ 1 nodes (points,numbers, etc.) x0 < x1 < ... < xn and a fixed arbitrary set of n + 1real numbers y0, y1, ..., yn, let us find a polynomial Ln(x) (here L isfrom Lagrange) of degree n such that

Ln(xi) = yi, i = 0, 1, ..., n,

i.e. the polynomial function Ln(x) is passing through the xOy-planepoints Mi(xi, yi), i = 0, 1, ..., n (see Fig.30).

F 30

Such a polynomial Ln(x) is called a Lagrange polynomial corre-sponding to the nodes x0 < x1 < ... < xn and to y0, y1, ..., yn.

T 27. Given x0 < x1 < ... < xn and y0, y1, ..., yn, thereis a unique Lagrange polynomial Ln(x) of degree n which is passingthrough the points Mi(xi, yi), i = 0, 1, ..., n.

P. Any polynomial Ln(x) of degree n can be represented as

Ln(x) = a0 + a1x+ ...+ anxn,

where a0, a1, ..., an are unknown coefficients. To determine Ln(x) meansto find these coefficients a0, a1, ..., an. Conditions Ln(xi) = yi, i =


0, 1, ..., n can be also written as a linear system in variables a0, a1, ..., an :

a0 + a1x0 + ...+ anxn0 = y0

a0 + a1x1 + ...+ anxn1 = y1

............................a0 + a1xn−1 + ...+ anx

nn−1 = yn−1

a0 + a1xn + ...+ anxnn = yn

.

The determinant of the system is a Vandermonde determinant equalto ∏

0≤i<j≤n(xj − xi), which is not zero because xi = xj for any i = j.

Thus the solution exists and it is unique.

To compute directly a0, a1, ..., an by solving this linear system is nota good idea because computation of large determinants is a difficult andnot always an exact methods (a big number of cumulative errors maylead to a very big final error!). To escape of this difficulty, the greatmathematician J. L. Lagrange (1736 — 1813) invented a nice alternativemethod. We know that the monomials 1, x, x2, ..., xn is a basis of thereal vector space Pn of all polynomials of degree less or equal to n. Thismeans that any polynomial P (x) of degree less or equal to n can beuniquely written in the form

P (x) = a0 + a1x+ ...+ anxn,

where x is the variable (it can take any value on R) and a0, a1, ..., anare real numbers which completely define our polynomial P. The greatidea of Lagrange was "to change" the initial basis 1, x, x2, ..., xn ofPn with another one more suitable for our situation. Let us define n+1polynomials of degree n in Pn :

(9.8)

ω0(x) = (x− x1)(x− x2)...(x− xn).................................................

ωi(x) = (x− x0)(x− x1)...(x− xi−1)(x− xi+1)...(x− xn).................................................

ωn(x) = (x− x0)(x− x1)...(x− xn−2)(x− xn−1).

Here x0 < x1 < ... < xn are our fixed nodes and let us remark that thepolynomial ω0(x) does not contain the factor (x− x0), ω1(x) does notcontain the factor (x− x1), ω2(x) does not contain the factor (x− x2)and so on. Let us also remark that

(9.9) ωi(xj) = 0 for any j = i and ωi(xi) = 0.

T 28. a) The set of polynomials

ω0(x), ω1(x), ..., ωi(x), ...ωn(x)


is a basis in Pn, the real vector space of all polynomials of degree lessor equal to n.

b) The unique Lagrange polynomial Ln(x) which is passing throughthe set of points Mi(xi, yi) (see theorem 27) can be represented as alinear combination of the polynomials ω0(x), ω1(x), ..., ωn(x):

(9.10) Ln(x) =y0

ω0(x0)ω0(x) +

y1ω1(x1)

ω1(x) + ...+yn

ωn(xn)ωn(x).

P. a) Since the dimension of Pn is n + 1 (1, x, x2, ..., xnis a basis of it!), we cannot have more than n + 1 linear independentelements in Pn (see any course in Linear Algebra, for instance see [Po1]or [Po2]). Thus, it is enough to prove that the set of polynomialsω0(x), ω1(x), ..., ωn(x) is linear independent. Indeed, let us take alinear combination of them:

(9.11) λ0ω0(x) + λ1ω1(x) + ...+ λnωn(x) = 0,

for any x ∈ R. Let us put x = x0 in (9.11). Since ω0(x0) = 0(x0, x1, ..., xn are distinct), ω1(x0) = 0, ..., ωn(x0) = 0, we get thatλ0 = 0. If we put now x = x1 in (9.11), we find λ1 = 0, etc. Finallywe find that all λ′s are zero, i.e. the set ω0(x), ω1(x), ..., ωn(x) islinear independent in Pn. Having n + 1 elements and dimension be-ing n + 1, any other polynomial Q cannot be outside the linear sub-space generated by ω0(x), ω1(x), ..., ωn(x). If we could find such apolynomial, we could enlarge the number of linear independent ele-ments to n + 2 which is not possible because the dimension is exactlyn+ 1 (see a course in Linear Algebra, for instance [Po1] or [Po2]). Soω0(x), ω1(x), ..., ωn(x) is also a generating system for Pn, i.e. this setof polynomials is a basis of Pn.

b) Since ω0(x), ω1(x), ..., ωn(x) is a basis, one can write Ln(x) as

(9.12) Ln(x) = µ0ω0(x) + µ1ω1(x) + ...+ µnωn(x).

To compute µ0, µ1, ..., µn, we successively make x = x0, x1, ..., xn in(9.12). For instance, if x = x0, we get

y0 = Ln(x0) = µ0ω0(x0),

because ω1(x), ..., ωn(x) are zero at x0. Thus, µ0 = y0ω0(x0)

. In the same

way we get that µ1 = y1ω1(x1)

, ..., µn = ynωn(xn)

, i.e. we just proved formula

(9.10).

E 44. (interpolation with parabolas) Let us interpolate threepoints M1(x1, y1), M2(x2, y2), M3(x3, y3) with a parabola, i.e. with a


Lagrange polynomial of degree 2. Let us use formula (9.10) and find

(9.13) L2(x) =y1

ω1(x1)ω1(x) +

y2ω2(x2)

ω2(x) +y3

ω3(x3)ω3(x),

where ω1(x) = (x− x2)(x− x3), ω2(x) = (x− x1)(x− x3) and ω3(x) =(x− x1)(x− x2).

E 5. Write L2(x) for x1 = 0.1, x2 = 0.5, x3 = 0.9 andy1 = 1, y2 = 1.5, y3 = 0.5. Draw the graphic of the resulting parabola!

Assume now that in Example 44 x2 is the midpoint of the segment[x1, x3]. Thus, x2 = x1+x3

2. Let us denote by h the differences

(9.14) x2 − x1 = x3 − x2 =x3 − x1

2= h.

Let us evaluate I = I[x1, x3; y1, y2, y3] =∫ x3x1L2(x)dx. Let us use now

formula (9.13) to compute this last integral. Since(9.15)∫ x3

x1

L2(x)dx = y1

∫ x3

x1

ω1(x)

ω1(x1)dx+y2

∫ x3

x1

ω2(x)

ω2(x2)dx+y3

∫ x3

x1

ω3(x)

ω3(x1)dx.

it is enough to compute the three integrals which appear on the rightside of the formula (9.15). For this, let us change the variable x witha new variable s :

x = x1 + sh.

Thus, ∫ x3

x1

ω1(x)

ω1(x1)dx =

h

2

∫ 2

0

(s− 1)(s− 2)ds =h

3,

∫ x3

x1

ω2(x)

ω2(x2)dx = −h

∫ 2

0

s(s− 2)ds =4h

3,

∫ x3

x1

ω3(x)

ω3(x1)dx =

h

2

∫ 2

0

s(s− 1)ds =h

3

and formula (9.15) becomes

(9.16)

∫ x3

x1

L2(x)dx = y1h

3+ y2

4h

3+ y3

h

3.

Let us consider now a continuous function f : [a, b] → R and an equidis-tant division a = x0 < x1 < x2 < ... < x2n = b of the interval [a, b] with2n nodes, where n is a nonzero natural number. Let h = xi−xi−1 = b−a

2n.

We take the following set of triple points: x0, x1, x2, x2, x3, x4, ...,x2n−2, x2n−1, x2n. For each x2i−2, x2i−1, x2i we approximate the re-striction of f to the interval [x2i−2, x2i] by the Lagrange polynomial

L(i)2 (x) which corresponds to the node points x2i−2, x2i−1, x2i and the


value y2i−2 = f(x2i−2), y2i−1 = f(x2i−1) and y2i = f(x2i) for i =

1, 2, ..., n. If one approximates∫ baf(x)dx with

S(f, n) =n∑

i=1

∫ 2i

x2i−2

L(i)2 (x)dx

and if one uses formula (9.16) to compute each term of this last sum,one gets ∫ b

a

f(x)dx ≈ b− a

6n[f(x0) + f(x2n)+

(9.17)+4(f(x1) + f(x3) + ...+ f(x2n−1)) + 2(f(x2) + f(x4) + ...+ f(x2n−2))].

This approximation formula is called Simpson formula and if f isof class C4 then the error can be estimated by

err =

∣∣∣∣∫ b

a

f(x)dx− S(f, n)

∣∣∣∣ ≤(b− a)5

2880M4,


∣∣f (IV )(x)∣∣ (see [GG]). For n = 2 look at the Fig. 31

to see how we approximate the graphic of f with two arcs of parabolas.

F 31

If we take an equidistant division a = x0 < x1 < x2 < ... < xn = band if we interpolate the arc Mi−1Mi, where Mi(xi, f(xi)), with theLagrange polynomial of degree 1, i.e. with the segment [Mi−1Mi], weget the above discussed trapezoids formula (see (9.6)).

E 6. Take n = 4 and f(x) = x2 defined on [0, 1]. Then∫ 1

0x2dx = 1

3. Use the three approximate formulas R(f, n), T (f, n) and


S(f, n) discussed above to approximate the integral∫ 1

0x2dx. Compare

the results with the exact computation, i.e. with 1/3.


1. Compute the following definite integrals:

a)∫ e1

ln2 xxdx; b)

∫ ln 2

0

√ex − 1dx; c)

∫ e1

lnxx2dx;

d)∫ π0

11+cos2 x

dx; e)∫ 2π

01

1+cos2 xdx;

2. Prove that if f : [−a, a] → R, a > 0, f is continuous and even,then

∫ a−a f(x)dx = 2

∫ a0f(x)dx.

3. Prove that if f : [−a, a] → R, a > 0, f is continuous and odd,then

∫ a−a f(x)dx = 0.

4. Prove that if f : R→ R is continuous and periodic of period T >

0 (this means that f(x+T ) = f(x) for any x in R, then∫ α+Tα

f(x)dx =∫ T0f(x)dx for any α ∈ R. In particular, prove that

∫ T2

−T2

f(x)dx =∫ T0f(x)dx .5. Compute the area bounded by the curves (help yourself by draw-

ing everything!):a)y = −x2, y = x2 and x = 2;b)x = y2 and x2 + y2 = 2, x > 0;

c) (x−5)2

16+ (y−5)2

4= 0;

d)xy = 1, y = 2x and y = 12x;

e)y = x2 − 4x+ 6 and y = −3x2 + 6;f)y = 2

5x and x = 2 − y − y2.

6. The curve y = sin x, x ∈ [0, π2] rotates around Ox-axis and then

around Oy-axis. Compute the two volumes of the revolutionary bodieswhich result.

7. Find the area bounded by the arc of the parabola y = 4x2, Ox-axis and the lines x = 0 and x = a (a > 0), using only the definitionof definite integral with Riemann sums.

8. Find the area bounded by the curvilinear trapezoid generatedby the hyperbola y = 1/x, x = a, x = b (0 < a < b) and Ox-axis.

9. Let f : [0, 1] → R be a continuous function defined on theinterval [a, b] with values in R. Prove that

limn→∞

1

n

n∑

k=1

f

(k

n

)=

∫ 1

0

f(x)dx.

Use this formula to compute the following limits:a) S = lim

n→∞

(1n2

+ 2n2

+ ...+ n−1n2

);


b) S = limn→∞

15+25+...+n5

n6;

c) S = limn→∞

(1n+1

+ 1n+2

+ ... + 1n+n

);

d) S = limn→∞

(n

n2+12+ n

n2+22+ ...+ n

n2+n2

);

e) S = limn→∞

√1+√

2+...+√n

n√n

.

10. Using the fact that F (x) =∫ xaf(t)dt is a primitive of f(x),

compute:

a) limx→1

∫ x1 ln tdt

x−1; b) lim

x→0

∫ x0

sin ttdt

x; c) lim

x→0

∫ x0

√1+t4dt

x3.

11. Find the graphic of the function f(x) =∫ x0

sin ttdt, x ∈ R.

12. Compute the limits limA→∞

∫ A0e−ax cos bx dx and

limA→∞

∫ A0e−ax sin bxdx.

13. Find the area bounded by the curves y = 2 − x2 and y3 = x2.14. Find the area bounded by the Ox-axis and the arc of the cycloid

x = a(t− sin t),y = a(1 − cos t)

, t ∈ [0, 2π], a > 0.

15. Find the area bounded by the curve: ρ = a(1+cos 2θ), θ ∈ [0, π](Draw it!).

16. Find the area bounded by the Bernoulli’s lemniscate: ρ2 =a2 cos 2θ, θ ∈ [−π

4, π

4] ∪ [3π

4, 5π

4].

17. Find the length of the curves: a) y = ln x, x ∈ [√

3,√

8]; b)x = 1

4y2 − 1

2ln y, y ∈ [1, e].

18. Find the lengths of the following curves:

a) the cycloid:

x = a(t− sin t),y = a(1 − cos t)

, t ∈ [0, 2π], a > 0,

b) the cardioid: ρ = a(1 + cos θ), θ ∈ [0, 2π] and

c) the astroide: x23 + y

23 = a

23 , a > 0.

19. Find the volume of the revolutionary ellipsoid determined by

the rotation of the ellipse x2

a2+ y2

b2= 1, a) around Ox-axis; b) around

Oy-axis.

20. Use the formula S = 2π∫ ?

?y√

1 + y′2dx to find the area ofthe parabolic mirror generated by the rotation of the arc of parabolax = 1

16ay2, y ∈ [0, 4a], around Ox-axis.

21. The curve y = a cosh xa(catenary), a > 0, x ∈ [0, a] rotates

around the Ox-axis and gives rise to a surface called catenoid. Findthe area of this catenoid.

22. Find the revolutionary areas obtained by the rotation aroundthe Ox-axis of the a) cycloid, b) cardioid and c) astroide from theproblem 18.


23. The circle x2 + (y − b)2 = a2, b > a, rotates around Ox-axis.Find the side area and the volume of the resulting solid (torus).

24. The parabola y = x2, x ∈ [0, 1] rotates itself around the firstbisectrix y = x. Find the side area and the volume of the resulting solid(Hint: use Riemann’s sums to find appropriate formulas or make firstof all a rotation of axes of 45).

25. Compute: a)∫ √3

1

√1+x2

x2dx; b)

∫ 2e

1|lnx− 1| dx;

c)∫ 1

−1max

[(13

)x, 3x

]dx;

26. Let us denote by In the following definite integral∫ π

20

sinn xdx,n = 1, 2, ... . Prove that In = n−1

nIn−2. Then compute I5 and I6.

27. Find the coordinates of the mass center of the plane domainbounded by the arc of parabola y2 = x, y ≥ 0, the Ox-axis and theline x = 2.

28. Find the matrix of the momentum of inertia for the arc of thehelix:

H :=

x = a cos ty = a sin tz = bt

, a, b > 0, t ∈ [0, 2π].

(Hint: I =

Ixx Ixy IxzIxy Iyy IyzIxz Iyz Izz

, where Ixx =

∫ 2π

0(y2 + z2)ds, Ixy =

∫ 2π

0z2ds, etc., i.e. the square of the distance of a pointM(x, y, z) of H

to the Ox-axis, to the Oxy-plane, etc. Here ds =√x′2 + y′2 + z′2dt is

the element of length on the curve H.29. Use the formula ds =

√x′2 + y′2dt to deduce an analogous

formula, ds =√ρ2 + ρ′2dθ, for a curve given in polar coordinates:

ρ = ρ(θ). Use this last formula to compute the length of the cardioid:ρ = a(1 + cos θ), θ ∈ [0, 2π], where a > 0.

30. The parabola y = x2

4, x ∈ [0, 3], rotates around Oy-axis. Find

the volume of the resultant solid.31. Find the coordinates of the mass center of the figures bounded

by the curves:

a) x2

a2+ y2

b2= 1, x, y ≥ 0, x = 0, y = 0; b) y = x2, y =

√x, x ≥ 0.


32. Use the formulas (see also the chapter with line integrals of thefirst type):

L =

∫ 2π

0

√x′2 + y′2dt,

xG =

∫ 2π

0x√x′2 + y′2dt

L,

yG =

∫ 2π

0y√x′2 + y′2dt

Lin order to compute the coordinates of the mass center of the line:


, t ∈ [0, 2π).

33. Let M1(0,12), M2(1, 1), M3(2,

12), M4(3, 1) and M5(4,

12) be 5

points in the xOy-plane. Write the Lagrange’s polynomial for thesepoints and compute the area bounded by it, the lines x = 0, x = 4,y = 0 and the graphic of this polynomial.

34. Use the trapezoid method with n = 5 to approximately compute∫ 4

0xdx. Compute this integral by Newton-Leibnizformula and observe the committed error.

CHAPTER 3

Improper (generalized) integrals

1. More on limits of functions of one variable

Usually, whenever we say that a sequence xn is convergent tob, b is considered to be a real number. Let us extend the notion ofconvergence to the "completed" real line R = R∪−∞,∞. If b = ∞,we say xn is convergent to b if for any real number M, there exists arank NM such that if n ≥ NM , then xn ≥ M. If b = −∞, we say thatxn is convergent to b if for any real number M, there exists a rankNM such that if n ≥ NM , then xn ≤M.

Let A be a nonempty subset of R. We say that b ∈ R = R ∪−∞,∞ is a limit point of A if there exists a nonconstant sequencean of elements of A which is convergent to b. For instance, 0 is a limitpoint of A =

1n

and ∞ is a limit point of A = n. For an interval I

(finite, infinite, closed, open or neither closed nor open) any point of Ior of ∂I, the boundary of I in R, is a limit point of I (prove it!).

D 9. (limit of a function at a given point) Let f : A→ Rbe a function defined on a subset A of R with real values and let b ∈ Rbe a limit point of A. We say that L = L(f, b) ∈ R is the limit of fat b if for any sequence xn , xn ∈ A, which is convergent to b, thesequence f(xn) is convergent to L. If L is finite, i.e. if L ∈ R, wesay that f has the finite limit L at b or that f has the limit L at b. Wenote this by L = lim

x→bf(x).

If in this definition we take only sequences xn , xn ∈ A, xn < b,xn → b then, if the limit Ll(f, b) = lim

xn→b,xn≤bf(xn) exist and it is the

same for any such sequence, we say that Ll(f, b) is the left limit of f atb. If we take sequences xn , xn ∈ A, xn > b, xn → b then, if the limitLr(f, b) = lim

xn→b,xn≥bf(xn) exists and it does not depend on the sequence

xn, we say that Lr(f, b) is the right limit of f at b. Sometimes it isnot possible to define one of these limits. For instance, if f : [0, 1] → Rand b = 0, it is not possible to define the left limit of f at 0. In thiscase we call Lr(f, 0) = L(f, b) the limit of f at 0. Ll(f, b) and Lr(f, b)are called the side limits of f at b. Let us use the above definitions to

105

106 3. IMPROPER (GENERALIZED) INTEGRALS

compute the side limits and the limit of f : R → R, f(x) = x2 + 1 atx = 2. If xn → 2, xn ≤ 2, then f(xn) = x2

n + 1 → 5. So Ll(f, 2) = 5. Itis easy to see that Lr(f, 2) = L(f, 2) = 5. But, if

f(x) =

x2, if x ≤ 23, if x > 2

, f : R→ R,

then Ll(f, 2) = 4, Lr(f, 2) = 3, and L(f, 2) does not exist because,generally speaking, any sequence zn which is convergent to 2 andwhich has an infinite number of terms on the left and on the right of

2, can be written like a union of two subsequences of it,z(1)kn

and

z(2)tn

, such that z

(1)kn

≤ 2 and z(2)tn ≥ 2. This means that f(z

(1)kn

) → 4

and f(z(2)tn ) → 3, thus the whole sequence f(zn) has not a unique

limit when n→ ∞. This is why we say that this f has not a limit at 2.The following theorem makes light on the relation between the side

limits and the limit itself.

T 29. Let A be a nonempty subset of R and let b be a limitpoint of A in R. Let us assume that both side limits of f at b exist andthat they are equal to M. Then f has a limit at b and this limit is equalto M.

P. Let xn be a sequence of elements of A which is conver-gent to b. If this sequence has either a finite terms on the left of b or afinite terms on the right of b, then the limit of the sequence f(xn)iseither Ll(f, b) or Lr(f, b) respectively. Since both these numbers areequal to M, the limit of f(xn) is M in this case. Assume now that

the sequence xn is the disjoint union of two subsequencesz(1)kn

and

z(2)tn

such that the terms of

z(1)kn

are on the left of b and the terms

ofz(2)tn

are on the right of b. Here "disjoint" means that these last se-

quences have no terms in common. Since the left limit at b and the right

limit at the same point are equal to M,f(z

(1)kn

)

andf(z

(2)tn )

are

both convergent toM. Let us prove that the whole sequence f(xn) isconvergent toM. For this let us take an ε-neighborhood (M−ε,M+ε)of M. There are two natural numbers N1 and N2 such that if kn ≥ N1

and tn ≥ N2 one has that f(z(1)kn

) ∈ (M−ε,M ] and f(z(2)tn ) ∈ [M,M+ε).

Let N = maxN1, N2, the greatest element of the set N1, N2. Now,if m ≥ N, then xm is either a z

(1)kn

with kn ≥ N1 or a z(2)tn with tn ≥ N2.

Thus f(xm) ∈ (M − ε,M + ε) in both cases. Hence, f(xm) → Mwhenever m→ ∞ and the proof is complete.

1. MORE ON LIMITS OF FUNCTIONS OF ONE VARIABLE 107

R 16. Since on the real line R we have two directions arounda given point b, on the left and on the right, we might be inclined togeneralize theorem 29 to the case of a function of many variables, sayof two variables, in the following sense. Let A be a nonempty domain(open and connected, see [Po] for instance) of the xOy-plane and letb = (b1, b2) be a limit point of A. Let f : A → R be a function oftwo variables defined on A with scalar real values. The temptation isto say that if the limit along any straight line which is passing throughb exists and it does not depend on the direction of these lines (thismeans that there exist a real number L such that if Mn = (xn, yn) isa sequence of points on a line y − b2 = m(x− b1) for all slopes m, i.e.yn − b2 = m(xn − b1) for any n, then f(xn, yn) has a limit at b). Butthis statement is wrong as follows from the example bellow. Let

f(x, y) =

y3√x+y , if y = − 3

√x

0, if y = − 3√x

,

f : R→ R. Take b = (0, 0) and an arbitrary line y = mx which passesthrough (0, 0). If (xn, yn) is such that yn = mxn, then

f(xn, yn) =mx

23n

1 +mx23n

→ 0,

when (xn, yn) → (0, 0). But, if (xn, yn) → (0, 0), xn = 0 and yn = 3√xn,

then

f(xn, yn) =1

2 0.

Thus 0 cannot be the limit of f at (0, 0). Hence, even f has the limit0 on any lines which is passing through (0, 0), it has limit 1/2 on thecubic curve y3 = x. It can be proved that a function f defined on aplane domain A has limit L at a point b of A if and only if it has thesame limit L along any smooth curve γ which is passing through b andis contained in A. This comes from the fact that any sequence of pointsMn(xn, yn) which converges to b can be interpolated by a smooth curveγ which contains the point b too. This last statement reduces to the fol-lowing fact. For any n = 2, 3, ..., three distinct points Pn−1(xn−1, yn−1),Pn(xn, yn), Pn+1(xn+1, yn+1) and two straight lines dn−1, dn+1 which ispassing through Pn−1 and Pn+1 respectively, there is a parametric curveof class C1 of the form

Γn :=

x(t) = xn−1 + An−1(t− n+ 1) + An(t− n + 1)2++An+1(t− n+ 1)3 + Cn+1(t− n+ 1)4,

y(t) = yn−1 +Bn−1(t− n+ 1) +Bn(t− n + 1)2++Bn+1(t− n + 1)3 +Dn+1(t− n + 1)4

,


t ∈ [n − 1, n + 1] such that it passes through our three points and istangent to dn−1, dn+1 at the points Pn−1 and Pn+1 respectively. Fort = n−1, x(n−1) = xn−1 and y(n−1) = yn−1. Since x

′(n−1) = An−1

and y′(n− 1) = Bn−1, one can find An−1and Bn−1 respectively as beingthe direction of dn−1. Let (ln+1,mn+1) the direction of the line dn+1.Then the coefficients An, An+1, Cn+1 can be determined by solving thefollowing linear system (in theses unknowns An, An+1, and Cn+1)

4An + 12An+1 + 32Cn+1 = l2 − An−1

An + An+1 + Cn+1 = xn − xn−1 − An−1

4An + 8An+1 + 16Cn+1 = xn+1 − xn−1 − 2An−1

.

The determinant of this linear system is equal to −16, so the systemhas a unique solution. This last system came from the conditions thatΓn be tangent to dn+1 at Pn+1, x(n) = xn and y(n) = yn. An anal-ogous system for y(x) will determine the coefficients Bn, Bn+1 andDn+1. Finally, in order to obtain an interpolating curve Γ of class C1

for a sequence Pn(xn, yn) which converges to b, such that Γ passesthrough b, we take a sequence of straight lines dn with its sequenceof slops mn convergent to a real number, say r and Γ to be the jointunion of all Γn, n = 2, 4, 6, ..., constructed above and the point b itselfwith the coordinates x(∞) and y(∞). When t runs from 1 up to ∞,t ∈ R, the corresponding point M(x(t), y(t)) goes from P1(x1, y1) up toP (b1, b2). Moreover, M(x(t), y(t)) is passing through P2, P3, ..., Pn, ...successively.

Practically, how do we prove that a function f : A→ R has a limitL at a limit point b ∈ R of A and how do we compute it? We simplytake an arbitrary sequence xn , xn ∈ A for all n, which converges tob and then ask ourselves if the sequence f(xn) is convergent to L.If we do not know the value of L in advance, we compute the limit off(xn) if it exists and, if this limit exists and if it does not depend ofthe sequence xn which converges to b, we must conclude that f hasthis last limit just computed at the point b.

For instance, let us study if the function f(x) = ln(x2 + 1) has alimit at x = 0. For this let xn → 0 be a sequence which converges to 0.Then, taking count of the usual elementary operations with sequences,we get that x2

n+1 → 1. Since the function ln : (0,∞) → R is continuousas being an elementary function (see [Po]), we get that the sequenceln(x2

n + 1) → ln 1 = 0. This last number does not depend on theconvergent to zero sequence xn , so the function f(x) = ln(x2 + 1)has the limit L = 0 at x = 0. In particular, the both side limits of f atzero are equal to zero.


The following two criterions are very useful in this chapter.

T 30. (ε-δ criterion) Let A be an interval in R and let b bea limit point of it in R. Let f : A→ R be a function defined on A withreal values. a) Assume that b if finite, i.e. it is in R. Then f has a limitL ∈ R at b if and only if for any ε > 0 (usually small) there exists a realnumber δ which depends on ε such that whenever x ∈ A∩ (b− δ, b+ δ),then f(x) ∈ (L − ε, L + ε), i.e. |f(x) − L| < ε. This means that if wego with x very close to b, then f(x) becomes very close to L. b) Assumenow that b = ∞. Then f has a limit L at b if and only if for any ε > 0(usually small) there exists a sufficiently large number δ > 0 whichdepends on ε such that whenever x ∈ (δ, b = ∞) (a neighborhood of∞),then f(x) ∈ (L − ε, L + ε), i.e. |f(x) − L| < ε. This means that if wepush x very close to ∞, then f(x) becomes very close to L. If b = −∞,then δ < 0 and −δ is large enough, i.e. in this last case δ is very closeto −∞. Moreover, the expression "x ∈ (δ, b = ∞) (a neighborhood of∞)" must be substituted with "x ∈ (−∞, δ) (a neighborhood of −∞)".

P. a) Suppose that f has a limit L ∈ R at b ∈ R, i.e. for allsequences xn → b one has that f(xn) → L. If for a fixed ε > 0 such anumber δ did not exist, then, for every natural number n = 1, 2, ..., wecould find a zn ∈ A∩ (b−1/n,b+1/n) such that f(zn) /∈ (L−ε, L+ ε).Thus zn → b and f(zn) does not converges to L, i.e. L cannot bea limit for f at b, a contradiction! Hence such a δ with the requiredproperties in the statement of the theorem must exist.

Conversely, assume that for any ε > 0 there exists a δ with theproperty described in the theorem. Take xn → b, xn ∈ A and take alsoa small ε > 0. For the corresponding δ (which depend on ε) one canfind a natural number N which depend on this δ such that if n ≥ N,then xn ∈ A∩ (b− δ, b+ δ). Use now the assumption and conclude thatfor n ≥ N, f(xn) ∈ (L− ε, L+ ε), i.e. f(xn) → L, so L is the limit off at the point b.

b) If b = ∞, let us assume that L is the limit of f at b = ∞, i.e. ifxn → ∞, xn ∈ A, then f(xn) → L. If one could find an ε > 0 for whichsuch a large number δ (depending on ε) with the property describedin theorem, b), did not exist, then, for this ε and for any n = 1, 2, ...we could construct a yn ∈ A∩ (n,∞) such that f(yn) /∈ (L− ε, L+ ε).Hence, yn → ∞ and f(yn) L, a contradiction! So such a δ mustexist.

Conversely, if for any small ε > 0 there exists a δ with the propertiesdescribed in theorem, b), take a sequence tn → b = ∞, tn ∈ A. Forsuch δ one can find a natural number M such that if n ≥ M, one hasthat tn ∈ A∩ (δ,∞). Let us use now the hypothesis and conclude that


f(tn) ∈ (L − ε, L + ε) for any n ≥ M. Thus f(tn) → L and the proofis completed.

Sometimes we cannot even guess the limit L in order to use theabove definition of the limit of f at b. For instance, let A = [0,∞),b = ∞ and f(x) =

∫ x0e−t

2dt, the famous Laplace error function which

is a basic mathematical tool in statistics. In higher mathematics itis proved that the function g(t) = e−t

2has no elementary primitives,

so we can never express f(x) as a composition of elementary func-tions (polynomial, rational, trigonometric, exponential and logarith-mic functions). Thus, if xn → ∞, how do we prove that the sequencef(xn) =

∫ xn0e−t

2dt

is convergent? How do we compute its limit?

The following criterion of Cauchy can decide if a limit of a function fat a given point b ∈ R exists or not, without guess a number L whichcould be a good candidate for the limit itself.

T 31. (Cauchy criterion) Let A be an interval of R and letb ∈ R be a limit point of it. Let f : A→ R be a function defined on A,with values in R. a) If b is finite, i.e. if b ∈ R, then f has a limit at b ifand only if for any small real number ε > 0 there is a real number δ > 0such that if two numbers x′ and x′′ are in A ∩ (b − δ, b + δ), then thedistance between f(x′) and f(x′′) is less than ε, i.e |f(x′) − f(x′′)| < ε.b) If b is not finite, say b = ∞, then f has a limit at b if and only if forany small real number ε > 0 there is a large real number δ > 0 such thatif two numbers x′ and x′′ are greater than δ, then the distance betweenf(x′) and f(x′′) is less than ε, i.e. |f(x′) − f(x′′)| < ε. For b = −∞we must replace this last δ with a negative, near to −∞, δ, i.e. in thiscase δ < 0 and −δ is very large. Moreover, the expression "x′ and x′′are greater than δ” must be substituted with "x′ and x′′ are less thanδ”.

P. a) Assume that L is the limit of f at b. Use now ε-δ cri-terion (see theorem 30, a)) for ε/2 instead of ε. Thus, there exists aδ > 0 such that if x ∈ A ∩ (b− ε/2, b + ε/2), then |f(x) − L| < ε/2.Take now this δ and for x′, x′′ ∈ A ∩ (b− ε/2, b + ε/2) one has that|f(x′) − L| < ε/2 and |f(x′′) − L| < ε/2. But

|f(x′) − f(x′′)| ≤ |f(x′) − L| + |f(x′′) − L| < ε/2 + ε/2 = ε.

Conversely, suppose that for any ε > 0, there is a real number δ > 0,such that if two numbers x′ and x′′ are in A ∩ (b − δ, b + δ), then thedistance between f(x′) and f(x′′) is less than ε, i.e |f(x′) − f(x′′)| < ε.We must prove that f has a limit L at b. For this, let xn → b, xn ∈ A,and let fix a small ε > 0. For this ε we take that δ > 0, defined by


the above hypothesis. Let N be a natural number such that if n ≥ N,then xn ∈ A∩ (b−δ, b+ δ). Applying our hypothesis to that fixed ε, weget that for any n,m ≥ N, one has that |f(xn) − f(xm)| < ε, i.e. thesequence f(xn) is a Cauchy sequence which has a limit L, dependingon the sequence xn which converges to b. If we take another sequenceyn → b and the same fixed ε with its corresponding δ, since xn, yn ∈A ∩ (b− δ, b+ δ) for any n ≥ N∗, common to both sequences, one hasthat |f(xn) − f(yn)| < ε, i.e. the sequences f(xn) and f(yn) haveone and the same limit L. Thus this L is the limit of f at b.

b) Assume now that b = ∞ and that f has the limit L at ∞. Useagain ε-δ criterion (see theorem 30, b)) for ε/2 instead of ε and continuethe same reasoning as in the finite case a).

Conversely, assume that for any small real number ε > 0 there isa large real number δ > 0 such that if two numbers x′ and x′′ aregreater than δ, then the distance between f(x′) and f(x′′) is less thanε, i.e. |f(x′) − f(x′′)| < ε. We must prove that f has limit at b = ∞.For this take a sequence xn → ∞, xn ∈ A and an arbitrary small realnumber ε > 0. For this ε take the above δ and take a natural numberN such that if n ≥ N, then xn > δ. Take now n,m ≥ N. Thus, xn,xm > δ and |f(xn) − f(xm)| < ε, i.e. the sequence f(xn) is a Cauchysequence. So it is a convergent to L sequence. Like in the proof of a)one can prove that this L does not depend on the sequence xn whichconverges to b. Hence L is the limit of f at b = ∞.

E 45. (Laplace error function) Let us use Cauchy criteriondescribed above to prove that the famous Laplace error function f(x) =∫ x0e−t

2dt, x ≥ 0, has a (finite) limit at b = ∞. Take for this x′ and

x′′ ≥ 0 and let us evaluate the distance between f(x′) and f(x′′) :

(1.1) |f(x′) − f(x′′)| =

∣∣∣∣∣

∫ x′′

x′e−t

2

dt

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′e−tdt

∣∣∣∣∣ =∣∣∣e−x′ − e−x

′′∣∣∣ ,

where x′, x′′ > 1. Take now a small ε > 0 and choose δ > 0 largeenough such that if x ≥ δ one has that e−x < ε/2. This is possiblebecause e−x → 0, when x → ∞. Coming back to formula (1.1) wefinally get that

|f(x′) − f(x′′)| < ε/2 + ε/2 = ε.

Thus Cauchy criterion works and so f has a limit at ∞. This limit isdenoted by

∫∞0e−t

2dt and its computation will be given in the next two

chapters. It is a first example of an improper (generalized) integral.

It is called in the literature the "Poisson integral" and its value is√π2.

Why an "improper integral"? Because the interval on which we perform


the integral computation of the function e−t2is not finite (bounded). In

the next section we shall systematically study such type of integrals.

2. Improper integrals of the first type

Up to now we have studied integrals on closed and bounded intervalsor on some finite unions of these, namely on subsets of the form A =∪ni=1[ai, bi], where ai < bi ≤ ai+1 < bi+1 for any i = 1, 2, ..., n−1. Let uscall such type of subsets, elementary subsets.We shall now extend theusual notion of Riemann integrals on such type of subsets to unboundedintervals or on a union between an elementary subset and one or twounbounded intervals of the form [a,∞) or (−∞, b] under the hypothesesthat any two of such subsets have in common at most a point. Forinstance,

B = (−∞,−5] ∪ [−5, 0] ∪ [1, 2] ∪ [5,∞)

is such a union. Since we can define∫ b−∞ f(x)dx by

∫∞−b f(−y)dy (make

the "change of variable" y = −x in your mind!), we can reduce every-thing to the definition of an integral of the following type

∫∞af(x)dx,

where f is a function defined on [a,∞). For∫∞−∞ f(x)dx we simply de-

fine it by limδ→∞

∫ δ−δ f(x)dx. If this last limit exists, then it is equal to

∫ 0

−∞ f(x)dx+∫∞0f(x)dx (see their definitions bellow). If the limit lim

δ→∞∫ δ−δ f(x)dx exists, we call it the principal value (pv) of the improper

integral∫∞−∞ f(x)dx. For instance, take the integral

∫∞−∞

1x2+1

dx :

pv

∫ ∞

−∞

1

x2 + 1dx = lim

δ→∞

∫ δ

−δ

1

x2 + 1dx = tan−1 ∞− tan−1(−∞) = π.

D 10. Let f : [a,∞) → R be a function defined on theunbounded interval (infinite interval) [a,∞) with real values such thatit is integrable on any subinterval of the form [a, x] ⊂ [a,∞). If thenew function F (x) =

∫ xaf(t)dt, F : [a,∞) → R has a limit L ∈ R at

b = ∞, we say that the symbol∫∞af(x)dx is convergent (or exists) and

its value is that L, i.e.∫ ∞

a

f(x)dx = limx→∞

∫ x

a

f(t)dt,

if this last limit exists. If the limit does not exist as a real number, wesay that the integral

∫∞af(x)dx is divergent or that it does not exist.

Such integrals like∫∞af(x)dx,

∫ b−∞ f(x)dx, or

∫∞−∞ f(x)dx are called

improper (generalized) integrals of the first type.

2. IMPROPER INTEGRALS OF THE FIRST TYPE 113

For instance,∫∞0

11+x2

dx is convergent because

limx→∞

∫ x

0

1

1 + t2dt = lim

x→∞[arctan x− arctan 0] =

π

2.

Thus,∫∞0

11+x2

dx = π2.

E 46. (Gabriel’s trumpet) Let us take a > 0 and f(x) = 1x

defined on the infinite interval [a,∞). Let us take A > a and let usrotate the arc of the graphic of f restricted to [a,A] around Ox-axis.The side surface of the solid obtained in this way is like a trumpet. Letus make A go to ∞. The "infinite" trumpet obtained in this way iscalled a Gabriel’s trumpet. The infinite area bounded by the graphic off(x), x ∈ [a,∞), the lines x = a and the Ox-axis (see Fig.1)

F 1

is ∫ ∞

a

1

xdx = lim

x→∞

∫ x

a

1

tdt = lim

x→∞[ln x− ln a] = ∞,

so this last improper integral is divergent. Let us compute the volumeof the revolutionary solid bounded by the Gabriel’s trumpet:

π

∫ ∞

a

1

x2dx = π lim

x→∞

∫ x

a

1

t2dt = π lim

x→∞

[−1

x+

1

a

]=π

a.

Thus the volume of this infinite solid is finite, namely πa. Hence the last

improper integral of the first type is convergent. If a > 0 goes to zero,the volume goes to ∞. Very strange! The area of a longitudinal centralsection of the Gabriel’s trumpet is infinite and its volume is finite!

T 32. (zero criterion) Let us consider the conditions andnotation from definition 10. If the integral

∫∞af(x)dx is convergent,

then there exists at least one sequence qn , qn → ∞, such that f(qn) →


0. In particular, if the integral∫∞af(x)dx is convergent, and if lim

x→∞f(x)

exists, then it must be zero. Thus, if the integral∫∞af(x)dx is conver-

gent, if f is continuous and if limx→∞

f(x) exists, then f must be bounded.

Hence, if limx→∞

f(x) exists and it is not equal to zero, then the improper

integral∫∞af(x)dx is divergent.

P. Let n0 be the least natural number greater then a. Then∫ ∞

a

f(x)dx =

∫ n0

a

f(x)dx+∞∑

n=n0

∫ n+1

n

f(x)dx.

But ∫ n+1

n

f(x)dx = f(qn)

∫ n+1

n

dx = f(qn),

where qn ∈ [n, n+1] (mean theorem). Since the series∑∞n=n0

∫ n+1

nf(x)dx

is convergent, its general term f(qn) converges to zero, when n → ∞.The other statements are simple consequences of the definition of alimit of a function at ∞ ( lim

x→∞f(x) = lim

n→∞f(qn) = 0 for any sequence

qn → ∞) and of the basic properties of sequences on a compact inter-val (see [Po], Weierstrass theorem). Indeed, if f were not bounded,there exists a sequence xn, xn ∈ [a,∞) for n = 1, 2, ..., such thatf(xn) → ∞ (say; you can easily put −∞ instead ∞). If xn itself isbounded, there is a subsequence xkn of xn with xkn → x0 ∈ [a,∞)(Cesàro’s Lemma, a particular case of Weierstrass theorem, see [Po]).The continuity of f implies that f(xkn) → f(x0). Since f(xkn) is asubsequence of f(xn), then f(xkn) → ∞. The uniqueness of the limitof a sequence implies that f(x0) = ∞, a pure contradiction! Thus thesequence xn is not bounded to ∞. Let xtn be a subsequence ofxn with xtn → ∞. Since lim

x→∞f(x) = 0, one has that f(xtn) → 0. But

f(xtn) → ∞, xtn being a subsequence of xn , thus we again obtaina contradiction (0 = ∞!). Hence f cannot be unbounded.

Here is an example of an unbounded function f(x) on [a,∞) suchthat

∫∞af(x)dx is convergent (contrary to a Riemann integral on a

finite and closed interval [a, b]!). Take N large enough such that N ≥ aand define f : [a,∞) → R, f(N + i) = N + i for any i = 0, 1, 2, ... andf(x) = 0 for any x /∈ N,N + 1, N + 2, .... Using only the definitionof an improper integral of the first type and theorem 19 we see that∫∞af(x)dx is convergent and it is zero, in spite of that fact that f is

unbounded: f(N + i) = N + i→ ∞, when i→ ∞.The following example is a standard example.


E 47. Let a be a positive real number. For any α ∈ R wedefine I(α) =

∫∞a

1xαdx. Then this integral is convergent if and only if

α > 1. Indeed, if α = 1, I(1) = lnx |∞a = ∞. Assume that α = 1. Then

F (x) =

∫ x

a

1

tαdt =

t−α+1

−α + 1

∣∣∣∣x

a

=x−α+1

−α + 1− a−α+1

−α + 1.

Since limx→∞

x−α+1 is finite and 0 in this case, if and only if −α+ 1 < 0,

i.e. α > 1.

There is a great number of similarities between the improper in-tegrals of the first type and the numerical series. First of all, for afunction f : [a,∞) → R+, such that for any x > a the "proper" inte-gral

∫ xaf(t)dt exists, let us write formally (for a fixed natural number

N ≥ a):

(2.1)

∫ ∞

a

f(x)dx =

∫ N

a

f(x)dx+∞∑

n=N

cn,

where cn =∫ n+1

nf(x)dx, n = N,N+1, ... . It is clear that the improper

integral is convergent if and only if the last numerical series on theright is convergent (sketch a proof!). If f has positive and also negativevalues "up to ∞”, this statement is not always true. Indeed, let f(x) =sin 2πx, x ∈ [0,∞). Since

F (x) =

∫ x

0

sin 2πtdt = − cos 2πt

2π

∣∣∣∣x

0

=1

2π[1 − cos 2x] ,

and since cosx has no limit at ∞, F (x) has no limit at ∞. Thus theimproper integral

∫∞0

sin 2πxdx is not convergent.

However, cn =∫ n+1

nsin 2πxdx = 0 and the series on the right in

formula (2.1) is equal to zero!We recall a nice theorem (theorem 21-the integral test, [Po]) which

is stated here in language of improper integrals.

T 33. (series test) Let f : [a,∞) → R+, be a function fdefined on an interval [a,∞), a ≥ 0, with nonnegative values. Assumealso that f is a continuous and a decreasing function. Let N be thefirst natural number greater or equal to a and let cn = f(n) for anyn = N,N + 1, ... . Then the improper integral

∫∞af(x)dx is convergent

if and only if the numerical series∑∞n=N cn is convergent.

Since the proof of this result was done in [Po], theorem 21, we omitit here.


For instance, the integral∫∞a

1x100+1

dx, a ≥ 0, is convergent because

the numerical series∑∞n=[a]+1

1n100+1

is convergent (apply the limit com-

parison test with the Riemann series for α = 100 > 1, etc.). To tryto compute directly a primitive for 1

x100+1is a crazy idea! And, if we

know that this integral is convergent but we cannot compute it ex-actly, where is the practical gain? There is a big one! If we know that

this integral is convergent we can approximate it with∫Ma

1x100+1

dx,whereM is a large natural number. But this last integral can be easilyapproximately computed by using some specialized software.

If we try to use approximate methods for computing a divergentimproper integral is another crazy idea! This means to obtain differentnumerical values which are randomly distributed to ∞. Thus thesevalues cannot approximate some unknown number!

This is way we need some tests in order to decide if an improperintegral is convergent or not. A basic test is obtained by applying theCauchy criterion for limits of functions (see theorem 31).

T 34. (Cauchy criterion for improper integrals I)Let f : [a,∞) → R be a function which is integrable on any finite

interval [a, x] for x ∈ [a,∞). Then∫∞af(x)dx is convergent if and only

if for any small ε > 0, there exists δ large enough and depending on ε,

such that if x′, x′′ ∈ [δ,∞), then∣∣∣∫ x′′x′f(x)dx

∣∣∣ < ε. This is equivalent

to say that∫ x′′x′f(x)dx→ 0, whenever x′, x′′ → ∞.

P. We simply apply theorem 31, b) to the function F (x) =∫ xaf(t)dt, etc.

E 48. Let us study the convergence of the integral I =∫∞1

arctanxx2

dx. Practically we want to prove that the limit of the func-

tion G(x′, x′′) =∣∣∣∫ x′′x′

arctanxx2

dx∣∣∣ is equal to zero, whenever x′, x′′ → ∞,

independently. Indeed, use the basic mean theorem (see theorem 22)and find∣∣∣∣∣

∫ x′′

x′

arctan x

x2dx

∣∣∣∣∣ = |arctan c|∣∣∣∣∣

∫ x′′

x′

1

x2dx

∣∣∣∣∣ ≤π

2

∣∣∣∣1

x′− 1

x′′

∣∣∣∣ → 0,

when x′, x′′ → ∞.

This last test is very used to derive other more useful in practicetests. In many cases the following test reduces the study of the im-proper integrals of the first type

∫∞af(x)dx to the case of the nonneg-

ative function |f(x)| . If the integral∫∞a

|f(x)| dx is convergent, we saythat

∫∞af(x)dx is absolutely convergent.


T 35. (absolute convergence implies convergence) Supposethat the integral

∫∞af(x)dx is absolutely convergent. Then it is also

convergent.

P. Since∫∞af(x)dx is absolutely convergent, one has that

the integral∫∞a

|f(x)| dx is convergent, so by using Cauchy criterionwe have that ∣∣∣∣∣

∫ x′′

x′|f(x)| dx

∣∣∣∣∣ → 0,

whenever x′, x′′ → ∞. Since∣∣∣∣∣

∫ x′′

x′f(x)dx

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′|f(x)| dx

∣∣∣∣∣ ,

one has that∣∣∣∫ x′′x′f(x)dx

∣∣∣ → 0, whenever x′, x′′ → ∞, i.e. our integral∫∞af(x)dx is also convergent.

For instance, let us prove that the integral I =∫∞1

sinxx2dx is con-

vergent. We shall prove that it is absolutely convergent. Indeed,∣∣∣∣∣

∫ x′′

x′

|sin x|x2

dx

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′

1

x2dx

∣∣∣∣∣ =

∣∣∣∣1

x′− 1

x′′

∣∣∣∣ → 0,

when x′ and x′′ go to ∞. Applying again Cauchy criterion, we get thatthe integral I is absolutely convergent.

We give now some convergence criteria for improper integrals of thefirst type of nonnegative functions.

T 36. (comparison test 1) Let f, g : [a,∞) → R+ be twofunctions defined on the interval [a,∞) with nonnegative values, inte-grable on any subinterval [a, x] ⊂ [a,∞) and such that f(x) ≤ g(x)for any x ∈ [a,∞). i) If

∫∞ag(x)dx is convergent, then

∫∞af(x)dx is

also convergent. ii) If∫∞af(x)dx is divergent, then

∫∞ag(x)dx is also

divergent.

P. We apply the Cauchy criterion in both cases.i) If


∣∣∣∣∣

∫ x′′

x′g(x)dx

∣∣∣∣∣ → 0,

when x′, x′′ → ∞. But∣∣∣∣∣

∫ x′′

x′f(x)dx

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′g(x)dx

∣∣∣∣∣ → 0,


so∣∣∣∫ x′′x′f(x)dx

∣∣∣ → 0, whenever x′, x′′ → ∞. Hence∫∞af(x)dx is con-

vergent.ii) If

∫∞ag(x)dx were convergent, from i) we would have that∫∞

af(x)dx is convergent, a contradiction. So

∫∞ag(x)dx must be

divergent.

For instance, since1

x8 + 7<

1

x8

and since∫∞1

1x8dx is convergent (α = 8 > 1 in example 47), this last

theorem tells us that∫∞1

1x8+7

dx is convergent. Since∫ ∞

0

1

x8 + 7dx =

∫ 1

0

1

x8 + 7dx+

∫ ∞

1

1

x8 + 7dx,

and since∫ 1

01

x8+7dx is a proper integral, one has that

∫∞0

1x8+7

dx is aconvergent integral. This is because

limx→∞

∫ x

0

1

t8 + 7dt = lim

x→∞

∫ 1

0

1

t8 + 7dt+ lim

x→∞

∫ x

1

1

t8 + 7dt =

=

∫ 1

0

1

t8 + 7dt +

∫ ∞

1

1

t8 + 7dt.

T 37. (limit comparison test) Let f, g : [a,∞) → R+ betwo functions defined on the interval [a,∞) with nonnegative values,integrable on any subinterval [a, x] ⊂ [a,∞). Assume in addition thatg(x) > 0 for any x ∈ [a,∞) and that the following limit "exists":

limx→∞

f(x)g(x)

= L ∈ R ∪ ∞.i) If L = 0,∞, then the improper integrals

∫∞af(x)dx and

∫∞ag(x)dx

have the same nature, i.e. they are simultaneously convergent or diver-gent.

ii) If L = 0 and∫∞ag(x)dx is convergent, then

∫∞af(x)dx is also

convergent.iii) If L = ∞ and

∫∞ag(x)dx is divergent, then

∫∞af(x)dx is also

divergent.

P. i) Take an ε > 0 sufficiently small such that L−ε > 0 (since

f and g have nonnegative values, L is nonnegative). Since limx→∞

f(x)g(x)

= L,

there exists a large enough δ > a such that if x ∈ [δ,∞), then

(2.2) L− ε <f(x)

g(x)< L + ε,


or

(2.3) (L− ε) g(x) < f(x) < (L+ ε)g(x).

If∫∞af(x)dx is convergent, then

∫∞δf(x)dx and

∫∞δ

f(x)L−εdx are also

convergent; since g(x) < f(x)L−ε , one has that

∫∞δg(x)dx is also convergent

(see theorem 36). Since∫ ∞

a

g(x)dx =

∫ δ

a

g(x)dx+

∫ ∞

δ

g(x)dx,

we have that∫∞ag(x)dx is convergent. If


(L+ ε)∫∞ag(x)dx and (L+ ε)

∫∞δg(x)dx are also convergent. Use

the last inequality of 2.3 and theorem 36 to see that∫∞δf(x)dx is also

convergent. Since

(2.4)

∫ ∞

a

f(x)dx =

∫ δ

a

f(x)dx+

∫ ∞

δ

f(x)dx,

one has that∫∞af(x)dx is convergent.

ii) If L = 0, in formula (2.3) we use only the last inequality to seethat if

∫∞ag(x)dx is convergent, then ε

∫∞ag(x)dx and ε

∫∞δg(x)dx are

also convergent so∫∞δf(x)dx is convergent. From formula (2.4) we get

that∫∞af(x)dx itself is convergent.

iii) If L = ∞, then for any fixedM > 0, there exists a large number

δ > a such that f(x)g(x)

> M if x ∈ [δ,∞). Since f(x) > Mg(x), x ∈ [δ,∞)

and since∫∞ag(x)dx and M

∫∞δg(x)dx are divergent, then

∫∞δf(x)dx

is divergent. Now we apply again formula (2.4) to see that∫∞af(x)dx

is divergent.

For instance, since

limx→∞

13√x2+1

1

x23

= 1

and since∫∞1

1

x23dx is divergent (α = 2/3 in example 47), this last

theorem says that∫∞1

13√x2+1

dx is divergent.

If in theorem 37 we take a > 0 and g(x) = 1xp, we get a very useful

test.

T 38. (test p) Let f : [a,∞) → R+ be a function definedon the interval [a,∞) with nonnegative values, integrable on any subin-terval [a, x] ⊂ [a,∞). Let

limx→∞

xpf(x) = L ∈ R ∪ ∞.


i) If L = 0,∞ and p > 1, then∫∞af(x)dx is convergent. If L = 0,∞

and p ≤ 1, then∫∞af(x)dx is divergent.

ii) If L = 0 and p > 1, then∫∞af(x)dx is convergent.

iii) If L = ∞ and p ≤ 1, then∫∞af(x)dx is divergent.

If L = 0,∞, we call the number k = −p, Abel’s degree of f(x).For instance, Abel degree of f(x) = x3 + 1 is equal to k = 3 becauselimx→∞

x−3(x3 + 1) = 1. The Abel degree is uniquely determined. Indeed,

limx→∞

xpf(x) = limx→∞

xp′f(x) = L = 0 implies that lim

x→∞xp−p

′= 1, or that

p = p′.Test p is very useful in apparently complicated problems. For in-

stance, is it possible to use an approximation like this:∫ ∞

0

√x+ 1 −√

x3√x2 + 1

dx ≈∫ 10000

0

√x+ 1 −√

x3√x2 + 1

dx?

To find an elementary primitive of√x+1−√x3√x2+1

is not a good idea! Let

us prove that this improper integral of the first type is convergent andthen everything will be clear.

For this, let us use test p (see theorem 38):

limx→∞

xp√x+ 1 −√

x3√x2 + 1

= limx→∞

xp(√x+ 1 +

√x)

3√x2 + 1

=

= limx→∞

xp

x12

(√1 + 1

x+ 1

)x233

√1 + 1

x2

=1

2limx→∞

xp

x12+ 23

= 0,∞

if and only if p = 76, which is greater than 1. Thus, our integral is

convergent (see theorem 38) and we can use that approximation. The

approximate integral∫ 10000

0

√x+1−√x3√x2+1

dx is a Riemann integral and it can

be easily approximated by one the usual method described in Chapter1.

Except Cauchy criterion, all the other tests given above are aboutimproper integrals of a nonnegative function. Here is a basic and use-ful test on improper integrals of functions which are not necessarilynonnegative.

T 39. (Dirichlet’s test) Let f, g be two continuous functionsdefined on [a,∞) with values in R. Assume that the Newton primitivefunction F (x) =

∫ xaf(t)dt is bounded on [a,∞), i.e. there exists a

number M > 0 such that |F (x)| ≤ M for any x ∈ [a,∞). We alsoassume that g is a function of class C1 on [a,∞) and lim

x→∞g(x) = 0.

Then∫∞af(x)g(x)dx is convergent.


P. We simply apply the Cauchy criterion (see theorem 34).∫ x′′

x′f(x)g(x)dx =

∫ x′′

x′F ′(x)g(x)dx

by parts= F (x)g(x) |x′′x′ −

−∫ x′′

x′F (x)g′(x)dx

mean formula= F (x′′)g(x′′) − F (x′)g(x′)−

−F (cx′,x′′) [g(x′′) − g(x′)] , where cx′,x′′ ∈ [x′, x′′].

Thus, ∣∣∣∣∣

∫ x′′

x′f(x)g(x)dx

∣∣∣∣∣ ≤

≤ |F (x′′)g(x′′)| + |F (x′)g(x′)| + |F (cx′,x′′)| [|g(x′′)| + |g(x′)|] ≤(2.5)M [|g(x′′)| + |g(x′)|] +M [|g(x′′)| + |g(x′)|] = 2M [|g(x′′)| + |g(x′)|] .

Take now a small ε > 0 and let δ > 0 be large enough such that ifx ∈ [δ,∞) then |g(x)| < ε

4M. This is true because lim

x→∞g(x) = 0. This δ

is the searched for δ in Cauchy criterion:∣∣∣∣∣

∫ x′′

x′f(x)g(x)dx

∣∣∣∣∣ ≤ 2M [|g(x′′)| + |g(x′)|] < 2M[ ε

4M+

ε

4M

]= ε,

if x′, x′′ ∈ [δ,∞).

E 49. (Dirichlet’s integral) If β is a real number, the fol-lowing integral

∫∞0

sinβxxdx is called the Dirichlet’s integral. We can

assume that β > 0. Let us prove that this integral is convergent. Firstof all let us remark that the function sinβx

xdefined initially only on

(0,∞) can be extended to a continuous function

h(x) =

sinβxx, if x ∈ (0,∞)β, if x = 0,

defined on [0,∞). Thus, we can substitute the initial integral with∫∞0h(x)dx. This last one is "identical" with the initial one because

does not matter what we put in x = 0, the value of the integral doesnot change if that one is convergent. Since

∫ ∞

0

sin βx

xdx =

∫ 1

0

sin βx

xdx+

∫ ∞

1

sinβx

xdx,

it will be enough to prove that the last integral∫∞1

sinβxxdx is convergent.

In order to prove this we apply the Dirichlet’s test by putting f(x) =sin βx, which has the Newton’s primitive F (x) =

∫ x1

sin βtdt = cosββ

−cos βxβ

bounded by 2βon [1,∞), and g(x) = 1

x, which is of class C1 and


decreasing to 0 at ∞. Hence the Dirichlet’s integral is convergent forany fixed β ∈ R.

E 50. (Fresnel’s integrals) The following integralsI =

∫∞0

sin x2dx and J =∫∞0

cosx2dx are called the Fresnel’s inte-grals. Let us prove that I is convergent (in fact both are convergent!).For this, let us write

∫ ∞

0

sin x2dx =

∫ 1

0

sin x2dx+

∫ ∞

1

sin x2dx.

It is enough to prove that∫∞1

sin x2dx is convergent. Let us denote

F (x) =∫ x1

sin t2dt and let us change the variable t =√u. So

∫ x

1

sin t2dt =1

2

∫ x2

1

1√u

sinudu.

Since any real number z ∈ [1,∞) can be written as z = x2 for an x ∈[1,∞), it is enough to see that the integral

∫∞1

1√u

sin udu is convergent.

Here we apply again the Dirichlet’s test for f(u) = sin u and g(u) = 1√u.

So the first Fresnel’s integral is convergent. We leave as an exercisefor the reader to prove (in the same way!) that the second Fresnel’sintegral is also convergent.

3. Improper integrals of the second type

Let f : [a, b] → R be a (Riemann) integrable function on the closedand bounded interval [a, b]. Theorem 13 says that in this case f mustbe a bounded function, i.e. there exists a real numberM > 0 such that|f(x)| ≤ M for any x ∈ [a, b]. The next result tries to say somethingon the reverse statement.

T 40. Let f : [a, b) → R be a bounded function on theinterval [a, b) which is integrable on any closed subinterval [a, x], wherex ∈ [a, b). Then f is integrable on the entire interval [a, b] and

(3.1)

∫ b

a

f(x)dx = limx→b,x<b

∫ x

a

f(t)dt.

If f is integrable on any subinterval [y, b], where y ∈ (a, b], then againf is integrable on the entire interval [a, b] and

(3.2)

∫ b

a

f(x)dx = limy→a,y>a

∫ b

y

f(t)dt

3. IMPROPER INTEGRALS OF THE SECOND TYPE 123

P. Since the proof for the second part of the statement is verysimilar to the proof of the first part, we shall prove only this last one.To prove the integrability of f on [a, b] we shall use Darboux Criterion(theorem 14). Since f is bounded on [a, b] one can choose M > 0 suchthat |f(x)| ≤ M on [a, b]. Let ε > 0 be a small positive real numberand let δ1 = ε/2M. Let δ > 0 such that δ ≤ δ1 and for any partition∆ = a = x0, x1, ..., xn−1 = b − δ1 of the interval [a, b − δ1] with‖∆‖ < δ, one has:

S∆ − s∆ =n−1∑

i=1

(Mi −mi) (xi − xi−1) < ε/2.

Here Mi = supx∈[xi−1,xi]

f(x) and mi = infx∈[xi−1,xi]

f(x). Let us consider now

the partition

(3.3) ∆′ = a = x0, x1, ..., xn−1, xn = b = ∆ ∪ bof the interval [a, b]. If we keep the above δ fixed, any partition ∆′′ of[a, b], ‖∆′′‖ < δ, can be represented as a partition ∆′ like in formula(3.3), with a δ′1 ≤ δ1, instead of δ1. Thus

S∆′ − s∆′ ≤ S∆ − s∆ + (Mn −mn)δ1 ≤ ε/2 +M · ε/2M = ε.

Applying Darboux Criterion we get that f is integrable and∫ b

a

f(x)dx = lim‖∆′‖→0

S∆′ = lim‖∆‖→0

S∆ =

= limδ1→0

∫ b−δ1

a

f(x)dx = limx→b,x<b

∫ x

a

f(t)dt.

For instance, applying this last result we find that the noncontinu-ous function f : [0, 1] → R,

f(x) =

x, if x ∈ [0, 1)

2, if x = 1,

is integrable and∫ 1

0

f(x)dx = limx→1,x<1

∫ x

0

tdt = limx→1,x<1

x2

2=

1

2.

We say that a function f : I → R, defined on an interval I withreal values is unbounded at a limit point b of I if there exists at leastone sequence xn with xn ∈ I, xn → b such that the sequence f(xn)is not bounded. For instance, f(x) = 1/x , x ∈ (0, 1] is not boundedat 0. We also say that 0 is a singular point for f. We can also write


limx→0,x>0

1x

= ∞, thus f(x) = 1/x is not bounded at 0. In our definition

we can take xn = 1/n.

D 11. Let [a, b] be a bounded and closed interval in Rand let f : [a, b) → R such that f is integrable on any subinterval [a, x],x ∈ [a, b) and f is unbounded at b.We say that f is integrable on [a, b] ifthe following limit lim

x→b,x<b

∫ xaf(t)dt exists, is finite and in this case we

denote it by∫ baf(t)dt. This last integral is called an improper integral

of the second type with singularity at b. We also say that the integral∫ baf(t)dt is convergent.

E 7. Let f : (a, b] → R be an integrable function on anyinterval [x, b], x ∈ (a, b] such that f is unbounded at a. Define the

convergence of∫ baf(t)dt in this last case.

For instance, f(x) = 1/x, x ∈ (0, 1] is unbounded at 0 and∫ 1

01xdx

is divergent because

limx→0,x>0

∫ 1

x

1

tdt = lim

x→0,x>0[ln 1 − ln x] = ∞.

But∫ 1

01√xdx is convergent because

limx→0,x>0

∫ 1

x

1√tdt = lim

x→0,x>02t

12

∣∣∣∣1

x

= limx→0,x>0

[2 − 2x

12

]= 2.

If f is unbounded at an interior point c of [a, b] but all the integrals∫ xaf(t)dt, x > a, x < c and

∫ byf(t)dt, y > c, y < b exist, then, by

definition,∫ baf(x)dx exists (is convergent) if both improper integrals

of the second type∫ caf(x)dx and

∫ bcf(x)dx exist (are convergent). In

this last case we write∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

For instance, I =∫ 1

−113√xdx is convergent because both integrals I1 =∫ 0

−113√xdx and I2 =

∫ 1

013√xdx are convergent. Indeed,

limx→0,x<0

∫ x

−1

13√tdt = lim

x→0,x<0

[3

2x23 − 3

2

]= −3

2,

and

limx→0,x>0

∫ 1

x

13√tdt = lim

x→0,x>0

[3

2− 3

2x23

]=

3

2.

Thus, I = I1 + I2 = 0.


In the case when at least one of the integrals∫ caf(x)dx or

∫ bcf(x)dx

is divergent (nonconvergent) one can define the principal value of

∫ b

a

f(x)dx

by the following formula

pv

∫ b

a

f(x)dxdef= lim

ε→0,ε>0

[∫ c−ε

a

f(x)dx+

∫ b

c+ε

f(x)dx

].

If this last limit exists, in practice one can work with it, but,...verycarefully, because its value is only a convention.

For instance,∫ 1

−11xdx is divergent because neither

∫ 0

−11xdx, nor∫ 1

01xdx is convergent.

However, the principal value of∫ 1

−11xdx is 0. Indeed,

∫ 1

−1

1

xdx = lim

ε→0,ε>0

[∫ −ε

−1

1

xdx+

∫ 1

ε

1

xdx

]=

= limε→0,ε>0

[ln ε− ln 1 + ln 1 − ln ε] = 0.

To use in practice the definition 11 is usually a very difficult taskbecause we must compute integrals of the type

∫ xaf(t)dt. In practical

problems we want to know if our improper integral of the second type∫ baf(x)dx with singularity at b is convergent or not. If it is convergent,

then we can approximate it with∫ b−εa

f(x)dx for a very small chosenpositive number ε. For instance, ε = 10−6. Now, to compute this last

Riemann integral∫ b−εa

f(x)dx one can use any approximate formulajust studied in Chapter 2. To decide the convergence of an improper

integral of the second type∫ baf(x)dx with singularity at b, we need

some "tests".We begin with a standard example.

E 51. Let I(α) =∫ ba

1(b−x)αdx be an improper integral of the

second type with singularity at b, where α is a fixed positive real number.Let us prove that our integral is convergent if and only if α < 1. Indeed,

(3.4) I(α) = limx→b,x<b

∫ x

a

1

(b− t)αdt.

But ∫ x

a

1

(b− t)αdt = −(b− x)−α+1

−α + 1+

(b− a)−α+1

−α + 1,


if α = 1, and ∫ x

a

1

(b− t)αdt = ln

b− a

b− x,

if α = 1. It is easy to see that the limit in (3.4) exists if and only if

α < 1. In this last case I(α) = (b−a)−α+1

−α+1.

If F (x) =∫ xaf(t)dt, x ∈ [a, b) is the area function of Newton for

function f which appears in the last definition, then∫ baf(t)dt is con-

vergent if and only if the function F : [a, b) → R has a (unique) finitelimit at the point b. Thus we can apply the general Cauchy Criterion(see theorem 31, a)) for the function F and obtain:

T 41. (Cauchy criterion for improper integrals II) Let f :[a, b) → R be a function which is integrable on any finite interval [a, x]

for x ∈ [a, b). Then∫ baf(x)dx is convergent if and only if for any small

ε > 0, there exists δ small enough and depending on ε, such that if

x′, x′′ ∈ (b− δ, b), then∣∣∣∫ x′′x′f(x)dx

∣∣∣ < ε. This is equivalent to say that∫ x′′x′f(x)dx→ 0, whenever x′, x′′ → b, x′, x′′ ∈ [a, b) and x′, x′′ < b.

E 52. Let us apply this last Cauchy criterion to see if theimproper integral of the second type

∫ 2

1sinx√x−1

dx is convergent or not.

Here x = 1 is the unique singularity of f(x) = sinx√x−1

on (1, 2]. Take an

ε > 0 and let us evaluate the quantity∣∣∣∫ x′′x′

sinx√x−1

dx∣∣∣ . Using the mean

formula (see theorem 22), we get∫ x′′

x′

sin x√x− 1

dx = sin c

∫ x′′

x′

1√x− 1

dx.

Thus, ∣∣∣∣∣

∫ x′′

x′

sin x√x− 1

dx

∣∣∣∣∣ ≤ 2√x′′ − 1 + 2

√x′ − 1 < ε,

if x′, x′′ ∈ (1, 1 + δ), with δ = ε2

16. Here a neighborhood of 1 in (1, 2] is

of the form (1, 1 + δ), with δ small enough.

The similarity with the improper integrals of the first type is nowvery clear. Indeed, it is enough to put instead of ∞, the real number band instead of the neighborhood (δ,∞) of infinite, with δ large enough,the "neighborhood" (b− δ, b) = [a, b) ∩ (b− δ, b+ δ) of b in A = [a, b).Even the other results relative to improper integrals of the first typecan be transferred almost word by word for improper integrals of thesecond type, to be clear, we prefer to present them here with proofs.


Cauchy criterion is very used to derive other more useful in practicetests. In many cases the following test reduces the study of the im-

proper integrals of the second type∫ baf(x)dx, with singularity at b, to

the case of the nonnegative function |f(x)| . If the integral∫ ba|f(x)| dx

is convergent, we say that∫ baf(x)dx is absolutely convergent.

T 42. (absolute convergence implies convergence) Suppose

that the integral of the second type,∫ baf(x)dx, with singularity at b, is

absolutely convergent. Then it is also convergent.

P. Since∫ baf(x)dx is absolutely convergent, one has that the

integral∫ ba|f(x)| dx is convergent thus, by using Cauchy criterion, we

have that ∣∣∣∣∣

∫ x′′

x′|f(x)| dx

∣∣∣∣∣ → 0,

whenever x′, x′′ → b. Since∣∣∣∣∣

∫ x′′

x′f(x)dx

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′|f(x)| dx

∣∣∣∣∣ ,

one has that∣∣∣∫ x′′x′f(x)dx

∣∣∣ → 0, whenever x′, x′′ → b, i.e. our integral∫∞af(x)dx is also convergent.

For instance, let us use this last result to prove that the integral

I =∫ 1

0sin( 4

√x)√x

dx, with the unique singularity x = 0, is convergent. We

shall prove that it is absolutely convergent. Indeed,∫ x′′

x′

|sin( 4√x)|√x

dx ≤∣∣∣∣∣

∫ x′′

x′

1√xdx

∣∣∣∣∣ =∣∣∣2√x′′ − 2

√x′∣∣∣ → 0,

when x′ and x′′ go to 0. Applying again Cauchy criterion, we get thatthe integral I is absolutely convergent.

Thus, first of all we give some criteria for improper integrals of thesecond type of nonnegative functions.

T 43. (comparison test 1) Let f, g : [a, b) → R+ be twofunctions defined on the interval [a, b) with nonnegative values, in-tegrable on any subinterval [a, x] ⊂ [a, b), unbounded at b and such

that f(x) ≤ g(x) for any x ∈ [a, b). i) If∫ bag(x)dx is convergent,

then∫ baf(x)dx is also convergent. ii) If

∫ baf(x)dx is divergent, then∫ b

ag(x)dx is also divergent.


P. We apply Cauchy criterion in both cases. i) If∫ bag(x)dx

is convergent, then ∣∣∣∣∣

∫ x′′

x′g(x)dx

∣∣∣∣∣ → 0,

when x′, x′′ → b, where x′, x′′ ∈ [a, b). But∣∣∣∣∣

∫ x′′

x′f(x)dx

∣∣∣∣∣ ≤∣∣∣∣∣

∫ x′′

x′g(x)dx

∣∣∣∣∣ → 0,

so∣∣∣∫ x′′x′f(x)dx

∣∣∣ → 0, whenever x′, x′′ → b. Hence∫ baf(x)dx is conver-

gent.

ii) If∫ bag(x)dxwere convergent, from i) we would have that

∫ baf(x)dx

is convergent, a contradiction. So∫ bag(x)dx must be divergent.

For instance, since

1√x(1 + x)

<1√x, for any x ∈ (0, 1],

and since∫ 1

01√xdx is convergent (α = 1/2 < 1 in example 51), this last

theorem tells us that∫ 1

01√

x(1+x)dx is convergent.

T 44. (limit comparison test) Let f, g : [a, b) → R+ be twofunctions defined on the interval [a, b) with nonnegative values, with b asthe unique singularity and integrable on any subinterval [a, x] ⊂ [a, b).Assume in addition that g(x) > 0 for any x ∈ [a, b) and that the

following limit "exists": limx→b

f(x)g(x)

= L ∈ R ∪ ∞.i) If L = 0,∞, then the improper integrals

∫ baf(x)dx and

∫ bag(x)dx

have the same nature, i.e. they are simultaneously convergent or diver-gent.

ii) If L = 0 and∫ bag(x)dx is convergent, then

∫ baf(x)dx is also

convergent.

iii) If L = ∞ and∫ bag(x)dx is divergent, then

∫ baf(x)dx is also

divergent.

P. i) Take an ε > 0 sufficiently small such that L−ε > 0 (since

f and g have nonnegative values, L is nonnegative). Since limx→b

f(x)g(x)

= L,

there exists a small enough δ > 0 such that a < b−δ and if x ∈ [b−δ, b),then

L− ε <f(x)

g(x)< L + ε,


or

(3.5) (L− ε) g(x) < f(x) < (L+ ε)g(x).

If∫ baf(x)dx is convergent, then

∫ bb−δ f(x)dx and

∫ bb−δ

f(x)L−εdx are also

convergent; since g(x) < f(x)L−ε , one has that

∫ bb−δ g(x)dx is also conver-

gent (see theorem 43). Since∫ b

a

g(x)dx =

∫ b−δ

a

g(x)dx+

∫ b

b−δg(x)dx,

we have that∫ bag(x)dx is convergent. If

∫ bag(x)dx is convergent, then

(L + ε)∫ bag(x)dx and (L + ε)

∫ bb−δ g(x)dx are also convergent. Use

the last inequality of (3.5) and theorem 43 to see that∫ bb−δ f(x)dx is

convergent. Since

(3.6)

∫ b

a

f(x)dx =

∫ b−δ

a

f(x)dx+

∫ b

b−δf(x)dx,

one has that∫ baf(x)dx is convergent.

ii) If L = 0, in formula (3.5) we use only the last inequality to see

that if∫ bag(x)dx is convergent, then ε

∫ bag(x)dx and ε

∫ bb−δ g(x)dx are

also convergent and∫ bb−δ f(x)dx is convergent. From formula (3.6) we

get that∫ baf(x)dx itself is convergent.

iii) If L = ∞, then for any fixed M > 0, there exists a small

number δ > 0 such that f(x)g(x)

> M if x ∈ [b−δ, b). Since f(x) > Mg(x),

x ∈ [b−δ, b) and since∫ bag(x)dx andM

∫ bb−δ g(x)dx are divergent, then∫ b

b−δ f(x)dx is also divergent. Now we apply again formula (3.6) to see

that∫ baf(x)dx is divergent.

For instance, since

limx→0,x>0

1√x(x+1)

1√x

= 1

and since∫ 2

01√xdx is convergent (α = 1/2 < 1 in example 51), this last

theorem says that∫ 2

01√

x(x+1)dx is convergent.

If in theorem 44 we take g(x) = 1(b−x)q , we get a very useful test.

T 45. (test q) Let f : [a, b) → R+ be a function defined onthe interval [a, b) with nonnegative values, with the unique singularity


b and integrable on any subinterval [a, x] ⊂ [a,∞). Let

limx→b,x<b

(b− x)qf(x) = L ∈ R ∪ ∞.

i) If L = 0,∞ and q < 1, then∫ baf(x)dx is convergent.

ii) If L = 0 and q < 1, then∫ baf(x)dx is also convergent.

iii) If L = ∞ and q ≥ 1, then∫ baf(x)dx is divergent.

Test q is very useful in apparently complicated problems. For in-stance, is it possible to use an approximation like this:

∫ 1

0

√x3 + 3

5√

1 − xdx ≈

∫ 0.99999

0

√x3 + 3

5√

1 − xdx?

To find an elementary primitive of√x3+35√1−x is not a good idea! Let us

prove that this improper integral of the second type is convergent andthen everything will be clear (see definition 11).

For this, let us use test q (see theorem 45):

limx→1,x<1

(1 − x)q√x3 + 3

5√

1 − x= L = 0,∞

if and only if q = 15. Then L = 2. Since q = 1/5, test q says that our

integral∫ 1

0

√x3+35√1−x dx is convergent and so we can use the indicated ap-

proximation. The approximate integral∫ 0.99999

0

√x3+35√1−x dx is a Riemann

integral and it can be easily approximated by one the usual methoddescribed in Chapter 2.

Except the Cauchy criterion, all the other tests given above areabout improper integrals of a nonnegative function. Here is a basic anduseful test on improper integrals of functions which are not necessarilynonnegative.

T 46. (Dirichlet’s test) Let f, g be two continuous functionsdefined on [a, b) with values in R. Suppose that f and g are unboundedat b. Assume that the Newton primitive function F (x) =

∫ xaf(t)dt is

bounded on [a, b), i.e. there exists a number M > 0 such that |F (x)| ≤M for any x ∈ [a, b). We also assume that g is a function of class C1

on [a, b) and limx→b

g(x) = 0. Then∫ baf(x)g(x)dx is convergent.

P. We simply apply Cauchy criterion (see theorem 41).∫ x′′

x′f(x)g(x)dx =

∫ x′′

x′F ′(x)g(x)dx

by parts= F (x)g(x) |x′′x′ −


−∫ x′′

x′F (x)g′(x)dx

mean formula= F (x′′)g(x′′) − F (x′)g(x′)−

−F (cx′,x′′) [g(x′′) − g(x′)] , where cx′,x′′ ∈ [x′, x′′].

Thus, ∣∣∣∣∣

∫ x′′

x′f(x)g(x)dx

∣∣∣∣∣ ≤

≤ |F (x′′)g(x′′)| + |F (x′)g(x′)| + |F (cx′,x′′)| [|g(x′′)| + |g(x′)|] ≤(3.7)M [|g(x′′)| + |g(x′)|] +M [|g(x′′)| + |g(x′)|] = 2M [|g(x′′)| + |g(x′)|] .

Take now a small ε > 0 and let δ > 0 be small enough such that ifx ∈ [b− δ, b) then |g(x)| < ε

4M. This is true because lim

x→bg(x) = 0. This

δ is the searched for δ in the Cauchy criterion:∣∣∣∣∣

∫ x′′

x′f(x)g(x)dx

∣∣∣∣∣ ≤ 2M [|g(x′′)| + |g(x′)|] < 2M[ ε

4M+

ε

4M

]= ε,

if x′, x′′ ∈ [b− δ, b).

Let us use this last theorem to decide if one can approximate the

improper integral of the second type,∫ 1

0sin 3√1−x√

1−x dx, with singularity

at 1 (why?), by the usual Riemann integral∫ 0.99999

0sin 3√1−x√

1−x dx. This

last one can be approximately computed by using one of the meth-ods described in Chapter 2. Thus we must show that the integral∫ 1

0sin 3√1−x√

1−x dx is convergent. Let us use the above Dirichlet’s test. For

this we take f(x) = 1√1−x and g(x) = sin 3

√1 − x. A primitive of f is

F (x) = −2√

1 − x, which is bounded on [0, 2) (|F (x)| ≤ 2). Since g isdecreasing to 0 when x → 1, Dirichlet’s test tells us that the integralis convergent.

Sometimes an improper integral can be a "combination" betweenimproper integrals of the first type and improper integrals of the secondtype, i.e. mixed improper integrals. For instance, let us see if theintegral I =

∫∞1

1(x2+1) 3

√x−1

dx is convergent or not. Here the integration

interval is unbounded and the integrand function is also unbounded atx = 1. Take a λ > 1 and write formally:

I =

∫ ∞

1

1

(x2 + 1) 3√x− 1

dx =

∫ λ

1

1

(x2 + 1) 3√x− 1

dx+

+

∫ ∞

λ

1

(x2 + 1) 3√x− 1

dx.


The first integral in sum is an improper integral of the second type. Wecan use test q for q = 1/3 < 1 to prove that it is convergent. The secondintegral in the above sum is an improper integral of the first type. It isalso convergent because we can apply test p for p = 2 + 1/3 = 7/3 > 1.Their sum will be by definition the value of I. It is easy to see thatthis value does not depend on the choice of λ > 1. Such an integralI is also called an improper integral. We must be very carefully withthe operations with improper integrals. For instance a sum between aconvergent integral and a divergent one is always a divergent integral.But, a sum between two divergent integrals is not always divergent asthe following example shows. Let I =

∫∞0

(sin x − sinx)dx = 0, but

both I1 =∫∞0

sin xdx and I2 =∫∞0

(− sin x)dx are divergent. Thus, thedecomposition I = I1 + I2 has no meaning at all!

We end this chapter with a more sophisticated example.

E 53. Let us see if the following integral I =∫∞0

sinx√x+sinx

dx

exists, i.e. if it is convergent. Let us see what happens on the interval[0,∞) with function f(x) = sinx√

x+sinx. It is easy to see that this function

could be unbounded only at x = 0. But

limx→0,x>0

sin x√x+ sin x

= limx→0,x>0

x− x3

3!+ ...

√x+ x− x3

3!+ ...

= 0,

so that our function is bounded on [0,∞) and the integral is an improperintegral of the first type, i.e. it has a singularity only at∞. Let us write

I = I1 + I2,

where I1 =∫ 2

0sinx√x+sinx

dx and I2 =∫∞2

sinx√x+sinx

dx. In order to prove

that I is convergent (divergent) it is enough to prove that I2 is conver-gent (divergent) because I1 is a proper integral, i.e. an usual Riemannintegral on [0, 2]. And any proper integral is convergent! Let us writenow

sin x√x+ sin x

=sinx√x

− sin2 x√x(√x+ sin x)

.

Applying Dirichlet’s test (theorem 39) to the integral∫∞2

sinx√xdx we get

that this one is convergent. Thus the nature of the integral I2 is thesame like the nature of the integral

∫∞2

sin2 x√x(√x+sinx)

dx. But this time

the integrand function sin2 x√x(√x+sinx)

is nonnegative and we can compare

it with an easier nonnegative function g(x) = sin2 x√x(√x+1)

≤ f(x) =sin2 x√

x(√x+sinx)

. It is enough to prove that∫∞2

sin2 x√x(√x+1)

dx is divergent to


conclude that∫∞2

sin2 x√x(√x+sinx)

dx is divergent and so I2 would be diver-

gent. Since the integral

−1

2

∫ ∞

2

1 − 2 sin2 x√x(√x+ 1)

dx = −1

2

∫ ∞

2

cos 2x√x(√x+ 1)

dx

is convergent (apply again Dirichlet’s test) and since the integral12

∫∞2

1√x(√x+1)

dx is divergent (apply test p for p = 1), from

∫ ∞

2

sin2 x√x(√x+ 1)

dx = −1

2

∫ ∞

2

1 − 2 sin2 x√x(√x+ 1)

dx+1

2

∫ ∞

2

1√x(√x+ 1)

dx,

we conclude that the integral∫∞2

sin2 x√x(√x+1)

dx is divergent. Thus I2 and

I are also divergent. This information is very important because it willbe a crazy idea to approximate with something a divergent integral. Youcan say "maybe the given integral has something wrong in its form ifsomebody insists to compute it!". For instance, if instead of

√x one put

4√x3 the new obtained integral

∫∞0

sinx4√x3+sinx

dx were convergent (prove

this following the same steps as above!).

We saw that there is a close connection between the improper inte-grals of the first type and the numerical series (see theorem 33). Onehas an analogous result for the improper integrals of the second type.

T 47. (series test 1) Let us consider the improper integral

of the second type I =∫ baf(x)dx, with the unique singularity at the

point b. We also assume that f is continuous on [a, b). Let us take asequence a = b0 < b1 < ... < bn < ... < b such that bn → b and let us

consider the numerical series S =∑∞n=0 un, where un =

∫ bn+1bn

f(x)dx.Then, if the integral I is convergent, the series S is also convergentand S = I. Moreover, if in addition f has nonnegative values on [a, b),then the converse statement is also true. Namely, if the series S isconvergent, then the integral I is also convergent and I = S.

We leave the proof of this theorem as an exercise for the reader.What happens with the above result if b = ∞. For instance, in order

to compute the series −2∑∞n=1

[1√n+1

− 1√n

], one take the function

f(x) = 1√1−x and compute

∫ 1

0f(x)dx = 2. Hence the series is conver-

gent and its sum is equal to 2. Here we take bn = 1 − 1n, etc.

Can we reduce the study and the computation of an improper in-

tegral of the second type∫ baf(x)dx with the unique singularity at b

to an improper integral of the first type? The answer is yes! Let ustake a small arbitrary number δ > 0 and let us perform the change of


variables: t = 1b−x , x = b− 1

t, in the integral I(δ) =

∫ b−δa

f(x)dx. Thus

I(δ) =∫ 1

δ1

b−a

f(b− 1

t

)· 1t2dt. Let us denote by g(t) the new function

of t,f(b− 1

t)

t2. Since lim

δ→0I(δ) is the same with lim

u→∞

∫ u1

b−ag(t)dt, the study

of∫ baf(x)dx was reduced to the study of

∫∞1

b−ag(t)dt. We leave as an

exercise to the reader to prove the test q by using this last reductionand the test p.

Here is a last remark which is useful in the following chapter wherewe need to consider improper integrals of both types at the same time.

R 17. Let b be a real number or ∞ (b ∈ R ∪ ∞) and leta a any fixed real number less then b. If [a, b) is the obviously definedinterval we say that V is a fundamental neighborhood of b in [a, b) ifV is of the form (b − δ, b), δ being any small nonnegative real numberwith 0 < δ < b − a, for b = ∞ and V = (M,∞) with any M > afor b = ∞. Let f : [a, b) → R be a function defined on [a, b) with realvalues such that it is (Riemann) integrable on any subinterval [a, c]with a < c < b. We say that b is the unique singular point of f ifeither b = ∞ and f is bounded at any point c ∈ [a,∞), or b = ∞and f is unbounded at b and bounded at any other point c ∈ [a, b). Let

f be as above with the unique singularity at b. The integral∫ baf(x)dx

is called a (simple) improper integral. It is convergent if and only ifthe limit lim

x→b,x>a

∫ xaf(t)dt exists. Its value is called the value of the

improper integral∫ baf(x)dx. All the other results which were given when

we separately discussed improper integrals of the first and of the secondtype can be reformulated in language of improper integrals defined like

in this remark. For instance, Cauchy criterion says that∫ baf(x)dx is

convergent if and only if for any small ε > 0, there exists a fundamentalneighborhood Vε of b in [a, b) such that whenever x′, x′′ ∈ Vε one has

that∣∣∣∫ x′′x′f(x)dx

∣∣∣ < ε. We invite the reader to check that this last

criterion is equivalent to each one of the Cauchy criterions given for theparticular situations: b = ∞ and b = ∞. We also leave as an exercisefor the reader to state the general form of the Dirichlet’s criterion forthe general definition given here in this remark.

By a proper integral we mean an usual Riemann integral∫ baf(x)dx,

where f : [a, b] → R is an integrable function on [a, b] ⊂ R. Herethe function f has no singular point. Moreover, being integrable it isbounded (see theorem 13). We say that this proper integral is con-vergent because the Cauchy criterion works in this case for the point


b. Indeed, let ε > 0 be a small number and let M = supx∈[a,b]

|f(x)| .

Let δ > 0 be small enough such that b − δ > a and δ < εM. Let x′,

x′′ ∈ [b− δ, b] for this choice of δ. Then∣∣∣∣∣

∫ x′′

x′f(x)dx

∣∣∣∣∣ ≤M |x′ − x′′| < M · εM

= ε.

Thus Cauchy criterion works in the case of a proper integral. Thisis why we say that a proper integral can be viewed as a convergentimproper integral of the second type.


1. Test the convergence of the following improper integrals:a)∫∞1

dx

2x+ 3√x2+1; b)

∫∞0

xdx√x5+1

; c)∫ 1

0dx

3√1−x4 ; d)∫ 2

1dxlnx

; e)∫∞π2

sinxx2dx;

f)∫∞0

x√x3+2x7+1

dx; g)∫ 3

1dx

3√x−1; h)

∫ 2

−1(x2+1)dx

(x+1) 3√x−1

; i)∫∞0

cos(x2)dx;

j)∫∞0x2 cos 2xdx; k)

∫∞0

arctgxx

dx; l)∫ 4

2dx

(x−2)5; m)

∫∞1

(x2+1)dx

x4 5√x−1

; n)∫ 1

0dxx lnx

;

o)∫∞t

dx√x(x−a)(x−b)

, where t > a > b > 0; p)∫∞0

(e−

a2

x2 − e−b2

x2

)dx.

2. Compute (in the convergence case!) the following improperintegrals:

a)∫∞0e−2x sin 3xdx; b)

∫∞0e−ax cos bxdx, a > 0; c)

∫∞0

dx(x2+a2)(x2+4)

;

d)∫∞0

arctanx(1+x2)3/2

dx; e)∫∞−∞

dx1+x2

; f)∫∞0

sin 2xdx g)∫∞0xe−x

2dx;

h)∫∞0x3e−kx

4dx, k > 0.

3. Can you use a computer to approximately compute the improperintegrals?:

a)∫∞1

dx

2x+ 3√x2+4x+5; b)

∫∞−1

dx

x2+ 3√x4+1;

c)∫ 1

0x−

12 (1 − x)−

23dx; d)

∫∞0x−

12e−xdx;

e)∫∞0e−x

8dx; f)

∫∞0e−4x4dx; g)

∫∞0

tan−1 x(4+x2)

dx; h)∫∞0

cosx2dx.

CHAPTER 4

Integrals with parameters

1. Proper integrals with parameters

E 54. (a motivation) A linear wire is a segment [a, b] witha density function f defined on it. We assume that physical propertiesof the wire are changing during time. Thus a = a(t), b = b(t) arefunctions of time and the density function f = f(x, t), where x ∈(a(t), b(t)) is a function of time and of the point x. When time t variesin an interval J, we suppose that the extremities a(t), b(t) of the wirebelong to a fixed interval K. At any fixed moment t ∈ J, the mass ofthe linear wire is:

(1.1) M(t) =

∫ b(t)

a(t)

f(x, t)dx,

if the function gt(x) = f(x, t) is integrable on [a(t), b(t)] for any t ∈J. We just obtained a new function M : J → R which describes thevariation of the mass of the wire as a function of time t. Since theintegral on the right in formula (1.1) depends on a "parameter" t, wesay that this last integral is "an integral with a parameter t”.

Here is a more general definition for the notion of an integral with nparameters t1, t2, ..., tn. Let J be a subset of Rn and let t = (t1, t2, ..., tn)be an arbitrary element of J. Let a, b : J → K ⊂ R ∪ ±∞ be twofunctions of n variables defined on J with values into a fixed interval Kof R ∪ ±∞ such that a(t) ≤ b(t) for any t in J . Let f : K × J → Rbe a (scalar) function of n + 1 variables (x, t1, t2, ..., tn) = (x, t) whichis integrable with respect to (w.r.t.) x on K for any fixed t in J. This

means that for any fixed t in J, the integral I(t) =∫ b(t)a(t)

f(x, t)dx exists

for any fixed vector t in J.

D 12. The function I : J → R defined by the formula

(1.2) I(t) =

∫ b(t)

a(t)

f(x, t)dx

137

138 4. INTEGRALS WITH PARAMETERS

is said to be an integral with n parameters t1, t2, ..., tn. If for any t ∈ J

this last integral∫ b(t)a(t)

f(x, t)dx is a proper integral we say that our

integral is a proper integral with parameters t1, t2, ..., tn.

Let us now consider some examples of integrals with parameters.The integral I(λ) =

∫ 1

0sinλxdx with a parameter λ is defined for any

λ ∈ R. Thus J = R in this case. Functions a(λ), b(λ) are constant inthis case, a(λ) = 0, and b(λ) = 1 for any λ ∈ R. Thus K = [0, 1] and soour integral is a proper integral with a parameter λ. Since a primitiveF (x) =

∫sinλxdx can be computed in this case: F (x) = − cosλx

λ, if

λ = 0 and F (x) = 0 if λ = 0. Hence

I(λ) = −cosλ

λ+

1

λ=

1 − cosλ

λ,

if λ = 0. If λ = 0, I(0) = 0. A natural question arises: Is the func-

tion I(λ) continuous at 0? Since limλ→0

I(λ) = 00

l′Hospital= lim

λ→0sinλ = 0,

the answer is yes, because I(0) = 0. This example is the easiest pos-sible case which can appear. Usually, the primitive with respect to xcannot be computed (expressed as an elementary function!) in I(t) =∫ b(t)a(t)

f(x, t)dx, so we need to use other "indirect" computation of this

integral. For instance, I(c) =∫∞ce−cx

2dx, for c > 0 is an improper

integral with a parameter c. Here J = (0,∞), K = (0,∞), a(c) = c,the identity function and b(c) = ∞. For any fixed c > 0 our integralis convergent (apply the test p for p = 2). Since any primitive of e−x

2

is not an elementary function (this is a very deep result in Math andcannot be proved in this course!), any primitive of e−cx

2, c > 0, is not

also an elementary function (why?). Thus we cannot directly computeI(c). In the next section we shall see how to compute such integralswith parameters (not exactly this last one! We shall only computeI ′(c) = e−c

3 ( 12c− 1

).

Let us consider in the following only proper integrals with one pa-rameter, because we can ignore all the others parameters while we sepa-

rately work with one of them. Thus, let us consider I(t) =∫ b(t)a(t)

f(x, t)dx,

t ∈ J ⊂ R, J an interval, a, b : J → K ⊂ R and f(x, t) is integrablew.r.t. the variable x on the interval K. The first natural question is ifthe function I is continuous on J if the functions a(t), b(t) and f(x, t)are continuous on J and respectively on K × J.

T 48. (continuity theorem) With the above notation, iffunctions a(t), b(t) and f(x, t) are continuous on J and respectively

1. PROPER INTEGRALS WITH PARAMETERS 139

on K × J, then the new function I : J → R, I(t) =∫ b(t)a(t)

f(x, t)dx is

also continuous on J.

P. Let us fix a point t0 in J and let us prove that the functionI is continuous at this fixed point t0. For this let us take an arbitrarysequence tn in J such that tn → t0. To prove the continuity of I att0 it is enough to prove that I(tn) → I(t0). For this last statement, letus evaluate the difference I(tn) − I(t0) :

I(tn) − I(t0) =

∫ b(tn)

a(tn)

f(x, tn)dx−∫ b(t0)

a(t0)

f(x, t0)dx =

(1.3)∫ a(t0)

a(tn)

f(x, tn)dx+

∫ b(t0)

a(t0)

f(x, tn)dx+

∫ b(tn)

b(t0)


a(t0)

f(x, t0)dx.

To evaluate the integrals∫ a(t0)a(tn)

f(x, tn)dx and∫ b(tn)

b(t0)f(x, tn)dx we apply

the mean formula (see theorem 22) and find:

(1.4)

∫ a(t0)

a(tn)

f(x, tn)dx = f(cn, tn) [a(t0) − a(tn)]

and

(1.5)

∫ b(tn)

b(t0)

f(x, tn)dx = f(dn, tn)[b(tn) − b(t0)],

where cn is between a(tn) and a(t0) and dn is between b(tn) and b(t0).Since functions a(t) and b(t) are continuous and tn → t0, we see thatthe sequence cn converges to a(t0) and the sequence dn tends tob(t0). Since f is a continuous function as a function of two variables,we see that f(cn, tn) → f(a(t0), t0) and f(dn, tn) → f(b(t0), t0). Buta(t0) − a(tn) → 0 and b(tn) − b(t0) → 0, so that

∫ a(t0)

a(tn)

f(x, tn)dx→ f(a(t0), t0) · 0 = 0

and ∫ b(tn)

b(t0)

f(x, tn)dx → f(b(t0), t0) · 0 = 0.

Thus, in formula (1.3) it remains to prove that(1.6)∫ b(t0)

a(t0)


a(t0)

f(x, t0)dx =

∫ b(t0)

a(t0)

[f(x, tn)− f(x, t0)]dx→ 0,


whenever n→ ∞. By taking a sufficiently rich division ∆ : t0 = a(t0) <t1 < ... < tk = b(t0)

of the interval [a(t0), b(t0)] and by using the uniform continuity off, one can easily prove that∫ b(t0)

a(t0)[f(x, tn) − f(x, t0)]dx→ 0.

Why this last theorem is important? For instance, let I(t) =∫ t2+2

0e−tx

2dx, t ∈ R and the problem is to compute L = lim

t→0I(t). Usu-

ally, we make t = 0 in the integral expression of I(t). But, making this,we just intrinsically applied the above continuity theorem. Indeed,L = lim

t→0I(t) = I(lim

t→0t) = I(0) =

∫ 2

0dx = 2. Of course, the continuity

conditions on a(t) = t2+2, b(t) = 0 and on f(x, t) = e−tx2are obviously

satisfied (in order to be able to apply the continuity theorem).In many practical problems we need to see if the function

I(t) =∫ b(t)a(t)

f(x, t)dx is differentiable or not and, if it is so, how do

we compute its derivative? The great German mathematician G. W.von Leibniz (1646-1716) discovered a very nice and useful formula forthe computation of I ′(t), the differential of function I(t).

We start with the case when both functions a, b : J → R areconstant, say a(t) = a0 and b(t) = b0 for any t in J. We also assumethat f(x, t) is differentiable with respect to t for any fixed x in K andthat its differential ∂f

∂t(x, t) is a continuous function w. r. t. t.

T 49. We assume that the functions a, b and f verify these

last properties. Then the integral I(t) =∫ b0a0f(x, t)dx is differentiable

and

(1.7) I ′(t) =

∫ b0

a0

∂f

∂t(x, t)dx

for any t ∈ J.

P. Let us fix t0 in J and let us evaluate the ratio

I(t) − I(t0)

t− t0=

∫ b0a0

[f(x, t) − f(x, t0)] dx

t− t0.

For any fixed x in [a0, b0] we define gx : [t, t0]± → R by gx(ζ) = f(x, ζ).

Here [t, t0]± = [t, t0] if t ≤ t0 and [t, t0]

± = [t0, t] if t0 ≤ t. Let us writeLagrange’s formula for the function gx on the interval [t, t0]

± (i.e. wemust consider 2 cases, t ≤ t0 and t ≥ t0 respectively). In both casesone has:

gx(t) − gx(t0) = g′x(cx,t)(t− t0),


where cx,t ∈ [t, t0]±. Thus

I(t) − I(t0)

t− t0=

∫ b0a0

∂f∂t

(x, cx,t)(t− t0)dx

(t− t0)=

∫ b0

a0

∂f

∂t(x, cx,t)dx.

Since ∂f∂t

(x, cx,t) is a new continuous function h(x, t) of two variables,we can apply the continuity theorem 48 to the proper integral

∫ b0

a0

∂f

∂t(x, cx,t)dx =

∫ b0

a0

h(x, t)dx

with parameter t and obtain

limt→t0

I(t) − I(t0)

t− t0=

∫ b0

a0

limt→t0

h(x, t)dx =

∫ b0

a0

∂f

∂t(x, t0)dx,

because cx,t → t0, whenever t → t0. Hence I(t) is differentiable at t0and formula (1.7) is true.

Let us use this last theorem to compute the derivative of the func-

tion I(t) =∫ π

2

πcos txxdx. It is known in a very high level of Mathematics

that in general no primitive of function g(x) = cos xx

can be an elemen-tary function (in literature the primitive

∫cos xxdx is called the integral

cosine). Thus, it cannot be computed with the usual Newton-Leibnizformula. Sometimes we are not interested in the mathematical expres-sion of I(t) but, for instance, we are interested to compute "its velocity"I ′(t). To do this, let us use formula 1.7

I ′(t) = −∫ π

2

π

sin txdx =1

t

[cosπt− cos

π

2t].

We now consider the case when one of functions a(t) or b(t) isconstant and the other is not constant but it is differentiable on J.Take for instance a(t) to be nonconstant and b(t) to be a fixed constantb0.We also assume that f(x, t) is continuous with respect to both of itsvariables and that ∂f

∂t(x, t) is continuous w.r t. t.

T 50. We assume that all of these last conditions are sat-

isfied for the integral of a parameter t, I(t) =∫ b(t)a(t)

f(x, t)dx, i.e. I(t) =∫ b0a(t)

f(x, t)dx in our case. Then function I(t) is differentiable and

(1.8) I ′(t) =

∫ b0

a(t)

∂f

∂t(x, t)dx− a′(t)f(a(t), t)

for any t ∈ J.


P. Let us fix a point t0 ∈ J and let us evaluate the ratio

I(t) − I(t0)

t− t0=

∫ b0a(t)

f(x, t)dx−∫ b0a(t0)

f(x, t0)dx

t− t0=

∫ a(t0)a(t)

[f(x, t)] dx

t− t0+

+

∫ b0a(t0)

[f(x, t) − f(x, t0)] dx

t− t0.

The limit of ∫ b0a(t0)

[f(x, t) − f(x, t0)] dx

t− t0when t→ t0 is exactly

∫ b0

a(t0)

∂f

∂t(x, t0)dx

(see the proof of theorem 49). Thus, it remains to prove that the

limit of∫ a(t0)a(t)

[f(x,t)]dx

t−t0 when t → t0 is equal to −a′(t0)f(a(t0), t0). To

do this, let us apply mean formula of theorem (22) to the integral∫ a(t0)a(t)

[f(x, t)] dx :

∫ a(t0)

a(t)

[f(x, t)] dx = f(dt, t) [a(t0) − a(t)] ,

where dt ∈ [a(t), a(t0)]± . Since a(t) is also continuous, a(t) → a(t0)

and so dt → a(t0), whenever t→ t0. Thus,

limt→t0

∫ a(t0)a(t)

[f(x, t)] dx

t− t0= f(a(t0), t0)· lim

t→t0

a(t0) − a(t)

t− t0= −a′(t0)f(a(t0), t0).

Finally we put together both limits computed above and obtain:

limt→t0

I(t) − I(t0)

t− t0= limt→t0

∫ a(t0)a(t)

[f(x, t)] dx

t− t0+limt→t0

∫ b0a(t0)

[f(x, t) − f(x, t0)] dx

t− t0=

−a′(t0)f(a(t0), t0) +

∫ b0

a(t0)

∂f

∂t(x, t0)dx,

i.e. we just obtained formula (1.8) for t = t0.

Let us apply this last formula to compute the following limit:

limt→π

2

∫ 1

sin tex

2dx

∫ 0

cos tex2dx

.


Let us see that we are in the nondeterminate case 00and so we can

apply l’Hôspital rule:

limt→π

2

∫ 1

sin tex

2dx

∫ 0

cos tex2dx

= limt→π

2

[∫ 1

sin tex

2dx

]′

[∫ 0

cos tex2dx

]′ = limt→π

2

(− cos t)esin2 t

(sin t) ecos2 t= 0.

Here we applied formula (1.8) to compute the both derivatives[∫ 1

sin tex

2dx

]′and

[∫ 0

cos tex

2dx

]′.

We are now ready to give the general formula of Leibniz for a properintegral with a parameter t.

T 51. (Leibniz formula) Let I(t) =∫ b(t)a(t)

f(x, t)dx be an

integral with a parameter t ∈ J, where both functions a(t) and b(t)are differentiable on J and f(x, t) is continuous and of class C1 withrespect to the variable t. Then the function I(t) is differentiable and thefollowing Leibniz formula is true:

(1.9) I ′(t) =

∫ b(t)

a(t)

∂f

∂t(x, t)dx+ b′(t)f(b(t), t) − a′(t)f(a(t), t).

P. First of all let us see that this formula generalizes bothformulas (1.7) and (1.8). We shall see that these last formulas areenough in order to deduce the general Leibniz formula (1.9). Let us fixa t0 ∈ J and let us write

I(t) =

∫ b(t)

a(t)

f(x, t)dx =

∫ a(t0)

a(t)

f(x, t)dx+

∫ b(t0)

a(t0)

f(x, t)dx+

∫ b(t)

b(t0)

f(x, t)dx.

It will be enough to prove that each integral-function of t on the rightof this last formula is differential and to separately find a formula forits derivative.

Indeed, from formula (1.8) we see that

(1.10)

[∫ a(t0)

a(t)

f(x, t)dx

]′=

∫ a(t0)

a(t)

∂f

∂t(x, t)dx− a′(t)f(a(t), t).

From formula (1.7) we find[∫ b(t0)

a(t0)

f(x, t)dx

]′=

∫ b(t0)

a(t0)

∂f

∂t(x, t)dx.

At last, from formula (1.8) one has:

(1.11)

[∫ b(t)

b(t0)

f(x, t)dx

]′= −

[∫ b(t0)

b(t)

f(x, t)dx

]′=


= −∫ b(t0)

b(t)

∂f

∂t(x, t)dx+ b′(t)f(b(t), t).

If we make the sum of both sides for these above three formulas we get

I ′(t) =

∫ a(t0)

a(t)

∂f

∂t(x, t)dx− a′(t)f(a(t), t) +

∫ b(t0)

a(t0)

∂f

∂t(x, t)dx−

−∫ b(t0)

b(t)

∂f

∂t(x, t)dx+ b′(t)f(b(t), t).

But∫ a(t0)

a(t)

∂f

∂t(x, t)dx+

∫ b(t0)

a(t0)

∂f

∂t(x, t)dx−

∫ b(t0)

b(t)

∂f

∂t(x, t)dx =

=

∫ a(t0)

a(t)

∂f

∂t(x, t)dx+

∫ b(t0)

a(t0)

∂f

∂t(x, t)dx+

∫ b(t)

b(t0)

∂f

∂t(x, t)dx =

=

∫ b(t)

a(t)

∂f

∂t(x, t)dx.

Hence

I ′(t) =

∫ b(t)

a(t)

∂f

∂t(x, t)dx+ b′(t)f(b(t), t) − a′(t)f(a(t), t).

Let us compute I ′(t) for I(t) =∫ t2t

sin txxdx, for t ∈ R. We simply

use Leibniz formula and find

I ′(t) =

∫ t2

t

cos txdx+ 2tsin t3

t− sin t2

t=

sin tx

t

∣∣∣∣t2

t

+ 2 sin t3 − sin t2

t=

=sin t3

t− 2

sin t2

t+ 2 sin t3.

The following result was discovered by the Italian mathematicianGuido Fubini (1879—1943) in a more general form. This is why it iscalled Fubini’s theorem. We state this result here in a particular form.This form is enough for us in the following applications.

T 52. (Fubini’s theorem) Let K = [a, b] and J = [c, d] betwo real closed intervals and let f : K×J → R be a continuous functionf(x, t) of two variables. Then

(1.12)

∫ b

a

(∫ d

c

f(x, t)dt

)dx =

∫ d

c

(∫ b

a

f(x, t)dx

)dt.


P. 1) First of all let us explain what formula (1.12) says. Let

us denote J(x) =∫ dcf(x, t)dt, a proper integral with a parameter x and

let denote by L(t) =∫ baf(x, t)dx, a proper integral with a parameter

t. Formula (1.12) says that∫ baJ(x)dx =

∫ dcL(t)dt, i.e. we may change

the order of integration.

For proving Fubini’s formula we put F (u) =∫ ua

(∫ dcf(x, t)dt

)dx

and G(u) =∫ dc

(∫ uaf(x, t)dx

)dt for any u ∈ [a, b]. If we succeed to

prove that F (u) = G(u) for any u ∈ [a, b], by making u = b, we wouldobtain formula (1.12). Let us carefully use Leibniz formula (1.9) tocompute F ′(u) and G′(u).

F ′(u) =

∫ d

c

f(u, t)dt, G′(u) =

∫ d

c

f(u, t)dt.

Thus, F ′(u) = G′(u) for any u ∈ [a, b]. Hence F (u) = G(u) +C, whereC is a constant w.r.t. u. Let us put u = a and find F (a) = G(a) + Cand, since F (a) = G(a) = 0, we find that C = 0, i.e. F (u) = G(u) andthe first proof is completed. Here is another proof. We present it atan intuitive level. But, by using the uniform continuity of f, one canimprove it up to a corect mathematical proof.

2) Let us consider two arbitrary divisions

∆x : a = x0 < x1 < ... < xn = b

and

∆t : c = t0 < t1 < ... < tm = d

of the intervals [a, b] and [c, d] respectively. Let us denote by M1 =∫ baJ(x)dx and by M2 =

∫ dcL(t)dt. We can approximate very well the

number M1 by Riemann sums of the form

(1.13)n∑

i=1

J(ξi)(xi − xi−1),

where ξi ∈ [xi−1, xi] for any i = 1, 2, ..., n. Here divisions ∆x andξi ∈ [xi−1, xi] are arbitrary and if the norm ‖∆x‖ → 0, then the aboveRiemann sums become closer and closer to M1. Let us fix now sucha division ∆x and a set of marking points ξi such that the corre-sponding Riemann sum to be as close as we want to M1. Each number

J(ξi) =∫ dcf(ξi, t)dt can be well approximate by Riemann sums of the

formm∑

j=1

f(ξi, ηj)(tj − tj−1).


Thus M1 can be well approximate with sums of the type

(1.14)n∑

i=1

m∑

j=1

f(ξi, ηj)(tj − tj−1)(xi − xi−1).

In the same manner we can prove that M2 =∫ dcL(t)dx can be well

approximate with double sums of the form

(1.15)m∑

j=1

n∑

i=1

f(ξi, ηj)(xi − xi−1)(tj − tj−1).

Since the sum and the product are commutative we see that the abovedouble sums are equal. Hence the numbers M1 and M2 can be wellapproximate by the one and the same set of numbers of the form

n∑

i=1

m∑

j=1

f(ξi, ηj)(tj − tj−1)(xi − xi−1).

Let us denote by A this set of numbers. Assume that M1 = M2. Then,taking ε such that 0 < ε < |M1−M2|

2, one sees that

(M1 − ε,M1 + ε) ∩ (M2 − ε,M2 + ε) = ∅.

But there exist an infinite number of elements in A which are containedin this intersection! Thus, our assumption that M1 = M2 is not true.This means that M1 = M2, i.e. Fubini’s formula.

To change the order of integration could be a real necessity in some

cases. For instance, the computation of∫ 2

1

(∫ 2

1ln txxdx

)dt in this or-

der, i.e. first of all we compute J(t) =∫ 2

1ln txxdx and then we compute∫ 2

1J(t)dt, is more complicated than to change the order of integra-

tion (Fubini’s formula), namely to compute∫ 2

1

(∫ 2

1ln txxdt)dx (convince

yourself!).

R 18. If f : D → R, where D is the rectangle [a, b] × [c, d],then a double sum of the form

m∑

j=1

n∑

i=1

f(ξi, ηj)(xi − xi−1)(tj − tj−1)

is called a Riemann double sum for the function f. If all this doublesums becomes closer and closer to a number I, we say that I is the

double integral of f on the rectangle D. Write I =

∫∫

D

f(x, t)dxdt.

2. IMPROPER INTEGRALS WITH PARAMETERS 147

Fubini’s formula says that if f is continuous, then this last numberalways exist and it can be computed as:

∫∫

D

f(x, t)dxdt =

∫ b

a

(∫ d

c

f(x, t)dt

)dx =

∫ d

c

(∫ b

a

f(x, t)dx

)dt.

Such formulas are called iterative formulas for the computation of adouble integral on a rectangle D = [a, b]× [c, d] (see more on this topicin chapter 6).

2. Improper integrals with parameters

A function f with n parameters t = (t1, t2, ..., tn) ∈ J ⊂ Rn, whereJ is a domain in Rn, is simply a function f(x, t) of n + 1 variables,f : K × J → R, where K is a real interval, finite or not. Let b be alimit point of K, i.e. b ∈ R ∪ ±∞ and it is a limit of at least aninfinite sequence of elements of K. We assume that for any fixed t inJ the following limit g(t) = lim

x→bf(x, t) exists. The obtained function

g : J → R is called the simple (pointwise) limit of f at b, alongthe set of parameters J. The limit function g is said to be uniformw.r.t. (along) the set of parameters J if for any small ε > 0, there is aneighborhood Vb of b in K (it is of the form (b− δ, b+ δ) ∩K, (M,∞)if b = ∞, or (−∞,M) if b = −∞) such that

|g(t) − f(x, t)| < ε

for any x ∈ Vb and for any t ∈ J.For instance f(x, t) = ext, x ∈ (0, 1], t ∈ J = (−3, 2] has the

uniform function limit g(t) = 1 at b = 0. Indeed, if ε > 0, |ext − 1| < εif

x < δ = min

1

2ln(1 + ε),−1

3ln(1 − ε)

for any t ∈ (−3, 2]. If instead of J = (−3, 2] we take J = R, we easilysee that g(t) = 1 is a simple limit of f(x, t) = ext at b = 0, but it is notuniform.

It is obvious that if a function f(x, t) of a variable x and with a setof n parameters t = (t1, t2, ..., tn) ∈ J ⊂ Rn, J a domain, x ∈ K, whereK is a real interval in R, has a uniform limit g(t) at b, w.r.t. J, thenit is also a simple limit at b.

Following the proof of theorem 31 it is not difficult to prove thefollowing analogous result.

T 53. (Cauchy criterion of the uniform limit) Let K be aninterval of R and let b ∈ R be a limit point of K. Let f : K × J → R


be a function with n parameters t = (t1, t2, ..., tn) ∈ J ⊂ Rn, where J isa domain in Rn. a) If b is finite, i.e. if b ∈ R, then f has an uniformlimit g(t) at b along the set of parameters J, if and only if for any smallreal number ε > 0 there is a real number δ > 0 such that if two numbersx′ and x′′ are in K ∩ (b − δ, b + δ), then the distance between f(x′, t)and f(x′′, t) is less than ε, i.e |f(x′, t) − f(x′′, t)| < ε for any t ∈ J.b) If b is not finite, say b = ∞, then f has a uniform limit at b alongthe set of parameters J if and only if for any small real number ε > 0there is a large real number M > 0 such that if two numbers x′ and x′′

are greater than M, then the distance between f(x′, t) and f(x′′, t) isless than ε, i.e. |f(x′, t) − f(x′′, t)| < ε for any t ∈ J. For b = −∞we must replace this last M with a negative, near to −∞, M, i.e. inthis case M < 0 and −M is very large. Moreover, the expression "x′

and x′′ are greater than M must be substituted with "x′ and x′′ are lessthan M".

E 55. Here is a more complicated example of how we applythis last criterion. Let us define f(x, t) =

∫ x0

arctan tuu(u+1)

du, x ∈ K =

[0,∞), t ∈ J = R. Let us study the limit limx→∞

f(x, t) = g(t). Here

b = ∞ is a limit point of R and it is the unique singular point of ourintegral because lim

u→0

arctan tuu(u+1)

= t, i.e. 0 is a false (illusory) singular

point. For any fixed t in R this limit is nothing else than the improperintegral

∫∞0

arctan tuu(u+1)

du. This integral is convergent because∫ ∞

0

arctan tu

u(u+ 1)du =

∫ 1

0

arctan tu

u(u+ 1)du+

∫ ∞

1

arctan tu

u(u+ 1)du,

∣∣∣∣arctan tu

u(u+ 1)

∣∣∣∣ ≤π

2

1

u(u+ 1)

and ∫ ∞

1

1

u(u+ 1)du

is convergent (apply test p with p = 2). Thus our limit exists for anyfixed t in R. Since g(t) =

∫∞0

arctan tuu(u+1)

du we cannot (for the moment)

compute g(t) and then to use the definition of the uniform limit todecide if this limit is or is not uniform w.r.t. t. Let us use Cauchycriterion 53 to prove that our limit is a uniform one. For this let us

evaluate∫ x′′x′

arctan tuu(u+1)

du for x′, x′′ very close to ∞ (large enough). Sim-

ply, it is sufficient to prove that∣∣∣∫ x′′x′

arctan tuu(u+1)

du∣∣∣ becomes smaller and

smaller when x′, x′′ become larger and larger independent (uniformly)with respect to the parameter t. Here we use a simple and efficient idea


of K. Weierstrass. Namely, let us put instead of∣∣∣ arctan tuu(u+1)

∣∣∣ a greaterfunction h(u) which does not contain the parameter t at all and then tryto prove the Cauchy criterion for the improper integral

∫∞1h(u)du. In-

deed, since∣∣∣ arctan tuu(u+1)

∣∣∣ ≤ π2

1u(u+1)

, h(u) = π2

1u(u+1)

and, since the improper

integral (without a parameter) π2

∫∞1

1u(u+1)

du is convergent (apply test

p with p = 2), so the Cauchy criterion condition works, one has thatfor any small ε > 0 there exists M > 0 such that if x′, x′′ > M we getπ2

∣∣∣∫ x′′x′

1u(u+1)

du∣∣∣ < ε. Thus, since

∣∣∣∣∣

∫ x′′

x′

arctan tu

u(u+ 1)du

∣∣∣∣∣ ≤∫ x′′

x′

∣∣∣∣arctan tu

u(u+ 1)

∣∣∣∣ du ≤ π

2

∣∣∣∣∣

∫ x′′

x′

1

u(u+ 1)du

∣∣∣∣∣ < ε

we obtain that∣∣∣∫ x′′x′

arctan tuu(u+1)

du∣∣∣ < ε for any x′, x′′ > M. Hence

g(t) =

∫ ∞

0

arctan tu

u(u+ 1)du =

∫ 1

0

arctan tu

u(u+ 1)du+

∫ ∞

1

arctan tu

u(u+ 1)du

is the uniform limit of f(x, t) relative to the parameter t.

T 54. (continuity theorem) Let f(x, t), f : K × J → R,where K is a real interval and J is the domain of parameters t =(t1, t2, ..., tn), J ⊂ Rn. Assume that f is continuous with respect to thevariable vector t and that f has a uniform limit g(t) at a limit pointb of K. Then g(t) = lim

x→bf(x, t) is a continuous function of its vector

variable t. Thus, for any fixed t0 ∈ J,

(2.1) limt→t0

[limx→b

f(x, t)]

= limx→b

f(x, t0).

P. Let us choose a fixed point t0 in J and let us prove thecontinuity of g at this point t0. For this, let us evaluate the difference(2.2)|g(t) − g(t0)| ≤ |g(t) − f(x, t)|+|f(x, t) − f(x, t0)|+|f(x, t0) − g(t0)| .

Since f has a uniform limit g(t) at b, for a fixed small ε > 0, there isa neighborhood Vb of b in K (see the above definition of this notion!)such that

(2.3) |g(t) − f(x, t)| < ε

3, and |f(x, t0) − g(t0)| <

ε

3


for any x ∈ Vb and for any t ∈ J. Now let us fix such an x0 ∈ Vb andwrite again the inequality (2.2) for x = x0:(2.4)|g(t) − g(t0)| ≤ |g(t) − f(x0, t)|+|f(x0, t) − f(x0, t0)|+|f(x0, t0) − g(t0)| .Since function t → f(x0, t) is continuous at t0, there is a small realnumber δ > 0 such that if ‖t− t0‖ < δ, we get

(2.5) |f(x0, t) − f(x0, t0)| <ε

3.

Let us use the three inequalities (2.3) and (2.5) in the evaluation (2.4)to finally obtain:

|g(t) − g(t0)| < ε

for any t ∈ J with ‖t− t0‖ < δ. But this exactly means the continuityof the function g at t0.

E 56. Let us take the function g(t) =∫∞0

arctan tuu(u+1)

du (this

is an improper integral with a parameter t, see the definition bellow)obtained as a uniform limit of f(x, t) =

∫ x0

arctan tuu(u+1)

du when x→ ∞ and

let us compute

limt→0

g(t) = limt→0

∫ ∞

0

arctan tu

u(u+ 1)du.

Since g is continuous (see the last theorem above), we can write 0 =g(0) = lim

t→0g(t). Thus this limit is equal to 0. We simply put x = 0

under the integral sign. But, we must be very careful and not to do thisbefore you did not prove that the limit is uniform!

With the supplementary condition that f(x, t) has a finite limit ata limit point t0 of J, one can extend formula (2.1) for this more generalcase. For instance, if a > 0, then

limt→∞

∫ ∞

a

arctan tu

u(u+ 1)du =

∫ ∞

a

arctan∞u(u+ 1)

du = −π2

lna

a + 1.

(What happens when a→ 0?).The above theorem is sometimes used to decide if a limit is uniform

or not. For instance, take f(x, t) = tx, where x ∈ R and t ∈ (0, 1]. Ifg(t) = lim

x→∞f(x, t), it it easy to see that

g(t) =

0, if t < 11, if t = 1

.

Since this function in not a continuous function, the above limit cannotbe uniform with respect to the parameter t. This is also a counterex-ample to the fact that the hypothesis on limit to be a uniform one isnecessary.


T 55. (integration theorem)Let K and J be real intervals,let b be a limit point of K and let f : K×J → R, f(x, t), be a functionwhich has a uniform limit g(t) at b, relative to the parameter t. Assumethat [a, c] is an interval contained in J and that f is continuous withrespect to t on [a, c]. Then g is integrable on [a, c] and

(2.6)

∫ c

a

g(t)dt = limx→b

∫ c

a

f(x, t)dt,

or

(2.7) limx→b

∫ c

a

f(x, t)dt =

∫ c

a

limx→b

f(x, t)dt.

P. Let ε > 0 be a small real number and let Vb be a neigh-borhood of b in K such that |f(x, t) − g(t)| < ε

c−a for any x ∈ Vb andfor any t ∈ J. Since g is continuous on [a, c] (see theorem 54) it isintegrable. Let us evaluate the difference∣∣∣∣∫ c

a

f(x, t)dt−∫ c

a

g(t)dt

∣∣∣∣ ≤∫ c

a

|f(x, t) − g(t)| dt < ε

c− a

∫ c

a

dt = ε.

Thus∣∣∫ caf(x, t)dt−

∫ cag(t)dt

∣∣ < ε for any x ∈ Vb and for any t ∈ J.

But this is equivalent to saying that∫ cag(t)dt = lim

x→b

∫ caf(x, t)dt.

This theorem is useful in practice. Let us compute for instancelimx→∞

∫ 1

0e−xt

2dt. We know that the primitive of e−xt

2as a function

of t cannot be computed because it is not an elementary function.Function f(x, t) = e−xt

2, x ∈ [0,∞), t ∈ [0, 1] has an uniform limit,

g(t) = 0, when x→ ∞. Let us apply now the last theorem and find

limx→∞

∫ 1

0

e−xt2

dt =

∫ 1

0

limx→∞

e−xt2

dt = 0.

The hypothesis on the limit to be uniform is necessary. For instance,let f(x, t) = xte−xt

2, t ∈ [0, 1] and x ∈ R. Let g(t) = lim

x→∞xte−xt

2= 0

for any fixed t in [0, 1]. If this limit were a uniform one, then formula(2.6) would be true. But,

limx→∞

∫ 1

0

xte−xt2

dt = limx→∞

−1

2e−xt

2

∣∣∣∣1

0

=1

2= 0.

Thus the limit cannot be uniform. To directly prove that this limitis not uniform w.r.t. t is not an easy task. Hence the condition ofuniformity in the last theorem is necessary.

T 56. (differentiation theorem) Let K and J be real inter-vals, let b be a limit point of K and let f : K × J → R, f(x, t), be a


function which has a simple limit g(t) at b. Assume that f is of classC1 on J and that the function ∂f

∂t(x, t) has a uniform limit h(t) at b.

Then g(t) is differentiable on J and g′(t) = h(t) for any t in J. Wealso can write this as[

limx→b

f(x, t)]′t

= limx→b

∂f

∂t(x, t).

P. For any u ∈ J we define

F (x, u) =

∫ u

a

∂f

∂t(x, t)dt = f(x, u) − f(x, a),

where a is a fixed point in J. Taking the uniform limits in both sideswe get

limx→b

∫ u

a

∂f

∂t(x, t)dt

th.55=

∫ u

a

limx→b

∂f

∂t(x, t)dt =

∫ u

a

h(t)dt =

= g(u) − g(a).

From the last equality we get that u→ g(u) is a differentiable function(it is expressed as an area function of Newton!) and g′(u) = h(u) forany u ∈ J (here we applied Leibniz formula for a proper integral).

We can now apply the above theory to our main topic, improperintegrals with parameters. What is such an integral? Assume that

in the symbolic expression of an integral I(t) =∫ b(t)a(t)

f(x, t)dx, with

a parameter t ∈ J, one of the functions a(t) or b(t) is a constant b ∈R∪±∞ (we just fixed b(t) to be a constant!) and for any t ∈ J, on the

interval [a(t), b]±, b is a singular point for the integral∫ ba(t)

f(x, t)dx and

it is unique with this property. Such an integral is called an improperintegral with a parameter t. The same definition can be given for manyparameters instead of one. For instance, I(t) =

∫∞t

arctan txx(x+1)

dx, t > 0,

or I(t) =∫ 1

0sin tx√1−xdx are improper integrals with a parameter t. In the

first example, b = ∞ and the improper integral is of the first type.In the second case, b = 1 and the improper integral is of the secondtype. To study such an integral, we simply need to study the nature

of a limit at the singular point b. Namely, let I(t) =∫ ba(t)

f(x, t)dx

with b as the unique singular point. In fact, I(t) = limx→b

F (x, t), where

F (x, t) =∫ xa(t)

f(u, t)du. If for any fixed t ∈ J, this limit exists, i.e. the

usual improper integral∫ ba(t)

f(x, t)dx is convergent, we say that our

improper integral with parameter t is simple convergent. If the limit

I(t) = limx→b

F (x, t) = limx→b

∫ x

a(t)

f(u, t)du


is a uniform limit at b (a limit point of K) w.r.t. parameter t, we saythat our improper integral is uniformly convergent. It is clear that auniformly convergent integral is also a simple convergent one. Let usgive a list with the above results, initial stated for limits of functions

with parameters, applied here to the fixed integral I(t) =∫ ba(t)

f(x, t)dx,

with a parameter t. We shall always assume that f(x, t) is defined onK × J, where K and J are intervals and that f is continuous withrespect to x on K. Recall that a : J → K is a function defined on Jwith values in K. The following theorem is simply a translation of theresults obtained in general for limits of functions with parameters in ourparticular case of limits of functions of the type F (x, t) =

∫ xa(t)

f(u, t)du,

with u in K, between a(t) and x ∈ K. This is why we do not give acomplete proof for the next theorem.

T 57. With the above notation and hypotheses, one has thefollowing statements:

a) The integral I(t) =∫ ba(t)

f(x, t)dx, with singularity b (a limit point

of K), is uniformly convergent if and only if for any small real number

ε > 0 there is a neighborhood Vb of b in K, such that∣∣∣∫ x′′x′f(x, t)dx

∣∣∣ < ε

for any x′, x′′ in Vb (Cauchy criterion).

b) If I(t) =∫ ba(t)

f(x, t)dx is uniformly convergent on J, then I(t)

is continuous on J (continuity theorem).

c) If I(t) =∫ ba(t)

f(x, t)dx is uniformly convergent on J and [a, c] ⊂K , then

∫ c

a

[∫ b

a(t)

f(x, t)dx

]dt =

∫ b

a(t)

[∫ c

a

f(x, t)dt

]dx.

(Fubini theorem). If a or c is the unique singularity of I(t) in K, theset of the limit points of K in R ∪ ±∞, and if all the integrals withparameters t or x are uniformly convergent, then the last formula isalso true.

d) Let us assume that the integral I(t) =∫ ba(t)

f(x, t)dx is convergent

and, in addition, that the integral H(t) =∫ ba(t)

∂f∂t

(x, t)dx is uniformly

convergent. Then I(t) is differentiable and:

(2.8) I ′(t) =

∫ b

a(t)

∂f

∂t(x, t)dx− a′(t)f(a(t), t),

for any t ∈ J (Leibniz formula).


e) (Weierstrass test) Let I(t) =∫ baf(x, t)dx, a ∈ R, be an improper

integral with a parameter t and with singularity b. If there exists afunction s(x), s : K → R such that |f(x, t)| ≤ s(x) for any x ∈ K,

t ∈ J and if the improper integral∫ bas(x)dx is convergent, then I(t) is

uniformly convergent.

P. Let us prove for instance Leibniz formula d).

Let F (x, t) =∫ xa(t)

f(u, t)du. Since the integral I(t) =∫ ba(t)

f(x, t)dx

is convergent, the function F (x, t) has as a limit at b, exactly the func-

tion I(t). Since the integral H(t) =∫ ba(t)

∂f∂t

(x, t)dx is uniformly conver-

gent, the function L(x, t) =∫ xa(t)

∂f∂t

(u, t)du has as a uniform limit the

function H(t). Applying theorem 56 we get that

I ′(t) = limx→b

∂F

∂t(x, t) = lim

x→b

[∫ x

a(t)

∂f

∂t(u, t)du− a′(t)f(a(t), t)

]=

=

∫ b

a(t)

∂f

∂t(x, t)du− a′(t)f(a(t), t).

Here we just applied Leibniz formula for the proper integral∫ xa(t)

f(u, t)du

with parameter t.Let us prove now Weierstrass test e). In order to prove that the

improper integral I(t) =∫ baf(x, t)dx with parameter t is uniformly

convergent, we use Cauchy criterion a) of the same theorem. Let ε > 0be a small real number and let Vb be a neighborhood of b in K such

that for any x′, x′′ ∈ Vb one has∣∣∣∫ x′′x′s(x)dx

∣∣∣ < ε (here we used the

fact that the improper integral∫ bas(x)dx is convergent). Since

∣∣∣∣∣

∫ x′′

x′f(x, t)dx

∣∣∣∣∣ ≤∫ x′′

x′|f(x, t)| dx ≤

∫ x′′

x′s(x)dx < ε,

we finally derive that∣∣∣∫ x′′x′f(x, t)dx

∣∣∣ < ε for any x′, x′′ ∈ Vb. Applying

again the Cauchy criterion a), we get that the integral I(t) is uniformlyconvergent.

Let us use Leibniz formula d) of this last theorem (for improper in-tegrals with a parameter) to compute the integral I(t) =

∫∞0

arctan txx(1+x2)

dx,

t > 0. We can easily see that this integral is uniformly convergent (forapplying Leibniz formula this is not necessarily!). To see this we write

∣∣∣∣arctan tx

x(1 + x2)

∣∣∣∣ ≤π

2

1

x(1 + x2).


Since the improper integral π2

∫∞1

1x(1+x2)

dx is convergent (apply p-

test with p = 3), Weierstrass test e) in the above theorem says that∫∞1

arctan txx(1+x2)

dx is uniformly convergent. Since limx→0

arctan txx(1+x2)

= t, the in-

tegral∫ 1

0arctan txx(1+x2)

dx is a proper integral. It is easy to see that a sum

between a proper integral and an uniformly convergent improper in-tegral is also an uniformly convergent improper integral (prove this!).Thus

I(t) =

∫ ∞

0

arctan tx

x(1 + x2)dx =

∫ 1

0

arctan tx

x(1 + x2)dx+

∫ ∞

1

arctan tx

x(1 + x2)dx

is an uniformly convergent integral. We also need that the integral

H(t) =

∫ ∞

0

∂

∂t

[arctan tx

x(1 + x2)

]dx =

∫ ∞

0

1

(1 + x2)(1 + t2x2)dx

of the derivative function with respect to t be uniformly convergent.Since ∫ ∞

0

1

(1 + x2)(1 + t2x2)dx =

∫ 1

0

1

(1 + x2)(1 + t2x2)dx+

+

∫ ∞

1

1

(1 + x2)(1 + t2x2)dx

and since∫ 1

01

(1+x2)(1+t2x2)dx is a proper integral for t = 0, it is sufficient

to prove that∫∞1

1(1+x2)(1+t2x2)

dx is uniformly convergent. Since

1

(1 + x2)(1 + t2x2)≤ 1

1 + x2

and since∫∞1

1(1+x2)

dx is convergent (test p for p = 2), applying again

Weierstrass test e) we get that∫∞1

1(1+x2)(1+t2x2)

dx is uniformly conver-

gent. Now we can apply Leibniz formula for a(t) = 0 and b = ∞ :

I ′(t) =

∫ ∞

0

1

(1 + x2)(1 + t2x2)dx.

But

1

(1 + x2)(1 + t2x2)=

1

1 − t2· 1

1 + x2− t2

1 − t2· 1

1 + t2x2

for any t = 1. Thus,

I ′(t) =1

1 − t2

∫ ∞

0

1

1 + x2dx− t2

1 − t2

∫ ∞

0

1

1 + t2x2dx =

=1

1 − t2arctan x|∞0 − t

1 − t2arctan tx |∞0 =

π

2

1

1 + t.


Taking the primitive on both sides in

I ′(t) =π

2

1

1 + t,

we get I(t) = π2

ln(1 + t) + C, where C does not depend on t. SinceI(0) = 0, we obtain C = 0. Thus, I(t) = π

2ln(1+t) for any t > 0, t = 1.

Since both functions I(t) and π2

ln(1 + t) are continuous with respectto t, the equality is true even for t = 1.

Let us use now Weierstrass test e) of this last theorem to provethe uniform convergence of the integral I(t) =

∫∞1

sin txx2

dx. For this, letε > 0 be a small real number and let M > 0 be a sufficiently large real

number such that if x′, x′′ > M then∫ x′′x′

1x2dx < ε (here we use the

fact that the integral∫∞1

1x2dx is convergent, see test p). But

∣∣∣∣∣

∫ x′′

x′

sin tx

x2dx

∣∣∣∣∣ ≤∫ x′′

x′

|sin tx|x2

dx ≤∫ x′′

x′

1

x2dx < ε

for any x′, x′′ > M. Thus, the Weierstrass test e) tells us that theintegral I(t) =

∫∞1

sin txx2

dx is uniformly convergent.Now we give some applications to the effective computation of some

classes of integrals with parameters.

E 57. Let us compute the value of the following integralwith two parameters α and β:

(2.9) I(α, β) =

∫ ∞

0

e−αx2 − e−βx

2

xdx, α, β > 0.

Since

limx→0

e−αx2 − e−βx

2

x=

0

0= limx→0

−2αxe−αx2 − 2βxe−βx

2

1= 0,

for any fixed α and β, the integral has only b = ∞ as a singularity.Considering successively I(α, β) as an integral with a parameter α andthen, with a parameter β respectively, we can apply Leibniz formulaseparately for α and β respectively. Let us prove the simple convergenceof I(α, β). Since

limx→∞

x2e−αx2

x= 0, lim

x→∞x2e

−βx2

x= 0,

if α, β > 0, the p-test with p = 2 gives us the simple convergence ofI(α, β). We need now the uniform convergence of

∂I

∂α=

∫ ∞

0

∂

∂α

[e−αx

2 − e−βx2

x

]dx = −

∫ ∞

0

xe−αx2

dx.


For any fixed α1 > 0, this last integral is uniformly convergent on theinterval J = [α1,∞). Indeed,∣∣∣xe−αx2

∣∣∣ = xe−αx2 ≤ xe−α1x

2

and∫∞0xe−α1x

2dx is convergent (apply test p for p = 2), thus applying

again Weierstrass test e) of theorem 57 we obtain that the integral−

∫∞0xe−αx

2dx is uniformly convergent w.r.t. α. In the same way we

can prove that the integral

∂I

∂β=

∫ ∞

0

∂

∂β

[e−αx

2 − e−βx2

x

]dx =

∫ ∞

0

xe−βx2

dx

is uniformly convergent on any interval [β1,∞) with respect to β ifβ1 > 0 is fixed. Thus

(2.10)∂I

∂α= −

∫ ∞

0

xe−αx2

dx =1

2αe−αx

2

∣∣∣∣∞

0

= − 1

2α

and

(2.11)∂I

∂β=

∫ ∞

0

xe−βx2

dx = − 1

2βe−βx

2

∣∣∣∣∞

0

=1

2β.

Integrating with respect to α in the equality (2.10) we get

I(α, β) = −1

2lnα + C(β).

Let us introduce this expression of I(α, β) in formula (2.11):

C ′(β) =1

2β,

so C(β) = 12

ln β + k, where k does not depend either on α or on β.Hence

I(α, β) = −1

2lnα +

1

2ln β + k =

1

2lnβ

α+ k.

We see that if α = β, I(α,α) = 0 = k. Finally, I(α, β) = 12

ln βα.

This last example can be generalized in the following form.

T 58. (Froullani’s integrals) Let f : [0,∞) → R be a con-tinuous bounded function with f(∞) = λ. For any 0 < a < b we define

I(a, b) =

∫ ∞

0

f(ax) − f(bx)

xdx.

a) Then the integral is simply convergent and

(2.12) I(a, b) = [f(0) − f(∞)] lnb

a.


b) If in addition f is of class C1 on [0,∞) and if for any ε > 0, M > 0there is a continuous function g(x), such that |f ′(xt)| ≤ g(x) for allx ∈ (0,∞), t ∈ [ε,M ], and

∫∞0g(x)dx is convergent, then I(a, b) is

uniformly convergent on any interval of the form [ε,M ] (i.e. a, b ∈[ε,M ]), where 0 < ε < M.

P. a) Let 0 < δ < ∆ <∞ and let us estimate the integral

(2.13)

∫ ∆

δ

f(ax) − f(bx)

xdx

ax=z=bx=z

∫ a∆

aδ

f(z)

zdz −

∫ b∆

bδ

f(z)

zdz.

Since∫ a∆

aδ

f(z)

zdz =

∫ bδ

aδ

f(z)

zdz +

∫ aδ

bδ

f(z)

zdz +

∫ a∆

aδ

f(z)

zdz

and∫ b∆

bδ

f(z)

zdz =

∫ aδ

bδ

f(z)

zdz +

∫ a∆

aδ

f(z)

zdz +

∫ b∆

a∆

f(z)

zdz,

by subtraction these last two equalities we get∫ a∆

aδ

f(z)

zdz −

∫ b∆

bδ

f(z)

zdz =

∫ bδ

aδ

f(z)

zdz −

∫ b∆

a∆

f(z)

zdz.

Let us apply now the mean formula of theorem 22 to both of these lastintegrals. We obtain

(2.14)

∫ ∆

δ

f(ax) − f(bx)

xdx = [f(ξ) − f(η)] ln

b

a,

where ξ ∈ [aδ, bδ] and η ∈ [a∆, b∆]. Let us make δ → 0 and ∆ → ∞in formula (2.14) we get

∫ ∞

0

f(ax) − f(bx)

xdx = [f(0) − f(∞)] ln

b

a.

b) Let us assume that f is of class C1 on [0,∞] and let us denote by ϕthe function: ϕ(t) = f(xt), t ∈ [a, b], for a fixed x ∈ (0,∞). We applyLagrange formula for this function ϕ on [a, b] :

ϕ(b) − ϕ(a) = ϕ′(c)(b− a),

or

f(bx) − f(ax) = f ′(cx) · x · (b− a),

where c ∈ (a, b). Thus, for 0 < ε < M,∣∣∣∣f(ax) − f(bx)

x

∣∣∣∣ ≤ (b− a)g(x).


Since (b − a)∫∞0g(x)dx is convergent, Weierstrass test c) of theorem

57 tells us that I(a, b) =∫∞0

f(ax)−f(bx)x

dx is uniformly convergent.

For instance,

I(a, b) =

∫ ∞

0

arctan ax− arctan bx

xdx =

π

2lna

b.

Since f(z) = arctan z is of class C1 on R,

f ′(xt) =1

1 + x2t2≤ 1

1 + x2ε2(= g(x))

for any t ∈ [ε,M ] with [a, b] ⊂ [ε,M ] and∫∞0

11+x2ε2

dx is convergent,we see that I(a, b) is uniform convergent on [ε,∞) for any fixed ε > 0.But not always one can use Froullani’s formula (2.12) even the integralis of the same form. For instance,

I(a, b) =

∫ ∞

0

cos ax− cos bx

xdx, 0 < a < b.

Here f(z) = cos z has no limit at z = ∞, so the above theorem cannotbe applied. To compute it, we consider another integral with parame-ters:

J(a, k) =

∫ ∞

0

1 − cos ax

xe−kxdx, a > 0, k > 0.

Since the integral ∫ 1

0

1 − cos ax

xe−kxdx

is a proper integral, the integral∫ ∞

1

1 − cos ax

xe−kxdx

is uniformly convergent with respect to a :∣∣∣∣1 − cos ax

xe−kx

∣∣∣∣ ≤1

xe−kx

and∫∞1

1xe−kxdx is convergent, see test p, for p = 2 and also

∫ ∞

0

e−kx sin axdx

is uniformly convergent w.r.t. a :∣∣e−kx sin ax

∣∣ ≤ e−kx


and∫∞0e−kxdx is convergent, we can write (Leibniz’s formula):

(2.15)∂J

∂a=

∫ ∞

0

e−kx sin axdx =a

a2 + k2(compute it two times by parts).

Thus

J(a, k) =

∫a

a2 + k2da =

1

2ln(a2 + k2) + C(k).

Making a → 0, because of the continuity of J(a, k) w.r.t. a (here weuse the uniform continuity of J relative to a and theorem 57 b)), onehas that 0 = ln k + C(k), or C(k) = − ln k. Thus

J(a, k) = ln

√a2 + k2

k.

If instead of a we put b, we get∫ ∞

0

1 − cos bx

xe−kxdx = ln

√b2 + k2

k.

But

I(a, b, k) =

∫ ∞

0

cos ax− cos bx

xe−kxdx =

= ln

√b2 + k2

k− ln

√a2 + k2

k= ln

√b2 + k2

√a2 + k2

for any a, b, k > 0. Since for any k ∈ [δ,∞), with δ > 0, the integralI(a, b, k) is uniformly convergent w.r.t. k

(

∣∣∣∣cos ax− cos bx

xe−kx

∣∣∣∣ ≤∣∣∣∣cos ax− cos bx

xe−δx

∣∣∣∣ and

∫ ∞

0

cos ax− cos bx

xe−δxdx is convergent),

fixing a k0 > 0 and taking 0 < δ < k0, we can apply again theorem 57and find that I(a, b, k) is continuous at k0 as a function of k. Now wecan make k → 0 in the equality

(2.16)

∫ ∞

0

cos ax− cos bx

xe−kxdx = ln

√b2 + k2

√a2 + k2

and finally find that∫∞0

cos ax−cos bxx

dx = ln bafor any a, b > 0. Is this

last integral uniform convergent with respect to the parameters a andb ? To use formula 2.16 it is not a good idea because the integralI(a, b, k) is uniformly convergent w.r.t. k only on any interval of the


type [δ,∞) and not on the entire interval (0,∞). We shall prove nowthat the integral

I(a, b) =

∫ ∞

0

cos ax− cos bx

xdx = ln

b

a

is not uniformly convergent w.r.t. a and b (simultaneously), wherea, b ∈ (0,∞). Suppose it is uniformly convergent. Applying theorem 57b) we get for a = 1

2n, b = 1

nand n→ ∞ that 0 = ln 2, a contradiction!

Thus our integral cannot be uniformly convergent on (0,∞). It is anopen problem for me if this integral is uniformly convergent on anyinterval of the type [δ,∞), where δ > 0. But on intervals of the type[δ,M ], 0 < δ < M? These informations could not help us to computethe integral by Leibniz formula because the integral

∫ ∞

0

∂

∂a

[cos ax− cos bx

x

]dx = −

∫ ∞

0

sin axdx

is not convergent!

E 58. (Dirichlet’s integral) The following integral with aparameter β,

I(β) =

∫ ∞

0

sin βx

xdx

is very useful in different branches of science and technique. It wasconsidered for the first time by the great German mathematician Jo-hann Peter Gustave Lejeune Dirichlet in the XIX-th century. To proveits existence, for any fixed β, he invented the Dirichlet test (see the-orem 39). In example 49 we just proved its simple convergence. Inorder to compute it we need it to be uniformly convergent. Even wehad succeeded to prove such thing, Leibniz formula leads us to compute

∫ ∞

0

∂

∂β

[sin βx

x

]dx =

∫ ∞

0

cosβxdx,

which is divergent! This is why Weierstrass invented an indirect methodto compute this integral. To force the convergence to ∞ (0 is not asingular point!), Dirichlet considered the following more complicatedintegral with two parameters α, β:

(2.17) D(α, β) =

∫ ∞

0

e−αxsin βx

xdx, α > 0, β ∈ R.

It is easy to see that we can reduce ourselves to the case β > 0. It is alsoclear that the integral D(α, β) is uniformly convergent if α ∈ [α0,∞),


β > 0, where α0 is a fixed positive real number as small as we want.Indeed, since ∣∣∣∣e−αx

sin βx

x

∣∣∣∣ ≤e−α0x

x,

since the integral∫∞1

e−α0x

xdx is convergent (test p for p = 2) and since

∫ ∞

0

e−αxsin βx

xdx =

∫ 1

0

e−αxsin βx

xdx+

∫ ∞

1

e−αxsinβx

xdx,

the first integral being a proper one, Weierstrass test e) of theorem 57tells us that the initial integral D(α, β) is uniformly convergent. Toapply Leibniz formula of differentiating under the integral sign withrespect to β (see theorem 57 d)) we also need to prove that the integral

T (β) =

∫ ∞

0

∂

∂β

[e−αx

sin βx

x

]dx =

∫ ∞

0

e−αx cosβxdx =α

β2 + α2

is uniformly convergent with respect to β. Indeed,∣∣e−αx cosβx

∣∣ ≤ e−αx

and∫∞0e−αxdx is convergent (= 1

α) implies the uniform convergence of

T (β) (Weierstrass test). Hence, Leibniz formula applied to D(α, β) asan improper integral with parameter β gives:

∂D

∂β=

∫ ∞

0

∂

∂β

[e−αx

sin βx

x

]dx =

α

β2 + α2.

Integrating with respect to β we get:

D(α, β) =

∫ ∞

0

e−αxsin βx

xdx = arctan

β

α+ C(α).

The uniform continuity w.r.t. β > 0 makes D(α, β) a continuous func-tion of β ((see theorem 57 b)). Thus D(α, 0) = 0 = C(α) for anyα > 0. Hence we finally obtain

(2.18) D(α, β) =

∫ ∞

0

e−αxsin βx

xdx = arctan

β

α.

The uniform continuity of D(α, β) with respect to α ∈ [α0,∞) for anyα0 > 0, makes D(α, β) continuous with respect to α > 0 (see theorem57 b)). Thus D(0, β) =

∫∞0

sinβxxdx = π

2. Now Dirichlet’s integral is

completely computed:

(2.19)

∫ ∞

0

sin βx

xdx =

π2, if β > 0

0, if β = 0−π

2, if β < 0.


Dirichlet’s integral is very useful in the computation of other integrals.For instance,

I =

∫ ∞

0

sin2 5x

x2dx = lim

x→∞

∫ x

0

sin2 5u

u2du = − lim

x→∞

∫ x

0

sin2 5u

(1

u

)′du =

(2.20) − limx→∞

[sin2 5u · 1

u

∣∣∣∣x

0

− 10

∫ x

0

sin 5u cos 5u

udu

].

Since

limu→0

sin2 5u

u= 5lim

u→0

sin 5u

5u· limu→0

sin 5u = 5 · 1 · 0 = 0,

sin2 5u · 1u

∣∣x0

= sin2 5xx

and 10∫ x0

sin 5u cos 5uu

du = 5∫ x0

sin 10uu

du. Thus thevalue of I from (2.20) becomes

I = − limx→∞

sin2 5x

x− 5

∫ ∞

0

sin 10u

udu.

Since∣∣∣ sin2 5x

x

∣∣∣ ≤ 1x, we find that the limit is equal to 0. Looking at the

Dirichlet’s integral we see that∫∞0

sin 10uu

du = π2(here β = 10). Thus,

I = −5π2.

E 59. (Euler-Poisson integral) The famous integral J =∫∞0e−x

2dx, which appears in Probability and Statistics whenever we

work with the Gauss-Laplace distribution and in many other places ofMathematics, Mechanics, etc. is called the Euler-Poisson integral. Tocompute it (the p-test for p = 2 assure us on the convergence!) let usintroduce a parameter u and a new variable t. Let us make a change ofvariables x↔ t, x = ut in J. We get J = u

∫∞0e−u

2t2dt. Thus

e−u2

J = ue−u2

∫ ∞

0

e−u2t2dt.

Let us integrate this last equality with respect to u and then let us changethe order of integration (see Fubini’s theorem for improper integrals,theorem 57 c); verify the uniform convergence!):

J

∫ ∞

0

e−u2

du = J2 =

∫ ∞

0

ue−u2

du ·∫ ∞

0

e−u2t2dt =

=

∫ ∞

0

(∫ ∞

0

e−u2(t2+1)u du

)dt =

∫ ∞

0

− 1

2(t2 + 1)e−(1+t2)u2

∣∣∣∞

0dt =

=1

2

∫ ∞

0

1

1 + t2dt =

1

2arctan t |∞0 =

π

4.


Hence

(2.21) J =

∫ ∞

0

e−x2

dx =

√π

2

If we put x =√tu, u being a new variable ant t > 0 a parameter, we

get

(2.22)

∫ ∞

0

e−tu2

du =1√t

√π

2.

E 60. (Fresnel’s integrals) The following integrals I1 =∫∞0

sin x2dx and I2 =∫∞0

cosx2dx are called Fresnel’s integrals. Letus see that I1 is convergent and then we shall compute it. Since I1 =

limA→∞

∫ A0

sin x2dx, let us evaluate∫ A0

sin x2dx. Let us change the vari-

able x with u = x2, i.e. x =√u. Thus, dx = 1

2√udu and the

last integral becomes∫ A20

sinu2√udu. In this last case u = 0 is a singu-

larity point for this integral. But the integral is convergent becauselim

u→0,u>0u12sinu2√u

= 0, q = 1/2 in the q-test, thus the integral is conver-

gent. Moreover I1 =∫∞0

sinu2√udu. It is also convergent at ∞. Indeed,

one can apply Dirichlet test (see theorem 39) and see that∫∞1

sinu2√udu is

convergent. We just proved that∫ 1

0sinu2√udu (A = 1) is convergent, so

I1 =

∫ 1

0

sinu

2√udu+

∫ ∞

1

sin u

2√udu

is also convergent. To compute I1 =∫∞0

sinu2√udu let us force the conver-

gence to∞ with a more complicated improper integral with a parameterk :

(2.23) J1(k) =

∫ ∞

0

sin u

2√ue−kudu.

It is easy to see that this integral is uniformly convergent on any in-

terval of the type [δ,∞) with δ > 0 (∣∣∣ sinu2√ue−ku

∣∣∣ ≤ e−δu√u,∫∞0

e−δu√udu is

convergent so we can apply Weierstrass test e) of theorem 57). Thisimplies that the function k → J1(k) is continuous on (0,∞) and itcan be extended by continuity with J1(0) =

∫∞0

sinu2√udu on [0,∞). Since

∫∞0e−ut

2dt = 1√

u

√π2(see formula (2.22)), we can put instead of 1√

uin

formula (2.23) the expression 2√π

∫∞0e−ut

2dt. Thus we get

J1(k) =

∫ ∞

0

sin u

2√ue−kudu =

1√π

∫ ∞

0

[e−ku sin u

∫ ∞

0

e−ut2

dt

]du =


=1√π

∫ ∞

0

[∫ ∞

0

sin u · e−u(k+t2)dt]du.

We now apply Fubini theorem (see theorem 57 c)) to change the orderof integration:

J1(k) =1√π

∫ ∞

0

[∫ ∞

0

sin u · e−u(k+t2)du]dt.

From formula (2.15),∫∞0e−kx sin axdx = a

a2+k2, we get

∫ ∞

0

sinu · e−u(k+t2)du =1

1 + (k + t2)2.

Thus

J1(k) =1√π

∫ ∞

0

1

1 + (k + t2)2dt.

The continuity of J1(k) on [0,∞) finally implies that

(2.24) I1 = J1(0) =

∫ ∞

0

sin u

2√udu =

1√π

∫ ∞

0

1

1 + t4dt.

To compute∫∞0

11+t4

dt we need the decomposition of 11+t4

into simplefractions. Since

t4 + 1 =(t2 + 1

)2 − 2t2 = (t2 + 1 − t√

2)(t2 + 1 + t√

2),

we get1

1 + t4=

At+B

t2 + 1 − t√

2+

Ct+D

t2 + 1 + t√

2,

where A = − 12√

2, C = 1

2√

2, B = D = 1

2. Hence

∫1

1 + t4dt =

A

2

∫(2t−

√2)

t2 + 1 − t√

2dt+

A+√

2B√2

∫1

t2 + 1 − t√

2dt+

+C

2

∫(2t +

√2)

t2 + 1 + t√

2dt+

B√

2 − C√2

∫1

t2 + 1 + t√

2dt =

=1

4√

2lnt2 + 1 + t

√2

t2 + 1 − t√

2− 1

4

∫1(

t− 1√2

)2

+ 12

d

(t− 1√

2

)+

+1

4

∫1(

t+ 1√2

)2

+ 12

d

(t +

1√2

)=

1

4√

2lnt2 + 1 + t

√2

t2 + 1 − t√

2+

+1

4·√

2 tan−1(√

2t− 1) +1

4·√

2 tan−1(√

2t + 1).


Thus,∫ ∞

0

1

1 + t4dt = 0 +

√2

4

[π2

+π

4+π

2− π

4

]=

π

2√

2.

Finally, I1 = 1√π· π

2√

2= 1

2

√π2. We leave as an exercise for the reader

to prove that I2 =∫∞0

cosx2dx is also 12

√π2.

R 19. Sometimes we meet the following situation. Let K bean interval [a, b] in R∪±∞ and let J = [t0, T ] be an interval in R∪∞ such that t0 ∈ R. Let f(x, t) be a function of two variables defined

on K×J such that it is of class C1 on K×J. Let I(t) =∫ baf(x, t)dx be

an improper integral with a parameter t with a or b, or both of them asits unique singular points. Assume that for any δ > 0, δ < T − t0, the

integrals I(t) and L(t) =∫ ba∂f∂t

(x, t)dx are uniformly convergent w.r.t.

t on [t0 +δ, T ].We also assume that I(t0) =∫ baf(x, t0)dx is convergent

and that L(t) =∫ ba∂f∂t

(x, t)dx is bounded in any neighborhood [t0, t0 +

δ] of t0. Then the function t → I(t) =∫ baf(x, t)dx is well defined

on [t0,∞) and it is continuous on this interval. Indeed, since I(t)is uniformly convergent on any interval of the type [t0 + δ, T ] it iscontinuous on (t0, T ] (see theorem 57 b)). Now we prove that we can"extend by continuity" this continuous function I(t), t ∈ (t0, T ] to the

entire interval [t0, T ] by putting I(t0) =∫ baf(x, t0)dx. Thus we want

to prove the continuity of I at t0 in [t0, T ]. Take a sequence tnn≥1,tn ∈ (t0, T ] and tn → t0. Let us evaluate the difference I(tn) − I(t0),when n → ∞. For any fixed x ∈ (a, b), let us apply Lagrange formulafor the function t→ f(x, t) on the interval [t0, tn] :

f(x, tn) − f(x, t0) =∂f

∂t(x, cx,tn)(tn − t0),

where cx,tn ∈ [t0, tn]. Thus,

I(tn) − I(t0) =

∫ b

a

[f(x, tn) − f(x, t0)] dx = (tn − t0)

∫ b

a

∂f

∂t(x, cx,tn)dx

is convergent to 0 when n → ∞ because∫ ba∂f∂t

(x, cx,tn)dx is boundedaround the point t0. Hence I(t) is also continuous at t = t0.

For instance, let I(t) =∫∞0

sinxxe−txdx, t ∈ [0,∞). It is not difficult

to prove that all conditions which appears in remark 19 are satisfiedfor t0 = 0. Thus we can say that I is also continuous at t = 0. Its valueI(0) =

∫∞0

sinxxdx = π

2(see Dirichlet’s integral 2.19).

E 61. Let us study and compute the following integral

3. EULER’S FUNCTIONS GAMMA AND BETA 167

I(α, β) =∫ 1

0xα−xβlnx

dx, α, β > −1. For this let us formally write

I(α, β) =

∫ 1

0

xα − 1

ln xdx−

∫ 1

0

xβ − 1

lnxdx.

It is sufficient to prove that any of this last two integrals are properintegrals (do it!). Thus I(α, β) is a proper integral and

∫ 1

0

xα − xβ

ln xdx =

∫ 1

0

(∫ α

β

xtdt

)dx =

∫ α

β

(∫ 1

0

xtdx

)dt =

=

∫ α

β

xt+1

t+ 1

∣∣∣∣1

0

dt =

∫ α

β

1

t+ 1dt = ln

α + 1

β + 1.

Here we just applied Fubini theorem 52, because all integrals whichappear can be considered proper integrals.

3. Euler’s functions gamma and beta

A polynomial function, P (x) = a0 +a1x+ ...+anxn, x ∈ R, a0, ...an

are fixed real numbers, a rational function R(x) = P (x)Q(x)

, where P (x),

Q(x) are polynomial functions, an exponential function f(x) = ax,a > 0, a = 1, a logarithmic function f(x) = loga x, x > 0, a > 0, a = 1,a power function f(x) = xα, x > 0, α ∈ R, a trigonometric functionf(x) = sin x, cosx, tanx, cot x, or their inverses, a composition ofthese previous ones, all of these functions are called elementary func-tions. We know that all these elementary functions are very useful inmany branches of applied mathematics. The first function which is notelementary but is very useful in many applications of mathematics isthe gamma function of L. Euler (1707 — 1783). It is also called Euler’sfunction of the second type. Its formal definition is the following:

D 13.

(3.1) Γ(x) =

∫ ∞

0

tx−1e−tdt, x ∈ (0,∞).

We see that this function is in fact an improper integral with a para-meter x. In this case we must begin with a proof of its existence. Weshall prove something more.

T 59. As an improper integral with a parameter x, thegamma function Γ(x) =

∫∞0tx−1e−tdt, x ∈ (0,∞) is uniformly conver-

gent on every closed subinterval [δ,M ] ⊂ (0,∞). In particular, it is


continuous and differentiable on (0,∞). Moreover, it is of class C∞ on(0,∞) and its derivative of order k is

(3.2) Γ(k)(x) =

∫ ∞

0

tx−1(ln t)ke−tdt, x ∈ (0,∞)

for any k = 0, 1, ... .

P. Since for k = 0 in formula (3.2) we obtain the defini-tion formula (3.1), we fix a k in N and we shall prove the statementof theorem for a more general case of improper integral Γ(k)(x) =∫∞0tx−1(ln t)ke−tdt, x ∈ (0,∞) with parameter t. Let us formally write:

Γ(k)(x) = Γ(k)1 (x) + Γ

(k)2 (x),

where

Γ(k)1 (x) =

∫ 1

0

tx−1(ln t)ke−tdt

and

Γ(k)2 (x) =

∫ ∞

1

tx−1(ln t)ke−tdt.

If we could prove that these two last improper integrals with parameterx are uniformly convergent on [δ,M ], their sum Γ(k)(x) would also be

uniformly convergent on the same interval. Let us start with Γ(k)1 (x). It

is an improper integral of the second type with a singular point t = 0.Since

∣∣tx−1(ln t)ke−t∣∣ ≤ tδ−1 |ln t|k e−t = tδ−1

(ln

1

t

)ke−t,

applying Weierstrass test e) of theorem 57, it will be enough to prove

the convergence of the integral∫ 1

0tδ−1

(ln 1

t

)ke−tdt. We can assume

that δ is sufficiently small (we prove the uniform convergence on alarger interval if we diminish δ!), for instance that 0 < δ < 1. Let us fixnow a number q such that 0 < q < 1, q + δ − 1 > 0 and another smallnumber ε with 0 < ε < q−1+δ

k.We did this such that the following limit

to be 0!

(3.3) limt→0,t>0

tqtδ−1

(ln

1

t

)ke−t = lim

t→0,t>0tqtδ−1t−kε

(ln 1

t

)k(

1t

)kε .

But we know that function ln z goes to ∞ whenever z → ∞ slowerthan any positive power of z, i.e. lim

z→∞ln zzε

= 0 (prove it again!). By

putting z = 1t, when t → 0, t > 0, we get that

(ln 1t )

k

( 1t )kε → 0, whenever


t→ 0, t > 0. Thus, in formula (3.3) we finally obtain:

limt→0,t>0

tqtδ−1

(ln

1

t

)ke−t = lim

t→0,t>0tq+δ−1−kε · 0 = 0,

because q + δ − 1 − kε > 0 (see here why we wanted ε to verify theinequalities: 0 < ε < q−1+δ

k!). Now, the q-test (see theorem 45) tells us

that the improper integral∫ 1

0tδ−1

(ln 1

t

)ke−tdt is convergent and that

Γ(k)1 (x) =

∫ 1

0tx−1(ln t)ke−tdt is uniformly convergent.

Let us prove now that Γ(k)2 (x) =

∫∞1tx−1(ln t)ke−tdt is also uni-

formly convergent. Since our integral is improper of the first type, weapply again Weierstrass test e) of theorem 57:

∣∣tx−1(ln t)ke−t∣∣ ≤ tM−1(ln t)ke−t.

It remains to prove that the improper integral∫∞1tM−1(ln t)ke−tdt is

convergent. Let us apply the p-test (see theorem 38):

limt→∞

tptM−1(ln t)ke−t = limt→∞

tptM−1tk

et

(ln t

t

)k= 0

for any p because the exponential function et goes faster to ∞ than anypower of t, i.e. lim

z→∞tα

et= 0 (prove it again!). Take for instance p = 2

and the test p tells us that the integral is convergent. Now the proofof the theorem is complete.

Two problems arise: 1) Is our improper integral Γ(x) =∫∞0tx−1e−tdt,

uniform convergent on an interval of the type (0,M ], for M > 0? Theanswer is no. Indeed, for any 0 < η < 1 one has:

(3.4)

∫ η

0

tx−1e−tdt ≥ e−η∫ η

0

tx−1dt = e−ηηx

x→ ∞,

whenever x→ 0. Thus, it is not possible for a given small ε > 0 to finda η, 0 < η < 1, such that

∣∣∣∣∫ η

0

tx−1e−tdt

∣∣∣∣ < ε

for any x ∈ (0,M ] because the quantity on the left goes to ∞ when xbecomes closer and closer to 0 (see formula (3.4)).

The second problem: 2) Is our integral Γ(x) =∫∞0tx−1e−tdt uni-

formly convergent on an interval of the type [M,∞), where M > 0?The answer is also no. Indeed, for any fixed N > 1,

∫ ∞

N

tx−1e−tdt ≥ Nx−1

∫ ∞

N

e−tdt = Nx−1e−N → ∞,


when x→ ∞. Thus, for a given small ε > 0 it is not possible to find asufficiently largeN > 1 such that

∫∞Ntx−1e−tdt < ε for any x ∈ [M,∞).

One of the most important property of Euler gamma function is thatone which refers to the "generalization" of the factorial function f(n) =n!, defined only on the set of natural numbers. Namely, Γ(n) = (n−1)!for any n ∈ N, n > 0. This formula will appear as a consequence of amore general recurrence property.

T 60. (basic properties of gamma)1) Γ(1) = 1.2) Γ(2) = 1.3) Γ(x) = (x−1)Γ(x−1) for any x > 1 (simple recurrence formula

or simple reduction formula).4) Γ(x) = (x−1)(x−2)...(x−n)Γ(x−n) for any fixed n ∈ N, n ≥ 1

and for any x > n (general recurrence formula or reduction formula)5) It is sufficient to know the values of Γ on the interval (0, 1].6) Γ(n) = (n− 1)! for any n ∈ N, n > 0 .7) Γ(x)Γ(1 − x) = π

sinπx, x ∈ (0, 1) (complementaries formula).

8) It is sufficient to know the values of Γ on the interval (0, 12] and

Γ(12) =

√π.

9)∫∞0yp−1e−tydy = Γ(p)

tp, t > 0, p > 0.

P. Property 7) cannot be proved here because its proof iseither too long or it is shorter but involves a lot of higher mathematics.Let us prove the other statements.

1) Γ(1) =∫∞0e−tdt = −e−t |∞0 = 1.

2)

Γ(2) =

∫ ∞

0

te−tdt =

∫ ∞

0

t(−e−t)′dt =by parts

= −te−t |∞0 +

∫ ∞

0

e−tdt = 0 + 1 = 1.

3)

Γ(x) =

∫ ∞

0

tx−1e−tdt =

∫ ∞

0

tx−1(−e−t)′dt =

= −tx−1e−t |∞0 +(x− 1)

∫ ∞

0

tx−2e−tdt = (x− 1)Γ(x− 1)

or

(3.5) Γ(x) = (x− 1)Γ(x− 1)

for any x > 1.


4) We simply apply the simple recurrence formula n times:

Γ(x) = (x− 1)Γ(x− 1) = (x− 1)(x− 2)Γ(x− 2) =

= (x− 1)(x− 2)(x− 3)Γ(x− 3) = ...

= (x− 1)(x− 2)...(x− n)Γ(x− n).

5) We know that Archimede’s axiom says that any x > 0 belongsto an interval of the type (n, n+ 1], where n ∈ N = 0, 1, 2, .... Thus,x− n ∈ (0, 1) and formula

(3.6) Γ(x) = (x− 1)(x− 2)...(x− n)Γ(x− n)

reduces the computation of Γ(x) to the computation of Γ(x−n), wherex− n ∈ (0, 1).

6) The formula Γ(n + 1) = n! immediately appears if we put informula (3.6) x = n + 1. Put then n − 1 instead of n in this lastobtained formula, etc.

8) In 5) we saw how to reduce the computation of Γ(x) to thecomputation of gamma at a subunitary value z = x− n. Assume thatz > 1

2. Then 1 − z < 1

2and, using the complementaries formula 7) we

get:

Γ(z) =π

Γ(1 − z) sin πz.

If we put x = 12in the complementaries formula, we get Γ(1

2) =

√π.

Another way to compute Γ(12) is to use directly the definition of Γ(1

2) :

Γ

(1

2

)=

∫ ∞

0

t−12e−tdt

t=z2= 2

∫ ∞

0

z−1e−z2

zdz =

= 2

∫ ∞

0

e−z2

dz = 2 ·√π

2=

√π,

because∫∞0e−z

2dz =

√π2

(Euler-Poisson integral-see example 59).

9) In Γ(p) =∫∞0zp−1e−zdz we put z = ty, where y is the new

variable and t is a positive parameter. Thus, Γ(p) = tp∫∞0yp−1e−tydy

or∫∞0yp−1e−tydy = Γ(p)

tp.

For instance, Γ(72) = 5

2Γ(5

2) = 5

2· 3

2· Γ(3

2) = 5

2· 3

2· 1

2Γ(1

2) = 15

√π

8.

E 62. (moment of order k) The following integralMk =

∫∞−∞ x

ke−x2dx is called in Statistics and in Mechanics a mo-

ment of order k (...of something!). If k is odd Mk = 0 (Why?). Ifk = 2n is even, then

M2n = 2

∫ ∞

0

x2ne−x2

dxx2=t=

∫ ∞

0

t(n+12)−1e−tdt = Γ(n+

1

2).


Let us use now the general recurrence formula and get:

M2n =

(n− 1

2

)(n− 3

2

)...

1

2Γ(

1

2) =

(2n− 1)(2n− 3)...1

2n−1·√π

2.

Let us give some information in order to sketch the graphic of Γ(x)(see Fig.1)

T 61. (derivatives and graphic for Γ(x)) Gamma functionΓ(x) =

∫∞0tx−1e−tdt, x ∈ (0,∞) has the following additional proper-

ties:1) Γ′′(x) > 0, i.e. the graphic of Γ is a convex one (keep water!-see

Fig.1)2) Γ has a unique local extremum point, namely a minimum point

xmin ∈ (1, 2) (see Fig.1)3) Oy-axis is a vertical asymptote (see Fig.1)4) lim

x→∞Γ(x) = ∞. Moreover, Γ(x) → ∞, whenever x → ∞ as fast

as n! → ∞, when n→ ∞ (see Fig.1).

P. 1) Formula (3.2) implies Γ′′(x) =∫∞0tx−1 (ln t)2 e−tdt > 0

for any x ∈ (0,∞). Thus the function Γ(x) is convex.2) Since Γ(1) = Γ(2) = 1 (see theorem 60), Rolle’s theorem (see for

instance [Po], Th.36) says that there exists a point xmin ∈ (1, 2) suchthat Γ′(xmin) = 0. Since Γ′′(x) > 0, xmin is a point of minimum for Γ.It is the unique extremum point for Γ. Otherwise, if there was anotherone z = xmin, then Γ′(z) = 0 (Fermat’s theorem-see for instance [Po],Th.35). Applying Rolle’s theorem to the function Γ′(x) on the intervalwith ends z and xmin respectively, we would find a point y in this lastinterval such that Γ′′(y) = 0, a contradiction, because Γ′′(x) > 0 forany x ∈ (0,∞). Thus xmin is the unique extremum point for Γ.

3) Let us compute

limx→0,x>0

Γ(x) = limx→0,x>0

Γ(x+ 1)

x=

1

+0= ∞.

Here we just used the recurrence formula Γ(x+1) = xΓ(x) (see theorem60) and the continuity of Γ (see theorem 59).

4) The continuity of Γ, just used above, and the basic formulaΓ(n) = (n− 1)! implies that

limx→∞

Γ(x) = limn→∞

Γ(n) = limn→∞

(n− 1)! = ∞.


F 1

Another fundamental non-elementary function is the beta functionof Euler or the first type Euler function. It is an improper integral withtwo parameters x and y. Its formal definition is:

D 14.

(3.7) B(x, y) =

∫ 1

0

tx−1(1 − t)y−1dt, x > 0, y > 0.

If x ≥ 1 and y ≥ 1 the beta function B(x, y) is a proper integralwith two parameters, i.e. it is convergent. The following result willclarify the other cases.

T 62. The improper integral B(x, y) =∫ 1

0tx−1(1− t)y−1dt,

x > 0, y > 0 is uniformly convergent on any strip of the form [p0,∞)×[q0,∞), where p0 > 0 and q0 > 0 (see Fig.2). In particular, the functionB(x, y) is continuous and even of class C∞ on (0,∞) × (0,∞).

P. We see that B(x, y) is an improper integral with two sin-gularities: t = 0 and t = 1 when x < 1 and y < 1. This is why weconsider the following decomposition (we separate these singularities):

B(x, y) = B0(x, y) +B1(x, y),

where

B0(x, y) =

∫ 12

0

tx−1(1 − t)y−1dt

and

B1(x, y) =

∫ 1

12

tx−1(1 − t)y−1dt.


If we succeed to prove that both B0(x, y) and B1(x, y) are uniformlyconvergent on [p0,∞) × [q0,∞) for any p0, q0 > 0, then B(x, y) itselfwould be uniformly convergent on [p0,∞) × [q0,∞). 1) Let us provethat B0(x, y) is uniformly convergent on [p0,∞) × [q0,∞). Since t = 0is the only singularity in this last case, if x < 1, we can consider p0 < 1and p0 ≤ x < 1 (if x ≥ 1 the integral is a proper one). Then

tx−1(1 − t)y−1 ≤ tp0−1(1 − t)y−1.

Let us apply now the Weierstrass test c) of theorem 57 and try to prove

that the integral∫ 1

20tp0−1(1 − t)y−1dt is convergent. For this we apply

the q-test (the singularity is t = 0):

limt→0,t>0

tqtp0−1(1 − t)y−1 = 1

for q = 1 − p0, which is < 1. Thus the integral is convergent andour integral B0(x, y) is uniformly convergent. 2) Let us now prove

that the improper integral B1(x, y) =∫ 112tx−1(1 − t)y−1dt is uniformly

convergent. This time the integral has as its unique singular point thepoint t = 1. We can consider that q0 < 1 (otherwise x ≥ 1 and ourintegral would be a proper one, i.e. a convergent one). So 0 < q0 <y < 1 and we can try to apply Weierstrass test c) of theorem 57. Since

tx−1(1 − t)y−1 ≤ tx−1(1 − t)q0−1,

it remains to prove that the improper integral∫ 112tx−1(1 − t)q0−1dt is

convergent (the singularity is t = 1). Indeed, applying the same q-testfor q = 1 − q0 we get

limt→1,t<1

(1 − t)1−q0tx−1(1 − t)q0−1 = 1.

Since 1 − q0 < 1, our last integral is convergent and consequently theintegral B1(x, y) is uniformly convergent. Hence B(x, y) = B0(x, y) +B1(x, y) is uniformly convergent on [p0,∞) × [q0,∞) for any p0, q0 >0. The same type of proof can be done for the uniform convergenceof

∫ 1

0∂∂x

[tx−1(1 − t)y−1] dt or for∫ 1

0∂∂y

[tx−1(1 − t)y−1] dt. Let us fix a

point (x0, y0) ∈ (0,∞) × (0,∞) and let us choose p0, q0 > 0 such thatx0 > p0 and y0 > q0. Thus, (x0, y0) ∈ [p0,∞) × [q0,∞) and, since thebeta function B(x, y) is uniformly convergent on [p0,∞)× [q0,∞), fromtheorem 57 b) and d), applied to B0(x, y) and to B1(x, y) separately,we see that B(x, y) is continuous, differentiable and even of class C∞

on (0,∞) × (0,∞) (work slowly everything!).

In the following result we put together the main properties of thefunction beta.


F 2

T 63. Let B(x, y) =∫ 1

0tx−1(1 − t)y−1dt, x > 0, y > 0 be

the beta function. Then1) B(x, y) = B(y, x) (symmetry).2) B(x, y + 1) = y

x+yB(x, y) (reduction formula).

3) B(x+ 1, y) = xx+y

B(x, y) (reduction formula).

4) B(x, y) =∫∞0

ux−1

(1+u)x+y du.

5) B(x, y) =∫∞0

uy−1

(1+u)x+y du.

6) B(x, y) = Γ(x)·Γ(y)Γ(x+y)

(Γ-reduction). This formula reduces the com-

putations with function B to computations with function Γ.

7) B(x, y) = 2∫ π

20

sin2x−1 u cos2y−1 udu (the trigonometric expres-sion).

P. First of all we want to remark that all the following com-putations are made in fact with proper integrals, then we take limitsto obtain the improper integrals formulations. But we omit to specifythis and we directly work with the improper form! Let us use thisextremely excessive method only to prove 1).

B(x, y) = limε→0,η→1

∫ η

ε

tx−1(1 − t)y−1dt =

u=1−t= − lim

ε→0,η→1

∫ 1−η

1−ε(1 − u)x−1uy−1du =


=

∫ 1

0

uy−1(1 − u)x−1du = B(y, x).

2)

B(x, y + 1) =

∫ 1

0

tx−1(1 − t)ydt =

∫ 1

0

(tx

x

)′(1 − t)ydt =

=

(tx

x

)(1 − t)y

∣∣∣∣1

0

+y

x

∫ 1

0

tx(1−t)y−1dt =y

x

∫ 1

0

tx−1(1+t−1)(1−t)y−1dt =

y

x

∫ 1

0

tx−1(1− t)y−1dt− y

x

∫ 1

0

tx−1(1− t)ydt =y

xB(x, y)− y

xB(x, y+ 1).

Thus [1 +

y

x

]B(x, y + 1) =

y

xB(x, y),

or

B(x, y + 1) =y

x+ yB(x, y).

3) We use now 1), 2) and again 1) to prove 3):

B(x+ 1, y) = B(y, x+ 1) =x

x+ yB(y, x) =

x

x+ yB(x, y).

4) In B(x, y) =∫ 1

0tx−1(1−t)y−1dt let us make the change of variable

t = u1+u

=

t =u

1 + u= 1 − 1

1 + u.

We simply get

B(x, y) =

∫ ∞

0

(u

1 + u

)x−1 (1

1 + u

)y−11

(1 + u)2du =

∫ ∞

0

ux−1

(1 + u)x+ydu.

5) If in the formula 4) we use the symmetry 1) we get:

B(x, y) = B(y, x) =

∫ ∞

0

uy−1

(1 + u)y+xdu,

i.e. formula 5).6) In formula

(3.8)

∫ ∞

0

yp−1e−tydy =Γ(p)

tp, t > 0, p > 0.

of theorem 60 9) we simply put t+ 1 instead of t and p+ q instead ofp (here q > 0). We get:

Γ(p+ q)

(t+ 1)p=

∫ ∞

0

yp+q−1e−(t+1)ydy.


Let us multiply this last equality by tp−1 and then let us integrate theobtained result with respect to t from 0 up to ∞:

(3.9) Γ(p+ q)

∫ ∞

0

tp−1

(t+ 1)p+qdt =

∫ ∞

0

tp−1

[∫ ∞

0

yp+q−1e−(t+1)ydy

]dt.

It is not so difficult to verify the conditions of theorem 57 c) and applyFubini’s result, i.e. we change the order of integration in the equal-ity (3.9). Moreover, we also use formula

∫∞0

tp−1

(t+1)p+q dt = B(p, q) just

obtained in 4):

Γ(p + q)B(p, q) =

∫ ∞

0

yp+q−1e−y[∫ ∞

0

tp−1e−tydt

]dy.

We use again formula (3.8) with t instead of y and y instead of t tocompute the brackets inside the right side of the last equality:

Γ(p + q)B(p, q) =

∫ ∞

0

yp+q−1e−yΓ(p)

ypdy =

= Γ(p)

∫ ∞

0

yq−1e−ydy = Γ(p)Γ(q),

or

B(p, q) =Γ(p)Γ(q)

Γ(p+ q).

7) In the definition formula B(x, y) =∫ 1

0tx−1(1 − t)y−1dt of the

function beta let us put t = sin2 u :

B(x, y) =

∫ π2

0

sin2x−2 u · cos2y−2 u · 2 sin u · cosudu =

= 2

∫ π2

0

sin2x−1 u cos2y−1 udu.

The function beta is useful in the computation of some types of

integrals. For instance, let us compute I =∫ π

2

0sin7 x 4

√cosxdx. It is

very easy to put in formula 7) of the last theorem 2x − 1 = 7 and2y−1 = 1

4. Thus x = 4 and y = 5

8. Hence I = 1

2B(4, 5

8). Let us use now

the Γ-reduction formula from the same last theorem and find:

I =1

2B(4,

5

8) =

1

2

Γ(4)Γ(58)

Γ(4 + 58)

=1

2

6 · Γ(58)(

3 + 58

) (2 + 5

8

) (1 + 5

8

) (58

)Γ(5

8)

=

=3 · 84

29 · 21 · 13 · 5.

Let us now compute J =∫∞0t14 (1 + t)−2dt.


Formula B(x, y) =∫∞0

ux−1

(1+u)x+y du from the above theorem can be

used with x− 1 = 14and x+ y = 2. Thus, x = 5

4and y = 3

4. Hence

J = B(5

4,3

4) =

Γ(54)Γ(3

4)

Γ(54

+ 34)

=14Γ(1

4) · Γ(3

4)

1.

Let us use the complementaries formula 7) of theorem 60 for x = 14and

find Γ(14) · Γ(3

4) = π

sin π4

= π√

22. Hence J = π

√2

8.

E 63. Another important integrals are I1(m) =∫ π

2

0sinm xdx

and I2(m) =∫ π

20

cosm xdx. Let us compute I1(m) (here m is a naturalnumber).

I1(m) =

∫ π2

0

sinm xdx =1

2B

(m+ 1

2,1

2

)=

1

2

Γ(m+1

2

)Γ(1

2)

Γ(m+2

2

) .

If m = 2k (even), then

I1(2k) =1

2

Γ(k + 12)Γ(1

2)

Γ(k + 1)=

1

2

(k − 12)(k − 3

2)...1

2

[Γ(1

2)]2

k!=

=π · 1 · 3 · 5 · ... · (2k − 1)

2k+1k!.

If m = 2k + 1 (odd), then

I1(2k + 1) =1

2

Γ(k + 1)Γ(12)

Γ(k + 1 + 12)

=1

2

k!Γ(12)

(k + 12)(k − 1

2)(k − 3

2)...1

2Γ(1

2)

=

=2kk!

1 · 3 · 5 · ... · (2k + 1).

We leave as an exercise to the reader to compute I2(m).

E 64. ("Traian Lalescu" national contest, 2006, Romania)

Let us compute Jn =∫ ba(x − a)n(b − x)ndx. First of all let us change

the variable x with u = x− a. Thus,

Jn =

∫ b−a

0

un(b− a− u)ndu = (b− a)n∫ b−a

0

un(

1 − u

b− a

)ndu.

Let us change again the variable u with t = ub−a . Hence

Jn = (b− a)2n+1

∫ 1

0

tn (1 − t)n dt = (b− a)2n+1B(n + 1, n + 1) =

= (b− a)2n+1Γ(n+ 1)Γ(n+ 1)

Γ(2n + 2)= (b− a)2n+1 (n!)2

(2n+ 1)!.



1. Prove that the function f(x) =∫∞xe−xy

2dy, x > 0 is well defined

(the improper integral is convergent). Prove that the improper integral∫∞xe−xy

2dy with a parameter x > 0 is uniformly convergent on any

interval of the type [x0,∞) with x0 > 0. Compute f ′(x) and f ′′(x)and prove that these improper integrals are uniformly convergent onintervals of the type [x0,∞) with x0 > 0.

2. Starting with equality∫∞0e−ptdt = 1

p, p > 0, compute

∫∞0tne−ptdt

for any n = 1, 2, ... (Justify all operations you do!).

3. Let u(x, y) =∫∞−∞

xf(z)x2+(y−z)2dz be an improper integral with two

parameters x, y. Study the convergence and the uniform convergenceof u(x, y). When can we compute ∂2u

∂x2and ∂2u

∂y2? Prove that ∆u = ∂2u

∂x2+

∂2u∂y2

= 0.

4. Starting with∫ 1

0xa−1dx = 1

a, a > 1 , find

∫ 1

0xn−1 ln3 xdx for

n = 2, 3, ... .5. Find an appropriate set of convergence for the following improper

integrals with parameters, verify the conditions of Leibniz formula andapply it to compute them.

a)∫∞0

arctan axx(1+x2)

dx;

b)∫ π

2

0arctan(λ tanx)

tanxdx, λ ≥ 0;

c)∫∞0

dx(x2+λ)n+1

(start with∫∞0

dx(x2+λ)

and compute its derivatives,

many times);

d)∫ 1

0ln(1−α2x2)x2√

1−x2 dx, |α| < 1;

e)∫ π0

ln(1+sin a cos x)cosx

dx;

f)∫∞0

e−ax−e−bx

xdx, a > 0, b > 0;

g)∫∞0

e−ax−e−bx

xsin(mx)dx, a, b > 0;

h)∫ π

2

0ln

(1+a cosx1−a cosx

)1

cos xdx, |a| < 1.

6. Prove that the famous Bessel’s function Jn(x) = 2π

∫ π0

cos(nt −x sin t)dt, n = 0, 1, 2, ... , is a solution of the ordinary differential equa-tion (Bessel’s equation): x2y′′ + xy′ + (x2 − n2)y = 0.

7. Compute:

a) limx→π

∫ sinx0 et

2dt∫ tan x

0 et2dt

; b) limx→∞

∫ x0 (tan−1 t)

2dt

√x2+1

; c)limx→0

∫ x2x e−t2dt∫ x2x3 e

−t2dt; d)lim

x→0

∫ x2sinx e

t3dt∫ x+11 et

2dt.

8. Use Dirichlet’s integral to compute:a)∫∞0

sin3 xxdx; b)

∫∞0

sin4 xxdx; c)

∫∞0

sin2 xx2

dx;

Hint: sin 3x = (sin x) (3 − 4 sin2 x); sin4 x = (1 − cos2 x) sin2 x;9. Use informations on Euler’s gamma and beta functions to com-

pute:


a)∫∞0t14 (1 + t)−2dt; b)

∫ π2

0sin3 x cos5 xdx;

c)∫ π

2

0sin3 x cos2 xdx; d)

∫ 1

0x−

23 (1 − x)−

13dx;

e)∫∞0x6e−x

2dx; f)

∫∞−∞ x

ne−x2dx, n a natural number;

g)∫∞0e−x

3dx; h)

∫ 1

0t3√

1 − tdt;

i)∫ 1

0(1 − x2)10dx; j)

∫ 1

01√

1−x3dx; k)∫∞0

x1+x3

dx; l)∫∞0

11+x3

dx;

m)∫∞0

dx(1+x2)2

; n)∫∞0

3√x1+x2

dx.

10) Compute:

a)(∫ t2te−tx

2dx

)′′∣∣∣∣t=1

; b)∫ π

40

ln(1+t·tan x)dx, t ≥ 0; c)∫ π

20

ln(cos2 x+

√2 sin2 x)dx (Hint: Compute first of all I(t) =

∫ π2

0ln(cos2 x+t sin2 x)dx,

then put t =√

2);

d)∫∞0

x2

1+x4dx;

11) Prove that equality: Γ(

12

)=

√π implies equality:

Γ(n + 1

2

)= 1·3·5·...·(2n−1)

2n

√π for any n ∈ N.

12) If A = Γ(16

)and B = Γ

(13

), compute

∫∞0

6√x(1 + x)−

43dx as

an expression of A and B.13) Reduce to computations with function Γ the following integral∫ π

2

0sin10 x cos12 xdx.

14) Prove that the equality: Γ(

12

)=

√π implies the equality:∫∞

0e−x

2dx =

√π2.

15) Prove that m =∫∞−∞ xf(x)dx and σ2 =

∫∞−∞(x − m)2f(x)dx,

where

f(x) =1

σ√

2πe−

(x−m)2

2σ2 ,

where σ > 0 andm is a fixed real number. This last function is called inStatistics the probability function of the Gauss-Laplace distribution . Itmodels for instance the distribution of errors in a particular sequencesof random measurements. m is called the mean and σ2 is called thevariance of this last distribution.

16) Compute I(λ) =∫ λ0

ln(1+λx)1+x2

dx, |λ| < 1 and then compute∫ 0.5

0ln(1+0.5x)

1+x2dx.

CHAPTER 5

Line integrals

1. The mass of a wire. Line integrals of the first type.

D 15. A 1-dimensional deformation or simply a defor-mation of an interval I ⊂ R = R ∪ ±∞ in the n-dimensionalspace Rn is a vector function −→r : I → Rn. For a t ∈ I, the vector−→r (t) = (x1(t), ..., xn(t)) is called the position vector of "the movingpoint" M(x1, x2, ..., xn) at the moment t". Such a deformation is alsocalled a parametric path in Rn because it is uniquely determined by then scalar functions x1 = x1(t), ..., xn = xn(t), where t is called a pa-rameter. A parametric representation (parametrization)of this path isusually given in the following way:

(1.1)

x1 = x1(t)...

xn = xn(t)

, t ∈ I.

The image −→r (I) in Rn is commonly called the deformation of I throughthe deformation functions x1 = x1(t), ..., xn = xn(t) or simply a curveC in Rn. This curve C is also called the support curve of the deforma-tion −→r . We say that the curve C is continuous if the vector function−→r is continuous. We say that C is smooth if −→r is of class C1 on I, thefunction −→r is injective (one-to-one) and "the velocity" −→r ′(t) = 0 forany t ∈ I. C is of class Ck on I if −→r is of class Ck on I. The curve Cis piecewise smooth if −→r is smooth, but a finite set t1, t2, ..., tm ⊂ I.If C is continuous these last points are discontinuity points for −→r , i.e.the curve is "interrupted" at these points. If C is continuous but itis not smooth because of the singular points t1, t2, ..., tm, we say thatthese last points are "corners" or "multiple points" for C. If n = 2, wedenote the parametric representation of a curve C in R2 by x = x(t),y = y(t), t ∈ I. In this last case we say that C is a plane curve (a 2D-curve). If n = 3, we denote the parametric representation of a curveC in R3 by x = x(t), y = y(t), z = z(t), t ∈ I. We say that C is aspace curve (a 3D-curve). In Fig.1 we see a piecewise smooth curve

181

182 5. LINE INTEGRALS

with singular points M1, M2 and M3. These are the unique "corners"of this continuous curve.

F 1

For instance, if n = 1 a continuous curve is an interval (why?). Letthe following parametric path be in plane:

(1.2)

x = R cos ty = R sin t

, t ∈ [0, 2π).

Thus, −→r : [0, 2π) → R, −→r (t) = (R cos t, R sin t) is the correspondingparametric path. Whenever t runs from 0 up to 2π (without reach-ing it; otherwise the point N(R, 0) would appear two times and ourcurve would not be smooth more-it would had a "double point"!) themoving point M(x(t), y(t)) will describe a complete circle of radius R,"oriented" in the direct trigonometric direction, or counterwise clocksense (see Fig.2).

Let now A(x1, y1) and B(x2, y2) be two points in the Cartesin planexOy (see Fig.3). The following parametrization:

(1.3)

x = x(t) = x1 + t(x2 − x1)y = y(t) = y1 + t(y2 − y1)

, t ∈ [0, 1]

describes exactly the "oriented" closed segment [AB].Indeed, if M(x(t), y(t)) is a moving point on [AB], for t = 0 it be-

comes A and for t = 1 it becomes B. It is an easy exercise of AnalyticalGeometry to see that any point on the segment [AB] is of this typefor a unique t ∈ [0, 1]. Since function −→r (t) = (x(t), y(t)) is of class C1

on [0, 1], our closed segment is a plane or 2D-smooth curve. A smoothcurve is also called a simple curve. Usually, in engineering, any curve

1. THE MASS OF A WIRE. LINE INTEGRALS OF THE FIRST TYPE. 183

F 2

C will be a finite union of simple curves Ci, i = 1, 2, ..., N, such thatCi ∩ Cj = ∅ or a point for i = j.

F 3

Let us come back to the circle (1.2), but instead of the interval[0, 2π) we consider a new interval [−π, π); this means that we start togo on the circle from M(−R, 0), in the counterclockwise direction, upto we come back to M. This is the same with changing the variable twith a new one v = t− π ∈ K = [−π, π). The new parametrization inthe variable v is:

x(v) = −R cos vy(v) = −R sin v

, t ∈ [0, 2π).

Now we change the variable v ∈ K = [−π, π) with another one u ∈J = [−∞,∞) such that tan v

2= u. Thus, the new parametrization of

the same circle is:

(1.4)

x(u) = −R1−u2

1+u2

y(u) = −R 2u1+u2

, u ∈ [−∞,∞) ⊂ R = R ∪ ±∞.


Here we have a new parametric path −→q : [−∞,∞) → R, −→q (u) =(x(u), y(u)), but the same curve, i.e. −→q ([−∞,∞)) = −→r ([0, 2π)). Whatis the relation between−→r and−→q ? If we describe the above substitutionswe get t = π + 2 arctan u. Hence the function λ : J → I, λ(u) =π+ 2 arctan u is a diffeomorphism, i.e. it is a bijection, it is of class C1

on J and its inverse λ−1(t) = tan t−π2

is also of class C1 on I. Finally,−→q (u) = −→r (λ(u)), i.e. −→q = −→r λ. Two paths −→r and −→q which can beobtained one from each other by a composition with a diffeomorphismare called equivalent. We see that the images of all equivalent pathsare identical. Thus they give rise to one and the same curve. In theAdvanced Mathematics the notion of a curve is an abstract object,namely the set of all equivalent parametric paths with a given one −→r .In fact one works with a representative or with a representation ofsuch an object, namely with a particular parametric path. Here we donot use such a sophisticated way to define a curve. We shall alwaysidentify the abstract curve with the image of a parametric path which isa representative of this curve. Thus a circle is an usual circle, a line is anusual line, etc. But, we always have to fix a particular parametrizationof such a "curve" to work with. Instead of writing all the parameterequations in (1.1) to define a parametric curve −→r : I → R, we simplywrite the couple (I,−→r ) and say:"Let the curve (or the deformation)(I,−→r )", etc.

D 16. Let (I,−→r ), −→r (t) = (x(t), y(t), z(t)),(or −→r (t) = (x(t), y(t))) be a 1-dimensional deformation in R3 (or

in R2) and let C = −→r (I) be its support or the effective curve.Let f : C → R be a scalar function defined on C. The triple (I,−→r , f)

is called a 3-D (a 2-D) wire or loaded curve with the density functionf on it.

This is a mathematical general model for a "thread" or a wire,loaded with a mass f(M) in each of its points (see Fig.4). In the fol-lowing we consider only piecewise smooth wires (I,−→r , f), i.e piecewisesmooth curves loaded with almost continuous density functions f, i.e.functions f which are continuous on C = −→r (I) except maybe a finitenumber of points on C.

For the moment we restrict ourselves to a smooth finite and closed(I = [a, b] is closed and finite) wire (I,−→r , f), where we denote by [AB]the arc C = −→r (I), A = −→r (a), B = −→r (b) (see Fig.4). We also assumethat f is continuous.

We know to compute the length l([AB]) (see formula (8.3), Ch.2)

of the arc [AB]. If the density function f were a constant c, then the


F 4

mass of the wire (I,−→r , f) would obviously be equal to c · l([AB]). Ifthe density function is continuous, it is also uniformly continuous (the

arc [AB] is closed and bounded and we can apply theorem 59, [Po]).

Thus, if x′,x′′ ∈ [AB] are very close, then f(x′) and f(x′′) are also veryclose (read again the definition of uniform continuity, definition 22 andexample 13 from [Po]). Let us consider a division ∆ :

a = t0 < t1 < ... < tn = b

of the segment I = [a, b] and let us look at the corresponding division

−→r (∆) : M0 = −→r (a),M1 = −→r (x1), ...,Mn = −→r (b),

of the arc [AB] (see Fig.4). If the norm ‖∆‖ of the division ∆ issmall enough then the uniform continuity of f implies that the length

of the arc [Mi−1Mi] are sufficiently small (as we want). Thus we canapproximate the density function f on such an arc with the value of

it at a fixed point Pi(x(ξi), y(ξi), z(ξi) ∈ [Mi−1Mi], where ξi ∈ [ti−1, ti]for any i = 1, 2, ..., n. The points ξi are called marking points forthe wire (I,−→r , f) and division ∆. We construct now a new "abstract"wire which approximate the initial one. Namely, we approximate the

arc [AB] with the union of the straight line segments [Mi−1Mi], i =1, 2, ..., n. Then we define a density function on this new curve (a polyg-onal line) by taking f(x) = f(Pi) = f(x(ξi), y(ξi), z(ξi)) for any x ∈[Mi−1Mi]. The mass of this new obtained wire (the polygonal line,


which is a continuous deformation of the same interval I = [a, b], to-gether the density function f) is:

(1.5) S((I,−→r , f); ∆; ξi) =n∑

i=1

f(x(ξi), y(ξi), z(ξi))∥∥∥−−−−−→Mi−1Mi

∥∥∥ .

Such a sum is said to be a Riemann sum for our wire (I,−→r , f), whichcorresponds to the division ∆ and the set of marking points ξi. Itis clear enough that we can define such sums even in the case of anarbitrary wire (I,−→r , f), not necessarily a smooth one.

D 17. We say that the general wire (I = [a, b],−→r , f) hasa mass H (a real number) if for any small positive real number ε > 0,there exists a small positive real number δ (which depends on ε) suchthat |H − S((I,−→r , f); ∆; ξi)| < ε for any division ∆ of [a, b] with‖∆‖ < δ and for any set of marking points ξi, ξi ∈ [ti−1, ti]. Thisis also equivalent to saying that H is a unique point with the follow-ing property: "There exists a sequence of divisions ∆n, ∆n ≺ ∆n+1

(the points of ∆n are between the points of ∆n+1) such that ‖∆n‖ → 0

and the sequence S((I,−→r , f); ∆n; ξ(n)i ) is convergent to H for any

choice of the set of marking points ξ(n)i for the division ∆n. This num-ber H is said to be the line (curvilinear) integral of the first type of thefunction f on the curve C = −→r ([a, b]).It is denoted by

∫Cf(x, y, z)ds,

were ds is called the element of length, i.e. the length of a "small"

curvilinear arc [Mi−1Mi] on C. If a wire (I = [a, b],−→r , f) has a massH we also say that the line integral

∫Cf(x, y, z)ds exists and it is equal

to H. When f = 1, the mass H coincides with the length l(C). Thus,in this last case, the length exists (C is rectifiable-see definition 8) andit is equal to

∫Cds.

We shall later give an example of a bounded wire even with f = 1which has no finite mass (in this case the mass is equal to its length)at all. Since in engineering we usually do not meet such an exoticsituation, we shall consider only piecewise smooth wires here, i.e. afinite union of smooth wires. Since taking masses is an additive process,we can restrict ourselves to the particular case of a smooth wire witha continuous density function. The following result give us a methodto compute the mass of such a wire.

T 64. Let (I = [a, b];−→r ; f), −→r (t) = (x(t), y(t), z(t), bea smooth wire with f a continuous function. Then the line integral∫Cf(x, y, z)ds of the first type exists (the mass of the wire exists) and


it can be computed by using the following formula:(1.6)∫

C

f(x, y, z)ds =

∫ b

a

f(x(t), y(t), z(t))√x′2(t) + y′2(t) + z′2(t)dt.

As we just know (see formula (8.3), Ch.2), ds =√x′2(t) + y′2(t) + z′2(t)dt.

P. The proof follows the idea of the proof of the formula (8.3,Ch.2). Let us evaluate the Riemann sum S((I,−→r , f); ∆; ξi) of for-mula (1.5):

S((I,−→r , f); ∆; ξi) =n∑

i=1

f(x(ξi), y(ξi), z(ξi))∥∥∥−−−−−→Mi−1Mi

∥∥∥ =

n∑

i=1

f(x(ξi), y(ξi), z(ξi))×

×√

[x(ti) − x(ti−1)]2 + [y(ti) − y(ti−1)]2 + [z(ti) − z(ti−1)]2.

=n∑

i=1

f(x(ξi), y(ξi), z(ξi))×

√x′2(ci)(ti − ti−1)2 + y′2(di)(ti − ti−1)2 + z′2(ei)(ti − ti−1)2,

where ci, di, ei ∈ [ti−1, ti] (we applied Lagrange formula for each func-tions x(t), y(t) and z(t) on the intervals [ti−1, ti]). Since the curveC is of class C1 on [a, b], functions x′(t), y′(t) and z′(t) are continu-ous, so that we can use freely the approximations: x′2(ci) ≈ x′(ξi),y′2(di) ≈ y′(ξi), z

′2(ei) ≈ z′(ξi). Thus, our last Riemann sum becomes:

(1.7) S((I,−→r , f); ∆; ξi) ≈n∑

i=1

f(x(ξi), y(ξi), z(ξi))√x′2(ξi) + y′2(ξi) + z′2(ξi)(ti − ti−1).

The approximation here is better and better if the norm of the di-visions ∆ is smaller and smaller. Hence these Riemann sums havea unique limit point H if and only if the sums on the right in for-mula (1.7) have such a point H. But the sums on the right are noth-ing else than Riemann sums for the usual definite Riemann integral∫ baf(x(t), y(t), z(t))

√x′2(t) + y′2(t) + z′2(t)dt. Thus, finally, this last

integral exists and it is equal to H. Hence H exists and the formula(1.6) is true.


We can summarize all of these in the following two formulas:

(1.8) mass([AB]) =

∫ b

a

f(x(t), y(t), z(t))√x′2(t) + y′2(t) + z′2(t)dt.

(1.9) length([AB]) =

∫ b

a

√x′2(t) + y′2(t) + z′2(t)dt.

If C = C1 ∪ ...∪Cn, where each curve C1, ..., Cn are smooth curvesand Ci ∩ Cj = ∅ or a point for i = j, then, by definition

∫

C

fds =

∫

C1

fds+

∫

C2

fds+ ...+

∫

Cn

fds

Moreover, the mapping f →∫Cfds is a linear mapping. It is easy to

see that if C is smooth and f is continuous then there exists a pointc ∈ [a, b] such that(1.10)∫

C

fds = f(x(c), y(c), z(c))

∫

C

ds = f(x(c), y(c), z(c)) · length(C).

This is called the mean formula for line integrals of the first type.We saw that any curve of class C1 on an interval [a, b] has a length,

i.e. it is rectifiable. Here is an example of a continuous curve which isnot rectifiable even I is a finite interval.

E 65. (Koch’s curve flake of snow) We shall construct asequence of curves (γn), each curve being of class C

1 but a finite numberof points (piecewise smooth curves) such that (γn) is uniformly conver-gent to a bounded curve (it can be embedded in a disc of a finite radius)γ which has an infinite length, i.e. it is not rectifiable. This last curveγ have no tangent line in any of its point. Let us construct the curveγ1, the polygonal line (A1A2A3A4A5) (see Fig.5).

Take a segment of a line, [AB] of length d > 0 and divide it into 3equal parts. Put A1 = A and A5 = B. Then construct A2, A3, A4 suchthat ∥∥∥−−−→A1A2

∥∥∥ =∥∥∥−−−→A2A3

∥∥∥ =∥∥∥−−−→A3A4

∥∥∥ =∥∥∥−−−→A4A5

∥∥∥ =d

3and ∥∥∥−−−→A1A3

∥∥∥ =∥∥∥−−−→A5A3

∥∥∥ =d

2

1

sinα,

where for instance α = π6. On each side AiAi+1, i = 1, 2, 3, 4, we make

the same construction as above by taking d3instead of d. Finally we

obtain the polygonal line (γ2) = (B1B2...B17) (see Fig.5). And so on.Let R1 = d

2tanα, α = π

6, be the height of the trapezoid (A1A2A4A5).

Since each trapezoid (B1B2B4B5), (B5B6B8B9), (B9B10B12B13) and


F 5

(B13B14B16B17) is similar to (A1A2A4A5), the corresponding angles αhave the same measure. Thus their heights are R2 = d

2·3 tanα. The

heights of the trapezoids which corresponds to (γ3) is R3 = d2·32 tanα.

For (γn) the corresponding R is Rn = d2·3n−1 tanα. Since R1 +R2 + ...+

Rn + ... = 34d tanα, (γ) = lim

n→∞(γn) is bounded. It is not so difficult

to see that the length of (γn) is equal to4n

3nd. Hence the length of (γ)

cannot be finite.

E 66. Let [OABO] be a loaded frame like in Fig.6,

F 6

with the density function f(x, y) = x2 + y2. a) Find the perimeterof the frame. b) Find the mass of it. c) Find the mass center of the arc

[AB]. d) Find the moment of inertia IOy of the segment [OB] relativeto the Oy-axis. Here is the solutions to these questions.


a) In spite of an easy elementary solution of this question, we shallapply here the above theory and formulas relative to the line integral ofthe first type. First of all we need to construct a parameterization ofthe frame. The support curve is not of class C1 because it has 3 corners

A,B and O. It is in fact a union of three smooth curves: [OA], [AB]and [OB]. Let us construct a parametrization for each of them.

(1.11) [OB] :

x = ty = t

, t ∈[0,

1√2

]

(1.12) [AB] :

x = cos ty = sin t

, t ∈[0,π

4

]

(1.13) [OA] :

x = ty = 0

, t ∈ [0, 1]

Now, length([OB]) =∫[OB]

ds, where ds =√x′2 + y′2dt =

√2dt. Thus,

length([OB]) =√

2

∫ 1√2

0

dt = 1.

Let us compute the length of the arc [AB]. In this case

ds =√

(− sin t)2 + cos2 tdt = dt.

Thus,

length([AB]) =

∫

[AB]

ds =

∫ π4

0

dt =π

4.

Now, the length of [OA] is obviously 1 and the perimeter of the frameis equal to 1 + π

4+ 1 = 2 + π

4.

b) Since

mass[OABO] = mass[OA] +mass[AB] +mass[OB] =

=

∫

[OB]

(x2 + y2)ds+

∫

[AB]

(x2 + y2)ds+

∫

[OA]

(x2 + y2)ds =

=

∫ 1√2

0

(t2 + t2)√

2dt+

∫ π4

0

(cos2 t+ sin2 t)dt+

∫ 1

0

(t2 + 02)dt =

=1

3+π

4+

1

3=

2

3+π

4.

c) First of all let us explain some formulas in connection with the mass

centre. Let G(xG, yG) be the mass centre of the loaded arc [AB]. Bydefinition, if we concentrate the entire mass of the arc in the point G,both static moments with respect to axes of this last obtained system isequal to static moments with respect to axes of the initial arc. If one


divide the arc into small pieces by the points M0 = A,M1, ...,Mn = Band if we consider the density function to be the same f(Pi) on each arc

[Mi−1Mi], where Pi(xi, yi) is a fixed point on this last arc, then the mo-

ment w.r.t. Ox-axis can be approximated by∑ni=1 yif(Pi)

∥∥∥−−−−−→Mi−1Mi

∥∥∥ ,i.e. the moment is equal to

∫[AB]

yf(x, y)ds. The moment w.r.t. Oy-

axis will be∫[AB]

xf(x, y)ds. Let us use the remarks made above relative

to the point G and write:

xG ·∫

[AB]

f(x, y)ds =

∫

[AB]

xf(x, y)ds

and

yG ·∫

[AB]

f(x, y)ds =

∫

[AB]

yf(x, y)ds.

Hence

(1.14) xG =

∫[AB]

xf(x, y)ds∫[AB]

f(x, y)ds, yG =

∫[AB]

yf(x, y)ds∫[AB]

f(x, y)ds.

Now we have to compute three line integrals. But the mass was justcomputed in b), mass = π

4. Let us compute the line integrals which

appear at the numerators:∫

[AB]

xf(x, y)ds =

∫

[AB]

x(x2 + y2)ds =

=

∫ π4

0

cos t(cos2 t+ sin2 t)dt = sin t |π40 =

1√2.

∫

[AB]

yf(x, y)ds =

∫

[AB]

y(x2 + y2)ds =

=

∫ π4

0

sin t(cos2 t+ sin2 t)dt = − cos t |π40 = 1 − 1√

2.

Hence

xG =

1√2π4

=2√

2

π, yG =

4 − 2√

2

π.

d) Recall that the moment of inertia of a point M(x, y) loaded with amass m w.r.t. Oy-axis is equal to mx2 (here x is the distance fromM to Oy-axis). By a process of "globalization" i.e. of "integration"(explain it like in the case of the mass centre!) we get

IOy =

∫

[OB]

x2fds =

∫ 1√2

0

t2(t2 + t2)√

2dt = 2√

2x5

5

∣∣∣∣1√2

0

=1

10.


Let us put together all the formulas which give us the coordinatesof mass centre of a space wire (I,−→r , f), where −→r (t) = (x(t), y(t), z(t)).Let C be the support curve of this wire. Then,

(1.15) xG =

∫Cxfds∫Cfds

; yG =

∫Cyfds∫Cfds

; zG =

∫Czfds∫Cfds

.

Let now denote Ix, Iy, Iz the moments of inertia of the above wire w.r.t.Ox-axis, Oy-axis and Oz-axis respectively. Let IO be the moment ofinertia of the same wire w.r.t. the origin O. Then

(1.16) Ix =

∫

C

(y2 + z2)fds; Iy =

∫

C

(x2 + z2)fds;

Iz =

∫

C

(y2 + x2)fds; IO =

∫

C

(x2 + y2 + z2)fds.

E 67. Let us consider the cylindrical helix

Γ :

x = a cos ty = a sin tz = bt

, t ∈ [0, 2π], a, b > 0.

Here ds =√x′2 + y′2 + z′2dt =

√a2 + b2dt. Let us rotate Γ around Oz-

axis and let us compute its moment of inertia. If nothing is mentionedon the density function, we tacitly consider it to be f(x, y, z) = 1. Thusone has

Iz =

∫

C

(y2 + x2)fds =

∫ 2π

0

a2√a2 + b2dt = 2πa2

√a2 + b2.

Another application of the line integral of the first type appearswhen we want to compute the area of some cylindrical surface with the

directrix arc [AB] in the xOy-plane, the generating lines parallel to Oz-axis and limited by the xOy-plane and a surface z = f(x, y) ≥ 0 (see

Fig.7). Let x = x(t), y = y(t) be a parametrization of the arc [AB].

We divide the arc [AB] into n arcs [Mi−1Mi]. Let Ci be an arbitrary

fixed point on the arc [Mi−1Mi] and let C ′i the point on the surfacez = f(x, y) which is projecting in Ci. The area of the cylindrical surface

(Mi−1MiM′iM

′i−1) with the directrix curve [Mi−1Mi] and generating

lines parallel to Oz-axis can be well approximated with the area of thespace rectangle [Mi−1MiM

′′iM

′′i−1] with base the segment [Mi−1Mi] and

the height equal to the length of the segment [CiC′i] (see Fig.7) This

area is equal to∥∥∥−−−−−→Mi−1Mi

∥∥∥ · f(Ci). Thus, the entire area can be well


approximated by the sum:

n∑

i=1

∥∥∥−−−−−→Mi−1Mi

∥∥∥ · f(Ci).

Looking at the definition 17 and at the formula 1.5 we obtain that the

area σ(ABB′A′) of the cylindrical surface bounded by [AB], [A′B′] andthe segments [AA′], [BB′] (see Fig.7) can be computed by using thefollowing formula:

(1.17) σ(ABB′A′) =

∫

[AB]

fds.

F 7

E 68. Let us find the area σ of the surface which are a partof the cylindrical surface with directrix curve the parabola y = 2x2 inxOy-plane, generator lines parallel to Oz-axis, delimited by the xOy-plane and the plane x+ y + z = 10 (see Fig.8). Since z = 10 − x− y,f(x, y) = 10−x−y in formula 1.17 and since x = t, y = 2t2, t ∈ [−1, 1]

is a parametrization of the curve [AOD], we get:

σ =

∫

[AOD]

(10 − x− y)ds =

∫ 1

−1

(10 − t− 2t2)√

1 + 16t2dt.

Since the function g(t) = t√

1 + 16t2 is an odd function on the symmet-

ric interval [−1, 1] w.r.t. the origin O, one has that∫ 1

−1t√

1 + 16t2dt =


0. Hence it remains to compute the following definite integral∫ 1

−1

(10 − 2t2)√

1 + 16t2dt = 2

∫ 1

0

(10 − 2t2)√

1 + 16t2dt,

etc.

F 8

2. Line integrals of the second type.

Let M be a moving point on a segment [a, b] from a to b in a field

of forces−→f (x) = f(x)

−→i , x ∈ [a, b].We know (see example 42) that the

workW−→f

[a, b] ofM is equal to∫ baf(x)dx. If the field

−→f is a space field

and if the point M is moving on an oriented space segment−→AB then,

in each position P ∈ [AB] of M, the effective force which produces the

work is the vector projection of−→f along the vector

−→AB, i.e. the vector

−→f (P )·−→AB‖−→AB‖

−→AB

‖−→AB‖ , where−→f (P ) · −→AB is the dot (scalar) product of the free

vectors−→f (P ) and

−→AB. Thus, if the length of the vector

−→AB is very

small we can approximate the work of M on [AB] by the dot product

(2.1)−→f (P ) · −→AB,

where P is a fixed point on [AB].

D 18. (orientation of an arc) Let us consider now a

smooth curve [AB] with a parametrization x = x(t), y = y(t) and

z = z(t), t ∈ [a, b] (see Fig.9). We say that the arc [AB] is oriented or

2. LINE INTEGRALS OF THE SECOND TYPE. 195

directly oriented if whenever t1 < t2 on [a, b], then length[AM1] <

length[AM2], where M1(x(t1), y(t1), z(t1)) is the point on the curvewhich corresponds to t1 ∈ [a, b] and M2(x(t2), y(t2), z(t2)) is the pointon the curve which corresponds to t2 ∈ [a, b]. This means that whenevert goes from a to b, the corresponding moving point M(x(t), y(t), z(t))goes from A to B on the curve, without turning back even for a verysmall "period of time". We always consider in our examples that a

curve [AB] has two orientations: "the direct one", from A to B and"the inverse one", from B to A.

It can be proved (see any elementary course in Differential geome-try) that any smooth curve Γ, realized as a deformation of an interval,

can be oriented, i.e. it has two orientations. We divide the arc [AB]

into n subarcs [Mi−1Mi]. The division A = M0,M1, ...,Mn = B isconsidered to be the image of a division ∆ : a = t0 < t1 < ... < tn = bof the segment [a, b] through the deformation −→r (t) = (x(t), y(t), z(t)),

t ∈ [a, b] of the segment [a, b] into the arc [AB] (see Fig.9).

F 9

If the norm of the division ∆ is very small, since the curve [AB]

is smooth, the lengths of the subarcs [Mi−1Mi] are very small. Thus

we can approximate the work of a continuous field of forces−→F (x, y, z)

= (P (x, y, z), Q(x, y, z), R(x, y, z)) defined on the curve [AB], along the


oriented arc [AB], by the following sum:

(2.2) S(−→F ,∆, ξi) =

n∑

i=1

−→F (Pi) ·

−−−−−→Mi−1Mi,

where Pi is a fixed point on the arc [Mi−1Mi], i.e. Pi(x(ξi), y(ξi), z(ξi)),where ξi, ξi ∈ [ti−1, ti], is a fixed set of marking points for the division

∆. Such a sum is called a Riemann sum for−→F , ∆ and ξi. Here the

work W−→F

( [Mi−1Mi])of−→F on [Mi−1Mi] was approximated with

−→F (Pi) ·−−−−−→

Mi−1Mi.

D 19. With the above notation and definitions, for ageneral continuous and oriented arc of a curve Γ = −→g ([a, b]), where−→g : [a, b] → R3(or R2), let us assume that there exists a real num-ber S with the following property: if ε > 0 is a small positive realnumber, then there exists another small positive real number δε > 0(depending on ε) such that if ∆ is a division of [a, b], one has that∣∣∣S − S(

−→F ,∆, ξi)

∣∣∣ < ε. This (unique) number S will be denoted by∮

Γ

−→F · d−→r or by

∮

Γ

Pdx+Qdy+Rdz and it is called the line integral of

the second type of−→F or of the differential form ω = Pdx+Qdy+Rdz

respectively. Here, formally, d−→r = (dx, dy, dz) and

∮

Γ

−→F · d−→r will be

interpreted as "the work" of the field of forces−→F along the oriented

curve Γ (considered with the direct orientation), or "the circulation"

of the field of forces−→F along the oriented curve Γ if the curve Γ is

"closed" (−→r (a) = −→r (b)). If we change the direct orientation withthe inverse one (write Γ−) the line integral will have a minus in front

of it, i.e.

∮

Γ−

−→F · d−→r def

= −∮

Γ

−→F · d−→r . Moreover, if Γ = ∪Ni=1Γi is

a finite union of nonoverlaping (they have in common at most onepoint) curves Γi, considered with the induced orientation of Γ, then∮

Γ

−→F · d−→r =

∑Ni=1

∮

Γi

−→F · d−→r (this is a consequence but I prefer to put

it in a definition!).


It is easy to see that the mapping−→F →

∮

Γ

−→F · d−→r for a fixed curve

Γ is a linear mapping. The basic problem is how to compute such a

line integral

∮

Γ

−→F ·d−→r . The following theorem will clarify this question.

T 65. Let us consider again the above orientated smooth

curve [AB] with a parametrization x = x(t), y = y(t) and z = z(t),t ∈ [a, b], with a division ∆ of the interval [a, b],with its Riemann sum,etc. (see Fig.9 and all the starting discussion). Then

(2.3)

∮

[AB]

−→F · d−→r =

∫ b

a

−→F (x(t), y(t), z(t)) · −→r ′(t)dt,

or

(2.4)

∮

[AB]

−→F · d−→r =

=

∫ b

a

[P (x(t), y(t), z(t))x′(t) +Q(x(t), y(t), z(t))y′(t)+

+R(x(t), y(t), z(t))z′(t)]dt,

where −→r (t) = (x(t), y(t), z(t)) = x(t)−→i + y(t)

−→j + z(t)

−→k .

P. First of all, in order to evaluate a Riemann sum of the

type (2.2), let us compute the dot product−→F (Pi) ·

−−−−−→Mi−1Mi.

−→F (Pi) ·

−−−−−→Mi−1Mi ≈ P (x(ξi), y(ξi), z(ξi)) [x(ti) − x(ti−1)] +

+Q(x(ξi), y(ξi), z(ξi)) [y(ti) − y(ti−1)] +(2.5)

+R(x(ξi), y(ξi), z(ξi)) [z(ti) − z(ti−1)] .(2.6)

Let us now apply Lagrange formula for functions x(t), y(t) and z(t)defined on the interval [ti−1, ti] :

x(ti) − x(ti−1) = x′(ci)(ti − ti−1),

where ci ∈ [ti−1, ti],

y(ti) − y(ti−1) = y′(di)(ti − ti−1),

where di ∈ [ti−1, ti] and

z(ti) − z(ti−1) = z′(ei)(ti − ti−1),


where ei ∈ [ti−1, ti]. Since the norm ‖∆‖ is very small, i.e. the lengthsof intervals [ti−1, ti] are small enough and since the functions x′(t), y′(t),z′(t) are continuous (the curve is of class C1 being smooth), we can wellapproximate x′(ci), y

′(di) and z′(ei) with x′(ξi), y′(ξi) and with z′(ξi)

respectively (ci, di and ei are very close to ξi and the previous functionsare continuous!). Thus formula (2.5) can be rewritten as

−→F (Pi) ·

−−−−−→Mi−1Mi =

= [P (x(ξi), y(ξi), z(ξi))x′(ξi) +Q(x(ξi), y(ξi), z(ξi))y

′(ξi)](ti − ti−1)+

+R(x(ξi), y(ξi), z(ξi))z′(ξi)(ti − ti−1).

Thus, the Riemann sum S(−→F ,∆, ξi) =

∑ni=1

−→F (Pi) ·

−−−−−→Mi−1Mi can be

approximated with

n∑

i=1

[P (x(ξi), y(ξi), z(ξi))x′(ξi) +Q(x(ξi), y(ξi), z(ξi))y

′(ξi)+

+R(x(ξi), y(ξi), z(ξi))z′(ξi)](ti − ti−1),

which is nothing else than a Riemann sum for the simple Riemannintegral

∫ b

a


+R(x(t), y(t), z(t))z′(t)]dt.

Taking limits as ‖∆‖ → 0, we get formula (2.4).Since d−→r = (x′(t), y′(t), z′(t))dt, formula (2.3) is an immediate con-

sequence of formula (2.4).

This last theorem says that in order to compute a line integral ofthe second type, we must find a suitable (not so complicated!) para-

meterization of the arc [AB], then we simply apply formula (2.4) toreduce the computation to a calculation of a Riemann definite integral.If the curve and the field of forces are all in the same plane xOy, weput z(t) = 0 and R(x, y, z) = 0 in formula (2.4), i.e. we get an easierformula: ∮

[AB]

−→F · d−→r =

∫ b

a

−→F (x(t), y(t)) · −→r (t)dt =

=

∫ b

a

[P (x(t), y(t))x′(t) +Q(x(t), y(t))y′(t)] dt.


E 69. Let−→F (x, y) = x2−→i + xy

−→j be a plane field of forces

(or, equivalently, let ω = x2dx + xydy be a plane differential form oforder I) and let Γ be the marked with arrows oriented curve in Fig.10(the arrows indicate the orientation!)

F 10

a) Find the work on the arc of the circle [BC]; b) Compute

∮

[AB]

−→F ·

d−→r ; c) Compute

∮

[CA]

Pdx+ Qdy;

d) Compute the circulation of the field−→F along the closed oriented

(like in Fig.10) curve [CABC]. Since our last curve is not smooth (ithas "corners"!) we must decompose it into 3 smooth nonoverlaping

curves: [CA], [AB] and [BC] respectively and then to write (see defin-ition 19):

(2.7)

∮

[CABC]

−→F · d−→r =

∮

[CA]

−→F · d−→r +

∮

[AB]

−→F · d−→r +

∮

[BC]

−→F · d−→r .

Let us find 3 separate parametrizations for the component curves [CA],

[AB] and [BC].

[CA] :

x = 2 + (−2)ty = 0 + (−1)t

, t ∈ [0, 1];


[AB] :

x = 0y = t

, t ∈ [−1, 2];

[BC] :

x = 2 cos ty = 2 sin t

, t ∈ [0,π

2]−,

i.e. t goes from π2to 0. Now we are ready to compute everything. a)

The work on the arc [BC] can be find as follows:∮

[BC]

−→F · d−→r =

∮

[BC]

Pdx+Qdy =

= −∫ π

2

0

[(2 cos t)2(2 cos t)′ + (2 cos t)(2 sin t)(2 sin t)′

]dt =

= −8

∫ π2

0

[− cos2 t sin t + cos2 t sin t

]dt = 0.

b)∮

[AB]

−→F · d−→r =

∫ 2

−1

[02 · 0 + 0 · tdt

]= 0.

c)∮

[CA]

Pdx+Qdy =

∫ 1

0

[(2 − 2t)2(−2) + (2 − 2t)(−t)(−1)

]dt = −7

3.

d) The work or the circulation on [CABC] is the sum of the resultsjust obtained in a), b) and c), i.e. −7

3.

Here is a nice application to Physics.

T 66. (the gain principle) Let W =

∮

Γ

−→F · d−→r be the

work of a moving point M of mass m in a continuous field of forces,along an oriented smooth curve Γ, realized as a deformation −→r (t) =(x(t), y(t), z(t)) of the interval [a, b]. Then W = T (B) − T (A), where

T = m2|−→v |2 is the kinetic energy of the moving system, −→v = −→r ′ is the

velocity vector and A, B are the end points of the trajectory Γ. Here thedifference T (B) − T (A) is called "the gain in kinetic energy". Thus,the gain principle in Physics says that the work done by a moving point

on a trajectory [AB] is equal to the gain in kinetic energy of the movingpoint.

3. INDEPENDENCE ON PATH. CONSERVATIVE FIELDS 201

P. Since the second law of dynamics says that−→F = m−→v ′, we

can apply formula (2.3) and successively write:

W =

∮

Γ

−→F · d−→r =

∫ b

a

−→F (−→r (t)) · −→v (t)dt = m

∫ b

a

−→v ′(t) · −→v (t)dt =

= m

∫ b

a

(−→v · −→v2

)′dt =

m

2|−→v |2

∣∣∣b

a= T (B) − T (A).

R 20. Maybe somebody asks what happens with a line in-

tegral

∮

Γ

−→F · d−→r if we change the parametrization of the given curve

Γ, i.e. if the parametric path −→r changes with an equivalent one−→r∗ .

To preserve the initial orientation we must assume that the change ofvariables Φ : [a∗, b∗] → [a, b] is an increasing one, i.e. Φ′(t) > 0. Then,−→r∗(t∗) = −→r (Φ(t∗)) and dt∗ = dt∗

dtdt. Thus,

∮

Γ

−→F (

−→r∗) · d−→r∗ =

∫ b∗

a∗

[−→F (

−→r∗) · d

−→r∗

dt∗

]dt∗ =

=

∫ b∗

a∗

[−→F (−→r (Φ(t∗))) · d

−→rdt

dt

dt∗

]dt∗ =

=

∫ b

a

−→F (−→r (t))

−→r′ (t)dt =

∮

Γ

−→F · d−→r .

Thus the value of the line integral does not change. It depends only onthe curve itself and not on its particular parametrization.

3. Independence on path. Conservative fields

Looking at formula (2.4) we see that the line integral

∮

[AB]

−→F · d−→r

depends on the curve Γ = [AB] which connects the points A and B

and on the field−→F . It does not depend on the parametrization itself,

because the change of variable formula in the usual definite integralgives us that the line integral does not change if we change −→r with anequivalent path −→r ′ (here −→r and −→r ′ both connect A and B).

D 20. Let D be a space (or plane) domain and let−→F :

D → R3(or R2) be a continuous field. We say that the integral

∮ −→F ·d−→r

(this is only a symbol here!) is independent on path if for any two


points A and B of D and for any (piecewise) smooth path γ ⊂ D,

which connects A and B, the line integral of the second type

∮

γ

−→F · d−→r

does not depend on the curve γ itself but only on the end points A andB of it. This means that if Γ is another path which connect A and B

one has that

∮

γ

−→F · d−→r =

∮

Γ

−→F · d−→r .

It is clear enough that this property characterizes the field−→F de-

fined on D. We shall define a class of fields which has the above prop-

erty, namely that the integral

∮ −→F · d−→r is independent on path.

D 21. A field−→F : D → R3 is said to be a conservative

(potential) field if there exists a scalar function U : D → R of class C1

such that gradU =−→F , i.e. if

−→F (x, y, z) = P (x, y, z)

−→i +Q(x, y, z)

−→j +

R(x, y, z)−→k then ∂U

∂x= P, ∂U

∂y= Q and ∂U

∂z= R. It is clear that

−→F is

conservative if and only if the differential form ω = Pdx+Qdy +Rdzhas a primitive U(x, y, z), i.e. a scalar function U(x, y, z) such thatdU = ω. We also say in this case that ω is exact or closed on D. The

function U is called the potential function or the primitive of−→F and

the function −U is called the potential energy of−→F .

For instance, the central field−→F = c−→r ("draw" it for c > 0 and

for c < 0), where −→r = x−→i + y

−→j + z

−→k is the position vector of the

pointM(x, y, z) and c is a real constant, is a conservative field because

U(x, y, z) = x2

2+ y2

2+ z2

2is a "potential" function for

−→F . If we add an

arbitrary constant K to this U(x, y, z) we also get a potential function

for−→F . It is clear enough that not all fields are conservative. For in-

stance, the plane field−→F (x, y) = 3x2−→i + xy

−→j is not a conservative

field on any domain of R2. Indeed, suppose that−→F is conservative on

a small disc B(a, ρ), with ρ > 0. Then there is a potential functionU(x, y) such that

(3.1)

∂U∂x

= 3x2,∂U∂y

= xy.

Let us integrate (find a primitive) the first equality w.r.t. x:∫∂U

∂xdx = U(x, y) +K(y) = x3 +H(y),

where K(y) and H(y) are constant w.r.t. x but, in general, functionsof y. Let us put C(y) = H(y)−K(y). Thus U(x, y) = x3 +C(y). With


this expression of U, let us come in the second equality of (3.1) andfind

C ′(y) = xy,

which is not possible, because C ′(y) cannot be a function of x. Thus,

our assumption is false, i.e.−→F (x, y) = 3x2−→i + xy

−→j cannot be a

conservative field.Such conservative fields appears in nature, in mechanics, astronomy,

material sciences, etc. For instance, the gravitational field−→G(x, y, z) =

−mg−→k , where m is a constant mass at any point M(x, y, z) and g =9.8m/s2 is the known gravitational acceleration constant (see Fig.11),is a conservative field.

F 11

Indeed, let us find the general form of all the potential functions

U(x, y, z) of−→G :

∂U∂x

= 0∂U∂y

= 0∂U∂z

= −mg.

Let us integrate the last equation w.r.t. z and put the constants whichappear only once:

U(x, y, z) = −mgz +K(x, y),

where K(x, y) is a constant w.r.t. z but, in general, it can depend on xand y because, in general, U may depend on x and y. Let us introduce


this last expression of U in the second equality and find:

∂K

∂y= 0.

Thus K is also a constant function w.r.t. y. Hence K(x, y) = K(x)only. So that U(x, y, z) = −mgz + K(x) and, finally, let us put thislast expression of U in the first equality and find:

K ′(x) = 0,

so that K is constant w.r.t. x, y and z respectively. Now, the general

form of a primitive function for−→G is

U(x, y, z) = −mgz +K.

This means that the potential energy of the gravitational field is ofthe form −U = mgz −K. Since for z = 0 (at the horizontal plane ofthe earth) the potential energy is equal to zero, K must be zero andwe refound a known formula from high school: the potential energy= mgh, where h is the height at which the mass m is.

A strange property of such conservative fields is that the work ofthem does not depend on path, or that the circulation of them on a"closed" curve (−→r (a) = −→r (b)) is always equal to zero. Here is themathematically stated result.

T 67. (Leibniz-Newton formula for line integrals) Let−→F :

D → R3 be a continuous conservative (potential) field and let Γ be anarc of an oriented smooth curve in D with its end points A and B. Let−→r (t) = (x(t), y(t), z(t)), t ∈ [a, b] be a parametrization of Γ. Then

a)(3.2)∮

Γ

−→F · d−→r = U(B) − U(A) = U(x(b), y(b), z(b)) − U(x(a), y(a), z(a)),

where U(x, y, z) is a potential function (a primitive) of−→F .

b) In particular,

∮

Γ

−→F · d−→r does not depend on the curve Γ which

connect the points A and B, i.e. the integral

∮ −→F ·d−→r does not depend

on path in D.

c) Another consequence is that two primitives U and V of−→F on D

differ one to each other by a constant only.


d) For any "closed" curve δ (−→r (b) = −→r (a)) of D one has that∮

δ

−→F · d−→r = 0. Conversely, if for any smooth closed curve δ of D we

have that

∮

δ

−→F ·d−→r = 0, then

∮ −→F ·d−→r does not depend on path in D.

P. a) Let U(x, y, z) be a primitive of−→F (x, y, z) = P (x, y, z)

−→i +Q(x, y, z)

−→j +R(x, y, z)

−→k .

Then ∂U∂x

= P, ∂U∂y

= Q and ∂U∂z

= R. Let us consider the composed

function g(t) = U(x(t), y(t), z(t)), t ∈ [a, b]. By using the chain rule(derivative of U along the curve Γ, see [Po], Th.68), let us compute itsderivative:

g′(t) =∂U

∂x(x(t), y(t), z(t))x′(t) +

∂U

∂y(x(t), y(t), z(t))y′(t)+

+∂U

∂z(x(t), y(t), z(t))z′(t) = P (x(t), y(t), z(t))x′(t)+

+Q(x(t), y(t), z(t))y′(t) +R(x(t), y(t), z(t))z′(t).

Let us use the basic formula (2.4) and write:∮

Γ

−→F · d−→r =

∫ b

a


+R(x(t), y(t), z(t))z′(t)]dt =

=

∫ b

a

g′(t)dt = g(b)− g(a) = U(x(b), y(b), z(b))−U(x(a), y(a), z(a)) =

= U(B) − U(A).

b) It is clear from a).

c) Let U , V be two primitives of−→F on D and let M0(x0, y0, z0) be

a fixed point of D. Let now M(x, y, z) be another (variable) point ofD. Since D is connected (D is a domain, i.e. open and connected) letus take a smooth arc γ in D which connect M0 and M, oriented "fromM0 to M". We apply now formula (3.2) for both primitives U and Vand find: ∮

Γ

−→F · d−→r = U(M) − U(M0) = V (M) − V (M0),

orU(M) − V (M) = U(M0) − V (M0) = K,

a constant, so that U(x, y, z) = V (x, y, z) + K and our theorem iscompletely proved.


d) In formula (3.2) we simply put x(a) = x(b), y(a) = y(b) andz(a) = z(b), etc. Assume now that on any "closed" smooth curve δ one

has that

∮

δ

−→F · d−→r = 0. Take now two points T and W in D and two

distinct smooth oriented curves Γ1, Γ2 which connect T and W . Then

the curve δ = Γ1 ∪ Γ−2 is "closed", so that

∮

Γ1∪Γ−2

−→F · d−→r = 0. But

∮

Γ1∪Γ−2

−→F · d−→r =

∮

Γ1

−→F · d−→r −

∮

Γ2

−→F · d−→r = 0,

i.e.

∮

Γ1

−→F ·d−→r =

∮

Γ2

−→F ·d−→r and our integral

∮ −→F ·d−→r does not depend

on path.

Now we are ready to make many applications. One of these isrelated to a large simplification of computations in a line integral ofthe second type.

E 70. Let γ be the "closed" oriented (as arrows indicate)curve

[OAmBnCO] of Fig.12.

F 12

Let us compute the circulation (or the work) of the plane field−→F (x, y) = 2xy

−→i + x2−→j along the curve γ. Because the curve is a

union of 4 smooth distinct curves, we ought to find a parametrizationfor each of them, compute 4 line integrals and then add the four ob-tained numbers (see example 69). All of this may be very boring and


tiresome. The best method is to check first of all if the field is conserv-ative or not. If we were lucky that the field to be conservative, then theline integrals does not depend on path and we can choose a very easypath which connect the starting and the end points of the arc. In ourcase the arc is closed, so that the integral is zero. Let us check now ifour field is conservative or not. For this, let us try to find U(x, y) suchthat

∂U

∂x= 2xy, and

∂U

∂y= x2.

We integrate the first equality w.r.t. x and find: U = x2y+K(y). Let usput this expression of U in the second equation and find: x2 +K ′(y) =x2. Thus, K ′(y) = 0 and K(y) is a constant K. Hence U(x, y) =x2y + K, where K is an arbitrary constant. Therefore our field isconservative and, as a consequence, the given line integral is zero (seetheorem 67).

But, how to test if a field is conservative or not without directchecking if a potential (primitive) function exists, like we did above?We begin with the following remark.

T 68. Let−→F : D → R3,−→

F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) be a conservative field

of class C1 on a space domain D. Then curl−→F =

−→0 , i.e. the field

−→F

is an irrotational field. If−→F : D → R2,

−→F (x, y) = (P (x, y, ), Q(x, y, )),

is a plane conservative field of class C1 on a plane domain D, then∂Q∂x

= ∂P∂yon D.

P. Let U(x, y, z) be a primitive function for the conservativefield −→

F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)).

Then ∂U∂x

= P, ∂U∂y

= Q and ∂U∂z

= R. We know that

curl−→F = ∇×−→

F =

(∂R

∂y− ∂Q

∂z

)−→i −

(∂R

∂x− ∂P

∂z

)−→j +

(∂Q

∂x− ∂P

∂y

)−→k .

If here instead of P, Q, R we put ∂U∂x, ∂U∂y

and ∂U∂z

respectively, we get∂R∂y

− ∂Q∂z

= ∂2U∂y∂z

− ∂2U∂z∂y

= 0 because U is a function of class C2 (since

P,Q,R are functions of class C1) and we can apply Schwarz’ theorem(see theorem 71, [Po]). Similarly we can easily show that ∂R

∂x− ∂P

∂z= 0

and ∂Q∂x

− ∂P∂y

= 0, i.e. curl−→F =

−→0 . In the plane field situation, we put

instead Q, ∂U∂y

and instead of P, ∂U∂x, so that ∂Q

∂x= ∂2U

∂x∂yand ∂P

∂y= ∂2U

∂y∂x.

Since u is of class C2, ∂2U∂x∂y

= ∂2U∂y∂x

, i.e. ∂Q∂x

= ∂P∂y.


We say that a plane domain D is simple connected if any contin-uous "closed" curve γ ⊂ D can be continuously contracted up to apoint in D. We say that a space domain D is simple connected if anycontinuous "closed" surface (the image of a sphere through a contin-uous deformation in R3) Σ can be continuously contracted up to apoint in D. Practically, if D has no "hole" it is simple connected. Forinstance, any disc in R2 is simple connected but R2(0, 0) is notsimple connected because it has a hole, the origin of the axes.

R 21. If D is a simple connected domain then theorem 68

has a reciprocal. Namely, if curl−→F =

−→0 or ∂Q

∂x= ∂P

∂yin the plane

field case, then−→F is a conservative field. We shall give an elegant

proof of this last result in the next chapter (for a plane domain, seeGreen’s formula) and, in another chapter (for a space domain), asan application to Stockes’ formula. We also sketch a proof for planedomains in remark 26. The condition that D be simple connected isessential as follows from the next example. Let D = R2(0, 0) be anot simple connected domain and let

−→F = (P,Q), where P = − y

x2+y2

and Q = xx2+y2

. It is easy to see that ∂Q∂x

= y2−x2(x2+y2)2

= ∂P∂ybut

∮

x2+y2=1

− y

x2 + y2dx+

x

x2 + y2dy =

∫ 2π

0

[sin2 t+ cos2 t]dt = 2π = 0.

Hence−→F cannot be conservative on D (see theorem 67 d)).

This last remark is very useful in practice. Let us compute

∮

γ

xdx+

ydy + zdz, where γ is the oriented path from Fig.13.

Since−→F = (x, y, z) is irrotational, i.e. curl

−→F =

−→0 , the integral

does not depend on path and its value is U(1, 5, 3) − U(0, 0, 0), where

U is a primitive of−→F . We know that such a primitive exists from the

last remark (D is the entire R2 so it is simple connected). Thus we candirectly compute it by integrating the system

∂U∂x

= x∂U∂y

= y∂U∂z

= z

.

It is easy to see that x2+y2+z2

2+ K, where K is a constant, is the

general solution of this system (use the same idea like in the case of the


F 13

gravitational field). Take U(x, y, z) = x2+y2+z2

2and compute U(1, 5, 3)−

U(0, 0, 0) = 352.

When a line integral of the second type

∮ −→F · d−→r does not depend

on path on a domain D, then instead of writing

∮

[AB]

−→F · d−→r , where

[AB] is a path in D which connects two points A and B of D, we

simply write

∮ B

A

−→F · d−→r . To compute such an integral we can use a

primitive of−→F or we can use any arc of a piecewise smooth curve γ

included in D which connect A and B in this order. Namely, we write∮ B

A

−→F · d−→r = U(B) − U(A) or

∮ B

A

−→F · d−→r =

∮

γ

−→F · d−→r . The usual

choice for γ is a segment of a line or a finite union of segments of lines,eventually parallel with the coordinate axes. For instance,

(3,0,−2)∮

(1,−1,2)

xdx+ ydy + zdz =

(3,−1,2)∮

(1,−1,2)

xdx+ ydy + zdz+

+

(3,0,2)∮

(3,−1,2)

xdx+ ydy + zdz +

(3,0−2)∮

(3,0,2)

xdx+ ydy + zdz =


=

∫ 3

1

xdx+

∫ 0

−1

ydy+

∫ −2

2

zdz =x2

2

∣∣∣∣3

1

+y2

2

∣∣∣∣0

−1

+z2

2

∣∣∣∣−2

2

= 4− 1

2+0 =

7

2.

R 22. The idea coming from this last example can be gen-eralized in order to construct a primitive for a conservative continuous

field−→F = (P,Q,R) on a domain D. In order to prove that a field

−→F

is conservative on a simple connected domain, the best is to use the

criterion just described in remark 21 namely, to verify if curl−→F =

−→0

(in the space case), or to verify that ∂Q∂x

= ∂P∂y(in the plane case). Let

us describe an effective construction for such a primitive U(x, y) in theplane case (for the space case the description is more tiresome). Firstof all let us choose a fixed point M0(x0, y0) in D. The primitive U weare going to construct will depend on the choice of this fixed point. Wejust know that all the primitives of a given field differ one to each otherby a constant. Thus, such a primitive is completely determined by itsvalue at a point. In our case we shall determine the primitive U(x, y)with the property: U(x0, y0) = 0. Take now another point M(c, d) ofD and try to define the value of U at this "moving" point M. Let[M0M1M2....M2n−1M2n], M2n = M, be a polygonal line in D, whichconnect M0 with M, such that the couples of segments

([M0M1], [M1M2]), ([M2M3], [M3M4]), ..., ([M2n−2M2n−1], [M2n−1M2n])

has the following property: in each couple ([M2j−2M2j−1], [M2j−1M2j]),j = 1, 2, ..., n, the first segment [M2j−2M2j−1] is parallel to Ox-axis andthe second [M2j−1M2j ] is parallel to Oy-axis. We start with the value ofU at M0 and shall find the value of it at M2 (we operate with the firstcouple). Then we use the value of U at M2 and, by the same procedure,we find the value of U at M4 and so on up to we succeed to find thevalue of U at M2n = M. Hence we reduced everything in describing theprocedure only for the first couple ([M0M1], [M1M2]). More exactly, it issufficient to construct U(x, y) in a small open disc B(M0, r) ⊂ D, r >0, with centre at M0 and radius r. Let P (s, u) ∈ B(M0, r) and let thecouple of segments ([M0P1], [P1P ]) such that P1(s, y0), i.e. the segment[M0P1] is parallel to Ox-axis and [P1P ] is parallel to Oy-axis. It isclear that the polygonal line [M0P1P ] is contained in the disc B(M0, r),i.e. it is contained in D too. Let us define now:

U(s, u) =

P∮

M0

Pdx+Qdy =

∮

[M0P1]

Pdx+Qdy +

∮

[P1P ]

Pdx+Qdy.

4. COMPUTING PLANE AREAS WITH LINE INTEGRALS 211

or

(3.3) U(s, u) =

∫ s

x0

P (x, y0)dx+

∫ u

y0

Q(s, y)dy.

We see that here we have integrals with parameters. Let us use the

conservativeness hypothesis of−→F = (P,Q) which implies ∂Q

∂x= ∂P

∂y

and the Leibniz formula (1.9), Ch.4, to compute (the existence is alsoassured!) ∂U

∂s(s, u) and ∂U

∂u(s, u) :

∂U

∂s(s, u) = P (s, y0) +

∫ u

y0

∂Q

∂s(s, y)dy = P (s, y0) +

∫ u

y0

∂P

∂y(s, y)dy =

P (s, y0) + P (s, u) − P (s, y0) = P (s, u).

Now,∂U

∂u(s, u) = Q(s, u).

Hence U is a primitive of−→F on that ball. In particular we have the

value of U at the point M2. We now go on with M2 instead of M0 andfind the value of U at M4, etc., up tp M2n. To be easier, we can choosefrom the beginning a finite set of overlapping balls inside D, the firstcontaining M0 and the last containing M = M2n. Then choose M1 inthe intersection of the first and of the second ball,M2 in the intersectionof the second and the third, etc.

4. Computing plane areas with line integrals

Let us compute the area of the plane domain D, bounded by theoriented piecewise smooth curve [ABCDA], which is its boundary ∂D(see Fig.14).

We see that the domain D bounded by the curve [ABCDA] issimple w.r.t. Ox-axis, i.e. if [a, b] is the projection of D on Ox-axis,then for any x0 ∈ (a, b) the straight lineX = x0 intersects the boundary∂D in at most two distinct points. The following result gives the areaof D as a line integral of the second type.

T 69. With the above notation and hypotheses we have thefollowing formula:

(4.1) area(D) = −∮

∂D

ydx

P. We know from a basic application of definite integral that

(4.2) area(D) =

∫ b

a

[g(x) − f(x)]dx =

∫ b

a

g(x)dx−∫ b

a

f(x)dx.


F 14

Since a parameterization of [CD] is: x = x, y = g(x), x ∈ [a, b]− (this

means that x goes from b to a) one has that −∮

[CD]

ydx =∫ bag(x)dx.

Since a parametrization of [AB] is: x = x, y = f(x), x ∈ [a, b], we

obtain that

∮

[AB]

ydx =∫ baf(x)dx. Now, if B(b, yB) and if C(b, yC), then

a parametrization of [BC] is: x = b, y = y ∈ [yB, yC ]. Hence

∮

[BC]

ydx =

0. Similarly,

∮

[DA]

ydx = 0, because x = a is fixed on [DA], so that dx = 0

on it. Finally, from these observations and from formula (4.2) we get:

−∮

∂(D)

ydx = −∮

[CD]

ydx−∮

[DA]

ydx−∮

[AB]

ydx−∮

[BC]

ydx =

=

∫ b

a

g(x)dx−∫ b

a

f(x)dx = area(D).

R 23. If the domain D is simple w.r.t. Oy-axis, i.e. if [c, d]is the projection of D on Oy-axis and if y0 ∈ (c, d), then the straight


line Y = y0 cuts the boundary ∂D of D in at most two points, then onecan similarly prove the following formula:

(4.3) area(D) =

∮

∂D

xdy.

If D is simple w.r.t. both axes then, by making the sum of formulas(4.1) and (4.3) we get a new "symmetric" formula":

(4.4) area(D) =1

2

∮

∂D

xdy − ydx,

the boundary being directly oriented, i.e. in the trigonometric direction,"from Ox to Oy" (see Fig.14).

The symmetric formula 4.4 is very useful in applications. For in-stance, let us compute the area bounded by the astroide x = a cos3 t,y = a sin3 t, t ∈ [0, 2π]. In this case the domain D is simple w.r.t. bothaxes. Since dx = −3a cos2 t sin tdt and dy = 3a sin2 t cos tdt, then

xdy − ydx =(3a2 sin2 t cos4 t + 3a2 sin4 t cos2 t

)dt =

= 3a2 sin2 t cos2 tdt =3

4a2 sin2 2tdt.

Thus

area(D) =1

2· 3

4a2

∫ 2π

0

sin2 2tdt =3

8a2

∫ 2π

0

1 − cos 4t

2dt =

3a2π

8.

R 24. If a bounded domain D is not simple, say with respectto the Ox-axis then, in our common examples, always it will be possibleto divide it into a finite number of nonoverlaping (which have no smalldisc in their intersections!) simple domains w.r.t. Ox-axis and thento apply the above formulas for each of them. Moreover, we can evencontinue the process of dividing the obtained domains up to we obtaindomains which are simple w.r.t. both axes. In Fig.15 we started witha domain D which is not simple w.r.t. Ox-axis and we divided it intothree nonoverlaping subdomain D1, D2 and D3, which are simple w.r.t.Ox-axis.

What is important to remark is that

∮

[A2A6]

ydx +

∮

[A6A2]

ydx = 0 and

∮

[A4A6]

ydx +

∮

[A6A4]

ydx = 0. Here, since in the drawing of Fig.15 the

segments on which we integrate are parallel to Oy-axis, each of these


F 15

integral is zero (dx = 0). In general, these dividing curves can not bealways chosen to be segments parallel with the Oy-axis. Hence, these

last two equalities come from the general relation:

∮

γ

−→F · d−→r +

∮

γ−

−→F ·

d−→r = 0. Let us use these observations to prove that formula (4.1) istrue even in the general case when the domain is not simple w.r.t. toOx-axis but it can be divided into a finite set of nonoverlaping simpledomains w.r.t. Ox-axis. Look at Fig.15 and follow the next reasoning:

area(D) = area(D1)+area(D2)+area(D3) = −∮

∂D1

ydx−∮

∂D2

ydx−∮

∂D3

ydx =

= −[

∮

[ A4A3A2]

ydx+

∮

[A2A6]

ydx+

∮

[A6A4]

ydx+

+

∮

[ A6A5A4]

ydx+

∮

[A4A6]

ydx+

∮

[ A2A1A6]

ydx+

∮

[A6A2]

ydx] =

−∮

[ A4A3A2]

ydx−∮

[ A2A1A6]

ydx−∮

[ A6A5A4]

ydx = −∮

∂D

ydx.

In this way we can extend all the formulas (4.1), (4.3) and (4.4) to thecase when the domain D has its boundary ∂D a closed direct oriented


piecewise smooth curve and it can be decomposed into a finite union ofsimple nonoverlaping domains. We can apply the same principle to adomain D with its boundary ∂D a polygonal line (polygonal domains).It is very easy to see that each such domain can be decomposed into a fi-nite union of nonoverlaping triangles. Since each triangle is a boundaryof a convex (so that simple) domain, the above three formulas 4.1, 4.3and 4.4 works also for such a "nonconvex" domain. Now, any domainD with its boundary a piecewise smooth curve can be well approximatedwith polygonal domains. Taking limits, we get that the three formulasworks even in this last more general case.

Hence, we can reduce everything to triangles. Since even such op-eration to divide a polygonal domain into triangles can be complicatedwe give below a general and wise formula to compute the area of abounded polygonal domain D.

E 71. Let us apply the formula (4.4) to find a generalformula for the computation of the area of the domain D, boundedby a polygonal direct oriented line [A0A1...An−1An], where Ai(xi, yi),i = 0, 1, ..., n and An = A0. Hence,

area(D) =1

2

n−1∑

j=0

∮

[AjAj+1]

xdy − ydx.

A natural parametrization of the oriented segment [AjAj+1] is x =xj + t(xj+1 − xj), y = yj + t(yj+1 − yj), where t ∈ [0, 1]. Thus,

xdy − ydx =

= [xj + t(xj+1 − xj)] (yj+1 − yj) − [yj + t(yj+1 − yj)] (xj+1 − xj) dt =

= (xjyj+1 − yjxj+1) dt =(−→OAj ×

−→OAj+1

)· −→k dt.

Hence,

area(D) =1

2

n−1∑

j=0

∫ 1

0

(−→OAj ×

−→OAj+1

)· −→k dt. =(4.5)

=1

2

[n−1∑

j=0

(−→OAj ×

−→OAj+1

)]· −→k .(4.6)

This last expression of the area does not depend on the origin O orof the coordinate system. Thus, a very nice consequence of the vectorformula 4.5 is the following.


T 70. Let A0, A1, ..., An−1, An = A0 andM be n+1 distinct

points in a plane (P ). Then the quantity∑n−1j=0

−−→MAj×

−−→MAj+1 is a real

number which does not depend on M. Its absolute value is equal to twotimes the area of the polygonal domain D bounded by the polygonal line[A0A1...An−1An].

R 25. Let D be a plane bounded domain with its boundary∂D a "closed" piecewise smooth direct oriented curve. For a moving

point M on the oriented curve ∂D we denote by −→r =−−→OM the po-

sition vector of M w.r.t. a fixed point O. Let us fix for a moment a

Cartesian plane frame O;−→i ,

−→j and the corresponding two axes Ox

and Oy. By d−→r we mean the vector d−→r = dx−→i + dy

−→j which can be

conceived as a small difference −→r ′ − −→r . If we compute −→r × d−→r we

obtain (xdy − ydx)−→k , where

−→k =

−→i × −→

j is an orthogonal versor onthe plane (P ) in which D is. Its sense is closely connected with the

orientation of ∂D by using the screw rule (O;−→i ,

−→j ,

−→k is a direct

Cartesian coordinate system:−→k =

−→i × −→

j ). Thus formula (4.4) canalso be written:

(4.7) area(D) =1

2

∮

∂D

(−→r × d−→r ) · −→k .

Since this formula is independent on the coordinate system O;−→i ,

−→j ,

we see what is intuitively clear that the area does not depend on afixed coordinate system. This is why formula (4.7) is useful in manyapplications.

5. Supplementary remarks on line integrals

We saw that a field−→F = P

−→i + Q

−→j + R

−→k ) is conservative if and

only if the differential form ω = Pdx + Qdy + Rdz is exact, i.e. if itcan be realized as the first differential of a primitive function U(x, y, z),namely if dU = ω, which is equivalent to saying that ∂U

∂x= P, ∂U

∂y= Q

and ∂U∂z

= R. Let us now prove the complicated result stated in remark21 for some special domains, namely starred domains. We say that the

differential form ω is closed if P,Q,R are of class C1 and curl−→F =

−→0

on D.

D 22. A subset S of R3 (or R2, or R) is said to be starredif there is a point A ∈ S such that for any other point M ∈ S the

5. SUPPLEMENTARY REMARKS ON LINE INTEGRALS 217

segment [AM ] is contained in S. This point A is called a center of thestarred set S. In particular any starred subset is a connected subset.

Any convex subset S is a starred subset, any fixed point A of Sbeing a center of it. In Fig.16 we see some examples of starred, convex,nonstarred, or nonconvex plane subsets.

F 16

In Fig.17 we see the same type of examples in space.

F 17

Recall that an direct oriented segment [AM ], A(a, b, c), M(x, y, z)of a line is a smooth deformation of the direct oriented interval [0, 1]

through the classical parametrization path −→r (t) = f1(t)−→i + f2(t)

−→j +


f3(t)−→k , t ∈ [0, 1], where

f1(t) = a+ (x− a)tf2(t) = b+ (y − b)tf3(t) = c+ (z − c)t

, t ∈ [0, 1].

T 71. Let D be a starred domain in R3 and let ω = Pdx+Qdy+Rdz be a closed differential form on D, i.e. P,Q,R are of class

C1 and curl−→F =

−→0 on D. Then ω is exact on D, i.e. there exists a

scalar function U(x, y, z) defined on D, such that its partial derivatives∂U∂x, ∂U∂y, ∂U∂zexist and ∂U

∂x= P, ∂U

∂y= Q and ∂U

∂z= R, i.e. U is a primitive

or a potential function for−→F . Here

−→F is a field with components P,Q

and R.

P. Let A(a, b, c) be a centre of the starred domain D and letM0(x0, y0, z0) be an arbitrary moving point of D. SinceD is starred andA is a centre of it, the segment [AM0] is contained in D. We define:

U(x0, y0, z0) =

∮

[AM0]

Pdx+Qdy +Rdz =

(5.1) =

∫ 1

0

[P (f1(t), f2(t), f3(t))(x0−a)+Q(f1(t), f2(t), f3(t))(y0−b)+

+R(f1(t), f2(t), f3(t))(z0 − c)]dt.

Don’t forget that

f1(t) = a+ (x0 − a)tf2(t) = b+ (y0 − b)tf3(t) = c + (z0 − c)t

, t ∈ [0, 1].

Let us apply Leibniz’ formula in the integral (5.1), with parameters x0,y0, z0, in order to compute the partial derivatives of U. The conditions

of theorem 51 are satisfied. We also use the condition curl−→F =

−→0 , i.e.

∂P∂y

= ∂Q∂x, ∂Q∂z

= ∂R∂y

and ∂R∂x

= ∂P∂z

:

∂U

∂x0

(x0, y0, z0) =

∫ 1

0

[P (f1(t), f2(t), f3(t))+∂P

∂x(f1(t), f2(t), f3(t))t(x0−a)+

+∂Q

∂x(f1(t), f2(t), f3(t))t(y0 − b) +

∂R

∂x(f1(t), f2(t), f3(t))t(z0 − c)]dt =

=

∫ 1

0

[P (f1(t), f2(t), f3(t)) +∂P

∂x(f1(t), f2(t), f3(t))t(x0 − a)+

+∂P

∂y(f1(t), f2(t), f3(t))t(y0 − b) +

∂P

∂z(f1(t), f2(t), f3(t))t(z0 − c)]dt =

5. SUPPLEMENTARY REMARKS ON LINE INTEGRALS 219

=

∫ 1

0

d

dt[tP (f1(t), f2(t), f3(t))] dt = tP (f1(t), f2(t), f3(t))|10 =

= P (f1(1), f2(1), f3(1)) = P (x0, y0, z0).

We leave the reader to use the same methods in order to prove that∂U∂y0

(x0, y0, z0) = Q(x0, y0, z0) and that ∂U∂z0

(x0, y0, z0) = R(x0, y0, z0).

If the segments [AB], [BC] and [CM0] are contained in D, then wecan integrate on the path [ABCM0] (see Fig.19) and we can define:

U(x0, y0, z0) =

∮

[ ABCM0]

Pdx+Qdy + Rdz.

The computations are easier in this case (do them!). Finally we get thesame function U defined on D (if A is fixed). For instance, let us usethis last idea to find a primitive of the form ω = yzdx+ zxdy + xydz.

Since curl−→F =

−→0 , there exists a primitive (it is clear that D = R3 is

a starred domain) U(x, y, z). Let us fix a centre A(0, 1, 2) (see Fig.18)

F 18

and let us compute the value of U at B(3, 1,−3) :

U(3, 1,−3) =

A1(3,1,2)∮

A(0,1,2)

yzdx+zxdy+xydz+

B(3,1,−3)∮

A1(3,1,2)

yzdx+zxdy+xydz =

=

∫ 3

0

1 · 2 · dx+

∫ −3

2

3 · 1 · dz = −9.


F 19

R 26. Let D be a plane simple connected domain and let−→F : D → R2,

−→F = (P,Q) be a plane field of class C1 on D such that

∂P∂y

= ∂Q∂xon D. Then the field is conservative, i.e. there exists a scalar

function U(x, y) such that gradU =−→F , or ∂U

∂x= P and ∂U

∂y= Q. Our

above experience tells us that this function must be of the form:

(5.2) U(x, y) =

∮

γ

Pdx+Qdy,

where γ is a smooth oriented curve contained in D and which connect afixed "for ever" point M0(x0, y0) of D and the moving point M(x, y) onD. Two problems arises: 1) to prove that the definition does not dependon the connecting curve γ and 2) to prove that ∂U

∂x= P and ∂U

∂y= Q.

The second problem is easy if we succeed to prove the first. Indeed,if the definition (5.2) does not depend on the path γ, then the integral∮Pdx+Qdy is independent on path, i.e. it is zero on any "closed" path

in D. And conversely! Let us fix a point L(a, b) in D and let us takean open disc B(L, r), r > 0, which is contained in D. Since this ball isa starred domain, applying the plane variant of theorem 71 we see that

the restriction of U to this ball is a primitive for−→F on D. In particular

∂U∂x

(a, b) = P (a, b) and ∂U∂y

(a, b) = Q(a, b). Hence we need only to prove


1). As we just remarked, it is enough to prove that

∮

Γ

Pdx+Qdy = 0 for

any smooth closed oriented curve Γ ⊂ D, if ∂P∂y

= ∂Q∂xon D. Since any

such curve Γ can be well and uniformly approximated with closed polyg-

onal lines, it is enough to prove

∮

[A0A1...An−1An]

Pdx+Qdy = 0 for any

closed oriented polygonal line [A0A1...An−1An], An = A0.We can dividethe domain bounded by this polygonal line in oriented triangles, suchthat the orientation of each triangle is uniquely determined (induced) bythe orientation of [A0A1...An−1An]. Hence it will be sufficient to provethat if ∂P

∂y= ∂Q

∂xon a domain Ω which contains a triangle [A0A1A2],

then

∮

[A0A1A2]

Pdx + Qdy = 0. Since always we can take a starred do-

main ∆ which is contained in D and which contains our given triangle[A0A1A2], we can apply theorem 71 to find a potential function U for−→F restricted to ∆. Now, theorem 67 says that

∮

[A0A1A2]

Pdx+Qdy = 0,

i.e. what we wanted to prove.


1. Compute the length of the curve y(x) = a cosh xa, x ∈ [−a, a],

a > 0.

2. Let Γ :

x = 3t+ 1y = 4t+ 1

, t ∈ [−1, 1]. Find its length and the

coordinates of its mass centre if the density function is f(x, y) = |x| .3. On the segment [AB], A(1, 1), B(3, 4) we consider a density

function f(x, y) = x2 + y2. Find its mass.

4. On the arc of a circle [ACB] with radius 2, A(2, 0), B(−2, 0)and C(0, 2) we consider a wire with density function f(x, y) = 1√

x2+y2.

Find the mass, the coordinates of the mass centre and the moment ofinertia w.r.t. Oy-axis.

5. The segment [0, 2π] is deformed into a helix:

x = a cos θy = a sin θz = bθ

,

where a, b > 0 and θ ∈ [0, 2π]. Assume that the density function isf(x, y, z) = 4. Compute the mass, the coordinates of the mass centreand the moment of inertia w.r.t. Oz-axis.

6. The wire Γ = M(x, y) : (x − 1)2 + y2 = 1, with the den-sity function f(x, y) = x rotates around Ox-axis. Find its moment ofinertia.


7. Let A(a, 0), B(0, b), n = [AB], be the segment AB and let m =

[AB], be the arc of the ellipse x2

a2+ y2

b2= 1 which connects A and B.

The metallic frame [AmBnA] is loaded with a density function f(x, y)at the point M(x, y), which is three times the value of the projectionof M on the Ox-axis. Find its mass centre.

8. Let A(1, 1), B(1,−1) and [AOB] be the arc of the parabola x =

y2. The frame [AOB] ∪ [AB] with the density function f(x, y) = x+ 1rotates around Ox-axis. Find its moment of inertia.

9. Let A(1, 1) and B(2, 0) be two points in the xOy-plane.Find I =

∫[OA]∪[AB]

(x3 + y3)ds.

10. Find∫Cxyds, where C is the square |x| + |y| = a > 0.

11. Find I =∫C

ds√x2+y2+4

, where C is the segment which connect

the points O(0, 0) and A(1, 2).

12. Find∫Cxyds, where C is the curve x2

a2+ y2

b2= 1, x, y ≥ 0.

13. Find∫C

dsx2+y2+z2

, where C :

x = 2 cos ty = 2 sin tz = bt

, t ∈ [0, 4π].

14. Find the length of the arc of a conic helix: x = 2et cos t,y = 2et sin t, z = 2et, between O(0, 0, 0) and A(2, 0, 2).

15. Find the length and the mass centre of the arc of a cycloid:


, t ∈ [0, π].

16. Find the mass of the ellipse-wire x2

a2+ y2

b2= 1 if its density

function is f(x, y) = |y| .

17. Find∫C

(x+ y)ds, where C is the arc:

x = t

y = 3t2√2

z = t3, t ∈ [0, 1].

18. Use Riemann sums to find the side area of the parabolic cylindery = 3

8x2, bounded by the planes z = 0, x = 0, z = x, y = 6.

19. Compute

∮

C

xdy, where C is the direct oriented (trigonometric

sense) curve which bounds the triangle realized by the coordinates axesand the line x

2+ y

3= 1.

20. Compute

∮

C

xydx + x2dy, where C is the arc of the parabola

y = 4x2 between O(0, 0) and A(1, 4), clockwise oriented.


21. Find the work of the field−→F (x, y) = xy

−→i + x2−→j on the fol-

lowing oriented curves:a) the rectangle frame [OABCO],whereA(0, 1), B(2, 1) andC(2, 0).

The orientation is O → A→ B → C →).b) the curve [OBAO], where [OB] and [AO] are segments of lines,

[BA] is an arc of a circle of radius 2, B(2, 0) and the angle ∡(AOB) isequal to 60. The orientation is O → B → A→ O.

c) the triangle frame [ABCA], where A(1, 2), B(3, 4) and C(5, 1).The orientation is A→ B → C → A.

d) the curve [OmAnO], where A(1, 1), [OmA] is the arc of theparabola y = x2 and [AnO] is the arc of the parabola x = y2. Theorientation is O → A→ O.

22. Find

∮

C

−→F · d−→r , where −→r (t) = t

−→i + t2

−→j + t3

−→k , t ∈ [0, 1] and

−→F (x, y, z) = (y, z, 2x). The orientation of C is the direct orientationinduced by the moving of t from 0 to 1.

23. Find the work of the field−→F (x, y, z) = xy

−→i + xz

−→j + yz

−→k on

the polygonal frame [ABCO], where A(3, 0, 0), B(3, 3, 0) and C(0, 3, 3).The orientation is A→ B → C → O. Is this field a conservative field?If it is so, compute again easier the work of the field!

24. Find α ∈ R such that the integral I(α) =

∮(αxy−1)dx+3x2dy

does not depend on path. Then compute

(2,3)∮

(1,1)

(αxy − 1)dx + 3x2dy for

this last found value of α.25. Prove that the following differential forms are exact on some

simple connected domains (describe them!) and then find their primi-tives.

a)dx+dy+dzx+y+z

; b)xdx+ydy+zdz√x2+y2+z2

; c)xdy+ydx√x2+y2

.

26. Use the formulas: area(D) =

∮

∂D

xdy = −∮

∂D

ydx = 12

∮

∂D

xdy−ydx

in order to find the areas of the figures bounded by the following closedcurves:

a)x2 + y2

4= 1; b)

x = a cos3 ty = a sin3 t

, t ∈ [0, 2π]. c)

x = t− sin ty = 1 − cos t

, t ∈[0, 2π], and the Ox-axis.

27. Find the length and the area bounded by


Γ :

x = R(2 cos t− cos 2t)y = R(2 cos t− sin 2t)

, t ∈ [0, 2π], R > 0.

28. Let Γ :

x = t cos ty = t sin tz = t

, t ∈ [0, π] be a wire with the density

function f(x, y, z) = x2 + y2. Find its static moment relative to xOy-plane.

29. Let Γ = M(x, y) : x2 + y2 = a2 and x + y + z = 0 with itsdirect orientation induced by the direct orientation of the plane xOy.

Compute the work of−→F = (1, x, z) on Γ.

30. Let P1 be the parabola x = y2 and let P2 be the unique parabolawhich contains the points B(3,−

√3), A(3,

√3) and C(2, 0). Find the

area bounded by these parabolas.31. Compute:

a)∫ (2,−1)

(−1,1)x(1 + x)dx− y(1 + y)dy; b)

∫ (−2,1,5)

(0,0,0)x2dx+

√ydy− z

√zdz;

32. Find V (x, y) if dV = (4x + 2y)dx + (2x − 6y)dy. Use this tosolve the differential equation y′ = 4x−6y

6y−2x.

CHAPTER 6

Double integrals

1. Double integrals on rectangles.

A lamina or a thin plate is a bounded plane closed domain (an openand connected bounded subset of R2together with its boundary)D witha density function f(x, y) defined on it. In practical applications f(x, y)is greater or equal than zero. But here we shall consider the densityfunction f : D → R to have arbitrary values and to be piecewise contin-uous. This means the f is bounded and continuous on D except maybea finite union of points or smooth curves (which have area zero!). Theboundary ∂D of D is considered to be a piecewise smooth curve (inparticular it has a zero area!). Let us recall the definition 7 of the areaof a plane figure. We say that a plane figure A has area (measure)σ(A) if for any ε > 0 there exist two elementary figures (finite unionof nonoverlaping (their intersections does not contain interior points!)simple rectangles [a, b] × [c, d]) Ei and Eo such that Ei ⊂ A ⊂ Eo,area(Eo)−area(Ei) < ε and area(Eo) ≤ σ(A) ≤ area(Ei). This meansthat we can well approximate the figure A with elementary figures frominside and from outside too. We know that the domain D which is thesupport of a lamina defined above has an area (because ∂D is piecewisesmooth!). We always in this chapter assume this and, because of thistype of approximation, we begin with a particular form of D, namelywhen D is a simple rectangle [a, b] × [c, d] and f : [a, b] × [c, d] → Ris a continuous function defined on it (any piecewise continuous func-tion can well be approximated with continuous functions!). Let therectangle (as a surface!) D = [ABCE] in Fig.1.

Let ∆x : a = x0 < x1 < ... < xn = b be an arbitrary divisionof [a, b] and let ∆y : c = y0 < y1 < ... < ym = d be an arbitrarydivision of [c, d]. Let Dij = [xi−1, xi] × [yj−1, yj] and let Pij(ci, dj) bea marking point in Dij for any i = 1, 2, ..., n and j = 1, 2, ...,m. Bydefinition the area of Dij is (xi − xi−1)(yj − yj−1) and D = ∪i,jDij.Let us assume now that ‖∆x‖ → 0 and ‖∆y‖ → 0, when n → ∞and m → ∞, i.e. area(Dij) → 0. Hence we can well approximate thedensity function f(x, y) on Dij with its value at the marked point Pij,i.e. f(x, y) ≈ f(ci, dj) for any (x, y) ∈ Dij. We can remark that ci

225

226 6. DOUBLE INTEGRALS

F 1

and dj are sets of marking points for ∆x and ∆y respectively. Thuswe can well approximate:

(1.1) mass(D) ≈n∑

i=1

m∑

j=1

f(ci, dj)area(Dij)def= Sf(∆x,∆y, (ci, dj))

Such a sum is called a double Riemann sum which corresponds todivisions ∆x,∆y, to marking points P (ci, dj) and to function f(x, y).We can also introduce double Darboux sums:

s∆x,∆y =n∑

i=1

m∑

j=1

mijarea(Dij) and S∆x,∆y =n∑

i=1

m∑

j=1

Mijarea(Dij),

where mij = inf(x,y)∈Dij

f(x, y) and Mij = sup(x,y)∈Dij

f(x, y).

It is not so difficult to state and to prove analogous results to thosegiven in chapter II. For instance, any double Riemann sum

Sf (∆x,∆y, (ci, dj))

is greater or equal than its corresponding s∆x,∆y and it is also less orequal than S∆x,∆y .

D 23. A function f(x, y) defined on a simple rectangleD is Riemann integrable on D if there exists a real number I such thatfor any ε > 0, there exists a small number δ (depending on ε) suchthat if ‖∆x‖ < δ and ‖∆y‖ < δ, then |I − Sf(∆x,∆y, (ci, dj))| < εfor any marking points (ci, dj) corresponding to the divisions ‖∆x‖and ‖∆y‖ . The number I is unique, it is called the double integral of

1. DOUBLE INTEGRALS ON RECTANGLES. 227

f on D and it is denoted by I =

∫∫

D

f(x, y)dxdy. Here dxdy is said

to be a (small) element of area of D.This is the virtual notation for(xi − xi−1)(yj − yj−1). In this particular case (the case of a rectangle)dxdy = dx · dy. We shall see that in general (for arbitrary domains)this is not true (for instance in the case of a disc).

We can give here the analogous theorems for the theorems 14 and16.

T 72. (Darboux criterion) f : D → R is Riemann inte-grable if and only if for any ε > 0, there exists δ > 0 such that whenever‖∆x‖ < δ and ‖∆y‖ < δ then one has that S∆x,∆y − s∆x,∆y < ε.

A direct consequence of this criterion is the following basic theorem.

T 73. Any continuous function f : D = [a, b] × [c, d] → Ris Riemann integrable. Moreover, any piecewise continuous function f(see its definition above) on a rectangle is also Riemann integrable.

The following theorem is a basic result for what we shall study herew.r.t. "multiple integrals", i.e. integrals in 2 or 3 dimensions.

T 74. (iterative formula) Let f : D → R be a continuousfunction on a rectangle D = [a, b] × [c, d]. Then f is integrable and(1.2)∫∫

D

f(x, y)dxdy =

∫ b

a

(∫ d

c

f(x, y)dy

)dx =

∫ d

c

(∫ b

a

f(x, y)dx

)dy.

These last formulas says that either we integrate f(x, y) w.r.t. y fromc to d and obtain a new function of x which is finally integrated froma to b (the first equality), or we firstly integrate f(x, y) w.r.t. x from ato b and obtain a new function of y which is finally integrated from cto d (the second equality).

P. We preserve the above notation in the next considera-

tions. Let J =

∫∫

D

f(x, y)dxdy, I1 =∫ ba

(∫ dcf(x, y)dy

)dx and I2 =

∫ dc

(∫ baf(x, y)dx

)dy. The number J exists because of theorem 73. The

integral I(x) =∫ dcf(x, y)dy is an integral with a parameter x. Thus

I1 =∫ baI(x)dx and we can well approximate it with Riemann sums of


the following type:

(1.3) I1 ≈n∑

i=1

I(ci)(xi − xi−1).

But each integral I(ci) =∫ dcf(ci, y)dy can be approximated by Rie-

mann sums of the form:

(1.4) I(ci) ≈m∑

j=1

f(ci, d(i)j )(yj − yj−1) ≈

m∑

j=1

f(ci, dj)(yj − yj−1),

because dj and d(i)j are both in [yj−1, yj], ‖∆y‖ → 0 and f is continuous.

Here ∆y is the union of all divisions ∆(i)y considered for each ci, i =

1, 2, ..., n. Now we come back to formula (1.3) with the approximationof I(ci) from formula (1.4) and find:

I1 ≈n∑

i=1

m∑

j=1

f(ci, dj)(yj − yj−1)(xi − xi−1) = Sf(∆x,∆y, (ci, dj)).

Thus I1 can be well approximated with double Riemann sums of thetype

Sf(∆x,∆y, (ci, dj)) Since f is Riemann integrable, all of these

sums have a unique limit point J =

∫∫

D

f(x, y)dxdy. Hence I1 = J.

Similarly we can prove (do it!) that J = I2, where I2 =∫ dcJ1(y)dy and

J1(y) =∫ baf(x, y)dx. Therefore, I1 = I2 = J.

E 72. Find the mass of the plate D = [0, 1] × [1, 2] if thedensity function is f(x, y) = x2y + xy2. We use the iterative formula(1.2):

mass(D) =

∫∫

D

(x2y + xy2)dxdy =

∫ 1

0

(∫ 2

1

(x2y + xy2)dy

)dx =

=

∫ 1

0

(x2y

2

2+ x

y3

3

)∣∣∣∣2

1

dx =

∫ 1

0

(x222

2+ x

23

3− x21

2− x

1

3

)dx =

=

∫ 1

0

(3

2x2 +

7

3x

)dx =

(3

2

x3

3+

7

3

x2

2

)∣∣∣∣1

0

=1

2+

7

6=

5

3.

The other variant of the formula is "easier" to work with.∫∫

D

(x2y + xy2)dxdy =

∫ 2

1

(∫ 1

0

(x2y + xy2)dx

)dy =


∫ 2

1

(x3

3y +

x2

2y2

∣∣∣∣1

0

)dy =

∫ 2

1

(y

3+y2

2

)dy =

(y2

6+y3

6

)∣∣∣∣2

1

=5

3.

E 73. Let us compute the mass centre coordinates of Dwhen the density function is a piecewise continuous function f(x, y).Look at Fig.1 and let us approximate the static moment relative to Oy-axis of the rectangle Dij = [xi−1, xi] × [yj−1, yj] by cif(ci, dj)area(Dij).Thus the "global" static moment of D relative to Oy-axis can be wellapproximated by

n∑

i=1

m∑

j=1

cif(ci, dj)(yj − yj−1)(xi − xi−1) = Sxf (∆x,∆y, (ci, dj)),

a Riemann sum of function xf(x, y). Hence the theoretical static mo-ment My of the lamina (D, f) relative to Oy-axis can be computed bythe formula:

(1.5) My =

∫∫

D

xf(x, y)dxdy.

In the same manner we can deduce a formula for the static momentMx of the lamina (D, f) w.r.t. Ox-axis:

(1.6) Mx =

∫∫

D

yf(x, y)dxdy.

The coordinates (xG, yG) of the mass centre G of the lamina (D, f) aredefined such that if we concentrate the mass of D at the point G, thenboth moments of this last system must coincide with the correspondentglobal moments, i.e.

My =

∫∫

D

xf(x, y)dxdy = xG ·mass(D),

and

Mx =

∫∫

D

yf(x, y)dxdy = yG ·mass(D).


Thus we just obtained the basic formulas:

(1.7) xG =

∫∫

D

xf(x, y)dxdy

∫∫

D

f(x, y)dxdy

, and yG =

∫∫

D

yf(x, y)dxdy

∫∫

D

f(x, y)dxdy

.

For instance, let us compute the mass centre of the square D = [−1, 1]×[0, 2] with the density function f(x, y) = y. First of all let us remarkthat D is symmetric w.r.t. Oy-axis. Moreover, the distribution of themass is symmetric w.r.t. the same axis: f(x, y) = f(−x, y) = y. Thus,G is on Oy-axis, i.e. xG = 0. In order to compute yG we need tocompute two integrals:

∫∫

D

f(x, y)dxdy =

∫∫

D

ydxdy =

∫ 1

−1

(∫ 2

0

ydy

)dx = 2

∫ 1

−1

dx = 4

and∫∫

D

yf(x, y)dxdy =

∫∫

D

y2dxdy =

∫ 1

−1

(∫ 2

0

y2dy

)dx =

8

3

∫ 1

−1

dx =16

3.

Hence yG =163

4= 4

3. We leave as an exercise to the reader to prove

that yG = 43is the mass centre of the wire x = 0, y = t ∈ [0, 2],

with the same density function f(x, y) = y. Use your "static" feelingto anticipate this result and to explain this feeling.

E 74. Let us deduce the moments of inertia IOx of a simplerectangle D with density function f(x, y), when D moves around Ox-axis. For a material point P (x, y) with a mass m, this moment ofinertia is equal to y2m, i.e. the mass m multiplied by the square of thedistance from P to the rotation axis Ox. Thus, we can well approximatethe moment of inertia of the rectangle Dij w.r.t. Ox-axis with thequantity d2

jf(ci, dj)area(Dij). Thus the global moment of inertia of Drelative to Ox-axis can be well approximated by

n∑

i=1

m∑

j=1

d2jf(ci, dj)(yj − yj−1)(xi − xi−1) = Sy2f (∆x,∆y, (ci, dj)).

Hence, we can write the formula:

(1.8) IOx =

∫∫

D

y2f(x, y)dxdy.


Analogously, we can deduce the following formula for the moment ofinertia of D w.r.t. Oy-axis:

(1.9) IOy =

∫∫

D

x2f(x, y)dxdy.

The sum IO = IOx+IOy =

∫∫

D

(x2+y2)f(x, y)dxdy is called the moment

of inertia w.r.t. O.Let us compute for instance the moment IOy of D = [0, 1] × [0, 1]

with the density function f(x, y) = xy.

IOy =

∫∫

D

x2f(x, y)dxdy =

∫ 1

0

(∫ 1

0

x3ydy

)dx =

∫ 1

0

x3

(∫ 1

0

ydy

)dx =

1

2

∫ 1

0

x3dx =1

2· 1

4=

1

8.

Let us also compute IOx for the rectangle of figure Fig.2. Wheneverwe have no density function given, we consider it to be the constantfunction f(x, y) = 1.

F 2

Thus we have:

IOx =

∫∫

D

y2dxdy =

∫ 1

−1

y2

(∫ 1

−1

dx

)dy = 2

y3

3

∣∣∣∣1

−1

=4

3.

E 75. We also can compute the volume of some type of3D-domaine called rectangular cylindrical solids. In Fig.3


F 3

we have a cylindrical surface with its directrix curve the rectangle[ABCD] in the xOy-plane and the generating lines parallel to the Oz-axis. We are interested to compute the volume of the solid Ω bounded bythis cylindrical surface, by xOy-plane and by the surface z = f(x, y) ≥0, i.e. the piece of this surface, [A′B′C ′D′]. Let us divide the basicrectangle [ABCD] into small sudomaines Dij = [xi−1, xi]× [yj−1, yj ] =[P1P2P3P4] and let us choose an arbitrary marking point Pij(ci, dj) ineach Dij , i = 1, 2, ..., n and j = 1, 2, ...,m. The volume of the cylindricalsolid which has as a basis the rectangle [P1P2P3P4] and as a "hat" thepiece [P ′1P

′2P

′3P

′4] of the surface z = f(x, y) can be well approximated

(when the area of Dij goes to 0) with the volume of a parallelepipedwith the basis the same rectangle [P1P2P3P4] and the height f(ci, dj),i.e. the entire volume of our rectangular cylindrical solid can be wellapproximated with double Riemann sums of the form:

n∑

i=1

m∑

j=1

f(ci, dj)(yj − yj−1)(xi − xi−1),

i.e. this last volume can be computed with the formula:

vol =

∫∫

Ω

f(x, y)dxdy.

2. DOUBLE INTEGRALS ON AN ARBITRARY BOUNDED DOMAIN 233

When f(x, y) has also negative values, we can use the more generalformula:

(1.10) vol =

∫∫

Ω

|f(x, y)| dxdy.

For instance, in Fig.4 we have a rectangular cylindrical body[ABCOA′B′C ′O′]. In this case the parallelepiped solid with the basis

the rectangle [ABCO] is cuted by the plane xa

+ yb

+ zc

= 1 in the upper

part. Hence z = c(1 − x

a− y

b

), a > 1, b > 1 and the volume of our

cylindrical body [ABCOA′B′C ′O′] can be computed as follows:

vol =

∫ 1

0

(∫ 1

0

[c− c

ax− c

by]dy

)dx =

∫ 1

0

[cy − c

axy − c

2by2

]∣∣∣1

0dx =

∫ 1

0

(c− c

ax− c

2b

)dx =

(cx− c

2ax2 − c

2bx)∣∣∣

1

0= c− c

2a− c

2b.

F 4

2. Double integrals on an arbitrary bounded domain

All the plane domains (connected subsets of R2) which we work within the following are closed (∂D ⊂ D) bounded and with a finite areaσ(D). This last general condition is not so easy to be verified in prac-tice. Since in engineering one needs only closed bounded domains Dwith a piecewise smooth boundary ∂D, we shall limit our study for suchdomains. It is not difficult to prove (see the beginning considerations


in the chapter with line integrals) that a piecewise smooth boundary (afinite union of nonoverlaping smooth curves) can be well and uniformlyapproximated by polygonal lines. Hence the area of D can be repre-sented as a limit of areas of figures bounded by polygonal lines. Sincesuch a figure can be divided into a union of nonoverlaping triangles andsince a triangle can be well approximated with simple rectangles (doit! or look in [Pal]), we finally deduce that any closed bounded domainD with a piecewise smooth boundary ∂D has an area (measure) σ(D)(see definition 7) and this area is a limit of areas of elementary figures(finite union of nonoverlaping simple rectangles). Hence, it is natural toextend the definition of the double integral on a simple rectangle to anelementary figure E = ∪Ni=1Di, where Di = [ai, bi]× [ci, di], Di∩Dj = ∅or a subset in R2 without interior points (points, segments of smoothlines, etc.), i = j, by the following formula:

∫∫

E

f(x, y)dxdy =N∑

i=1

∫∫

Di

f(x, y)dxdy.

A "theoretical" idea is to continue to extend the definition of the doubleintegral to a closed bounded domain D with piecewise smooth ∂D (⇒area(∂D) = 0), which can be well approximated from inside and from

outside with two sequences E(n)ins and E

(n)out of elementary figures, by the

following formula:∫∫

D

f(x, y)dxdy = limn→∞

∫ ∫

E(n)ins

f(x, y)dxdy = limn→∞

∫ ∫

E(n)out

f(x, y)dxdy.

Here E(n)ins ⊂ D ⊂ E

(n)out, and lim

n→∞σ(E

(n)ins) = σ(D) = lim

n→∞σ(E

(n)out). Prac-

tically this definition can not be used at all! Thus, we shall use ourprevious experience to take again the definition of the double integralon the class of plane domains D just mentioned above. By diam(D),the diameter of a bounded domain D, we mean the following number:

diam(D) = sup∥∥∥

−−−→MM ′

∥∥∥ : M,M ′ ∈ D.

For instance, the diameter of a closed disc B[A, r] is 2r, i.e. its usualdiameter. The diameter of a rectangle Ω is equal to the length of oneof its diagonal. The diameter of a parallelogram is equal to the lengthof its greatest diagonal. And so on! A division ∆ of the domain D(like above!) is a finite set of domains Di , i = 1, 2, ..., n, of the sametype like D, such that ∪Di = D and Di ∩ Dj = ∅ or a figure ofzero area (in fact points or piecewise smooth curves!) for i = j (see


Fig.5). The norm of a division ∆ is the following nonnegative realnumber: ‖∆‖ = maxdiam(Di) : i = 1, 2, ..., n. A set of markingpoints Pi(ci, di) , i = 1, 2, ..., n for a division ∆ is a set of some fixedpoints Pi(ci, di) ∈ Di for any i = 1, 2, ..., n. For a function f : D → R,for a division ∆ of D and for a set of marking points Pi(ci, di) wedefine the corresponding Riemann sum:

(2.1) Sf(∆, Pi(ci, di)) =n∑

i=1

f((ci, di))area(Di).

F 5

D 24. A function f : D → R is (Riemann) integrable

on D if there exists a real number I (denoted by

∫∫

D

f(x, y)dxdy) such

that for any small ε > 0, there exists a small δ (depending on ε) withthe following property: if ∆ is a division of D with ‖∆‖ < δ, then|I − Sf(∆, Pi(ci, di))| < ε for any set of marking points Pi(ci, di)of ∆. This means that we can well approximate the number I withRiemann sums of the form Sf(∆, Pi(ci, di)), when ‖∆‖ → 0.

We shall state without a proof some results which are analogouswith the corresponding results for the simple Riemann integral intro-duced and studied in chapter II. It will not be so difficult for the readerto imitate the proofs given there. If some difficulties will appear, lookin [Pal] for complete proofs.

T 75. Let D be a plane domain like above and let f : D →R be a Riemann integrable function. Then f must be bounded.


This last result says that the class of integrable functions is includedin the class of bounded functions. This is why any function consideredby us in the following is assumed to be bounded.

Let D be like above, let ∆ = Di, i = 1, 2, ..., n be a division of Dand let f be a bounded function defined on D with values in R. Letmi = inff(x, y) : (x, y) ∈ Di and let Mi = supf(x, y) : (x, y) ∈ Difor any i = 1, 2, ..., n. Then

s∆(f) =n∑

i=1

mi · area(Di)

is called the inferior Darboux sum of f relative to ∆ and

S∆(f) =n∑

i=1

Mi · area(Di)

is said to be the superior Darboux sum of f relative to ∆. It is clearthat

m · area(D) ≤ s∆(f) ≤ Sf(∆, Pi(ci, di)) ≤ S∆(f) ≤M · area(D),

for any set of marking points Pi(ci, di) of the division ∆, where m =inff(x, y) : (x, y) ∈ D and M = supf(x, y) : (x, y) ∈ D. Since theset s∆(f) : ∆ goes on the set of all divisions ofD is upper bounded byM ·area(D), it has a least upper bound I∗(f). Since the set S∆(f) : ∆goes on the set of all divisions of D is lower bounded by m · area(D),it has a greatest lower bound I∗(f). In general, I∗(f) ≤ I∗(f).

T 76. (Darboux criterion) Let D be like above and let f :D → R be a bounded function. Then f is integrable on D if andonly if I∗(f) = I∗(f), i.e. if and only if for any ε > 0 there exists aδε > 0 such that if ‖∆‖ < δε then S∆(f) − s∆(f) < ε. In this last case

I∗(f) = I∗(f) =

∫∫

D

f(x, y)dxdy.

This criterion is very useful whenever we want to prove that a spec-ified class of functions is integrable.

T 77. Let D be a closed bounded plane domain with ∂D apiecewise smooth curve and let f : D → R be a continuous functiondefined on D. Then f is integrable on D. If f is continuous on a sub-set DA of D such that A has an area equal to zero, then f is alsointegrable on D.

P. We prove only the first statement of the theorem. SinceD is closed and bounded, D is a compact subset of R2. Since f is


continuous, it is uniformly continuous (see [Po], theorem 59). In orderto prove that f is integrable we shall use the above Darboux criterion(see theorem 76). For this, let us take a small ε > 0. The uniformcontinuity of f implies that there exists a small δ > 0 (depending on ε)such that if x,x′ ∈ D and ‖x− x′‖ < δ, then |f(x) − f(x′)| < ε

area(D).

Let us take a division ∆ = Di, i = 1, 2, ..., n of D with ‖∆‖ < δ.Since eachDi is a compact subset of R2 and since f is continuous, thereexist x(i) and x′(i) in Di with

mi = inff(x, y) : (x, y) ∈ Di = f(x(i))

and

Mi = supf(x, y) : (x, y) ∈ Di = f(x′(i))

(see [Po], theorem 58). Hence

S∆(f) − s∆(f) =n∑

i=1

[f(x′(i)) − f(x(i))

]· area(Di) ≤

≤n∑

i=1

∣∣f(x′(i)) − f(x(i))∣∣ · area(Di) <

ε

area(D)

n∑

i=1

area(Di) = ε,

because ‖∆‖ < δ implies diam(Di) < δ and so,∥∥x′(i) − x(i)

∥∥ < δ.Thus we can apply the uniform continuity of f and finally we get thatS∆(f) − s∆(f) < ε, i.e. that f is integrable.

R 27. Up to now we used closed bounded domains Di withpiecewise smooth boundaries, as subdomains of a division ∆ of a fixeddomain D of the same type. Since any such subdomain Di can bewell approximated with elementary figures (finite union of nonoverlap-ing simple rectangles) or with finite unions of nonoverlaping triangles,discs, parallelograms, etc., we can substitute these Di in definition 24 orin theorem 76 with triangles, discs, parallelograms, etc. It is clear thatin general it is not possible to cover a domain D with nonoverlapingdiscs, for instance. But, for any small ε > 0, there exists such a unionof discs Uε ⊂ D, such that area(DUε) < ε. Hence we can work withthe following alternative equivalent definition for an integrable functionf. We say that f : D → R is (Riemann) integrable on D if there ex-ists a real number I such that for any small ε > 0 and η > 0 thereexists a union Uη = Bi of nonoverlaping discs, such that Uη ⊂ D,area(DUη) < η and |I −∑n

i=1 f((ci, di))area(Bi)| < ε for any set ofmarking points Pi(ci, di), Pi(ci, di) ∈ Bi, i = 1, 2, ..., n. Similarly wecan work with triangles, parallelograms, general rectangles, etc., insteadof discs.


The following result put together some basic properties of the dou-ble integral.

T 78. a) the mapping f

∫∫

D

f(x, y)dxdy is a linear

mapping defined on the vector space Int(D) of all integrable functionsdefined on D. This means that∫∫

D

[αf(x, y) + βg(x, y)]dxdy = α

∫∫

D

f(x, y)dxdy + β

∫∫

D

g(x, y)dxdy.

b) the mass of a lamina (D, f) can be computed as follows:

mass(D) =

∫∫

D

f(x, y)dxdy.

In particular the area of D is equal to

∫∫

D

dxdy.

c) if f ≤ g on D, then

∫∫

D

f(x, y)dxdy ≤∫∫

D

g(x, y)dxdy. In par-

ticular, if f ≥ 0 on D, then

∫∫

D

f(x, y)dxdy ≥ 0.

d)

∣∣∣∣∣∣

∫∫

D

f(x, y)dxdy

∣∣∣∣∣∣≤

∫∫

D

|f(x, y)| dxdy.

e)

∫∫

D

f(x, y)dxdy = f(x0, y0) ·area(D), where f is continuous and

(x0, y0) is a point of D (the mean formula). This formula says thatany nonhomogeneous plate has the same mass as a homogenous platewith the same domain D and the constant density equal to the valueof the initial density function at a certain point of D. In particular,

if f is continuous, f(x, y) ≥ 0 and

∫∫

D

f(x, y)dxdy > 0, then f is

not zero at least on a small disc contained in D. Moreover, if f ≥ 0

and

∫∫

D

f(x, y)dxdy = 0, then the set A of points (x, y) for which

f(x, y) > 0 has area equal to zero (the proof of this statement is moredifficult).


f) if ∆ = Di, i = 1, 2, ..., n is a division of D, then∫∫

D

f(x, y)dxdy =n∑

i=1

∫∫

Di

f(x, y)dxdy.

Since we have no effective method to compute a double integral upto the present time, we postpone the work with some examples. Thuswe continue our study by presenting a basic method for computingdouble integrals, namely the "iterative method". The name comes fromthe fact that by using this method we shall reduce the computation ofa double integral to the successive (iterative) calculation of two simpleRiemann integrals. We begin with some prerequisites.

We say that a domain D like above (closed, bounded and ∂D is apiecewise smooth curve) is simple w.r.t. Ox-axis if any straight linex = x0, parallel toOy-axis, x ∈ (a, b), where [a, b] = prOx(D), intersectsthe boundary ∂D in one or two distinct points (see Fig.6). The readercan easily observe that the domain D from Fig.6 is also simple w.r.t.Oy-axis. Let [c, d] = prOy(D) and let Ω = [a, b] × [c, d] = [LMNP ] be"the least" simple rectangle which contains D (see Fig.6). We write

the boundary ∂D as a union of two arcs: ∂D = [ABUC] ∪ [CEV A].

Assume that a parametrization of [ABUC] is:

x = x, x ∈ [a, b]

y = α(x), a

parametrization of [CEV A] is:

x = x, x ∈ [a, b]

y = β(x)and that α(x) ≤

β(x) for any x ∈ [a, b].

F 6


With these notation and definitions one has the following basicresult.

T 79. (iterative general formula) Let D be a simple domainw.r.t. Ox-axis like above and let f : D → R be a piecewise continuousfunction (bounded and continuous on D except a subset of zero area)defined on D. Then

(2.2)

∫∫

D

f(x, y)dxdy =

∫ b

a

(∫ β(x)

α(x)

f(x, y)dy

)dx.

Here I(x) =∫ β(x)α(x)

f(x, y)dy is an integral with a parameter x in its

general form (see formula (1.2)). This formula says that in orderto compute a double integral on this particular type of domain D, wefix an x on [a, b], the projection of D on Ox-axis, cut the domain Dwith the vertical line which passes through x and obtain the segment[UV ], where U(x, α(x)) and V (x, β(x)), compute the simple integral

I(x) =∫ β(x)α(x)

f(x, y)dy with a parameter x, along this segment and fi-

nally compute the next simple integral∫ baI(x)dx. In fact, we just com-

puted the mass I(x) of the segment [UV ], which depends on x, then we

"made the sum"∫ baI(x)dx of all these masses I(x).

P. We intend to use the similar iterative formula for a rec-tangle (see formula (1.2)) for an extension-function f(x, y) : Ω =[LMNP ] → R of f : D → R to Ω. We simply extend f from D tothe rectangle Ω = [a, b] × [c, d] by putting zero in all the points ofΩD :

f(x, y) =

f(x, y), if (x, y) ∈ D0, if (x, y) ∈ ΩD

.

Since this function is continuous on Ω except maybe the points of ∂D,

which has zero area, we can apply theorem 77 and find that f is inte-grable on Ω and∫∫

D

f(x, y)dxdy =

∫∫

Ω

f(x, y)dxdy −∫ ∫

ΩD

0 · dxdy =

∫∫

Ω

f(x, y)dxdy.

But now we can apply the iterative formula (1.2) for the rectangleΩ = [a, b] × [c, d] and find:

∫∫

Ω

f(x, y)dxdy =

∫ b

a

(∫ d

c

f(x, y)dy

)dx =


=

∫ b

a

(∫ α(x)

c

f(x, y)dy +

∫ β(x)

α(x)

f(x, y)dy +

∫ d

β(x)

f(x, y)dy

)dx =

=

∫ b

a

(∫ β(x)

α(x)

f(x, y)dy

)dx.

For instance, let us compute the mass of the plate D with thedensity function f(x, y) = 3xy2, where D is the domain bounded bythe line y = x and the parabola y = x2. Since x ∈ [0, 1] and sincex ≥ x2, when x ∈ [0, 1], we get:

mass(D) =

∫∫

D

3xy2dxdy =

∫ 1

0

x

(∫ x

x23y2dy

)dx =

∫ 1

0

x · y3∣∣xx2dx =

∫ 1

0

x · (x3 − x6)dx =x5

5− x8

8

∣∣∣∣1

0

=1

5− 1

8=

3

40.

R 28. If the domain D is not simple w.r.t. Ox-axis but itis simple w.r.t. Oy-axis (see Fig.7) we obtain the analogous formula:

(2.3)

∫∫

D

f(x, y)dxdy =

∫ d

c

(∫ γ(y)

δ(y)

f(x, y)dx

)dy,

where x = δ(y), y = y is the parametrization of the arc [ XW4SGmY ]

and x = γ(y), y = y is the parametrization of the arc [ YW1TW2nHW3X](Prove this last formula!).

Let for instance the domain D bounded by the following curves:x = y2 + 1, x = 1, and x = 5. Let us compute the coordinates of itsmass centre G(xG, yG).Whenever one gives us no density function, thismeans that the plate D is tacitly considered to be homogenous withthe constant density 1. Moreover, D is symmetric w.r.t. Ox-axis, sothat yG = 0 (prove it by pure calculation!). Since a double integral ona general domain D like above can be well approximated by sums ofdouble integrals on simple rectangles, all the formulas from (1.7) and(1.8) can be used for our nonrectangular domain D. Hence, we need tocompute two double integrals:

mass(D) = area(D) =

∫∫

D

dxdy = 2

∫ 5

1

(∫ √x−1

0

dy

)dx =


F 7

= 2

∫ 5

1

√x− 1dx = 2

(x− 1)32

32

∣∣∣∣∣

5

1

=4

3· 4

32 =

32

3.

Let us compute now the static moment w.r.t. Oy-axis, by using formula(2.3) (decomposing D into two nonoverlaping subdomains, computealso this moment by using formula (2.2)):

MOy(D) =

∫∫

D

xdxdy =

∫ 2

−2

(∫ y2+1

1

xdx

)dy =

=1

2

∫ 2

−2

[(y2 + 1)2 − 1]dy =

∫ 2

0

[y4 + 2y2]dy =24 · 11

15.

Thus,

xG =

∫∫

D

xdxdy

mass(D)=

24·1115323

=33

30.

Prove that this xG is the same like the x-coordinates of the mass centreof the domain D′ bounded by y = 2, x = y2 + 1 and by x = 1.

Let us now compute the moment of inertia IOx(D) of the plate Dfrom Fig.8 with the density function f(x, y) = 2x. The formula is thesame formula for rectangles (see formula (1.8)) extended to a general


domain D (one can deduce this formula directly. Do it!):

IOx(D) =

∫∫

D

y2f(x, y)dxdy =

∫∫

D

2xy2dxdy =

=

∫ 2

0

2x

(∫ 2√

2x

x2y2dy

)dx =

=2

3

∫ 2

0

(16√

2x32 − x6)dx =

2

3

[32√

2

5x52 − x7

7

]∣∣∣∣∣

2

0

=768

35.

F 8

In example 75 we deduced a formula (1.10) to compute the volumeof a cylindrical solid bounded by a rectangular domain D and "above"by a piece of a surface z = f(x, y). The same type of reasoning can beextended for a general domainD like above and the formula remains the

same: vol =

∫∫

D

|f(x, y)| dxdy. Let us apply this formula to compute

the volume of the cylindrical solid (generated by lines parallel to Oz-axis) with the basis the domain D bounded by the coordinate axesOx, Oy, and the line x + y = 1, and limited "above" by the surfacez = x2 + y2. Hence

vol =

∫∫

D

(x2 + y2)dxdy =

∫ 1

0

(∫ 1−x

0

(x2 + y2)dy

)dx =


∫ 1

0

(x2y +

y3

3

∣∣∣∣1−x

0

)dx =

∫ 1

0

(x2(1 − x) +

(1 − x)3

3

)dx =

[x3

3− x4

4− (1 − x)4

12

]∣∣∣∣1

0

=1

3− 1

4+

1

12=

1

6.

E 8. Let D be a simple domain w.r.t. Ox-axis and let [a, b]be its projection on Ox-axis. For any x ∈ [a, b] let Mx(x, α(x)) andNx(x, β(x)), α(x) ≤ β(x) be the points at which the vertical line X = xcuts ∂D, the boundary of D. Let G(x, θ(x)) be the mass centre (thedensity function is 1) of the segment [MxNx]. Let (γ, f) be the wirex = x, y = θ(x), and f(x, θ(x)) = length[MxNx], where x ∈ [a, b] andlet H be its mass centre. Prove that H is the mass centre of D.

E 9. State and comment a Lebesgue type criterion for dou-ble integrals.

E 10. Let f, g : D → R be two piecewise continuous func-tions defined on a domain D like above. Let A = (x, y) ∈ D :f(x, y) = g(x, y). We assume that A has zero area. Prove that∫∫

D

f(x, y)dxdy =

∫∫

D

g(x, y)dxdy.

3. Green formula. Applications.

The following formula connects a line integral of the second typewith a double integral. Such formulas are important because one canreduce the computations from a dimension to an inferior one.

T 80. (Green formula) Let D be a closed bounded domainwith ∂D a direct oriented (trigonometric direction) piecewise smoothand "closed" curve. We also assume that the domain D is simple w.r.t.

both axes. Let−→F (x, y) = P (x, y)

−→i +Q(x, y)

−→j be a plane field of class

C1 defined on D. Then

(3.1)

∮

∂D

Pdx+Qdy =

∫∫

D

(∂Q

∂x− ∂P

∂y

)dxdy.

This formula is called the Green formula.

P. In Fig.9we assume the following parametrizations:

(3.2) arc[ACB] :

x = x

y = ϕ(x), x ∈ [a, b],

3. GREEN FORMULA. APPLICATIONS. 245

F 9

(3.3) arc[BDA] :

x = x

y = ψ(x), x ∈ [a, b]−,

(3.4) arc[DAC] :

x = α(y)y = y

, y ∈ [c, d]−,

(3.5) arc[CBD] :

x = β(y)y = y

, y ∈ [c, d].

a) We firstly prove that −∫∫

D

∂P∂ydxdy =

∮

∂D

Pdx. Indeed,

−∫∫

D

∂P

∂ydxdy = −

∫ b

a

(∫ ψ(x)

ϕ(x)

∂P

∂y(x, y)dy

)dx =

=

∫ b

a

P (x, ϕ(x))dx−∫ b

a

P (x, ψ(x))dx =

=

∮

arc[ACB]

Pdx+

∮

arc[BDA]

Pdx =

∮

∂D

Pdx.

b) We prove now that

∫∫

D

∂Q∂xdxdy =

∮

∂D

Qdx. Indeed,

∫∫

D

∂Q

∂xdxdy =

∫ d

c

(∫ β(y)

α(y)

∂Q

∂x(x, y)dx

)dy =


∫ d

c

Q(β(y), y))dy −∫ d

c

Q(α(y), y))dy =

∮

arc[CBD]

Qdy +

∮

arc[DAC]

Qdy =

∮

∂D

Qdy.

We now add the two equalities obtained in a) and b) and we finally getthe Green formula.

R 29. If our domain D is not simple with respect to oneaxis or to both, we break it into a finite number of subdomains whichare simple w.r.t. both axis and they preserve the other properties of D.Then we apply Green formula for each of these last domains. Let usdo this for the domain D of Fig.10,

F 10

bounded by the arc [ABCDEFGHA] :∫∫

D

(∂Q

∂x− ∂P

∂y

)dxdy =

∫∫

D1

(∂Q

∂x− ∂P

∂y

)dxdy+

+

∫∫

D2

(∂Q

∂x− ∂P

∂y

)dxdy +

∫∫

D3

(∂Q

∂x− ∂P

∂y

)dxdy =

Green=

∮

arc[GHABCEG]

Pdx+Qdy+

∮

arc[EFGE]

Pdx+Qdy+

∮

arc[CDEC]

Pdx+Qdy =


=

∮

arc[GHABC]

Pdx+Qdy +

∮

arc[CE]

Pdx+Qdy +

∮

arc[EG]

Pdx+Qdy+

(3.6) +

∮

arc[EFG]

Pdx+Qdy +

∮

arc[GE]

Pdx+Qdy+

+

∮

arc[CDE]

Pdx+Qdy +

∮

arc[EC]

Pdx+Qdy.

Since

∮

arc[CE]

Pdx+Qdy+

∮

arc[EC]

Pdx+Qdy = 0 and

∮

arc[EG]

Pdx+Qdy+

∮

arc[GE]

Pdx+Qdy = 0, in formula (3.6) it remains

∫∫

D

(∂Q

∂x− ∂P

∂y

)dxdy =

∮

arc[GHABC]

Pdx+Qdy+

∮

arc[CDE]

Pdx+Qdy+

+

∮

arc[EFG]

Pdx+Qdy =

∮

∂D

Pdx+Qdy,

i.e. we obtained the Green formula for the domain D.

E 76. (formulas for areas again). Let us put in theorem80 or in remark 29 Q(x, y) = x and P (x, y) = −y. We get again theknown formula for the computation of an area by using line integralsof the second type:

area(D) =

∫∫

D

dxdy =1

2

∮

∂D

xdy − ydx.

If we put now Q(x, y) = x and P (x, y) = 0, we get

area(D) =

∫∫

D

dxdy =

∮

∂D

xdy.

If we put P (x, y) = −y and Q(x, y) = 0, we get

area(D) =

∫∫

D

dxdy = −∮

∂D

ydx.


E 77. Let in addition D be a simple connected domain.Then the condition ∂Q

∂x= ∂P

∂yimplies the fact that the line integral

∮

arc[AB]

Pdx + Qdy depends only on the points A and B but not on the

geometrical form of the arc [AB] which connect these points A andB. Indeed, if we take a direct oriented closed piecewise smooth curveγ ⊂ D and if we denote by Dγ ⊂ D the subdomain of D bounded by γ(here we need that D be simple connected), by applying Green formulafor Dγ with ∂Dγ = γ, we get:

∮

γ

Pdx+Qdy =

∫∫

Dγ

(∂Q

∂x− ∂P

∂y

)dxdy = 0.

Thus, from remark 26 we get that

∮

arc[AB]

Pdx+Qdy is independent on

the path which connect A and B and, consequently, the field−→F = (P,Q)

is conservative.

E 78. (a practical method for computing the static momentsfor plane domains) In Mechanics we meet the following situation. LetD be a plane domain with a complicated boundary ∂D. Moreover, wedo not know exactly to describe this boundary; it is only supplied by alarge number of points

A0(x0, y0), A1(x1, y1), ..., An(xn, yn)

which are situated on ∂D (see Fig.11).

F 11


Assume that we have to compute the k-th static moment of D w.r.t.

Oy-axis, k = 0, 1, 2, ..., i.e. the number

∫∫

D

xkdxdy. But, since for

Q(x, y) = xk+1

k+1, ∂Q∂x

= xk, we can take Q(x, y) = xk+1

k+1and P = 0 in the

Green formula:∫∫

D

xkdxdy =

∮

∂D

xk+1

k + 1dy.

If n is large enough and if the lengths∥∥∥−−−−→AjAj+1

∥∥∥ are small enough, wecan well approximate ∂D with the polygonal line [A0A1...An−1AnA0].Thus

∫∫

D

xkdxdy =

∮

∂D

xk+1

k + 1dy ≈(3.7)

≈ 1

k + 1

n∑

i=1

∮

[Ai−1Ai]

xk+1dy

+

1

k + 1

∮

[AnA0]

xk+1dy.(3.8)

Let us compute the line integral

∮

[Ai−1Ai]

xk+1dy. A natural parametriza-

tion of the direct oriented segment [Ai−1Ai] is the following:

[Ai−1Ai] :

x = xi−1 + t(xi − xi−1)y = yi−1 + t(yi − yi−1)

, t ∈ [0, 1].

Thus∮

[Ai−1Ai]

xk+1dy =

∫ 1

0

[xi−1 + t(xi − xi−1)]k+1 (yi − yi−1)dt =

=yi − yi−1

xi − xi−1

· 1

k + 2[xi−1 + t(xi − xi−1)]

k+2∣∣∣1

0=

=1

k + 2(yi − yi−1)

xk+2i − xk+2

i−1

xi − xi−1.

Hence

(3.9)

∫∫

D

xkdxdy ≈


≈ 1

(k + 1)(k + 2)

[(y0 − yn)

xk+20 − xk+2

n

x0 − xn+

n∑

i=1

(yi − yi−1)xk+2i − xk+2

i−1

xi − xi−1

].

This last formula is important in practice. For instance, if D is boundedby a polygonal line [A0A1...An−1AnA0] and if k = 0, we get

(3.10) area(D) =1

2

[(y0 − yn)(x0 + xn) +

n∑

i=1

(yi − yi−1)(xi + xi−1)

]

For instance, if n = 2 (a triangle), we get a known formula forcomputing the area of a triangle

area(∆A0A1A2) =1

2

∣∣∣∣∣∣

x0 y0 1x1 y1 1x2 y2 1

∣∣∣∣∣∣.

Here we do not need to put the absolute value for the determinant be-cause the direct trigonometric direction of the sequence of points A0,A1, A2 assures the nonnegativity of this determinant. We propose tothe reader to find nice geometrical or mechanical interpretations forformulas (3.9) and (3.10).

We can even continue to find "nice" formulas. Let them×m matrix

Iij(D) =

∫∫

D

xiyjf(x, y)dxdy

of the generalized moment of inertia for a plate (D, f), where f is thedensity function. If one can well approximate this f by a polynomialP (x, y) in two variables, we can reduce the problem of the calculationof such moments to the computation of double integrals of the following

type

∫∫

D

xiyjdxdy for i, j ∈ 0, 1, 2, .... Applying again Green formula

we get∫∫

D

xiyjdxdy =

∮

∂D

xi+1

i + 1yjdy.

We leave as an exercise to the reader to compute this last integral for adomain D bounded by the above polygonal line [A0A1...An−1AnA0] andto discuss the obtained formula.

4. CHANGE OF VARIABLES IN DOUBLE INTEGRALS. POLAR COORDINATES.251

E 79. Let us use the Green formula (see theorem 80) to

compute the line integral I =

∮

arc[OABO]

xydx+ x2dy (see Fig.12). Thus,

I =

∫∫

D

(2x− x)dxdy =

∫ 1

0

x

(∫ 2x

x

dy

)dx =

∫ 1

0

x2dx =1

3.

F 12

4. Change of variables in double integrals. Polarcoordinates.

When we have to compute a double integral

∫∫

D

f(x, y)dxdy we

need an analytical expression of f and a parameterization of the bound-ary ∂D of the domain D. It is clear that if these ones are complicated,our integral can not easily be computed.

E 80. For instance, let us try to find the mass of the discD = (x, y) : x2 + y2 ≤ 9 with the density function f(x, y) = x2y2.

mass(D) =

∫∫

D

x2y2dxdy =

∫ 3

−3

x2

(∫ √9−x2

−√

9−x2y2dy

)dx =


=2

3

∫ 3

−3

x2 y3∣∣√

9−x20

dx =2

3

∫ 3

−3

x2(9 − x2)√

9 − x2dx.

But, to compute this last simple integral it is not so immediately. Weshall see later that the best idea is to change the disc D with a rectan-gular domain because usually to integrate on rectangles is easier thento integrate on discs. Here is some theoretical support for this strongmethod of computing double integrals.

Let us start with a smooth (of class C1) deformation (or a smoothparametrization sheet) −→g : Ω → D of a closed bounded plane domainΩ onto another closed bounded domain D ("onto" means that −→g (Ω) =D). We also assume that ∂Ω and ∂D are piecewise smooth curves (thusΩ and D have areas!) and that −→g is a diffeomorphism on Ω, i.e. itis invertible and its inverse −→g −1 is also smooth, i.e. of class C1 on D.

If −→g (u, v) = x(u, v)−→i + y(u, v)

−→j , where x(u, v) and y(u, v) are the

component functions of the field −→g , then we simply write:

−→g :

x = x(u, v)y = y(u, v)

, (u, v) ∈ Ω.

Here we must be careful because the same letter x is used for thevariable x and, at the same time, like a name of a function x(u, v) ofvariables u and v. The same is for the letter y. We intentionally didthis because we wanted to show that the variable x was changed witha function x(u, v) of variables u and v and the variable y was changedwith a function y(u, v) of the new variables u and v. This change of the"old" variables x and y with the "new" ones u and v is called a changeof variables. More exactly, the deformation −→g : Ω → D is called achange of variables if it is a diffeomorphism (see Fig.13)

Let us partially divide the domain Ω into rectangles Ωij = [ui−1, ui]×[vj−1, vj ], i = 1, 2, ..., n and j = 1, 2, ...,m. In each rectangle Ωij wechoose a fixed marked point Pij(ξi, ηj). A small rectangle [P0P1P2P3] =[ui−1, ui] × [vj−1, vj] is deformed by −→g into a curvilinear parallelogram[M0M1M2M3] (see Fig.13 and Fig.14). First of all we want to estimatethe area of this last curvilinear parallelogram [M0M1M2M3] in lan-guage of the area of the rectangle [P0P1P2P3] and of the deformationfunctions x(u, v) and y(u, v).

T 81. area([M0M1M2M3]) ≈∣∣J(ξi, ηj)

∣∣·area([P0P1P2P3]),where J(ξi, ηj) is the value of the Jacobian

J(u, v) = det

(∂x∂u

(u, v) ∂x∂v

(u, v)∂y∂u

(u, v) ∂y∂v

(u, v)

)


F 13

of −→g (u, v) computed at the point Pij(ξi, ηj). Here the sign "≈" means"well approximation". This approximation is better and better if thenorms of the divisions generated by ui and vj are smaller andsmaller, i.e. if ui − ui−1 → 0 and vj − vj−1 → 0 for any i and j.

The above formula is usually written in a formal way:

(4.1) dxdy = |J(u, v)| dudv.We say that the initial element of area dudv (which is the area ofa small rectangle!), after the deformation process generated by −→g , is"dilated" and becomes |J(u, v)| dudv. The dilation (or contraction) co-efficient is exactly |J(u, v)| in a small neighborhood of the point (u, v).This means that the changes of areas are perfectly controlled by theJacobian matrix of the deformation −→g .

P. We approximate the area of the curvilinear parallelogram[M0M1M2M3] with the area of the parallelogram [M0M1M2M

∗3 ],

generated by the vectors−−−−→M0M1 and

−−−−→M0M2 (see Fig.14).

But the area of this last parallelogram is equal to∥∥∥−−−−→M0M1 ×

−−−−→M0M2

∥∥∥ .Let compute

−−−−→M0M1 ×

−−−−→M0M2. Firstly we have:

−−−−→M0M1 = [x(ui, vj−1) − x(ui−1, vj−1)]

−→i +[y(ui, vj−1) − y(ui−1, vj−1)]

−→j =

=

[∂x

∂u(ci, vj−1)

−→i +

∂y

∂u(c′i, vj−1)

−→j

](ui − ui−1).

Here we used Lagrange formula for functions t x(t, vj−1) and t y(t, vj−1) on the interval [ui−1, ui]. Since the length of this last intervalis small enough, since both functions just defined are of class C1, since


F 14

ci, c′i, ξi ∈ [ui−1, ui] and since ηj ∈ [vj−1, vj ], we can further make the

following approximation:

(4.2)−−−−→M0M1 ≈

[∂x

∂u(ξi, ηj)

−→i +

∂y

∂u(ξi, ηj)

−→j

](ui − ui−1).

Secondly one gets:−−−−→M0M2 = [x(ui−1, vj) − x(ui−1, vj−1)]

−→i +[y(ui−1, vj) − y(ui−1, vj−1)]

−→j =

=

[∂x

∂v(ui−1, dj)

−→i +

∂y

∂v(ui−1, d

′j)−→j

](vj − vj−1).

Here we used Lagrange formula for functions s x(ui−1, s) and s y(ui−1, s) on the interval [vj−1, vj]. Since the length of this last intervalis small enough, since both functions just defined are of class C1, sincedj , d

′j , ηj ∈ [vj−1, vj ] and since ξi ∈ [ui−1, ui], we can further make the

following approximation:

(4.3)−−−−→M0M2 ≈

[∂x

∂v(ξi, ηj)

−→i +

∂y

∂v(ξi, ηj)

−→j

](vj − vj−1).

Hence −−−−→M0M1 ×

−−−−→M0M2 ≈

≈[∂x

∂u(ξi, ηj) ·

∂y

∂v(ξi, ηj) −

∂x

∂v(ξi, ηj) ·

∂y

∂u(ξi, ηj)

](ui−ui−1)(vj−vj−1)

−→k .

Thus ∥∥∥−−−−→M0M1 ×−−−−→M0M2

∥∥∥ ≈


≈∣∣∣∣∂x

∂u(ξi, ηj) ·

∂y

∂v(ξi, ηj) −

∂x

∂v(ξi, ηj) ·

∂y

∂u(ξi, ηj)

∣∣∣∣ area([P0P1P2P3]) =

=∣∣J(ξi, ηj)

∣∣ area([P0P1P2P3]).

Since area([M0M1M2M3]) ≈∥∥∥−−−−→M0M1 ×

−−−−→M0M2

∥∥∥ , we finally get:

area([M0M1M2M3]) ≈∣∣J(ξi, ηj)

∣∣ area([P0P1P2P3]),

i.e. the statement of the theorem.

T 82. (change of variables formula) Let us preserve theabove notation and hypotheses. Let f : D → R be a piecewise continu-ous function defined on D. Then

(4.4)

∫∫

D

f(x, y)dxdy =

∫∫

Ω

f(x(u, v), y(u, v)) |J(u, v)| dudv.

P. Let us observe that the union of the nonoverlaping subdo-mains Dij = −→g (Ωij), where Ωij = [ui−1, ui] × [vj−1, vj], i = 1, 2, ..., nand j = 1, 2, ...,m, well approximate (when the diameters diam(Ωij) ofΩij go to zero) the entire domain D (see Fig.13). Thus, using theorem81 and the definition of double integrals as the limit of double Riemannsums, we get:

∫∫

D

f(x, y)dxdy ≈n∑

i=1

m∑

j=1

f(Mij)area(Dij) ≈

≈n∑

i=1

m∑

j=1

f(x(ξi, ηj), y(ξi, ηj)) ·∣∣J(ξi, ηj)

∣∣ · (ui − ui−1)(vj − vj−1) ≈

≈∫∫

Ω

f(x(u, v), y(u, v)) |J(u, v)| dudv.

Hence the two fixed numbers

∫∫

D

f(x, y)dxdy and

∫∫

Ω

f(x(u, v), y(u, v)) |J(u, v)| dudv become closer and closer! There-

fore they must coincide, i.e.∫∫

D

f(x, y)dxdy =

∫∫

Ω

f(x(u, v), y(u, v)) |J(u, v)| dudv,

what we wanted to prove.


Formula (4.4) is called "change of variables formula" and it is ex-tremely useful in practice.

The most popular change of variables are the change of the Carte-sian coordinates (x, y) with the polar coordinates (ρ, θ) :

(4.5)

x = ρ cos θy = ρ sin θ

,

where ρ goes from 0 (without it!) up to ∞ and θ ∈ [0, 2π). The defor-mation induced by this change of variables is −→g : (0,∞) × [0, 2π) →R2(0, 0), −→g (ρ, θ) = (ρ cos θ, ρ sin θ). It is important because it de-forms the interior of a rectangle Ω = (0, R] × [0, 2π), R > 0, into theinterior of the disc B((0, 0), R) = (x, y) : x2 + y2 ≤ R2. Since somefinite length smooth curves do not count for us during the double inte-gration process, we will simply say that any disc in a fixed plane is the

image after a deformation

x = x0 + ρ cos θy = y0 + ρ sin θ

, ρ ∈ (0, R), θ ∈ [0, 2π),

of a rectangle from the (ρ, θ)-plane. Here (x0, y0) is the centre of thedisc and R is its radius. The first one (see formula (4.5)) is a particularcase of this last more general polar coordinates change. The Jacobianof the polar coordinates change is equal to

det

(cos θ −ρ sin θsin θ ρ cos θ

)= ρ = 0.

(see also [Po], Ch.11, Section 6). Hence formula (4.1) becomes:

(4.6) dxdy = ρdρdθ.

For instance, let consider again the double integral of example 80 andlet us change the Cartesian variables (x, y) with the polar coordinates(ρ, θ) (see formula (4.5)). Here dxdy = ρdρdθ. Thus, formula (4.4) canbe applied:

∫ ∫

D: x2+y2≤9

x2y2dxdy =

∫ 3

0

(∫ 2π

0

ρ4 cos2 θ sin2 θdθ

)ρdρ =

=

∫ 3

0

ρ5(∫ 2π

0

cos2 θ sin2 θdθ

)dρ =

(1

4

∫ 2π

0

sin2 2θdθ

)∫ 3

0

ρ5dρ =

=

(∫ 2π

0

1 − cos 4θ

2dθ

)· 35

4=

35π

4.

We can remark how easy become the calculations if we changed thevariables!


E 81. Let us compute the inertia moment of the ellipsex2

a2+ y2

b2≤ 1 w.r.t. Oy-axis. Since nothing is said on the density

function, we take f(x, y) = 1. If one tries to compute the double integral

which appear here,

∫ ∫

D: x2

a2+ y2

b2≤1

x2dxdy, by using only iterative formulas,

one will see that the calculations becomes very difficult. Let us try touse the following "general polar coordinates" change of variables:

x = aρ cos θy = bρ sin θ

, ρ ∈ (0, 1], θ ∈ [0, 2π).

It is easy to see that in this last case J = abρ and our integral becomes(we simply apply change of variables formula (4.4):

∫ ∫

D: x2

a2+ y2

b2≤1

x2dxdy =

∫ 1

0

(∫ 2π

0

ρ2 cos2 θdθ

)abρdρ =

= ab

∫ 1

0

ρ3

(∫ 2π

0

cos2 θdθ

)dρ = ab

(∫ 1

0

ρ3dρ

)(∫ 2π

0

cos2 θdθ

)=

= ab1

4

∫ 2π

0

1 + cos 2θ

2dθ =

πab

4.

E 82. The area of the ellipse x2

a2+ y2

b2≤ 1 can be immediately

computed by using the above change of variables.

area =

∫ ∫

D: x2

a2+ y2

b2≤1

dxdy =

∫ 1

0

(∫ 2π

0

dθ

)abρdρ = πab.

In particular, the area of a disc of radius R is equal to πR2.

E 83. Sometimes the disc has not its centre in the originof the axes. For instance, let us compute the mass centre of the discD : x2 + y2 + 4y ≤ 0 if the density function is f(x, y) =

√x2 + y2.

Since the equation of the disc can be written as x2 + (y + 2)2 ≤ 4, wesee that the centre of the disc is in C(0,−2) and its radius is equal to2. It is symmetric w.r.t. Oy-axis, geometrically and also from the pointof view of statics (the mass is symmetrically expanded w.r.t. Oy-axis).


Hence xG = 0. Let us compute

yG =

∫∫

D

y√x2 + y2dxdy

∫∫

D

√x2 + y2dxdy

.

To compute these two double integrals we shall use the common polarcoordinates (because the circle ∂D contains the origin! Otherwise thismethod fails!). Let us write the Cartesian equation of the disc in thelanguage of polar coordinates: ρ2 cos2 θ + ρ2 sin2 θ + 4ρ sin θ ≤ 0, orρ+ 4 sin θ ≤ 0, or 0 < ρ ≤ −4 sin θ. This time, for any fixed value of θ,ρ goes between 0 and −4 sin θ (Draw and look at it!). Now θ itself goesbetween π and 2π. Hence∫∫

D

√x2 + y2dxdy =

∫ 2π

π

(∫ −4 sin θ

0

ρ · ρdρ)dθ =

∫ 2π

π

ρ3

3

∣∣∣∣−4 sin θ

0

dθ =

= −64

3

∫ 2π

π

sin3 θdθ =64

3

∫ 2π

π

(1 − cos2 θ)d(cos θ) =

=64

3

(cos θ − cos3 θ

3

)∣∣∣∣2π

π

=256

9.

Now, we look at the last calculations and make some slight modificationsin order to compute the double integral in the numerator:

∫∫

D

y√x2 + y2dxdy =

∫ 2π

π

(∫ −4 sin θ

0

ρ2 sin θ · ρdρ)dθ =

∫ 2π

π

sin θ

(ρ4

4

∣∣∣∣−4 sin θ

0

)dθ =

256

4

∫ 2π

π

sin5 θdθ =

= −64

∫ 2π

π

(1 − cos2 θ)2d(cos θ) =

= −64

∫ 2π

π

(1 − 2 cos2 θ + cos4 θ)d(cos θ) =

−64

[cos θ − 2

3cos3 θ +

1

5cos5 θ

]∣∣∣∣2π

π

= −64 · 16

15.

Thus yG =−64· 16

152569

= −125.


E 84. Sometimes the centre of the disc is not in the originof the axes and the boundary of the disc does not passes through theorigin. For instance, let us compute the mass of the disc D : (x −1)2 + (y + 3)2 ≤ 1 if the density function is f(x, y) = x. Let us use thefollowing change of variables:

x = 1 + ρ cos θy = −3 + ρ sin θ

, ρ ∈ (0, 1), θ ∈ [0, 2π).

Thus,

mass(D) =

∫∫

D

xdxdy =

∫ 1

0

(∫ 2π

0

(1 + ρ cos θ) dθ

)ρdρ = π.

E 85. Let us compute for instance the mass of the plate[A0A1A2A3] with the density function f(x, y) = y (see Fig.15).

F 15

Since−−−→A0A1 = 10

−→i + 3

−→j and

−−−→A2A3 = 10

−→i + 3

−→j as free vectors,

we conclude that the domain D = [A0A1A2A3] is a parallelogram. But,

to compute the mass of D, i.e. the double integral

∫∫

D

ydxdy with the

iterative formulas, it is not so easy. Let us try to change this "com-plicated" domain into a rectangle, on which it will be easy to integrate.We need some knowledge from a Linear Algebra course. Let us startwith the square Ω = [0, 1] × [0, 1] and let us try to find a "translated"linear mapping −→g : R2 → R2 which "without the translation part"is an isomorphism of vector spaces and such that −→g (Ω) = D. Thus


−→g (u, v) = (au+bv+e, cu+dv+h) (because of the linearity of the com-ponent functions of −→g without the translation) and we only need to de-termine the six numbers a, b, c, d, e, h and then to prove that −→g (Ω) = D.We even force that the corners of the square Ω = [0, 1] × [0, 1] to goexactly to the "corresponding" corners of the parallelogram (our feelingis that −→g is a dilation of axes + a translation + a rotation + a lin-ear deformation of angles!). Hence, the conditions: −→g (0, 0) = (3, 2),−→g (1, 0) = (13, 5), −→g (0, 1) = (4, 6) and −→g (1, 1) = (14, 9) (see Fig.6.15)supply us with the values of a, b, c, d, e, h. In fact we do not need thelast equality because such a function −→g carries collinear vectors intocollinear vectors (Why?). Namely, a = 10, b = 1, c = 3, d = 4, e = 3and h = 2. Thus −→g (u, v) = (10u+ v+ 3, 3u+ 4v+ 2). Since the matrix

T =

(10 13 4

)of the linear part (u, v) → (au+bv, cu+dv) of −→g is in-

vertible, we see that −→g is a diffeomorphism from Ω to D, in particular−→g is a smooth deformation (without "defects") of Ω into D. Moreover,it is also reversible, i.e. "one can come back in the same smoothly man-ner"). The Jacobian of −→g is exactly the determinant of the matrix T.Theorem 81 says that during the deformation process induced by −→g thearea of Ω increases detT = 36 times. But why −→g (Ω) = D? Let us takea pointM ∈ [0, 1]×[0, 1] and let us write its position vector w.r.t. a fixed

Cartesian system uO′v :−−→O′M = α

−→i′ + β

−→j′ where 0 ≤ α, β ≤ 1. We

need to prove that −→g (M) ∈ D, i.e. that (10α+β+3, 3α+4β+2) ∈ D,or, equivalently, that V = (10α + β, 3α + 4β), as a free vector in thexOy-Cartesian system can be decomposed as a linear combination of

vectors−−−→A0A1 = 10

−→i + 3

−→j and

−−−→A0A2 =

−→i + 4

−→j with subunitary

coefficients. Indeed, let us write:

(10α+ β)−→i + (3α + 4β)

−→j = λ(10

−→i + 3

−→j ) + η(

−→i + 4

−→j ),

or 10α + β = 10λ+ η3α + 4β = 3λ+ 4η

.

Thus λ = α ∈ [0, 1] and η = β ∈ [0, 1], i.e. −→g (M) ∈ D. Now we areready to compute the mass of the plate D. The Jacobian is equal to 36,thus

mass(D) =

∫∫

D

ydxdy =

∫ ∫

[0,1]×[0,1]

(3u+ 4v + 2) · 36dudv =

= 36

∫ 1

0

(∫ 1

0

(3u+ 4v + 2)dv

)du = 36

∫ 1

0

(3u+ 4) du =


= 36

(3

2+ 4

)= 18 · 11 = 18 · (10 + 1) = 198.

Until this moment we considered only double integrals on boundeddomains. A similar theory can be done for double integrals on un-bounded domains (see also the improper integrals of the first type).

For instance, let us compute again (we computed it with the helpof an integral with a parameter) and almost immediately the famousPoisson’s integral I =

∫∞0e−x

2dx.

I2 =

∫ ∞

0

e−x2

dx ·∫ ∞

0

e−y2

dy =

∫ ∫

D: x≥0,y≥0

e−x2−y2dxdy =

x=ρ cos θ=

y=ρ sin θ

∫ ∞

0

ρe−ρ2

(∫ π2

0

dθ

)dρ =

π

2

(−1

2e−ρ

2

)∣∣∣∣∞

0

=π

4.

Thus I =√π2. The integral

∫ ∫

D: x≥0,y≥0

e−x2−y2dxdy is an improper dou-

ble integral of the first type (the function is bounded but the domainis unbounded). Formula (4.5) and the theory of improper simple inte-grals are sufficient for the practical problems related to improper doubleintegrals.

For instance, let us consider the following improper double integral

of the second type J =

∫ ∫

D: x2+y2≤1

13√x2+y2

dxdy. In this case the domain

is bounded but the function f(x, y) = 13√x2+y2

is not bounded at (0, 0).

Like in the situation of an improper simple integral of the second type,we isolate the singularity by a neighborhood (a small disc in the 2-Dcase) Dε : x2 + y2 ≤ ε2, for a small ε > 0 and write:

J = limε→0

∫ ∫

DDε: ε2≤x2+y2≤1

13√x2 + y2

dxdy =

x=ρ cos θ=

y=ρ sin θlimε→0

∫ 1

ε

(∫ 2π

0

ρ−23ρdθ

)dρ = 2π lim

ε→0

[ρ43

43

]∣∣∣∣∣

1

ε

=3

2π.

Here we used the formula dxdy = ρdρdθ. Pay attention, do not forgetρ, the Jacobian of the deformation −→g , i.e. the dilation coefficient ofareas when one passes from rectangles to discs!


R 30. Sometimes we encounter the following type of prob-lems. We know that the area of a domain D is computed as follows:

(4.7) σ(D) =

∫ √154

−√154

(∫ 3−√

4−y2

1−√

1−y2dx

)dy.

a) The first problem is to describe the domain D. b) The second problemis to change the order of integration in formula (4.7). Looking at the

iterated integrals of (4.7) we see that while y goes between −√

154and√

154, for any fixed value of y in this interval, x ∈ [1 −

√1 − y2, 3 −√

4 − y2], i.e.

1 −√

1 − y2 ≤ x ≤ 3 −√

4 − y2.

This double inequality can be decomposed into two inequalities: (x −1)2 + y2 ≤ 1, i.e. the disc with centre at C1(1, 0) and radius r = 1and (x − 3)2 + y2 ≥ 4, i.e. the exterior of the open disc with centre atC2(3, 0) and radius R = 2. To change the order of integration it is notso easy. First of all we see that the domain D described by the last twoinequalities:

D :

(x− 1)2 + y2 ≤ 1(x− 3)2 + y2 ≥ 4

is simple w.r.t. the Oy-axis but it is not simple w.r.t. the Ox-axis.However, it is symmetric w.r.t. Ox-axis. Hence the area of D is twotimes the area of the superior part of D. i.e. the domain D∗ :

D∗ :

(x− 1)2 + y2 ≤ 1(x− 3)2 + y2 ≥ 4

, y ≥ 0.

Even if this last domain is not simple with respect to the Ox-axis (makea drawing and see this!). In fact D∗ can be decomposed into a unionof two nonoverlaping domains D∗

1 and D∗2 corresponding to x ∈ [0, 1]

and to x ∈ [1, 54] respectively (prove this!). By making explicit the

expression of y as a function of x in the equalities (x − 1)2 + y2 = 1and (x− 3)2 + y2 = 4 we finally get:

σ(D) = 2σ(D∗) = 2

[∫ 1

0

(∫ √2x−x2

0

dy

)dx+

∫ 54

1

(∫ √2x−x2

√6x−x2−5

dy

)dx

].

Hence we succeeded to change the order of integration in formula (4.7).A nice challenge for the reader is to obtain the same thing but, withoutusing the explicit drawing of D, i.e. to use only the algebraic descriptionof D!



1. Compute the following double integrals:

a)

∫ ∫

x∈[0,1], y∈[0,1]

ex+ydxdy; b)

∫ ∫

x∈[0,∞), y∈[1,2]

e−xydxdy;

c)

∫∫

D

xy2dxdy, where D is the domain bounded by y = 2x and

y = x2; d)

∫∫

D

x2y2dxdy, where D is the finite domain bounded by

x = 0, y = 0 and 2x− 3y + 1 = 0.2. Compute the mass of the lamina D bounded by y = x2 and

y = 2√x, where x ∈ [0, 1], if the density function is f(x, y) =

√xy.

3. Compute the static moments of the plate D :

x2 + y2 ≤ 4

3y ≥ x2

relative to both axes and then find the coordinates of its mass centre.4. Find the mass and the mass centre of the disc D : x2 + y2 ≤ 25

with the density function f(x, y) = 1 + |y| .5. Find the mass and the inertia moment w.r.t. O of the homoge-

nous ellipse x2

a2+ y2

4a2≤ 1 if f(x, y) = 3kg/m2.

6. Find the inertia moment of the plate D : x2 + y2 − 3x ≤ 0 w.r.t.Ox-axis if the density function is f(x, y) = 1√

x2+y2+1.

7. Compute

∫∫

D

ln(x2+y2)x2+y2

dxdy, for D : 1 ≤ x2 + y2 ≤ e2. But for

D′ : 0 ≤ x2 + y2 ≤ e2?8. Find the static moment of D : x2 + y2 ≤ 2, x ≤ y w.r.t. Oy-axis.9. Compute the mass of the disc x2 + y2 − x− y ≤ 0 if the density

function is f(x, y) = x+ y.10. Let A(1,−

√3), B(1,

√3) and C(−2, 0) be three points in the

xOy-plane. Compute the work of the field−→F (x, y) = x3−→i −y3−→j along

the direct oriented (trigonometric way!) curve [ABCA], by using Greenformula.

11. Compute directly and then by using Green formula the staticmoment of order 3 w.r.t. 0 of the square-plate bounded by the squareABCD, where A(0,−1), B(1, 0), C(0, 1) and D(−1, 0). (like usuallytake f = 1!).

12. Use Green formula (or any other formula) to compute


∮

arc[ABCOA]+

ydx− xdy if A(4,−2), B(4, 0), C(4, 2), [COA] is an arc

of the parabola x = y2, [AB] is a segment of a straight line and arc[BC]is an arc of the circle (x− 4)2 + (y − 1)2 = 1, x ≥ 4 which connect Band C.

13. Let A(i, 2i+ 1), i = 1, 2, ..., n and An+1(n+ 1, 0). Use exam-

ple 78 like a model to compute

∫∫

D

xdxdy,

∫∫

D

ydxdy and

∫∫

D

xydxdy,

where D is the domain bounded by the closed polygonal line[OA1A2...AnAn+1O].14. Find the areas bounded by the following curves:

a)

y = sin xy = 0

, x ∈ [0, 2π]; b) y2 = ax, x2 = by, a > 0, b > 0; c)

|x| + |y| = 1; d)

x2 + y2 = R2

x2

a2+ y2

b2≤ 1

(discussion on a, b, R).

15. Use an appropriate change of variables to compute

∫∫

D

xydxdy,

where:a) D is the parallelogram generated by the vectors

−→OA and

−−→OB,

with A(1, 2) and B(2, 1).b) D is the domain bounded by the lines: y = x, y = 2x, x+ y = 1

and x+ y = 4.c) D is the domain bounded by the curves: x = y2, x = 4y2, x = 1,

y > 0 and x = 4.

CHAPTER 7

Triple integrals

1. What is a triple integral on a parallelepiped?

Double integrals

∫∫

D

f(x, y)dxdy are connected with plane domains

D, i.e. with 2-dimensional objects. When we compute such an integralwe always finally come to the computation of two simple integrals.

If one substitutes plane domains with domainsD in the 3-dimensionalspace R3, one gets a more sophisticated notion, namely the notion of a

triple integral

∫∫∫

D

f(x, y, z)dxdydz. Here dxdydz is called an element

of volume and f(x, y, z) stands for "the density" function, etc. Sinceall the theory of such triple integrals is nothing else then a slight gener-alization of the theory of double integrals, we do not insist more on it.Thus some proofs will be omitted in this chapter, because the readercan easily recover them by a simple imitation of the correspondingproofs made by us in the case of the double or simple integrals.

D 25. By a space domain D we mean a subset D of R3

which is closed, bounded, connected and its boundary ∂D is a piecewisesmooth surface of class C1 (see bellow the definition of this notion orlook in any book of Differential Geometry). A couple Ω = (D, f) ofa space domain D and a piecewise continuous function f : D → R issaid to be a solid in R3. Here a piecewise continuous function f definedon D is a function defined on D, which is continuous except a subsetA of D such that vol(A) = 0 (we bellow recall the notion of a volumeor a measure in R3. See also the general definition 7). We say that apiecewise smooth surface of class C1 is simply the image −→g (W ) of afield −→g : W → R3, where W is a plane domain, such that −→g is of classC1 except a subset B of W of zero area (i.e. area(B) = 0). If f = 1 wealso call D a solid. Sometimes, even we do not specify anything aboutf, we call such a space domain D a solid. The function f is called adensity function because it associates to any point M(x, y, z) of D, anumber f(x, y, z), "the density of the solid at M".

265

266 7. TRIPLE INTEGRALS

For other definitions on topological notions in R3 we send the readerto any course of Differential Calculus (Multivariable Analysis), for in-stance to [Po].

A parallelepiped domain D is a space domain of the form D =[a, b]×[c, d]×[e, g]. In Fig.1 we see such a domain Ω = [ABCEA′B′C ′E′]and its projections on the coordinates planes xOy, yOz and zOx.

F 1

Let now D = [a, b] × [c, d] × [e, g] be a parallelepiped space domainand let us consider the following divisions ∆x : a = x0 < x1 < ... <xn = b, ∆y : c = y0 < y1 < ... < ym = d, ∆z : e = z0 < z1 <... < zp = g of the segments [a, b], [c, d] and [e, g] respectively. Let∆ = ∆x × ∆y × ∆z = Dijk = [xi−1, xi] × [yj−1, yj ] × [zk−1, zk] be the3-dimensional division of D generated by ∆x, ∆y and ∆z (see Fig.2).

Let

Pijk(ξi, ηj, θk), ξi ∈ [xi−1, xi], ηj ∈ [yj−1, yj ], θk ∈ [zk−1, zk]

(i.e. Pijk ∈ Dijk) be a fixed set of marking points of the division ∆ andlet

‖∆‖ = maxi,j,k

diam(Dijk)be the norm of the division ∆. Let f : D → R be a density functiondefined on D. If ‖∆‖ is small enough, we can well approximate themass of the solid (Dijk, f |Dijk

) by the number f(ξi, ηj , θk) · vol(Dijk),i.e. with the density at the fixed point Pijk multiplied by the volume

1. WHAT IS A TRIPLE INTEGRAL ON A PARALLELEPIPED? 267

F 2

of Dijk. Thus, the entire mass of D can be well approximated by thefollowing triple Riemann sum:

(1.1) Sf(∆; Pijk(ξi, ηj, θk)) =n∑

i=1

m∑

j=1

p∑

k=1

f(ξi, ηj, θk) · vol(Dijk),

which corresponds to the density function f, to the division ∆ and tothe set of marking points Pijk, i = 1, 2, ..., n, j = 1, 2, ...,m, k =1, 2, ..., p.

D 26. We say that a function f : D = [a, b]× [c, d]× [e, g]→ R is (Riemann) integrable on D is there exists a real number I suchthat for any ε > 0, there exists a real number δε (depending on ε) withthe property that

∣∣I − Sf(∆; Pijk(ξi, ηj , θk))∣∣ < ε

for any division ∆ with ‖∆‖ < δε and for any set of marking points

Pijk(ξi, ηj , θk) of∆. This number I is denoted by

∫∫∫

D

f(x, y, z)dxdydz

and it is called the triple integral of f on the domain D. Here dxdydz iscalled an element of volume and it symbolizes a small volume vol(Dijk).In general, dxdydz is not equal to the product dx · dy · dz. But here itis so, because the volume of a small parallelepiped Dijk is "virtually"


equal to dx · dy · dz. We see that I =

∫∫∫

D

f(x, y, z)dxdydz is the mass

of the solid (D, f) and vol(D) =

∫∫∫

D

dxdydz because for f = 1 each

Riemann sum Sf (∆; Pijk(ξi, ηj , θk)) is equal to vol(D).

It is easy to see that the above defined number I is unique.Introducing Darboux sums like in the case of the double or simple

integrals, it is not difficult to prove a Darboux type criterion and thefollowing basic result.

T 83. (iterative formulas for parallelepipeds) LetD = [a, b]×[c, d] × [e, g] be a parallelepiped space domain and let f : D → R be apiecewise continuous (see the definition above) function defined on D.Then f is Riemann integrable on D and

(1.2)

∫∫∫

D

f(x, y, z)dxdydz =

∫ b

a

(∫ d

c

(∫ g

e

f(x, y, z)dz

)dy

)dx.

Moreover, the order of integration in this formula can be changed. Thisformula says that first of all we compute the simple integral with twoparameters x and y, J(x, y) =

∫ gef(x, y, z)dz, then we compute the sim-

ple integral K(x) =∫ dcJ(x, y)dy with one parameter x, and finally we

compute∫ baK(x)dx and find the value of

∫∫∫

D

f(x, y, z)dxdydz. This

means that the computation of a triple integral on a parallelepiped do-main reduces to the calculation of three simple integrals, step by step,i.e. in an iterative style.

P. For simplifying the reasoning, we assume that f is contin-uous on D. Let us denote

T =

∫ b

a

(∫ d

c

(∫ g

e

f(x, y, z)dz

)dy

)dx

and

I =

∫∫∫

D

f(x, y, z)dxdydz.

Since, by definition, I exists if it is the unique limit point of the Rie-mann sums Sf (∆; Pijk(ξi, ηj , θk)), we shall prove that T itself is theunique limit point of such type of sums. So I exists and it will be equalto T. Let us preserve the notation from the statement of the theorem.

1. WHAT IS A TRIPLE INTEGRAL ON A PARALLELEPIPED? 269

Since J(x, y) =∫ gef(x, y, z)dz can be well approximated by Riemann

sums of the form:

J(x, y) ≈p∑

k=1

f(x, y, θk)(zk − zk−1),

we see that K(x) =∫ dcJ(x, y)dy can be well approximated by sums of

the following type:

(1.3) K(x) ≈p∑

k=1

(∫ d

c

f(x, y, θk)dy

)(zk − zk−1).

Each integral∫ dcf(x, y, θk)dy can be well approximated by Riemann

sums of the type:∫ d

c

f(x, y, θk)dy ≈mk∑

j=1

f(x, η(k)j , θk)(y

(k)j − y

(k)j−1).

We put together all the divisions ∆(k)y : c = y

(k)0 < ... < y

(k)mk = d, k =

1, 2, ..., p and form a bigger division ∆y : c = y0 < y1 < ... < ym = d.Hence

K(x) ≈p∑

k=1

m∑

j=1

f(x, ηj , θk)(yj − yj−1)(zk − zk−1).

Now, T =∫ baK(x)dx can be well approximated by sums of the type:

T ≈p∑

k=1

m∑

j=1

(∫ b

a

f(x, ηj , θk)dx

)(yj − yj−1)(zk − zk−1).

But ∫ b

a

f(x, ηj, θk)dx ≈njk∑

i=1

f(ξ(jk), ηj, θk)(x(jk)i − x

(jk)i−1 ).

By putting together all the divisions ∆(jk)x : a = x

(jk)0 < ... < x

(jk)njk = b,

j = 1, 2, ...,m, k = 1, 2, ..., p we get that

T ≈p∑

k=1

m∑

j=1

n∑

i=1

f(ξi, ηj, θk)(xi − xi−1)(yj − yj−1)(zk − zk−1) =

=n∑

i=1

m∑

j=1

p∑

k=1

f(ξi, ηj, θk) · vol(Dijk),

because of the commutative and associative properties of the usual ad-dition of real numbers. Hence T can be well approximated by Riemannsums of the type (1.1), i.e. I = T.


E 86. Let us compute the mass of the unitary cube D =[0, 1]× [0, 1]× [0, 1] with the density function f(x, y, z) = 3xyz+1. Theiterative method described above will be used.

mass(D) =

∫∫∫

D

(3xyz + 1)dxdydz =

=

∫ 1

0

(∫ 1

0

(∫ 1

0

(3xyz + 1)dz

)dy

)dx =

=

∫ 1

0

(∫ 1

0

[3xyz2

2+ z

]∣∣∣∣z=1

z=0

dy

)dx =

∫ 1

0

(∫ 1

0

[3xy

2+ 1

]dy

)dx =

∫ 1

0

[3xy2

4+ y

]∣∣∣∣y=1

y=0

dx =

=

∫ 1

0

[3x

4+ 1

]dx =

[3x2

8+ x

]∣∣∣∣x=1

x=0

=11

8.

Formulas for mass centres, static moments relative to coordinatesplanes and inertia moments of solids can be immediately written by asimple analogy with the similar formulas for laminas. Let (D, f) be asolid, where D is a parallelepiped space domain. Then the coordinatesof the mass centre G of D are:

(1.4) xG =

∫∫∫

D

xf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

, yG =

∫∫∫

D

yf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

,

zG =

∫∫∫

D

zf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

.

If the solid (D, f) is rotating around Oz-axis say, then the inertia mo-

ment Iz =

∫∫∫

D

(x2 + y2)f(x, y, z)dxdydz, etc. All of these formulas

can be deduced exactly like in the case of double integrals.

2. TRIPLE INTEGRALS ON A GENERAL DOMAIN. 271

2. Triple integrals on a general domain.

A general domain is a closed, connected and bounded domain Din R3 with its boundary ∂D a piecewise smooth surface (see definition25). We recall the notion of a volume or measure associated to adomain D. By an elementary domain in R3 we mean a finite unionof parallelepiped domains Di = [ai, bi] × [ci, di] × [ei, gi], i = 1, 2, ..., nsuch that if i = j, the intersection Di ∩ Dj is a subset of R3 withoutinterior points, i.e. it is not possible to find an open ball B(M0, r) =M(x, y, z) :

∥∥∥−−−→M0M∥∥∥ < r

which is included in Di∩Dj. Look in Fig.3

such an elementary domain Ω.

F 3

The volume of such an elementary domain is equal, by definition,to the sum of all volumes of the component parallelepipeds. Let usrecall definition 7 of a volume (measure in the 3-dimensional case).

D 27. A subset D of R3 has a volume vol(D) ∈ R+ if forany small real number ε > 0 there exist two elementary domains Eintand Eext such that:

1) Eint ⊂ D ⊂ Eext,2) vol(Eext) − vol(Eint) < ε.3) vol(Eint) ≤ vol(D) ≤ vol(Eext).

This means that a subset D ⊂ R3 has a volume vol(D), a nonneg-ative real number, if and only if there exist two towers of elementary

domains E(j)int and E(j)

ext such that(2.1)

E(1)int ⊂ E

(2)int ⊂ ... ⊂ E

(k)int ⊂ ... ⊂ D ⊂ ... ⊂ E

(m)ext ⊂ ... ⊂ E

(2)ext ⊂ E

(1)ext,


vol(E(m)ext ) ց vol(D) and vol(E

(k)int) ր vol(D) asm and k are convergent

to ∞ (become larger and larger). Since vol(E(m)ext ) − vol(E

(m)int ) → 0 as

m → ∞, vol(D) is the unique real number which is contained in all

intervals [vol(E(m)int ), vol(E

(m)ext )], m = 1, 2, .... (see Cantor’s axiom in

[Po]).Using this definition, it is not difficult to prove that a finite set

of points, of segments of finite length smooth curves, or of smoothsurfaces, have volume zero. In particular, the boundary ∂D of a generaldomain D has volume zero. As a consequence, a general domain D hasa finite volume, vol(D). It is also easy to see that a bounded subset Aof R3, which has no interior points, has a volume vol(A) and this lastone is equal to zero.

The volume function D vol(D) is an additive function, i.e.vol(D1∪D2) = vol(D1)+vol(D2)−vol(D1∩D2) and it is an increasingfunction, i.e. D ⊂ D′ implies vol(D) ≤ vol(D′).

Let now (D, f) be a solid, i.e. D is a general domain and f :D → R is a piecewise continuous function. A division (partition)∆ of D is a finite set of general subdomains Di, i = 1, 2, ..., n ofD such that ∪ni=1Di = D and for i = j, Di ∩ Dj is a subset withno interior points (the empty set, a finite number of points, smoothcurves, smooth surfaces, etc.). The norm of ∆ is the nonnegative realnumber ‖∆‖ = max diam(Di) : i = 1, 2, ..., n , where diam(Di) =

sup∥∥∥

−−−→MM ′

∥∥∥ : M,M ′ ∈ Di. We usually denote ∆ = Di. A set of

marking points Pi(ξi, ηi, θi), Pi ∈ Di, i = 1, 2, ..., n, is usually associ-ated to a division ∆. Let

Sf (∆, Pi(ξi, ηi, θi)) =n∑

i=1

f(ξi, ηi, θi)vol(Di)

be the Riemann sum which corresponds to the division ∆ and to theset of marking points Pi(ξi, ηi, θi), i = 1, 2, ..., n.

D 28. We say that a function f : D → R is (Riemann)

integrable if there exists a real number I denoted by

∫∫∫

D

f(x, y, z)dxdydz,

such that if ε > 0, there is a δ > 0 which depends on ε, with

|I − Sf(∆, Pi(ξi, ηi, θi))| < ε

for any division ∆ with ‖∆‖ < δ and for any set Pi(ξi, ηi, θi) of mark-ing points of the division ∆. This means that we can well approximatethe number I with Riemann sums of the form Sf (∆, Pi(ξi, ηi, θi)),


when ‖∆‖ → 0. I =

∫∫∫

D

f(x, y, z)dxdydz is said to be the triple inte-

gral of f on the domain D and dxdydz is called an element of volumeof D.

This number I is uniqueAll the properties of double integrals can be extended to triple

integrals.

T 84. Let D be a space domain and let f : D → R be aRiemann integrable function. Then f must be bounded.

P. Let us fix a value for ε, say ε = 1, in the above definition28. This last definition says that there exists at least one division∆ = Di, i = 1, 2, ..., n such that |I − Sf(∆, Pi(ξi, ηi, θi))| < 1 forany set of marking points Pi of the division ∆. A consequence of thislast inequality is that the Riemann sums Sf(∆, Pi(ξi, ηi, θi)) remainsbounded when we arbitrarily change the set of marking points Pi,Pi ∈ Di, i = 1, 2, ..., n. If f were unbounded, it were unbounded at leastin a subdomain Di0 . This means that there exists a sequence of points

P (m)i0

, P (m)i0

∈ Di0, m = 1, 2, ... with∣∣∣f(P

(m)i0

)∣∣∣ → ∞ when m → ∞.

We see now that in the corresponding Riemann sums,

Sf (∆, P (m)i (ξ

(m)i , η

(m)i , θ

(m)i )) =

n∑

i=1

f(ξ(m)i , η

(m)i , θ

(m)i )vol(Di),

where all the points P(m)i = Pi are fixed, except P

(m)i0

which is mov-

ing, the i0-terms, f(ξ(m)i0, η

(m)i0, θ

(m)i0

)vol(Di0) is unbounded to ∞ or to

−∞. Hence, the set of all Riemann sums Sf(∆, P (m)i (ξ

(m)i , η

(m)i , θ

(m)i ))

cannot remain bounded. A contradiction with the above observation!Thus, f must be bounded.

This last result says that the class of integrable functions is includedin the class of bounded functions. This is why any function consideredby us in the following is assumed to be bounded.

Let D be a domain with a volume vol(D), let ∆ = Di, i =1, 2, ..., n be a division of D and let f be a bounded function definedon D with values in R. Let mi = inff(x, y, z) : (x, y, z) ∈ Di and letMi = supf(x, y, z) : (x, y, z) ∈ Di for any i = 1, 2, ..., n. Then

s∆(f) =n∑

i=1

mi · vol(Di)


is called the inferior Darboux sum of f relative to ∆ and

S∆(f) =n∑

i=1

Mi · vol(Di)

is said to be the superior Darboux sum of f relative to ∆. It is clearthat

m · vol(D) ≤ s∆(f) ≤ Sf(∆, Pi(ξi, ηi, θi)) ≤ S∆(f) ≤M · vol(D),

for any set of marking points Pi(ξi, ηi, θi) of the division ∆, wherem = inff(x, y, z) : (x, y, z) ∈ D and M = supf(x, y, z) : (x, y, z) ∈D. Since the set s∆(f) : ∆ goes on the set of all divisions of Dis upper bounded by M · vol(D), it has a least upper bound I∗(f).Since the set S∆(f) : ∆ goes on the set of all divisions of D is lowerbounded by m ·vol(D), it has a greatest lower bound I∗(f). In general,I∗(f) ≤ I∗(f).

T 85. (Darboux criterion) Let D be a domain with a vol-ume vol(D) and let f : D → R be a bounded function. Then f is inte-grable on D if and only if I∗(f) = I∗(f), i.e. if and only if for any ε > 0there exists a δε > 0 such that if ‖∆‖ < δε then S∆(f)− s∆(f) < ε. In

this last case I∗(f) = I∗(f) =

∫∫∫

D

f(x, y, z)dxdydz.

This criterion is very useful whenever we want to prove that a spec-ified class of functions are integrable.

T 86. Let D be a closed bounded space domain with ∂D apiecewise smooth surface and let f : D → R be a continuous functiondefined on D. Then f is integrable on D. If f is continuous on a subsetDA of D such that A has a volume equal to zero, then f is alsointegrable on D.

P. We prove only the first statement of the theorem. SinceD is closed and bounded, D is a compact subset of R3. Since f iscontinuous, it is uniformly continuous (see [Po], theorem 59). In orderto prove that f is integrable we shall use the above Darboux criterion85. For this, let us take a small ε > 0. The uniform continuity of fimplies that there exists a small δ > 0 (depending on ε) such that ifx,x′ ∈ D and ‖x− x′‖ < δ, then |f(x) − f(x′)| < ε

vol(D). Let us take a

division ∆ = Di, i = 1, 2, ..., n of D with ‖∆‖ < δ. Since each Di isa compact subset of R3 and since f is continuous, there exist x(i) andx′(i) in Di with

mi = inff(x, y, z) : (x, y, z) ∈ Di = f(x(i))


and

Mi = supf(x, y, z) : (x, y, z) ∈ Di = f(x′(i))

(see [Po], theorem 58). Hence

S∆(f) − s∆(f) =n∑

i=1

[f(x′(i)) − f(x(i))

]· vol(Di) ≤

≤n∑

i=1

∣∣f(x′(i)) − f(x(i))∣∣ · vol(Di) <

ε

vol(D)

n∑

i=1

vol(Di) = ε,

because ‖∆‖ < δ implies diam(Di) < δ and so,∥∥x′(i) − x(i)

∥∥ < δ.Thus we can apply the uniform continuity of f and finally we get thatS∆(f) − s∆(f) < ε, i.e. that f is integrable.

R 31. Up to now we used closed bounded domains Di withpiecewise smooth boundaries, as subdomains of a division ∆ of a fixeddomain D of the same type. Since any such subdomain Di can be wellapproximated with elementary domains (finite union of nonoverlapingparallelepipeds) or with finite unions of nonoverlaping tethrahedrons,balls, prisms, etc., we can substitute these Di in definition 28 or intheorem 85 with tethrahedrons, balls, prisms, etc. It is clear that ingeneral it is not possible to cover a domain D with nonoverlaping balls,for instance. But, for any small ε > 0, there exists such a union ofballs Uε ⊂ D, such that vol(DUε) < ε. Hence we can work with thefollowing alternative equivalent definition for an integrable function f.We say that f : D → R is (Riemann) integrable on D if there ex-ists a real number I such that for any small ε > 0 and η > 0 thereexists a union Uη = Bi of nonoverlaping balls, such that Uη ⊂ D,vol(DUη) < η and |I −∑n

i=1 f(ξi, ηi, θi)vol(Bi)| < ε for any set ofmarking points Pi(ξi, ηi, θi), Pi(ξi, ηi, θi) ∈ Bi, i = 1, 2, ..., n. Simi-larly we can work with tethrahedrons, prisms, general parallelepipeds,etc., instead of balls. This is why if one has a tower of domains like inthe formula (2.1),

E(1)int ⊂ E

(2)int ⊂ ... ⊂ E

(k)int ⊂ ... ⊂ D,

such that vol(E(k)int) ր vol(D), then the approximation formulas can be

used:

(2.2)

∫∫∫

D

f(x, y, z)dxdydz = limk→∞

∫∫ ∫

E(k)int

f(x, y, z)dxdydz,


or

(2.3)

∫∫∫

D

f(x, y, z)dxdydz ≈∫∫ ∫

E(k)int

f(x, y, z)dxdydz

for k large enough. In the engineering practice this last formula ismostly used.

The following result put together some basic properties of the tripleintegrals.

T 87. a) the mapping f

∫∫∫

D

f(x, y, z)dxdydz is a lin-

ear mapping defined on the vector space Int(D) of all integrable func-tions defined on D. This means that

∫∫∫

D

[αf(x, y, z) + βg(x, y, z)]dxdydz =

= α

∫∫∫

D

f(x, y, z)dxdydz + β

∫∫∫

D

g(x, y, z)dxdydz.

b) the mass of a solid (D, f) can be computed as follows:

mass(D) =

∫∫∫

D

f(x, y, z)dxdydz.

In particular the volume of D is equal to

∫∫∫

D

dxdydz.

c) if f ≤ g on D, then

∫∫∫

D

f(x, y, z)dxdydz ≤∫∫∫

D

g(x, y, z)dxdydz.

In particular, if f ≥ 0 on D, then

∫∫∫

D

f(x, y, z)dxdydz ≥ 0.

d)

∣∣∣∣∣∣

∫∫∫

D

f(x, y, z)dxdydz

∣∣∣∣∣∣≤

∫∫∫

D

|f(x, y, z)| dxdydz.


e)∫∫∫

D

f(x, y, z)dxdydz = f(x0, y0, z0) · vol(D),

where f is continuous and (x0, y0, z0) is a point of D (mean formula).This formula says that any nonhomogeneous solid has the same mass asa homogenous solid with the same domain D and the constant densityequal to the value of the initial density function at a certain point ofD. In particular, if f is continuous, f(x, y, z) ≥ 0 and

∫∫∫

D

f(x, y, z)dxdydz > 0,

then f is not zero on a small ball which is contained in D. Moreover,if f ≥ 0 and

∫∫∫

D

f(x, y, z)dxdydz = 0,

then the set A of points (x, y, z) for which f(x, y, z) > 0 has the volumeequal to zero (the proof of this statement is more difficult).

f) if ∆ = Di, i = 1, 2, ..., n is a division of D, then∫∫∫

D

f(x, y, z)dxdydz =n∑

i=1

∫∫∫

Di

f(x, y, z)dxdydz.

All the other applications which we met in the case of a tripleintegral on parallelepipeds can be deduced here for a general domainD. For instance, the coordinates of the mass centre of a solid (D, f)are:

(2.4) xG =

∫∫∫

D

xf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

, yG =

∫∫∫

D

yf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

,

zG =

∫∫∫

D

zf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

.


If our solid (D, f) is rotating around the Oz-axis for instance, theinertia moment can be computed as follows:

Iz =

∫∫∫

D

(x2 + y2)f(x, y, z)dxdydz,

because the square of the distance of a point M(x, y, z) up to the Oz-axis is equal to x2 + y2. The reader can easily deduce other formulaeif one knows such a formula for a point with a "mass" at it. Forinstance, a solid (D, f) is moving around a straight line (d) and thedistance of a pointM(x, y, z) of this solid up to the line (d) is a functionH(x, y, z). Then, the inertia moment of this single point is equal toH2(x, y, z) · f(x, y, z). Now we "globalize" this last relation and obtaina general formula for the inertia moment of the entire solid, when it isrotating around the line (d) :

I(d) =

∫∫∫

D

H2(x, y, z) · f(x, y, z)dxdydz.

Since we have no effective method to compute a triple integral up tothe present time, we postpone to work with some examples. Thus wecontinue our study by presenting the basic method of computation ofthe triple integrals, namely the "iterative method". The name comesfrom the fact that by using this method we shall reduce the computa-tion of a triple integral to the successive (iterative) calculation of threesimple Riemann integrals.

3. Iterative formulas for a general space domain

Let us begin with some prerequisites.We say that a surface (S) is explicitly (described) with respect to z

if it has a parametrization of the following type:

(S) :

x = xy = y

z = z(x, y), (x, y) ∈ Ω = prxOy(S),

the projection of (S) on the xOy-plane (see Fig.4).A domain D is simple w.r.t. Oz-axis if there exist two continuous

functions z = ϕ(x, y) and z = ψ(x, y) such that:

D = M(x, y, z) : ϕ(x, y) ≤ z ≤ ψ(x, y), where (x, y) ∈ Ω ⊂ xOy ,i.e. D is the "continuous" union of all segments

[M1(x, y, ϕ(x, y)),M2(x, y, ψ(x, y))],

3. ITERATIVE FORMULAS FOR A GENERAL SPACE DOMAIN 279

F 4

when (x, y) runs on a plane domain Ω contained in the xOy-plane (seeFig.5). This means that D is a cylindrical domain delimited "up anddown" by the explicitly described surfaces z = ϕ(x, y) and z = ψ(x, y).

F 5


T 88. (a general iterative formula I) Let D be a generaldomain (see its definition at the beginning of this section) in R3 andlet (D, f) be a solid (f : D → R is a piecewise continuous function).We also assume that D is simple w.r.t. Oz-axis. Then:

(3.1)

∫∫∫

D

f(x, y, z)dxdydz =

∫∫

Ω

[∫ ψ(x,y)

ϕ(x,y)

f(x, y, z)dz

]dxdy.

This formula says that first of all we compute a simple integral

J(x, y) =

∫ ψ(x,y)

ϕ(x,y)

f(x, y, z)dz,

with two parameters x and y, on the segment

[M1(x, y, ϕ(x, y)),M2(x, y, ψ(x, y))],

relative to the variable z. Then, we compute a double integral∫∫

Ω

J(x, y)dxdy of the just obtained function J(x, y), on the domain

Ω, the projection of D on the xOy-plane (see Fig.5).

P. In order not to lose ourselves in sophisticated technicalconsiderations and to escape the essential, we prove this result onlywhen f is continuous. The idea of the proof is like the idea used forproving a similar formula for a double integral (see theorem 79). Letus construct the following three segments: [a, b] = prOx(D), [c, d] =prOy(D) and [e, g] = prOz(D), the projections of the domain D on thethree coordinates axes. Let us denote by E = [a, b] × [c, d] × [e, g], theparallelepiped constructed with these last segments. It is clear thatD ⊂ E (see Fig.5). It is "clear" that if we extend the density functionf : D → R to E by putting the density zero at all the points of Ewhich are not in D, the mass of D will be the same as the mass of E.

Indeed, let us define f : E → R,

f(x, y, z) =

f(x, y, z), if (x, y, z) ∈ D,

0, if (x, y, z) ∈ ED.

Since f is continuous on D, f is discontinuous (but bounded!) at moston ∂D, which is a finite union of smooth bounded surfaces. Hence, the

set of points at which f is discontinuous has volume zero. This means

that f is integrable on E and we can apply to its triple integral the


iterative formula (1.2):∫∫∫

D

f(x, y, z)dxdydz =

∫∫∫

E

f(x, y, z)dxdydz =

=

∫ b

a

(∫ d

c

(∫ g

e

f(x, y, z)dz

)dy

)dx =

=

∫ ∫

[a,b]×[c,d]

(

∫ ϕ(x,y)

e

f(x, y, z)dz +

∫ ψ(x,y)

ϕ(x,y)

f(x, y, z)dz+

+

∫ g

ψ(x,y)

f(x, y, z)dz)dxdy =

∫ ∫

[a,b]×[c,d]

(∫ ψ(x,y)

ϕ(x,y)

f(x, y, z)dz

)dxdy =

=

∫∫

Ω

(∫ ψ(x,y)

ϕ(x,y)

f(x, y, z)dz

)dxdy,

i.e. we just obtained the iterative formula (3.1). Here we used the

fact that f(x, y, z) = 0 at all the points of the segment [A,B] whichare outside of the segment [M1M2] and, at all the points of E whichare projected on the xOy-plane in the points of the region [a, b] ×[c, d]Ω (see Fig.5). For the other points (x, y, z) of D, f(x, y, z) =f(x, y, z).

E 87. Let Ω = M(x, y, z) : x2 +y2 ≤ z2, 0 ≤ z ≤ 2 be theconic domain from Fig.6 and let f(x, y, z) = 3z be a density functiondefined on Ω. Let us compute the mass of the solid (Ω, f).

We must be careful here because Ω is not more a plane domain likein the above theorem. Hence ϕ(x, y) =

√x2 + y2 and ψ(x, y) = 2 and

for Ω of the above theorem one has Ωxy = prxOy(D) (see Fig.6).Thus, using formula 3.1, we get:

mass(Ω) =

∫∫∫

Ω

3zdxdydz =

∫ ∫

x2+y2≤4

(∫ 2

√x2+y2

3zdz

)dxdy =

3

2

∫ ∫

x2+y2≤4

z2∣∣2√

x2+y2dxdy =

3

2

∫ ∫

x2+y2≤4

[4 − (x2 + y2)

]dxdy =

= 6

∫ ∫

x2+y2≤4

dxdy − 3

2

∫ ∫

x2+y2≤4

(x2 + y2)dxdy.


F 6

The first double integral is nothing else then 6 times the area of a discwith radius 2, thus its value is equal to 24π. Let us use polar coordinateschange of variables to compute the second double integral:

∫ ∫

x2+y2≤4

(x2 + y2)dxdy =

∫ 2

0

(∫ 2π

0

ρ2 · ρdθ)dρ =

=

∫ 2

0

ρ3(∫ 2π

0

dθ

)dρ = 2π

ρ4

4

∣∣∣∣2

0

= 8π.

Hence,

mass(Ω) = 24π − 3

2· 8π = 12π.

We see that in the above iterative formula first of all one computesa simple integral and then one computes a double integral. We shallsee in the following that in some cases it is possible to compute firstlya double integral and then a simple one. Moreover, in the next resultthe general domain D which appeared in the statement of theorem 88needs not to be simple.

T 89. (a general iterative formula II) Let Ω be a generaldomain and let (Ω, f) be a solid such that [e, g] = prOz(Ω), S = Ω ∩Z = z, where z ∈ [e, g], is the section of Ω realized by the plane Z = zand Ωz is the projection of S on the xOy-plane (see Fig.7). Then,

(3.2)

∫∫∫

Ω

f(x, y, z)dxdydz =

∫ g

e

∫∫

Ωz

f(x, y, z)dxdy

dz.


Here, first of all we compute a double integral I(z) =

∫∫

Ωz

f(x, y, z)dxdy

with a parameter z and then we calculate a simple integral∫ geI(z)dz

of the last computed function I(z).

F 7

P. To avoid some nonessential technical tricks, we assumethat f : Ω → R is a continuous function. Let E = [ABCDA′B′C ′D′] =[a, b] × [c, d] × [e, g] be a parallelepiped which contains Ω and

f(x, y, z) =

f(x, y, z), if (x, y, z) ∈ Ω,

0, if (x, y, z) ∈ EΩ.

be a density function on E which extend f up to E. It is easy to see

that the mass of (E, f) is equal to the mass of (Ω, f). Applying theiterative formula (1.2) for the parallelepiped domain E, we get:

∫∫∫

Ω

f(x, y, z)dxdydz =

∫∫∫

E

f(x, y, z)dxdydz =

∫ g

e

∫ ∫

[a,b]×[c,d]

f(x, y, z)dxdy

dz =

∫ g

e

∫∫

Ωz

f(x, y, z)dxdy

dz,

because f(x, y, z) = 0, when (x, y) is in [a, b] × [c, d] but not in Ωz and

f(x, y, z) = f(x, y, z), when (x, y) ∈ Ωz.


E 88. The solid (Ω, f), where Ω : x2 + y2 ≤ z2, 0 ≤ z ≤ 2and f(x, y, z) = z2 is rotating around Oz-axis. Let us compute theinertia moment of it w.r.t. to Oz-axis (see Fig.8). The projection ofthe domain Ω on the Oz-axis is the segment [0, 2]. For any z0 ∈ [0, 2],the projection of the section of Ω determined by the plane Z = z0 is thedisc Ωz0 : x2 + y2 ≤ z20 . Let us apply now the formula (3.2) to compute

the triple integral Iz =

∫∫∫

Ω

z2(x2 + y2)dxdydz.

Iz =

∫∫∫

Ω

z2(x2 + y2)dxdydz =

∫ 2

0

z2

∫ ∫

x2+y2≤z2

(x2 + y2)dxdy

dz =

=

∫ 2

0

z2[∫ z

0

ρ3

(∫ 2π

0

dθ

)dρ

]dz = 2π

∫ 2

0

z2 · z4

4dz =

π

2· z

7

7

∣∣∣∣2

0

=64π

7.

F 8

Both iterative formulas (3.1) and (3.2) refers to theOz-axis. We caneasily imagine the other four types of formulas which refer to Oy-axisand to Ox-axis.

E 89. For instance, let us find the coordinates of the masscentre of the domain D : x2 + z2 ≤ 2y, 0 ≤ y ≤ 8 and the densityfunction f(x, y, z) = y. The matter is about a circular paraboloidaldomain with Oy as an axe of symmetry. The symmetry here is alsoa static symmetry because the mass is symmetrically distributed w.r.t.


the Oy-axis. Indeed, f(−x, y,−z) = y = f(x, y, z), i.e. the density ofthe point M(x, y, z) is the same as the density of its symmetric pointM ′(−x, y,−z) w.r.t. Oy-axis. Thus, the mass centre G is on the Oy-axis, i.e. xG = zG = 0. Let us compute

yG =

∫∫∫

D

y · ydxdydz

∫∫∫

D

ydxdydz

.

Let us firstly compute I =

∫∫∫

D

ydxdydz, the mass of the solid. Let us

use the first iterative formula (3.1). The projection of the domain onthe zOx-plane is the disc: x2 + z2 ≤ 16, with radius equal to 4. Fora fixed point P (x, 0, z) in this last disc, all the points of D which areprojecting in P are the points of the space segment

[M1(x,x2 + z2

2, z),M2(x, 8, z)].

Thus, ∫∫∫

D

ydxdydz =

∫ ∫

x2+z2≤16

(∫ 8

x2+z2

2

ydy

)dxdz =

=1

2

∫ ∫

x2+z2≤16

[y2 |8x2+z2

2

]dxdz =

1

2

∫ ∫

x2+z2≤16

[64 − (x2 + z2)2

4

]dxdz =

= 32

∫ ∫

x2+z2≤16

dxdz − 1

8

∫ ∫

x2+z2≤16

(x2 + z2)2dxdz.

The first integral is 32 times the area of a disc of radius 4, i.e. it isequal to 32 × π × 16. Let us compute the second double integral∫ ∫

x2+z2≤16

(x2 + z2)2dxdz =

∫ 4

0

(∫ 2π

0

ρ4 · ρdθ)dρ = 2π

ρ6

6

∣∣∣∣4

0

= π46

3.

Hence, the mass of D is equal to 32 × π × 16 − π 45

6= 45π

3.

Let us now compute the static moment w.r.t. xOz-plane,

J =

∫∫∫

D

y2dxdydz =

∫ ∫

x2+z2≤16

(∫ 8

x2+z2

2

y2dy

)dxdz =


=1

3

∫ ∫

x2+z2≤16

[y3 |8x2+z2

2

]dxdz =

1

3

∫ ∫

x2+z2≤16

[83 − (x2 + z2)3

8

]dxdz =

=83

3

∫ ∫

x2+z2≤16

dxdz − 1

24

∫ ∫

x2+z2≤16

(x2 + z2)3dxdz =

=83

3· 16π − 1

24

∫ 4

0

(∫ 2π

0

ρ6 · ρdθ)dρ =

=83

3· 16π − 1

24· 2π · ρ

8

8

∣∣∣∣4

0

=83

3· 16π − 1

24· 2π · 48

8=

=1

24 · 8

(85 · 16π − 48 · 2π

)= 2π · 45.

Hence, yG = 2π·4545π3

= 6.

4. Change of variables in a triple integral

Let us consider two general space domain Ω and D, both in R3, thefirst described relative to a Cartesian coordinate system ou, ov, owand the second relative to the usual Cartesian coordinate system

Ox,Oy,Oz. Recall that a general space domain D is a closed,bounded and connected subset D of R3 such that its boundary ∂D is apiecewise smooth surface. A smooth deformation from Ω to D is a vec-tor function −→g : Ω → D, −→g (u, v, w) = (x(u, v, w), y(u, v, w), z(u, v, w))which is a diffeomorphism, i.e. it is a bijection and −→g and its inverse−→g −1 : D → Ω are vector functions (fields) of class C1. Here x, y, zare independent variables, but x(u, v, w), y(u, v, w), z(u, v, w) are threefunctions of three independent variables u, v and w. We prefer not tointroduce new notation for the name of these three last functions (likeϕ(u, v, w), ψ(u, v.w), θ(u, v, w)). A change of the "old" variables x, yand z with the "new" ones u, v and w is nothing else then such a smoothdeformation or a "regular transformation", another more mathematicalterm for the same notion.

First of all we are interested how changes the volume of a small (itsdiameter is small enough!) parallelepiped domain [P0P1P4P3P2P

′1P

′4P

′3]

of the domain Ω, which contains a fixed marking point P (ξ, η, θ),through the deformation −→g . Let [M0M1M4M3M2M

′1M

′4M

′3] be the im-

age−→g ([P0P1P4P3P2P′1P

′4P

′3]) of [P0P1P4P3P2P

′1P

′4P

′3] through −→g (see

Fig.9).If P0(u0, v0, w0), P1(u0, v1, w0), P2(u0, v0, w1), P3(u1, v0, w0),

4. CHANGE OF VARIABLES IN A TRIPLE INTEGRAL 287

F 9

M0(x0, y0, z0), M1(x1, y1, z1), M2(x2, y2, z2) and M3(x3, y3, z3) then,

(4.1)

x0 = x(u0, v0, w0)y0 = y(u0, v0, w0)z0 = z(u0, v0, w0)

,


,


,


.

T 90. (the change of an elementary volume) With theabove notation and hypotheses, one has that:

vol([M0M1M4M3M2M′1M

′4M

′3]) ≈(4.2) ∣∣J−→g (ξ, η, θ)

∣∣ vol([P0P1P4P3P2P′1P

′4P

′3]) =(4.3)

=∣∣J−→g (ξ, η, θ)

∣∣ (u1 − u0)(v1 − v0)(w1 − w0).

The approximation means that the volume of the image of the paral-lelepiped

[P0P1P4P3P2P′1P

′4P

′3]

through the deformation −→g is closer and closer to the quantity∣∣J−→g (ξ, η, θ)∣∣ (u1 − u0)(v1 − v0)(w1 − w0)


as soon as the diameter of the parallelepiped is smaller and smaller.Hence, we can look at the quantity

∣∣J−→g (ξ, η, θ)∣∣ as to a local "dilation"

or "contraction" coefficient of our deformation −→g . It controls the mod-ification of small volumes during the deformation process.

P. The following sequence of consecutive approximations willgive us the final approximation of the formula (4.2).(4.4)

vol([M0M1M4M3M2M′1M

′4M

′3]) ≈ vol

([−−−→M0M1,

−−−→M0M2,

−−−→M0M3

]),

where[−−−→M0M1,

−−−→M0M2,

−−−→M0M3

]is the skew general parallelepiped con-

structed with the vectors−−−→M0M1,

−−−→M0M2,

−−−→M0M3. Let us remember, from

any course of Vector Calculus, the basic formula of calculation of sucha volume. It is in fact the absolute value of the mixed product

(−−−→M0M1,

−−−→M0M2,

−−−→M0M3),

which is computed as the value of the following determinant:

(4.5) det

x1 − x0 y1 − y0 z1 − z0x2 − x0 y2 − y0 z2 − z0x3 − x0 y3 − y0 z3 − z0

.

Since x1 = x(u0, v1, w0) and x0 = x(u0, v0, w0), let us consider thefunction v x(u0, v, w0) and let us apply to it Lagrange formula (see[Po], Corollary 5, pag.88) on the interval [v0, v1] :

x1 − x0 =∂x

∂v(u0, c, w0) · (v1 − v0),

where c ∈ (v0, v1). Since the diameter of the parallelepiped[P0P1P4P3P2P

′1P

′4P

′3] is very small, the point (u0, c, w0) is very closed

to P (ξ, η, θ). Since the functions x(u, v, w), y(u, v, w), z(u, v, w) are ofclass C1, the function ∂x

∂v(u, v, w) is continuous. Hence ∂x

∂v(u0, c, w0) is

sufficiently close to ∂x∂v

(ξ, η, θ). Finally we get the approximation:

(4.6) x1 − x0 ≈∂x

∂v(ξ, η, θ) · (v1 − v0)

The same type of reasoning lead us to the other approximations:

y1 − y0 ≈ ∂y

∂v(ξ, η, θ) · (v1 − v0),

z1 − z0 ≈ ∂z

∂v(ξ, η, θ) · (v1 − v0),


x2 − x0 ≈ ∂x

∂w(ξ, η, θ) · (w1 − w0),

y2 − y0 ≈ ∂y

∂w(ξ, η, θ) · (w1 − w0),

z2 − z0 ≈ ∂z

∂w(ξ, η, θ) · (w1 − w0),

x3 − x0 ≈ ∂x

∂u(ξ, η, θ) · (u1 − u0),

y3 − y0 ≈ ∂y

∂u(ξ, η, θ) · (u1 − u0),

z3 − z0 ≈ ∂z

∂u(ξ, η, θ) · (u1 − u0).

With these approximations, we come back to the above determinant(4.5) and find:

det

x1 − x0 y1 − y0 z1 − z0x2 − x0 y2 − y0 z2 − z0x3 − x0 y3 − y0 z3 − z0

≈

≈ det

∂x∂v

(ξ, η, θ) ∂y∂v

(ξ, η, θ) ∂z∂v

(ξ, η, θ)∂x∂w

(ξ, η, θ) ∂y∂w

(ξ, η, θ) ∂z∂w

(ξ, η, θ)∂x∂u

(ξ, η, θ) ∂y∂u

(ξ, η, θ) ∂z∂u

(ξ, η, θ)

(u1−u0)(v1−v0)(w1−w0).

Since the determinant of a transposed matrix is the same like the de-terminant of the initial matrix and since by changing the rows betweenthem, the absolute value of the obtained determinant does not change,we finally get that the absolute value of this last value is nothing elsethen ∣∣J−→g (ξ, η, θ)

∣∣ (u1 − u0)(v1 − v0)(w1 − w0),

i.e. the formula which appeared in the statement of the theorem. Letus recall the definition of the Jacobian of −→g :

(4.7) J−→g (ξ, η, θ) = det

∂x∂u

(ξ, η, θ) ∂x∂v

(ξ, η, θ) ∂x∂w

(ξ, η, θ)∂y∂u

(ξ, η, θ) ∂y∂v

(ξ, η, θ) ∂y∂w

(ξ, η, θ)∂z∂u

(ξ, η, θ) ∂z∂v

(ξ, η, θ) ∂z∂w

(ξ, η, θ)

Let us use this basic result in order to prove the next theorem. Weshall again use the following simple observation: if two real numbers Iand J are limit points for one and the same subset A of real numbersand if A has a unique limit point, then I = J. Instead of saying that Iis a limit point of A, we can say that I can be well approximated withelements of A, etc.


T 91. (general change of variables in triple integrals) Withthe above notation and hypotheses one has:

(4.8)

∫∫∫

D

f(x, y, z)dxdydz =

=

∫∫∫

Ω

f(x(u, v, w), y(u, v, w), z(u, v, w)) ·∣∣J−→g (u, v, w)

∣∣ dudvdw.

P. We can well approximate the domain Ω with a union ofnonoverlaping (which have no interior common points) parallelepipedsΩijk = [ui−1, ui]×[vj−1, vj ]×[wk−1, wk], i = 1, 2, ..., n, j = 1, 2, ...,m, k =1, 2, ..., p. LetDijk = −→g (Ωijk) be the image of Ωijk after the deformation−→g . Since −→g is a diffeomorphism, D can also be well approximatedby the set of subdomains Dijk, i = 1, 2, ..., n, j = 1, 2, ...,m, k =1, 2, ..., p.

Let Pijk(ξi, ηj , θk), Pijk ∈ Ωijk, be a set of marking points in Ω(Pay attention, Ωijk is not a division of Ω). We just remarked above

(see remark 31) that the triple integral I =

∫∫∫

D

f(x, y, z)dxdydz can

be well approximated by sums of the following type (here

Mijk(x(ξi, ηj, θk), y(ξi, ηj, θk), z(ξi, ηj , θk)) ∈ Dijk,

i = 1, 2, ..., n, j = 1, 2, ...,m, k = 1, 2, ..., p

is a set of marking points in D):(4.9)

I ≈n∑

i=1

m∑

j=1

p∑

k=1

f(x(ξi, ηj, θk), y(ξi, ηj , θk), z(ξi, ηj, θk)) · vol(Dijk).

But, using formula (4.2), we get: vol(Dijk) ≈∣∣J−→g (ξi, ηj , θk)

∣∣ (ui −ui−1)(vj − vj−1)(wk − wk−1). Thus, coming back to formula (4.9), weget:

I ≈n∑

i=1

m∑

j=1

p∑

k=1

f(x(ξi, ηj, θk), y(ξi, ηj, θk), z(ξi, ηj , θk))×

×∣∣J−→g (ξi, ηj , θk)

∣∣ (ui − ui−1)(vj − vj−1)(wk − wk−1),

which is a sum that well approximates the triple integral

J =

∫∫∫

Ω


∣∣ dudvdw.


Since I and J are fixed numbers and since I ≈ J as well as we want,we must conclude that I = J, i.e. the required equality in formula(4.8).

R 32. When we want to use the change of variables formula(4.8), first of all we must try to realize our domain D as the image of asimpler domain Ω, a parallelepiped if this is possible, through a smoothdeformation −→g . Even −→g : Ω → D is not more a difeomorphism,but it is so whenever we restrict it to an open subset Ω′ ⊂ Ω withvol(ΩΩ′) = 0 and the subset D −→g (Ω′) has volume zero, the changeof variables formula continues to work. Indeed,

∫∫∫

D

f(x, y, z)dxdydz =

∫∫ ∫

D−→g (Ω′)

f(x, y, z)dxdydz+

+

∫∫ ∫

−→g (Ω′)

f(x, y, z)dxdydz = 0 +

∫∫ ∫

−→g (Ω′)

f(x, y, z)dxdydz =

=

∫∫∫

Ω′


∣∣ dudvdw =

=

∫∫∫

Ω


∣∣ dudvdw.

Here we used an immediate result (prove it!): "The integral (simple,double, triple, etc.) on a subset of measure (length, area, volume, etc.)zero is always equal to zero. Indeed, look carefully to any general Rie-mann sum,...it is zero!

R 33. Moreover, during the proof of the above theorem 91we did not used the fact that D or Ω are closed. Even some parts oftheir boundaries ∂D or ∂Ω are missing, the formula continues to work.This will be the situation in some future applications, when either theJacobian J−→g is zero at some points of ∂Ω, or −→g is not a bijection or adiffeomorphism around some points of Ω (see bellow the change of theCartesian coordinates x, y, z into the spherical coordinates ρ, ϕ, θ).

E 90. Let D be the skew parallelepiped [OACBDA′C ′B′]

generated by the vectors−→OA =

−→i + 3

−→j +

−→k ,

−−→OB = −3

−→i +

−→k and−−→

OD =−→i +

−→j + 4

−→k (see Fig.10).

Assume that D is loaded with a density function f(x, y, z) = x +y + z. Let us compute the mass of D, i.e. the triple integral I =


F 10

∫∫∫

D

f(x, y, z)dxdydz. We see that any of the iterative methods fails

in this case because of the "complicated" geometrical form of D (it canalso be divided into "many" simple subdomains w.r.t. Oz-axis). Let ususe the change of variables method described above. For this, we tryto find a linear transformation −→g : R3 → R3 such that the image ofthe cube Ω = [0, 1] × [0, 1] × [0, 1] through −→g be exactly our domainD. Any course of Linear Algebra tells us that there exists one and onlyone linear transformation (an isomorphism or automorphism in fact!)which changes the basis (1, 0, 0), (0, 1, 0), (0, 0, 1) into another basis−→OA,−−→OB,−−→OD. Here we just identified the arithmetical vector spaceR3 with the geometrical 3-D space V3 of all free space vectors. It iseasy to write the matrix M−→g of this linear transformation w.r.t. thecanonical basis (1, 0, 0), (0, 1, 0), (0, 0, 1) of R3 :

M−→g =

1 −3 13 0 11 1 4

.

Thus, −→g acts on the column vector (u, v, w)t ∈ R3 as follows:

−→g

uvw

=

1 −3 13 0 11 1 4

uvw

=

u− 3v + w3u+ w

u+ v + 4w

.

Hence −→g (u, v, w) = (u− 3v+w, 3u+w, u+ v+ 4w). Why −→g (Ω) = D?To answer to this question we can use the linearity of −→g . Indeed, anelement (u, v, w) ∈ R3 is in Ω = [0, 1] × [0, 1] × [0, 1] if and only if


u, v, w ∈ [0, 1]. But

−→g (u, v, w) = u−→g (1, 0, 0) + v−→g (0, 1, 0) + w−→g (0, 0, 1) =

= u−→OA+ v

−−→OB + w

−−→OD ∈ D,

since u, v, w ∈ [0, 1] (look at the parallelogram rule of adding vectors!).

Thus, −→g (Ω) ⊂ D. Since any vector−−→OM, with M ∈ D can be written

as a linear combination of the form:−−→OM = u

−→OA+ v

−−→OB + w

−−→OD,

for u, v, w ∈ [0, 1], one has that −→g is also surjective. The determinantof M−→g is equal to 35 = 0, so that −→g is an isomorphism. Since x =x(u, v, w) = u−3v+w, y = y(u, v, w) = 3u+w and z = z(u, v, w) = u+v+4w, −→g is also of class C1. Its inverse is also a linear transformation,so that it is of class C1 too. The Jacobian of −→g is detM−→g = 35. Nowwe can apply change of variables formula 4.8 and find:

mass(D) =

∫∫∫

D

(x+ y + z)dxdydz =

= 35

∫∫ ∫

[0,1]×[0,1]×[0,1]

(u− 3v + w + 3u+ w + u+ v + 4w)dudvdw =

=

∫ 1

0

(∫ 1

0

(∫ 1

0

(5u− 2v + 6w) du

)dv

)dw =

=

∫ 1

0

(∫ 1

0

(5

2u2 − 2uv + 6uw

∣∣∣∣1

0

)dv

)dw =

=

∫ 1

0

(∫ 1

0

(5

2− 2v + 6w

)dv

)dw =

=

∫ 1

0

(5

2v − v2 + 6wv

∣∣∣∣1

0

)dw =

∫ 1

0

(3

2+ 6w

)dw =

=3

2w + 3w2

∣∣∣∣1

0

=3

2+ 3 =

9

2.

E 91. (spherical coordinates) Let us recall a very importanttype of change of variables, namely the change of the Cartesian coordi-nates x, y, z into the "spherical coordinates", ρ, ϕ, θ (see Fig.11). Hereρ is the distance of a point M(x, y, z) up to the origin O, ϕ is the angle

between the Ox-axis and the projection−−→OM ′ of the position vector

−−→OM,


F 11

on the xOy-plane and, θ is the angle between the Oz-axis and the po-

sition vector−−→OM (see Fig.11). Since

∥∥∥−−→OM ′

∥∥∥ =∥∥∥−−→OM

∥∥∥ cos(π2− θ

)=∥∥∥−−→OM

∥∥∥ sin θ and since x =∥∥∥−−→OM ′

∥∥∥ cosϕ and y =∥∥∥−−→OM ′

∥∥∥ sinϕ, one has

the following connections between the Cartesian coordinates (x, y, z)and the spherical coordinates (ρ, ϕ, θ) :

(4.10)

x = x(ρ, ϕ, θ) = ρ sin θ cosϕ,y = y(ρ, ϕ, θ) = ρ sin θ sinϕz = z(ρ, ϕ, θ) = ρ cos θ

.

Since the Jacobian J−→g (ρ, ϕ, θ) of this transformation must not be zero,i.e.

J−→g (ρ, ϕ, θ) = det

sin θ cosϕ −ρ sin θ sinϕ ρ cos θ cosϕsin θ sinϕ ρ sin θ cosϕ ρ cos θ sinϕ

cos θ 0 −ρ sin θ

=

= −ρ2 sin θ = 0,

the maximal definition domain of this regular transformation is: ρ ∈(0,∞), ϕ ∈ [0, 2π) and θ ∈ (0, π). We see that the maximal definitiondomain of −→g is the infinite parallelepiped Ω = [0,∞) × [0, 2π] × [0, π]without some of its exterior sides. The image of −→g is the whole spaceR3 without the Oz-axis. During the triple integration process all ofthese "gaps" of dimension 1 or 2 have zero volume, so that they means"nothing" (see also remark 32). For instance, if we have to compute


the integral I =

∫∫∫

D

f(x, y, z)dxdydz, where D : x2 + y2 + z2 ≤ R2 is

the closed ball of radius R > 0 and with centre at O, we need in generalto use the change of variables formula by using spherical coordinates.Let D∗ = DOz-axis, let ΩR = [0, R] × [0, 2π] × [0, π] and let Ω∗

R =(0, R] × [0, 2π) × (0, π). Then, using the remark 33, we get

(4.11)

∫∫∫

D

f(x, y, z)dxdydz =

∫∫∫

D∗

f(x, y, z)dxdydz =

(4.12) =

∫∫∫

Ω∗R

f(ρ sin θ cosϕ, ρ sin θ sinϕ, ρ cos θ) · ρ2 sin θdρdϕdθ

(4.13) =

∫∫∫

ΩR

f(ρ sin θ cosϕ, ρ sin θ sinϕ, ρ cos θ) · ρ2 sin θdρdϕdθ.

Hence,

(4.14)

∫∫∫

D

f(x, y, z)dxdydz =

=

∫∫∫

ΩR

f(ρ sin θ cosϕ, ρ sin θ sinϕ, ρ cos θ) · ρ2 sin θdρdϕdθ,

or

(4.15)

∫∫ ∫

D: x2+y2+z2≤R2

f(x, y, z)dxdydz =

=

∫ R

0

ρ2[∫ 2π

0

(∫ π

0

f(ρ sin θ cosϕ, ρ sin θ sinϕ, ρ cos θ) sin θdθ

)dϕ

]dρ.

Let us apply now formula (4.15) in order to compute the mass ofthe ball D : x2 +y2 +z2 ≤ 9 with the density function f(x, y, z) = 4 |z|:

mass(D) = 4

∫∫ ∫

D: x2+y2+z2≤9

|z| dxdydz =

= 4

∫ 3

0

ρ2

[∫ 2π

0

(∫ π

0

ρ |cos θ| sin θdθ)dϕ

]dρ =


= 4

∫ 3

0

ρ3[∫ 2π

0

dϕ

](∫ π

0

|cos θ| sin θdθ)dρ =

= 4 · 2π

(∫ 3

0

ρ3dρ

)(∫ π

0

|cos θ| sin θdθ)

=

= 8πρ4

4

∣∣∣∣3

0

(∫ π2

0

cos θ sin θdθ −∫ π

π2

cos θ sin θdθ

)=

= 162π

(∫ π2

0

sin θd(sin θ) −∫ π

π2

sin θd(sin θ)

)=

= 162π

(sin2 θ

2

∣∣∣∣π2

0

− sin2 θ

2

∣∣∣∣π

π2

)= 162π.

Not always the too much processed formula (4.15) is useful. For in-stance, let C(0, 0, a) be a point on the Oz-axis, a > 0 and let D be theclosed ball with centre at a and with radius a (see Fig.12).

F 12

Let us compute the inertia moment Iz =

∫∫∫

D

(x2 + y2)dxdydz of

the solid (D, 1) (or of D with the density function understood to bef(x, y, z) = 1). Since the sphere ∂D contains the origin O, it is agood idea to use the change of variables formula, by changing the oldCartesian coordinates (x, y, z) with the spherical coordinates (ρ, ϕ, θ).It is clear that ifM(x, y, z) ∈ D, then the corresponding ϕ and θ verifythe obvious conditions: ϕ ∈ [0, 2π] and θ ∈ [0, π

2] (see Fig.12). The

Cartesian relation which describes D is: x2 + y2 + (z − a)2 ≤ a2, or


x2 + y2 + z2 − 2az ≤ 0. If instead x, y, z we put their correspondingexpressions as functions of ρ, ϕ, θ from (4.10), we get: 0 ≤ ρ ≤ 2a cos θ,i.e. ρ ∈ [0, 2a cos θ]. Thus, the general formula (4.8) becomes:

∫∫∫

D

(x2+y2)dxdydz =

∫ 2π

0

[∫ π2

0

(∫ 2a cos θ

0

ρ2 sin2 θ · ρ2 sin θdρ

)dθ

]dϕ =

=

∫ 2π

0

[∫ π2

0

sin3 θ

(∫ 2a cos θ

0

ρ4dρ

)dθ

]dϕ =

=

(∫ 2π

0

dϕ

)[∫ π2

0

sin3 θ

(∫ 2a cos θ

0

ρ4dρ

)dθ

]=

= 2π

[∫ π2

0

sin3 θ

(∫ 2a cos θ

0

ρ4dρ

)dθ

]=

(4.16) = 2π

[∫ π2

0

sin3 θ

(ρ5

5

∣∣∣∣2a cos θ

0

)dθ

]=

26a5π

5

∫ π2

0

sin3 θ cos5 θdθ.

Now we have a good opportunity to use formula 7) of theorem 63:

B(x, y) = 2∫ π

2

0sin2x−1 u cos2y−1 udu (the trigonometric expression of

the Euler’s beta function), in order to compute the last integral in4.16.

∫ π2

0

sin3 θ cos5 θdθ =1

2B(2, 3) =

1

2

Γ(2)Γ(3)

Γ(5)=

1

2

1 · 2

4!=

1

24.

Hence, ∫∫∫

D

(x2 + y2)dxdydz =26a5π

5· 1

24=

8πa5

15.

E 92. (cylindrical coordinates) Let us come back to Fig.11

and associate to any point M(x, y, z) the number r =∥∥∥−−→OM ′

∥∥∥ ∈ [0,∞),

the distance of M ′ (the projection of M on the xOy-plane) up to theorigin O, the angle ϕ ∈ [0, 2π] and the height z ∈ R of M. The associ-ation (x, y, z) (r, ϕ, z) gives rise to a new change of variables. Here(r, ϕ, z) are called cylindrical coordinates of M and the correspondingchange of variables is described by the following functions:

(4.17)

x = x(r, ϕ, z) = r cosϕy = y(r, ϕ, z) = r sinϕz = z(r, ϕ, z) = z

,


where r ∈ (0,∞), ϕ ∈ [0, 2π) and z ∈ R. The slight modification in thedefinition domain of the vector function −→g (r, ϕ, z) = (r cosϕ, r sinϕ, z)appeared because we want this function to be a smooth deformation(here the Jacobian is equal to r, which must be not zero). As in thecase of spherical coordinates, what we got out (r = 0, i.e. the Oz-axis, ϕ = 2π, i.e. the zOx-plane) are 1or 2-dimension pieces, so notimportant during the triple integration process. The name "cylindricalcoordinates" comes from the fact that the parallelepiped Ω = [0, R] ×[0, 2π]× [0, h] is deformed through the above cylindrical deformation −→ginto the cylindrical domain D : x2 + y2 ≤ R2, 0 ≤ z ≤ h. This lastone is not "in fact’ a smooth deformation, but the exception points areneglected during a process of triple integration, because a set of suchpoints has always zero volume. This is why we usually force things anddo not care of the true definition of a deformation!

Let us compute for instance the coordinates of the mass centre ofthe cylindrical solid (D, f), where D : y2 + z2 ≤ 16, 3 ≤ x ≤ 5 andf(x, y, z) = 2x + 1. Since the domain is symmetric (like a geometricalobject!) w.r.t. Ox-axis and since the distribution of masses is alsosymmetric w.r.t. to the same line Ox, the coordinates of the masscentreG of the solid are (xG, 0, 0). Hence, we compute only (see formula(2.4)) two triple integrals.

xG =

∫∫∫

D

xf(x, y, z)dxdydz

∫∫∫

D

f(x, y, z)dxdydz

.

Since the matter is about a cylindrical domain, we apply the changeof variable formula (4.8), but not with exactly the functions whichappear in formula (4.17). We have to make some small modificationsin this formula, but not in the essential feature of it. In the initialdescription of the cylindrical coordinates, the Oz-axis was a "pivot". Inour case "the pivot" is the Ox-axis, so that the cylindrical coordinatesin our case are (r, ϕ, x) for any pointM(x, y, z) (Why?-make a draw andexplain which are r and ϕ in this last situation!). Thus, the appropriateformulas are:

x = xy = r cosϕz = r sinϕ

,


where r ∈ [0, 4], ϕ ∈ [0, 2π] and x ∈ [3, 5]. Now we are really ready touse the change of variables formula (4.8) with the Jacobian equal to r.

mass(D) =

∫∫∫

D

(2x+ 1)dxdydz = 2

∫∫∫

D

xdxdydz +

∫∫∫

D

dxdydz =

= 2

∫∫∫

D

xdxdydz + vol(D) = 32π + 2

∫∫∫

D

xdxdydz =

= 32π +

∫ 4

0

r

(∫ 2π

0

[∫ 5

3

2xdx

]dϕ

)dr =

= 32π +

∫ 4

0

r

(∫ 2π

0

[x2 |53

]dϕ

)dr = 32π + 16

∫ 4

0

r

(∫ 2π

0

dϕ

)dr =

= 32π + 32π · 8 = 288π.

Let us compute now

∫∫∫

D

xf(x, y, z)dxdydz, i.e. the static moment

w.r.t. yOz-plane.∫∫∫

D

x(2x+ 1)dxdydz = 2

∫∫∫

D

x2dxdydz +

∫∫∫

D

xdxdydz.

But the last triple integral on the right was just computed above, sothat:∫∫∫

D

x(2x+ 1)dxdydz = 128π + 2

∫ 4

0

r

(∫ 2π

0

[∫ 5

3

x2dx

]dϕ

)dr =

= 128π +2

3

∫ 4

0

r

(∫ 2π

0

[53 − 33

]dϕ

)dr = 128π +

3136

3π =

3520π

3.

Finally, xG = 3520π3·288π ≈ 4.074 ∈ [3, 5].

The next result is very important in Mechanics. It can be done forsimple, double, triple or surface integrals (these last ones will be studiedin the next chapter). We shall give it here only for triple integrals. Weleave as an exercise for the reader to state and to prove such a resultfor simple, double or surface integrals.

T 92. (the reduction theorem in Statics) Let

(D1, f1), (D2, f2), ..., (Dn, fn)

be n general nonoverlaping solids in R3 and let

G1(x1, y1, z1), G2(x2, y2, z2), ..., Gn(xn, yn, zn)


be their corresponding mass centres. Then the mass centre G(xG, yG, zG)of the system of solids (D, f), where

D = D1 ∪D2 ∪ ... ∪Dnand f(x, y, z) = fk(x, y, z) if (x, y, z) ∈ Dk, has the coordinates xG, yGand zG computed with the formulas:

xG =x1mass(D1) + x2mass(D2) + ... + xnmass(Dn)

mass(D),(4.18)

yG =y1mass(D1) + y2mass(D2) + ...+ ynmass(Dn)

mass(D),

zG =z1mass(D1) + z2mass(D2) + ...+ znmass(Dn)

mass(D),

where obviously mass(D) = mass(D1) + mass(D2) + ... + mass(Dn).The above fractions, say the first one, is called the weighted average ofthe numbers x1, x2, ..., xn, with the weights mass(D1), ...,mass(Dn)respectively.

P. To prove this theorem, we simply use the main propertiesof the triple integrals, described in theorem 87 and the definition of themass centre of a solid (see formulas (2.4)). We also prove the formulafor xG only, because the other two can be proved in exactly the samemanner. Thus,

xG =

∫∫∫

D

xf(x, y, z)dxdydz

mass(D)=

=

∫∫∫

D1

xf(x, y, z)dxdydz + ...+

∫∫∫

Dn

xf(x, y, z)dxdydz

mass(D)=

=

∫∫∫

D1

xf1(x,y,z)dxdydz

mass(D1)·mass(D1) + ...+

∫∫∫

Dn

xfn(x,y,z)dxdydz

mass(Dn)·mass(Dn)

mass(D)=

=x1mass(D1) + x2mass(D2) + ...+ xnmass(Dn)

mass(D).


E 93. (a system of two solids) Let us apply this basic result(theorem 92) to compute the coordinates xG, yG and zG, the coordinatesof the mass centre for the system D, formed with two homogenous solidsD1, D2 (considered with the density function f = 1), where D1 is the

"north" hemisphere of the ellipsoid: x2

a2+ y2

b2+ z2

c2≤ 1, z ≥ 0 and D2 is

the "south" hemisphere of the ball: x2+(y−2b)2+z2 ≤ b2, z ≤ 0, wherea, b, c > 0. To compute the coordinates of the mass centres for D1 andD2 respectively, we need appropriate changes of variables, inspired ofthe usage of the spherical coordinates. For D1 we consider the "generalspherical coordinates", ρ, ϕ, θ :

D1 :

x = x(ρ, ϕ, θ) = ρa sin θ cosϕ,y = y(ρ, ϕ, θ) = ρb sin θ sinϕz = z(ρ, ϕ, θ) = ρc cos θ.

,

where ρ ∈ [0, 1], ϕ ∈ [0, 2π] and θ ∈ [0, π2]. The Jacobian of this trans-

formation is equal to −abcρ2 sin θ (prove this!). For instance, let uscompute the volume of D1 (which is equal to its mass because the den-sity function is 1) by using formula 4.8:

vol(D1) =

∫∫∫

D1

dxdydz = abc

∫ 1

0

ρ2

[∫ 2π

0

(∫ π2

0

sin θdθ

)dϕ

]dρ =

= abc

(∫ 1

0

ρ2dρ

)(∫ 2π

0

dϕ

)(∫ π2

0

sin θdθ

)= abc · 1

3· 2π · 1 =

2πabc

3.

Thus, the volume of the entire ellipsoid x2

a2+ y2

b2+ z2

c2≤ 1 is equal to 4πabc

3.

If a = b = c = R, we get the volume of a sphere of radius R : 4πR3

3. We

shall need bellow the volume of a half of a ball of radius b : vol = 2πb3

3.

Let us compute z1 =

∫∫∫

D1

zdxdydz

∫∫∫

D1

dxdydz

, the z-coordinate of G1, the mass

centre of D1. Since Oz is a symmetry axis for D1, y1 = x1 = 0.∫∫∫

D1

zdxdydz = abc2∫ 1

0

ρ3

[∫ 2π

0

(∫ π2

0

sin θ cos θdθ

)dϕ

]dρ =

abc2

4· 2π · sin2 θ

2

∣∣∣∣π2

0

=πabc2

4.


Hence z1 =πabc2

44πabc3

= 3c16.

Since D2 is symmetric w.r.t. the line x = 0, y = 2b, x2 = 0, y2 = 2b,

so that we need only to compute z2 =

∫∫∫

D2

zdxdydz

∫∫∫

D2

dxdydz

. Here G2, the mass

centre of D2 has the coordinates x2, y2 and z2. An appropriate changeof variables for D2 are:

x = x(ρ, ϕ, θ) = ρ sin θ cosϕ,y = y(ρ, ϕ, θ) = 2b+ ρ sin θ sinϕ

z = z(ρ, ϕ, θ) = ρ cos θ.,

where ρ ∈ [0, b], ϕ ∈ [0, 2π] and θ ∈ [π2, π]. The Jacobian is obviously

J = −ρ2 sin θ, because the function y was modified by the addition of aconstant. Hence,

∫∫∫

D2

zdxdydz =

∫ b

0

ρ3

[∫ 2π

0

(∫ π

π2

sin θ cos θdθ

)dϕ

]dρ =

=

(∫ π

π2

sin θ cos θdθ

)· 2π · b

4

4= −πb

4

4.

Thus z2 =−πb4

42πb3

3

= −3b8.

Let us use now formula (4.18) to compute xG, yG and zG.

xG =0 · vol(D1) + 0 · vol(D2)

vol(D)= 0

yG =0 · vol(D1) + 2b · vol(D2)

vol(D)=

2b · 2πb3

32πabc

3+ 2πb3

3

=2b3

ac + b2

zG =3c16

· vol(D1) − 3b8· vol(D2)

vol(D)=

3c16

· 2πabc3

− 3b8· 2πb3

32πabc

3+ 2πb3

3

=3

16

ac2 − b3

ac+ b2.

Many other practical properties of the mass centres can be provedby simple applications of the basic properties of triple integrals. Forinstance, for two nonoverlaping solids (D1, f1), (D2, f2) with their masscentresG1(x1, y1, z1) andG2(x2, y2, z2), the mass centreG of the system(D, f), where D = D1 ∪D2, is on the segment which joins G1 and G2

(prove it!). What happens when G1 = G2?



1. Find the mass of the following solids with their correspondingdensity function f(x, y, z).

a) D = [0, 1] × [0, 1] × [0, 1], f(x, y, z) = xy sin(πz); b) D = [0, 1] ×[0, 2] × [−2, 3], f(x, y, z) = 3z;

c) x2+y2 ≤ 1, 1 ≤ z ≤ 2, f(x, y, z) =√x2 + y2; d) x2+y2+z2 ≤ 1,

f(x, y, z) =√

1 + (x2 + y2 + z2)3/2;e) x2 + y2 ≤ 2x, y ≥ 0, z ≥ 0, z ≤ a, where a > 0, f(x, y, z) =

z√x2 + y2; f) the domain bounded by x2 + y2 = R2, z = 0, z = 1,

y ≤ x√

3, x, y, z ≥ 0, f(x, y, z) = 1; g) the tetrahedron bounded byx = 0, y = 0, z = 0, x+ y + z = 1, f(x, y, z) = (1 + x+ y + z)−3.

2. Find the volumes of the following domains:a) the domain D bounded by y2 + z2

4= 2x, x = 1; b) x2 + y2 + z2 −

4z ≤ 0; c) (x−1)2

a2+ y2

a2+ (z+4)2

b2≤ 1;

d) x2+y2

a≤ z ≤ a, a > 0, 0 ≤ y ≤ x; e) x2 + y2 + z2 ≤ z; f)

1 ≤ x ≤ 2, 1 ≤ y ≤ x, 1 ≤ z ≤ xy.3. Find the coordinates of the mass centre for the following solids:a) x2 + y2 + z2 ≤ 1, f(x, y, z) =

√x2 + y2 + z2; b) x2 + y2 ≤ z2,

0 ≤ z ≤ 1; f(x, y, z) =√x2 + y2;

c) x2 + y2 ≤ 1, 1 ≤ z ≤ 2, f(x, y, z) = 2z;4. Find the coordinates of the mass centre of the following domains,

bounded by:a) x2 + y2 = z, 0 ≤ z ≤ 1; b) x2 + y2 = z, x + y + z = 0; c)

y2 + 2z2 = 4x, x = 2;5. Find the inertia moment of the following solids w.r.t. Ox-axis:a) x2 +y2 +z2 ≤ 9, f = 1; b) (x−1)2 +y2 +z2 ≤ 1; c) x2 +y2 ≤ 2x,

x ≥ 0, y ≥ 0, 0 ≤ z ≤ 2, f = 2z;d) [0, 1] × [0, 2] × [0, 3]; e) x2 + z2 ≤ y2, 0 ≤ y ≤ 2, f = 2y +

1; f) the general parallelepiped generated by−→OA,

−−→OB,

−→OC, where

A(1, 3, 1), B(−2, 1, 1) and C(0,−1, 2); g) the tetrahedron [OABV ],where A(−1, 3, 2), B(−3, 1, 4) and V (1, 1, 1).

6. Find the mass centre forD = D1∪D2∪D3, whereD1 : x2+y2 ≤ 9,0 ≤ z ≤ 3, f1(x, y, z) = z,

D2 : (x−5)2 +y2 +z2 ≤ 4, f2(x, y, z) = x and D3 : [10, 11]× [5, 6]×[0, 1], f3(x, y, z) = z.

CHAPTER 8

Surface integrals

1. Deformation of a 2D-domain into a space surface

Let uov be a plane Cartesian coordinate system in R2 and let Oxyzbe a Cartesian coordinate system in R3. Let D be a plane generaldomain (bounded, closed, connected and ∂D is a piecewise smoothcurve) and let −→g : D → R3 be a continuous injective and piecewise ofclass C1 field. Moreover, we assume that at "almost" every pointM ofS = −→g (D) there exists a unique tangent plane to S. "Almost" meansthat the set A of points on S at which we have not a unique tangentplane has zero area (or area(g−1(A) = 0). We always assume that S isclosed as a topological object in R3, i.e. that R3S is an open subsetof R3. We recall that a subset A of R3 is said to be open (in R3) iffor any point P of it there exists an open ball B(P, r), r > 0 of R3

with B(P, r) ⊂ A. Such a field −→g as above is said to be a (piecewisesmooth) deformation of D into the (space) surface S = −→g (D). Since−→g (u, v) = (x(u, v), y(u, v), z(u, v), where x(u, v), y(u, v) and z(u, v) arepiecewise functions of class C1 of independent variables u and v, wealso represent −→g or S in a parametric way:

(1.1) −→g :

x = x(u, v)y = y(u, v)z = z(u, v)

, (u, v) ∈ D .

Here x of x(u, v) is the name of a function of two variables u and v.For instance, x(u, v) = u2 − v2, etc. By an abusive language we shalluse (x, y, z) either as free (independent) variables in R3, or as namesof the three functions of u and v like in formula (1.1). The threeequalities in formula (1.1) are called the parametric equations of thesurface S = −→g (D) and the surface itself S is also called the support ofthe deformation −→g . Such a deformation −→g is also called a parametricsheet. The same surface S can have many parametric representations.For instance, the plane 2x−3y+z−1 = 0 can have the following three

305

306 8. SURFACE INTEGRALS

parametric representations:

−→g 1 :

x = xy = y

z = −2x+ 3y + 1(x, y) ∈ R2.

Here D = R2, u = x, v = y, x(u, v) = u, y(u, v) = v and z(u, v) =−2u+ 3v + 1. Another parametrization of the same plane is:

−→g 2 :

x = xy = (2x+ z − 1) /3

z = z(x, z) ∈ R2.

It is now clear how do we construct the last one:

−→g 1 :

x = (1 − z + 3y) /2y = yz = z

(y, z) ∈ R2.

Hence, we can simply call a surface any deformation −→g : D → R3 oronly the support S = −→g (D) of this last one. In practice, we give asurface by its parametric equations or by an implicit form of it, S =(x, y, z) : F (x, y, z) = 0, where F is of class C1 and

gradF (x, y, z) =

(∂F

∂x(x, y, z),

∂F

∂y(x, y, z),

∂F

∂z(x, y, z)

)= −→

0 ,

i.e. in a neighborhood of each point M(a, b, c) of this surface (i.e.(a, b, c) verifies the equation F (x, y, z) = 0) we can apply the implicitfunction theorem (see [Po], theorem 82) and we can represent the sur-face in one of the following three forms:

S :

x = xy = y

z = z(x, y), (x, y) ∈ D ,S :

x = xy = y(x, z)z = z

, (x, z) ∈ D ,

S :

x = x(y, z)y = yz = z

, (y, z) ∈ D .

For instance, the unity sphere S = (x, y, z) : x2 +y2 +z2 = 1 is givenhere as an implicit equation. If we want to express it as z = z(x, y)about a pointM0(a, b, c), the partial differential

∂F∂z

(a, b, c) must be notzero, i.e. 2c = 0, or c = 0. So we can do this about all pointsM0(a, b, c)on the sphere S, but the points on the circle x2 + y2 = 1, z = 0.Since z = ±

√1 − x2 − y2, round such a point M0(a, b, c) z cannot

be defined as a univalent (univalued) function. About such a pointwe can represent the equation of the sphere S as an explicit function

1. DEFORMATION OF A 2D-DOMAIN INTO A SPACE SURFACE 307

w.r.t. x or y. Indeed, if c = 0, at least one of a or b must be not zero(otherwise 0 + 0 + 0 = 1!), say b > 0. Then the continuous expression±√

1 − x2 − z2 must be positive on a neighborhood U (find it!) of(a, b, 0), so that y =

√1 − x2 − z2 on U.

Let us recall now some facts from the differential geometry of a

surface S :

x = x(u, v)y = y(u, v)z = z(u, v)

, (u, v) ∈ D . Let P0(u0, v0) be a fixed point

in the plane domainD (see Fig.1) and let denote the above deformation−→g (u, v) by −→r (u, v) = x(u, v)

−→i + y(u, v)

−→j + z(u, v)

−→k =

−−→OM0, the

position vector of M0(x0, y0, z0), where x0 = x(u0, v0), y0 = y(u0, v0),z0 = z(u0, v0). To easier understand what is M0, we put −→r (P0) = M0,etc. (see Fig.1). Thus, a running point M on the surface S is theimage through −→r of a running point P (u, v) of the plane domain D.This means that M(x(u, v), y(u, v), z(u, v)) is such a running point onS. Usually, we associate to any point M0(x(u0, v0), y(u0, v0), z(u0, v0))of S two important vectors which generate the tangent plane TM0 atM0 to S. These are:

−→r u(P0) = xu(u0, v0)−→i + yu(u0, v0)

−→j + zu(u0, v0)

−→k and(1.2)

−→r v(P0) = xv(u0, v0)−→i + yv(u0, v0)

−→j + zv(u0, v0)

−→k .

Our hypotheses on S say that the tangent plane TM0 exists and itis unique, so that the two vectors −→r u(P0) and −→r v(P0) which give its

direction−→N (P0) = −→r u(P0) × −→r v(P0) cannot be collinear. The versor

−→n (P0) =−→r u(P0)×−→r v(P0)

‖−→r u(P0)×−→r v(P0)‖ is called the normal versor at M0 to S. Here

xu = ∂x∂u, xv = ∂x

∂v, −→r u = ∂−→r

∂u, ... etc. Thus, the equation of the tangent

plane TM0 is:

det

X − x0 Y − y0 Z − z0xu(u0, v0) yu(u0, v0) zu(u0, v0)xv(u0, v0) yv(u0, v0) zv(u0, v0)

= 0.

Let us introduce three new functions A(u, v), B(u, v) and C(u, v):−→r u(u, v) × −→r v(u, v) = A(u, v)

−→i + B(u, v)

−→j + C(u, v)

−→k . Hence the

normal versor −→n (u, v), at a running pointM(x(u, v), y(u, v), z(u, v)) ofS, has the following expression:

(1.3) −→n (u, v) =A(u, v)√

A2(u, v) +B2(u, v) + C2(u, v)

−→i +

+B(u, v)√

A2(u, v) +B2(u, v) + C2(u, v)

−→j +


+C(u, v)√

A2(u, v) +B2(u, v) + C2(u, v)

−→k .

(see Fig.1).

F 1

We are interested now how the area of a small rectangle [P0P1P2P3],where P0(u0, v0), P1(u1, v0), P2(u0, v1) and P3(u1, v1) changes when wedeform this rectangle through the deformation −→r (u, v) up to the sur-face [M0M1M2M3] (see Fig.1). The idea is the same like the idea usedin proving theorem 81.

T 93. With the above notation and hypotheses, one has

(1.4) area[M0M1M2M3] ≈≈

√A2(u0, v0) +B2(u0, v0) + C2(u0, v0) · area[P0P1P2P3],

the approximation being better and better as the diameter of [P0P1P2P3]is smaller and smaller. Hence, the number√

A2(u0, v0) +B2(u0, v0) + C2(u0, v0)

acts as a dilation or contraction coefficient of the elementary area

area[P0P1P2P3] = (u1 − u0)(v1 − v0).

Formula (1.4) can also symbolically be written:

(1.5) dσ =√A2 +B2 + C2dudv,


where dσ is called an element of area on the surface S. The pointP0(u0, v0) can be substituted with any other fixed point P (ξ, η) of therectangle

[P0P1P2P3], i.e.

(1.6) area[M0M1M2M3] ≈

≈√A2(ξ, η) +B2(ξ, η) + C2(ξ, η) · area[P0P1P2P3].

P. Let us approximate the area of the 3D-curvilinear paral-lelogram

[M0M1M2M3] with the area of the parallelogram [M0M1M2M′3] gen-

erated by the vectors−−−−→M0M1 and

−−−−→M0M2 (see Fig.1), i.e. with the num-

ber

(1.7)∥∥∥−−−−→M0M1 ×

−−−−→M0M2

∥∥∥ =

∥∥∥∥∥∥det

−→i

−→j

−→k

x1 − x0 y1 − y0 z1 − z0x2 − x0 y2 − y0 z2 − z0

∥∥∥∥∥∥,

where x1 = x(u1, v0), y1 = y(u1, v0), z1 = z(u1, v0), x2 = x(u0, v1),y1 = y(u0, v1), z1 = z(u0, v1). But

x1 − x0 = x(u1, v0) − x(u0, v0) =

=Lagrange

∂x

∂u(c1, v0) · (u1 − u0) ≈

∂x

∂u(u0, v0) · (u1 − u0),

because c1 ∈ (u0, u1) and u1 − u0 is small enough. Thus, in the sameway, we can deduce the following approximations:

x1 − x0 ≈ ∂x

∂u(u0, v0) · (u1 − u0) ≈

∂x

∂u(ξ, η) · (u1 − u0)

y1 − y0 ≈ ∂y

∂u(u0, v0) · (u1 − u0) ≈

∂y

∂u(ξ, η) · (u1 − u0)

z1 − z0 ≈ ∂z

∂u(u0, v0) · (u1 − u0) ≈

∂z

∂u(ξ, η) · (u1 − u0)

x2 − x0 ≈ ∂x

∂v(u0, v0) · (v1 − v0) ≈

∂x

∂v(ξ, η) · (v1 − v0)

y2 − y0 ≈ ∂y

∂v(u0, v0) · (v1 − v0) ≈

∂y

∂v(ξ, η) · (v1 − v0)

z2 − z0 ≈ ∂z

∂v(u0, v0) · (v1 − v0) ≈

∂z

∂v(ξ, η) · (v1 − v0).

Let us come back to formula (1.7) and obtain:∥∥∥−−−−→M0M1 ×

−−−−→M0M2

∥∥∥ ≈


≈

∥∥∥∥∥∥det

−→i

−→j

−→k

∂x∂u

∂y∂u

∂z∂u

∂x∂v

∂y∂v

∂z∂v

∥∥∥∥∥∥

∣∣∣∣∣∣(u0,v0)

(u1 − u0)(v1 − v0) =

√A2(u0, v0) +B2(u0, v0) + C2(u0, v0)(u1 − u0)(v1 − v0) ≈≈

√A2(ξ, η) +B2(ξ, η) + C2(ξ, η)(u1 − u0)(v1 − v0),

orarea[M0M1M2M3] ≈

≈√A2(u0, v0) +B2(u0, v0) + C2(u0, v0) · area[P0P1P2P3] ≈√A2(ξ, η) +B2(ξ, η) + C2(ξ, η) · area[P0P1P2P3].

Let us compute the element of area dσ on different surfaces S withthe help of formula (1.5).

E 94. (cylindrical deformation) Let D = [0, 2π] × [0, h]

(see Fig.2) and let −→r (u, v) = R cosu−→i + R sin u

−→j + v

−→k , u ∈ [0, 2π],

v ∈ [0, h] be a deformation defined on D. Since the image −→r (D) is thecircular cylinder of radius R and height h (see Fig.2),

F 2

this last deformation is said to be a cylindrical deformation. Theparametric equations of this cylinder are:

(1.8) −→g :

x = x(u, v) = R cosuy = y(u, v) = R sin uz = z(u, v) = v

, (u, v) ∈ D = [0, 2π] × [0, h].


Let us compute the element of area dσ on this cylinder. We need onlyto compute A, B, and C as the coordinates functions of −→r u(u, v) ×−→r v(u, v). But

−→r u = −R sin u−→i +R cosu

−→j + 0 · −→k

−→r v = 0 · −→i + 0 · −→j +−→k .

Thus,

−→r u×−→r v = det

−→i

−→j

−→k

−R sin u R cosu 00 0 1

= R cosu

−→i +R sin u

−→j +0

−→k .

So that√A2 +B2 + C2 = R and dσ = Rdudv. This means that if R

is large enough, then the areas on the cylinder x2 + y2 = R2 are verymuch modified. The dilation coefficient of the areas is R.

E 95. (conical deformation) Let D = [0, 2π] × [0, h] (see

Fig.3) and let −→r (u, v) = Rhv cos u

−→i + R

hv sin u

−→j + v

−→k , u ∈ [0, 2π] and

v ∈ [0, h] be a deformation of the rectangle D onto a cone (see Fig.3).

F 3

The Cartesian equation of this last cone is x2+y2 = R2

h2z2, z ∈ [0, h].

Let us compute the expression√A2 +B2 + C2.

−→r u ×−→r v = det

−→i

−→j

−→k

−Rhv sinu R

hv cosu 0

Rh

cosu Rh

sin u 1

=

=R

hv cosu

−→i +

R

hv sin u

−→j − R2

h2v−→k .


Thus

√A2 +B2 + C2 =

√R2

h2v2 +

R4

h4v2 =

R

hv

√1 +

R2

h2.

Hence dσ = Rhv√

1 + R2

h2dudv.

E 96. (spherical deformation) Let D = [0, 2π] × [0, π] (seeFig.4)

F 4

and let −→r (ϕ, θ) = R sin θ cosϕ−→i +R sin θ sinϕ

−→j + R cos θ

−→k , ϕ ∈

[0, 2π] and θ ∈ [0, π] be a deformation of the rectangle D onto the spherex2 + y2 + z2 = R2. In this case,

−→r ϕ ×−→r θ = det

−→i

−→j

−→k

−R sin θ sinϕ R sin θ cosϕ 0R cos θ cosϕ R cos θ sinϕ −R sin θ

=

= −R2 sin2 θ cosϕ−→i − R2 sin2 θ sinϕ

−→j − R2 sin θ cos θ

−→k .

Hence√A2 +B2 + C2 = R2 sin θ.

In the case of a surface S which is explicitly given w.r.t. z, z =z(x, y), the formulas can be putted in another more convenable form.Indeed, the parametrization of S is:

x = xy = y

z = z(x, y), (x, y) ∈ D.

2. SURFACE INTEGRAL OF THE FIRST TYPE. THE MASS OF A 3D-LAMINA.313

Here x stands for u and y stands for v. Let us compute A, B and C inthis particular case:

−→r x ×−→r y = det

−→i

−→j

−→k

1 0 ∂z∂x

0 1 ∂z∂y

= −∂z

∂x

−→i − ∂z

∂y

−→j +

−→k

Hence

(1.9) A = −∂z∂x, B = −∂z

∂y, C = 1.

Therefore

(1.10) dσ =

√(∂z

∂x

)2

+

(∂z

∂y

)2

+ 1dxdy

in this case.

2. Surface integral of the first type. The mass of a3D-lamina.

LetD be a (general) plane domain like above (closed, bounded, con-nected and ∂D is a piecewise smooth) in the uov-plane. Let −→r : D →R3, −→r (u, v) = x(u, v)

−→i + y(u, v)

−→j + z(u, v)

−→k be a space piecewise

smooth deformation of D onto a surface S = −→r (D). Let f : S → Rbe a piecewise continuous function of variables x, y and z, defined onS. It will be called a density function for S. The pair (S, f) will becalled a 3D-lamina ( 3-D plate). Let ∆ = Di, i = 1, 2, ..., n be adivision of nonoverlaping subdomains of the same type like D and let

∆∗ = −→r (Di)def= Si, i = 1, 2, ..., n be the corresponding division of

the space surface S. Let Pi(ξi, ηi), Pi(ξi, ηi) ∈ Di be a set of mark-ing points for the division ∆ and let Mi(x(ξi, ηi), y(ξi, ηi), z(ξi, ηi)),i = 1, 2, ...n be the set of marking points of ∆∗ which are the imagesof Pi through the deformation −→r . The norm of the division ∆ is likeusual the number ‖∆‖ = maxdiam(Di) : i = 1, 2, ..., n. It is clearthat ‖∆‖ → 0 implies that ‖∆∗‖ → 0. A Riemann sum for f,∆∗ andMi is defined as follows:

(2.1) Sf(∆∗, Mi) =

n∑

i=1

f(x(ξi, ηi), y(ξi, ηi), z(ξi, ηi)) · area(Si).

A natural interpretation of this last sum is the following. The entiremass mass(S) of the 3D-lamina (S, f) can be well approximated withthe sum of all masses of the small homogenous 3D-laminas (Si, fi),where for the density function fi we take the constant function with


the same value f(x(ξi, ηi), y(ξi, ηi), z(ξi, ηi)) at any point of Si, i.e. weapproximated on Si the variable density f with its value at a fixedpoint Mi of Si. Thus the Riemann sum (2.1) is nothing else then thesum of the masses of all these new constructed laminas.

D 29. Let us preserve the above notation and definitions.We say that the function f is Riemann integrable on S if there exists areal number I such that for any small ε > 0, there exists a small δ > 0(which depends on ε) with the following property: if ‖∆∗‖ < δ, then|I − Sf(∆

∗, Mi)| < ε for any P (ξi, ηi) ∈ Di. Hence I can be wellapproximated with Riemann sums of the type 2.1. This I can also beinterpreted as the mass mass(S), the mass of the 3D-lamina (S, f), i.e.the mass of S with the density function f (when f ≥ 0). The number I

is denoted by

∫∫

S

f(x, y, z)dσ and we call it the surface integral of the

first type of f on the surface S. In particular, for f = 1, we obtain the

very useful formula: area(S) =

∫∫

S

dσ. Like above, dσ is said to be the

element of area on S and it can be interpreted (a symbol in fact!) asan area of a very small surface Si, when ‖∆∗‖ → 0.

This number I is unique.By using only this definition, it is possible to prove (prove them by

imitating the similar proofs for double integrals!) the following mainproperties of the surface integral of the first type:

1) The mapping f

∫∫

S

f(x, y, z)dσ is a linear mapping, i.e.

(2.2)∫∫

S

[αf(x, y, z)+βg(x, y, z)]dσ = α

∫∫

S

f(x, y, z)dσ+β

∫∫

S

g(x, y, z)dσ.

2)

(2.3)

∫ ∫

S=∪Ni=1Si

f(x, y, z)dσ =N∑

i=1

∫∫

Si

f(x, y, z)dσ,

where Si are nonoverlaping surfaces.3)

(2.4) f ≤ g implies

∫∫

S

f(x, y, z)dσ ≤∫∫

S

g(x, y, z)dσ,


in particular if f ≥ 0, then

∫∫

S

f(x, y, z)dσ ≥ 0.

4) (a mean theorem)

(2.5)

∫∫

S

f(x, y, z)dσ = f(ξ, η, θ) · area(S),

where M(ξ, η, θ) is a point on S and f is assumed to be continuous.5)

(2.6) area(S) =

∫∫

S

dσ

6)

mass(S) =

∫∫

S

f(x, y, z)dσ,

where f is a density function on S.7) The coordinates of the mass centre of the 3D-lamina (S, f) are

calculated with the help of the following formulas:(2.7)

xG =

∫∫

S

xf(x, y, z)dσ

∫∫

S

f(x, y, z)dσ

, yG =

∫∫

S

yf(x, y, z)dσ

∫∫

S

f(x, y, z)dσ

, zG =

∫∫

S

zf(x, y, z)dσ

∫∫

S

f(x, y, z)dσ

.

8) Let (d) be a straight line in space and let g(x, y, z) be thedistance-function of a point M(x, y, z) up to (d). Let S be a surface inspace and let (S, f) be a 3D-lamina. Then the moment of inertia of(S, f) w.r.t. to (d) (when S rotates around (d)) is:

(2.8) I(d) =

∫∫

S

g2(x, y, z)f(x, y, z)dσ.

In particular, if (d) is the Oz-axis, then

Iz =

∫∫

S

(x2 + y2) · f(x, y, z)dσ.

Now we shall see how to practically compute such surface integralsof the first type. They are in fact the 2-dimensional variants of the lineintegrals of the first type (see definition 17).


T 94. Let (S, f) be a 3D-lamina as above (D is a generalplane domain, −→r : D → R3 is a piecewise smooth deformation of Donto S, f is piecewise continuous density function on S). Then we can

reduce the computation of I =

∫∫

S

f(x, y, z)dσ to the computation of a

double integral on the initial undeformed domain D :

(2.9)

∫∫

S

f(x, y, z)dσ =

=

∫∫

D

f(x(u, v), y(u, v), z(u, v))√A(u, v)2 +B(u, v)2 + C(u, v)2dudv.

We simply substituted the independent variables x, y, z with the defor-mation functions x(u, v), y(u, v), z(u, v) and finally we putted insteadof dσ its expression from formula 1.5.

P. The proof is based on the same idea like the leading ideaof all the proofs for obtaining change of variables formulas. We wellapproximate D with a finite union ∪Nj=1Rj ⊂ D of nonoverlaping sim-ple rectangles Ri. Let R∗i = −→r (Ri) be the image of Ri through thedeformation −→r . Because of the hypotheses on D and on the deforma-tion −→r of D onto S, the union ∪Nj=1R

∗j well approximate S and the

curvilinear parallelograms R∗j are nonoverlaping figures. Thus,(2.10)∫∫

S

f(x, y, z)dσ ≈N∑

j=1

f(x(ξj, ηj), y(ξj , ηj), z(ξj , ηj)) · area(R∗j),

where Pj(ξj , ηj) ∈ Rj is a set of marking points in D. But theorem93 applied to area(R∗j ) transforms the last sum of the formula (2.10)into a sum which approximates the double integral of (2.9):

N∑

j=1

f(x(ξj , ηj), y(ξj, ηj), z(ξj, ηj)) · area(R∗j) ≈

≈N∑

j=1

f(x(ξj, ηj), y(ξj , ηj), z(ξj, ηj))×

×√A(ξj, ηj)

2 +B(ξj, ηj)2 + C(ξj , ηj)

2 · area(Rj) ≈


≈∫∫

D


Hence,∫∫

S

f(x, y, z)dσ ≈

≈∫∫

D


Since both these numbers are fixed numbers, they must coincide, i.e.the equality (2.9) is completely proved.

We can now compute different quantities which are expressed bysurface integrals of the first type.

E 97. Let S : x2 + y2 = 2z, 0 ≤ z ≤ 2 be a paraboloid (seeFig.5).

F 5

It is realized as the image of disc x2+y2 ≤ 4 of radius 2 and centre at

O (see Fig.5), through the deformation −→r (x, y) = x−→i +y

−→j + x2+y2

2

−→k .

Since a parametrization for S is explicitly expressed w.r.t. z, z(x, y) =x2+y2

2, we can apply formula (1.10) and find

(2.11) dσ =

√(∂z

∂x

)2

+

(∂z

∂y

)2

+ 1dxdy =√x2 + y2 + 1dxdy.


a) Let us compute area(S).

area(S) =

∫∫

S

dσ =

∫ ∫

D:x2+y2≤4

√x2 + y2 + 1dxdy =

x=ρ cos θ=

x=ρ sin θ

∫ 2

0

ρ√ρ2 + 1

(∫ 2π

0

dθ

)dρ =

= π

∫ 2

0

(ρ2 + 1

) 12 d(ρ2 + 1) = π

(ρ2 + 1)32

32

∣∣∣∣∣

2

0

=2π

3

(53/2 − 1

).

b) Find its mass, if the density function is f(x, y, z) = 3z.

mass(D) =

∫∫

S

f(x, y, z)dσ =3

2

∫ ∫

D:x2+y2≤4

(x2+y2)√x2 + y2 + 1dxdy =

=3

2· 2π

∫ 2

0

ρ3√ρ2 + 1dρ

ρ2+1=t2= 3π

∫ √5

1

t2(t2 − 1)dt = etc.

c) Find the moment of inertia of (S, f), f = 3z, around the Ox-axis.

Ix = 3

∫∫

S

x2 + y2

2

(y2 +

(x2 + y2)2

4

)√

1 + x2 + y2dxdy =

=3

8

∫ 2

0

ρ3√ρ2 + 1

(∫ 2π

0

[4ρ2 sin2 θ + ρ4

]dθ

)dρ =

(2.12) =3

2

∫ 2

0

ρ5√ρ2 + 1

(∫ 2π

0

sin2 θdθ

)dρ+

3π

4

∫ 2

0

ρ7√ρ2 + 1dρ.

But ∫ 2π

0

sin2 θdθ =

∫ 2π

0

1 − cos 2θ

2dθ = π,

so that 2.12 becomes

Ix =3π

2

∫ 2

0

ρ5√ρ2 + 1dρ +

3π

4

∫ 2

0

ρ7√ρ2 + 1dρ = etc.

3. FLUX OF A VECTOR FIELD THROUGH AN ORIENTED SURFACE. 319

F 6

3. Flux of a vector field through an oriented surface.

Let us consider a plane (P ) and a fixed point O (the origin) in it.Let (dx) be an oriented line in (P ) which passes through O. Let (dy)be another oriented line in (P ) which also passes through O and isperpendicular on (dx) at O (see Fig.6).

In fact we have two possibilities to orientate (by fixing a nonzerovector or "an arrow" with the support line parallel to (dy)!) this last

line (dy). One is given by a vector−−→OB (see Fig.6) and the other is de-

fined by the vector−→OC = − −−→

OB. If we put a screw at O, perpendicular

on (P ) and, if we rotate it from−→OA to

−−→OB around O on the "smallest

path" (= 90), the direction of its advancement gives rise to anotheroriented line (dz) which is perpendicular on (dx) and (dy). This means

that the direction of (dz) is given by the vector−−→OD =

−→OA×−−→

OB, i.e. wejust obtained "the up" part of the plane (P ). "The down" part of (P )

is given by − −−→OD. Let

−→i ,

−→j and

−→k be the versors of

−→OA,

−−→OB and

−−→OD

respectively (see Fig.6). If one chooses the other orientation of (dy),

given by−→OC and, if one rotates the screw from

−→OA to

−→OC, it advances

on the other direction of−→k , i.e. on the direction of −−→

k . The Cartesian

coordinate system O;−→i ,

−→j ,

−→k is said to be direct or right oriented

and the other O;−→i ,

−→j ,

−→−k is called inverse or left oriented. Thus

our plane (P ) has two "faces": one which corresponds to−→k (as a free

vector if O runs on the plane) and the other which corresponds to −−→k .

This implies that if we move continuously O along a closed oriented

(see definition 18) smooth curve (γ) ⊂ (P ), the vector−→k (fixed in O)

is moving on this curve and comes back at the same initial position.


We can observe that if we continuously move the plane Cartesian co-

ordinates system O;−→i ,

−→j in the plane (P ), we never can reach the

position of the Cartesian coordinate system O;−→i ,−−→

j . This is whywe call O;

−→i ,

−→j a direct oriented plane Cartesian coordinate system

and the other, O;−→i ,−−→

j , an inverse orientated Cartesian coordinatesystem.

The definition of a direct or inverse orientation of a Cartesian co-ordinate system does not depend on the position of the point O, but

on the basis of vectors −→i ,−→j ,−→k . This is why the true mathemati-cal model of "orientation" is connected to the bases of the 3D-vectorspace of free vectors. We say that two basis of vectors v1, v2, v3 andw1, w2, w3 have the same orientation if the matrix C = (cij), con-structed as follows:

w1 = c11v1 + c21v2 + c31v3

w2 = c12v1 + c22v2 + c32v3

w3 = c13v1 + c23v2 + c33v3,

(called the transition matrix from the basis v1, v2, v3 to the basisw1, w2, w3) has its determinant detC a positive real number. Theinfinite set of all the bases in V3, the real vector space of all 3D-freevectors, can be divided into two nonoverlaping (their intersection isempty) subsets A and A−. Two bases v1, v2, v3 and w1, w2, w3 be-longs to the same subset if the transition matrix from one to anotherhas a positive determinant. Let denote by A the subset (or class) which

contains the basis −→i ,−→j ,−→k and by A− the subset (or class) which

contains the basis −→i ,−→j ,−→−k. We leave as an exercise for the readerto prove that A∪A− contains all the bases of V3 and that A∩A− = ∅,the empty set. We call A the direct orientation of the space V3 andA− the inverse orientation of V3. The same can be done for V2, the realvector space of the 2D-free vector space (i.e. the plane free vectors).Thus, whenever we fix a Cartesian coordinate system uov in a plane(P ), we just fixed a direct orientation:"from u to v". What is this?

The direction of the ou-axis is given by a versor−→i′ . The direction of

ov-axis is given by a versor−→j′ . By definition the class A of orientation

into which the base −→i′ ,−→j′ belongs is said to be the direct orientationof the Cartesian system uov. "From u to v" means that we take firstly−→i′ then, secondly, we take

−→j′ . We simply say that uov is an oriented

plane Cartesian coordinate system (here the direct orientation is "fromu to v” and the inverse one is "from v to u") (see Fig.7).The closed


curve in the uov-plane of Fig.7 is direct oriented because "its direc-tion" or "its arrow" is "from u to v”, the trigonometric direction or thecounterclockwise direction. Let D be a plane general domain (closed,bounded, connected and ∂D an oriented piecewise smooth curve). Be-ing in the plane uov, we say that D is oriented by this plane. Moreexactly, ∂D is directly oriented as we can see in Fig.7. If we walk on∂D in the direction of ∂D, the plane domain D must remain on theleft side. This last quality makes D to be an oriented domain.

Let −→g (u, v) = x(u, v)−→i + y(u, v)

−→j + z(u, v)

−→k be a smooth de-

formation (see its exact definition in Section 1, this chapter) definedon an oriented (like above!) general domain D, with values in V3 (orin R3). Let (S) = −→g (D) be the surface defined by this deformation.Let P0(u0, v0) be a point in D and let the direct orientated "coordinatecurves" γu : v = v0 and γv : u = u0 which are passing through P0

(see Fig.7). Let M0 = −→g (P0) and let Γu and Γv be the images on thesurface (S) of γu and γv, respectively. We assume that the curves Γuand Γv are also oriented with the same orientation like that one trans-ported from γu and γv, respectively (see Fig.7). Γu and Γv are calledthe coordinate curves on (S) which passe through M0.Thus,

(3.1) Γu :

x = x(u, v0)y = y(u, v0)z = z(u, v0)

, u ∈ prou−axis(D);

Γv :

x = x(u0, v)y = y(u0, v)z = z(u0, v)

, v ∈ prov−axis(D).

Hence, the "velocity" along Γu is

−→r u =∂−→r∂u

=

(∂x

∂u,∂y

∂u,∂z

∂u

)

and the "velocity" along Γv is

−→r v =∂−→r∂v

=

(∂x

∂v,∂y

∂v,∂z

∂v

).

We associate to the point M0 ∈ (S) its normal versor

−→n (M0) =−→r u(u0, v0) ×−→r v(u0, v0)

‖−→r u(u0, v0) ×−→r v(u0, v0)‖on (S) (see Fig.7). Here, in the definition of −→n (M0), we have incor-porated the orientation of D namely, we see that in the cross productof the numerator we take −→r u(M0), then

−→r v(M0), i.e. "from u to v”.Since −→n (M0) is a vesor, it uniquely defines a point M∗

0 (its extremity


if −→n (M0) =−−−→OM∗

0 ) on the unit sphere (US) : x2 + y2 + z2 = 1 in theOxyz-space (the space into which (S) is considered). Let us leave M0

to run freely on (S) and let us denote it by M. We obtain in this waya new function M M∗, which associates to a point M of (S), apoint M∗ on (US). Let us call this function the orientation functionand denote it by Or : (S) → (US), Or(M) = M∗.

F 7

D 30. Let us preserve all of these notation and hypotheseslike above. If the vector function Or : (S) → (US), Or(M) = M∗, isa continuous function, we say that our surface (S) is orientable. Thisfunction defines the direct (right) orientation ("from u to v") or thedirect face of (S). The function Or− : (S) → (US), Or−(M) = −M∗,the symmetric of M w.r.t. the origin O, is also continuous if Or iscontinuous and it defines the inverse (left) orientation ("from v to u")or the reverse face of (S). The continuity of Or means that if a pointMstarts from a fixed pointM0 ∈ (S) and runs on any continuous "closed"curve γ ⊂ (S) which containsM0, then the normal vector

−→n (M), whichstarts from −→n (M0), after covering the entire curve γ, it comes back toits initial position −→n (M0). Then we say that (S) has two distinct faces,i.e. it is orientable. Moreover, we also see that the orientation functionOr is continuous if the deformation −→g : D → −→g (D) = (S) has acontinuous inverse and if Or −→g : D → (US) is continuous. This isthe case for instance when we can find an explicit parameterization of−→g and Or −→g is continuous.

E 98. For instance, a plane (P ) : ax + by + cz + d = 0is an orientable surface. Indeed, say a = 0, a > 0 and let us use the


following explicit w.r.t. x parametrization

x = − by+cz+da

y = yz = z

, (y, z) ∈ R2.

Let us consider R2 with the oriented yOz-Cartesian coordinate system("from y to z”). Since

−→r y ×−→r z = det

−→i

−→j

−→k

− ba

1 0− ca

0 1

=

−→i +

b

a

−→j +

c

a

−→k ,

then −→n (M) =−→i + b

a

−→j + c

a

−→k√

1+ b2

a2+ c2

a2

= a−→i +b

−→j +c

−→k√

a2+b2+c2. Thus, the orientation function

becomes:(−by + cz + d

a, y, z

)→

(a√

a2 + b2 + c2,

b√a2 + b2 + c2

,c√

a2 + b2 + c2

),

i.e. a constant function, which always is continuous. Therefore, anyplane has two faces, i.e. it is orientable.

Not all surfaces are orientable, i.e. they have two faces. An easy toconstruct surface which has only one face is Möbius surface (see Fig.8).This surface is obtained by taking a rectangle strip [ABCD] and bygluing [AB] with [BC] such that A ≡ C and B ≡ D. Let (γ) be the"closed" curve which was initially the midline in the rectangle [ABCD](see Fig.8). Let us continuously and increasingly move a normal versor−→n along (γ) by starting from M. We see that whenever we come backtoM the initial versor −→n becomes −−→n . Thus, the orientation functionM

−→n (M) M∗ ∈ (US) is not continuous. Hence this surfacecannot be orientable, i.e. it has only one face (one cannot separatestwo distinct faces of it!).

A natural question arises. Is it possible to find a "special" parame-trization for this Möbius surface such that it be orientable? It can beproved that this last possibility cannot appear. To be or not to be ori-entable does not depend on a particular parameterization of a surfaceS. But, it is possible to change the direct face of (S) with the inverseone, if we change the initial parametrization of (S) with another one.For instance, let us consider the same plane (P ) : ax+ by+ cz+ d = 0,a > 0, as in example 98 with the following parametrization:

x = −−bu+cv+da

y = −uz = v

, (u, v) ∈ R2.


F 8

It is easy to see that−→n (M) = −a−→i +b

−→j +c

−→k√

a2+b2+c2, i.e. the normal versor of the

other side. Let (S) be a surface with two equivalent parametrizations−→g : D → (S) and −→g ′ : D′ → (S). Then there exists a diffeomorphismφ : D′ → D such that −→g ′ = −→g φ. It is easy to prove that −→g and−→g ′ defines one and the same face if and only if the Jacobian of φ is apositive number.

In general, if a surface (S) has an explicit parametrization, sayw.r.t. z,

(S) :

x = xy = y

z = z(x, y), (x, y) ∈ D ⊂ xOy,

then A = − ∂z∂x, B = − ∂z

∂yand C = 1 (see formula (1.9)). Since

(3.2) −→n (M) = cosα−→i + cosβ

−→j + cos γ

−→k =

=A(u, v)√

A(u, v)2 +B(u, v)2 + C(u, v)2−→i +

+B(u, v)√

A(u, v)2 +B(u, v)2 + C(u, v)2−→j +

+C(u, v)√

A(u, v)2 +B(u, v)2 + C(u, v)2−→k ,

then cos γ = 1√A2+B2+1

> 0 i.e. γ ∈ [0, π2). Many times, this last

information is sufficient to determine the direct face (which correspondsto the direct orientation) of the surface (S).


E 99. Let (S) :

x = xy = y

z =√x2 + y2

, x2 + y2 ≤ R2 be the

conic surface of Fig.9.

F 9

We are interested to find the mathematical description of the normalversors −→n which define the exterior side of this conic surface (see −→nin Fig.9). The angle γ between −→n and Oz is greater then 90, so thatcos γ < 0, i.e.

cos γ = − 1√A(u, v)2 +B(u, v)2 + 1

.

Hence, we must put a minus in front of cosα, cosβ and cos γ, in theexpression of −→n (see (3.2)). Therefore, in our case, for anyM(x, y, z) ∈(S),

−→n (M) =

(x√

2√x2 + y2

,y√

2√x2 + y2

,− 1√2

).

R 34. It is not difficult to see that if D is a smooth domain(closed, bounded, connected and ∂D is a smooth curve) and if −→r : D →−→r (D) = (S) is a smooth deformation of D onto the surface (S) (−→r isof class C1, injective and −→r u ×−→r v = 0 on D), then (S) is orientable.

Let now D be a closed, bounded and connected subset of the uov-plane with its boundary ∂D a piecewise directed oriented curve in uov.Let D = ∪ni=1Di be a division of D with nonoverlaping (Di ∩Dj has a


zero area for i = j) oriented subdomains of the same type like D (seeFig.10). We see in this figure that the intersection curve Di ∩Dj (if itexist like a curve!) is always double oriented. Moreover, the orientationof each Di (i.e. the orientation of ∂Di) is completely determined bythe orientation of D (i.e. of ∂D).

F 10

Let −→r (u, v) = x(u, v)−→i + y(u, v)

−→j + z(u, v)

−→k , −→r : D → −→g (D) =

(S) be a piecewise smooth deformation and let Pi(ui, vi) ∈ Di, i =1, 2, ..., n be a set of marking points of the division Di of D. ThenSi = −→g (Di), i = 1, 2, ..., n is a division of (S). If

‖Di‖ = maxdiam(Di) : i = 1, 2, ..., nbecomes smaller and smaller, then ‖Si‖ becomes smaller and smaller,except maybe a subset of points of zero area of D. Let Mi = −→r (Pi),i.e. Mi(x(ui, vi), y(ui, vi), z(ui, vi)) for i = 1, 2, ..., n. We assume that(S) and, as a consequence, all (Si) are orientable surfaces. Let

−→n (Mi) =−→r u(ui, vi) ×−→r v(ui, vi)

‖−→r u(ui, vi) ×−→r v(ui, vi)‖=

(3.3) =A(ui, vi)√

A(ui, vi)2 +B(ui, vi)2 + C(ui, vi)2−→i +


+B(ui, vi)√

A(ui, vi)2 +B(ui, vi)2 + C(ui, vi)2−→j +

+C(ui, vi)√

A(ui, vi)2 +B(ui, vi)2 + C(ui, vi)2−→k ,

be the normal versor at Mi, which belongs to the direct orientation

of (S). Let−→F : (S) → R3,

−→F (x, y, z) = P (x, y, z)

−→i + Q(x, y, z)

−→j +

R(x, y, z)−→k , be a field of vectors of class C1 almost everywhere on

(S), i.e. the subset of (S) on which−→F is not of class C1 has zero

area. The impact of the force−→F (M) with the surface (S) at the point

M ∈ (S) is measured by the projection of this force−→F (M) on the

normal versor −→n (M), i.e.−→F (M) ·−→n (M) (dot product). This quantity

is called the flux of the vector−→F (M) through the surface (S), at the

point M. If the diameter of (Si) is small enough, we can approximateit with a plane surface, which can be viewed as a part of the tangent

plane of (S) at the fixed point Mi ∈ (Si). We also approximate−→F (M)

with the constant vector−→F (Mi) for any pointM ∈ (Si). Thus, the flux−→

F (M) ·−→n (M) atM can be approximated with the flux−→F (Mi) ·−→n (Mi)

at Mi. Therefore, the total flux on (Si) can be approximated with[−→F (Mi) · −→n (Mi)

]· area(Si) (here the first ” · ” means the dot product

but the second ” · ” means the usual multiplication between two realnumbers). This last number is the volume of a right cylinder with the

basis (Si) and the height−→F (Mi)·−→n (Mi). Hence, the global flux Φ(S)(

−→F )

of the field−→F through the oriented (S) can be well approximated with

the following Riemann sum:

(3.4) S−→F ,(S)

(Si, Mi) =n∑

i=1

[−→F (Mi) · −→n (Mi)

]· area(Si).

D 31.−→F is integrable on the direct oriented surface (S)

if there exists a real number I such that for any small real numberε > 0, one can find another small real number δ > 0 (which de-pends on ε) such that if Di is a division of D with ‖Di‖ < δ,

then∣∣∣I − S−→

F ,(S)(Si, Mi)

∣∣∣ < ε for any set of marking points Pi,Pi ∈ Di. I is called the surface integral of the second type of

−→F on

the oriented surface (S). It is denoted by

∮ ∮

(S)+

−→F · −→n dσ. The sign "+"


means that the integral is considered on the direct face of (S). When-

ever we write

∮ ∮

(S)

−→F · −→n dσ, we must understand that the matter is

about the direct face of (S).

Hence, we can well approximate I with Riemann sums of the form(3.4). This last number I can be interpreted as the "global" flux of

the field−→F through the oriented surface (S), considered with its direct

face (or with its direct orientation "from u to v"). The surface integralof the second type is the analogous notion in the 2-dimensional case ofthe line integral of the second type for oriented curves (this last lineintegral is an 1-dimentional object).

Using only this definition, we can easily prove (do it!) the followingbasic properties of the surface integrals of the second type.

1)

∮ ∮

(S)−

−→F · −→n dσ = −

∮ ∮

(S)+

−→F · −→n dσ, where (S)− means the surface

(S) with its inverse orientation ("from v to u"). To see this one canchange −→n (Mi) with −−→n (Mi) in the above Riemann sums.

2) The mapping−→F

∮ ∮

(S)

−→F · −→n dσ is a linear mapping, i.e.

∮ ∮

(S)

(α−→F + β

−→G)· −→n dσ = α

∮ ∮

(S)

−→F · −→n dσ + β

∮ ∮

(S)

−→G · −→n dσ.

3) ∮ ∮

(S)∪(T )

−→F · −→n dσ =

∮ ∮

(S)

−→F · −→n dσ +

∮ ∮

(T )

−→F · −→n dσ,

where (T ) is another oriented surface such that on the common curve∂(S) ∩ ∂(T ) the orientation induced by the orientation of (T ) mustbe the inverse orientation of the same intersection curve viewed withthe orientation induced by the orientation of (S) (see the situation oftwo neighboring oriented subdomains in Fig.10 or in Fig.13). If thesurfaces (S) and (T ) have no small smooth curve in common, thenthe orientation of (T ) cannot be connected with the orientation of (S)by anything else, so that in the union (S) ∪ (T ) they keep their ownorientation.

Let us now try to prove the existence and to compute the surface

integral of the second type

∮ ∮

(S)

−→F · −→n dσ. For this we approximate the


area of (Si) in formula (3.4) by using formula (1.5):

(3.5) area(Si) ≈√A(ui, vi)2 +B(ui, vi)2 + C(ui, vi)2area(Di).

We can also prove such a formula by the use of formula (2.9) and themean theorem for double integrals:

area(Si) =

∫ ∫

(Si)

dσ =

∫∫

Di

√A(u, v)2 +B(u, v)2 + C(u, v)2dudv =

=√A(u∗, v∗)2 +B(u∗, v∗)2 + C(u∗, v∗)2

∫∫

Di

dudv ≈

≈√A(ui, vi)2 +B(ui, vi)2 + C(ui, vi)2area(Di),

because both points Pi(ui, vi) and P ∗i (u∗, v∗) are in the same smalldomain Di. Let us come back in formula (3.4) with this expression ofthe area(Si) and with the expression of −→n (Mi) from formula (3.3):

n∑

i=1

[−→F (Mi) · −→n (Mi)

]· area(Si) ≈

≈n∑

i=1

[P (Mi)A(Mi) +Q(Mi)B(Mi) +R(Mi)C(Mi)] area(Di).

But this last sum is a Riemann sum of the double integral∫∫

D

[P (x(u, v), y(u, v), z(u, v))A(u, v)+Q(x(u, v), y(u, v), z(u, v))B(u, v)+

+R(x(u, v), y(u, v), z(u, v))C(u, v)]dudv.

Since this last double integral and I, the surface integral of the secondtype∮ ∮

(S)

−→F · −→n dσ, can be well approximated by the same Riemann

sums, we must conclude that they must be equal, i.e.∮ ∮

(S)

−→F · −→n dσ =

(3.6)

=

∫∫

D

[P (x(u, v), y(u, v), z(u, v))A(u, v)+Q(x(u, v), y(u, v), z(u, v))B(u, v)+

+R(x(u, v), y(u, v), z(u, v))C(u, v)]dudv.


Thus we just reduced the computation of a surface integral of the sec-ond type to the calculation of a double integral. If we want to changethe face of (S), we must put a minus sign in front of the above doubleintegral. We can also express (prove it!) the surface integral of thesecond type as a surface integral of the first type:(3.7)∮ ∮

(S)

−→F ·−→n dσ =

∫∫

(S)

[P (x, y, z) cosα+Q(x, y, z) cosβ +R(x, y, z) cos γ] dσ,

where cosα, cosβ and cos γ are the coordinates of the normal versor−→n (M) at the point M(x, y, z). This is not a practical formula. Butit is useful in order to find a new expression for the surface integral∮ ∮

(S)

−→F · −→n dσ. If we look at Fig.11

F 11

we see that the area-projection prxOydσ of an orientated small ele-ment dσ (for instance the area of an (Si))of the surface (S) on the xOy-plane is a direct oriented area in this plane, usually denoted by dxdy. Itis clear enough that the order dx then dy in the symbol dxdy is essential,because it reflects the direct orientation "from x to y" in the xOy-plane.Hence dxdy = −dydx (dxdy is the "up" area and dydx is the "down"


area-see the arrows in Fig.11). For instance, przOydσ = dzdy, becausethe orientation of the boundary of this projection is "from z to y" (seeFig.11). Since the area-projections dxdy ≈ cos γdσ, dydz ≈ cosαdσand dzdx ≈ cosβdσ (see also remark 36), from formula (3.7) (takingagain Riemann sums, etc. ... do it!), we get:

(3.8)

∮ ∮

(S)

−→F · −→n dσ =

∮ ∮

(S)

Pdydz +Qdzdx+Rdxdy,

this last being a formal expression, not used in practice. The expressionω = Pdydz +Qdzdx+Rdxdy is said to be a differential form of order2 in three variables. Do not forget the order: dzdx and not dxdz!

E 100. Let us compute I =

∮ ∮

(S)

xdydz+ ydzdx, where (S)

is the interior face of the paraboloid: z = x2 + y2, 0 ≤ z ≤ 4 (seeFig.12).

F 12

In this case P (x, y, z) = x, Q(x, y, z) = y and R(x, y, z) = 0, i.e.−→F = (x, y, 0). Since the surface has an explicit parametrization w.r.t.z :

x = xy = y

z = x2 + y2, x2 + y2 ≤ 4,


we can use formula (3.2) and find A = − ∂z∂x

= −2x, B = −∂z∂y

= −2y,

C = 1, γ ∈ [0, π2],

−→n (x, y, z) =

(−2x√

1 + 4x2 + 4y2,

−2y√1 + 4x2 + 4y2

,1√

1 + 4x2 + 4y2

).

Let us use now the basic formula (3.6):

∮ ∮

(S)

xdydz + ydzdx = −2

∫ ∫

D:x2+y2≤4

(x2 + y2)dxdy =

= −2

∫ 2

0

ρ3(∫ 2π

0

dθ

)dρ = −4π

∫ 2

0

ρ3dρ = −16π.

R 35. If (S) is not of class C1 but it is piecewisely of classC1 (see Fig.13), then we give an orientation to each smooth piecesof (S) such that on the common intersection curves, these last curvesmust have 2 distinct type of orientation (each of them coming fromthe orientation of the neighboring subdomains). In fact, the orienta-tion of ∂(S) completely determines the orientation of all the componentsubdomains.

F 13

E 101. Let us compute the flux of the field−→F = (0, 0, z)

on the exterior faces of the tetrahedron [OABC] of Fig.14. Since thesurface is a union of 4 distinct surfaces with different parametrizations,


we successively compute:

I4 =

∮ ∮

[ABC]

(0, 0, z) · −→n4dσ.

Since

F 14

[ABC] :

x = xy = y

z = 1 − x− y, x+ y ≤ 1,

A = 1, , B = 1, C = 1, so

I0 =

∫ ∫

x+y≤1

(1 − x− y)dxdy =

∫ 1

0

(∫ 1−x

0

(1 − x− y)dy

)dx =

∫ 1

0

(y − xy − y2

2

)∣∣∣∣1−x

0

dx =

∫ 1

0

(x2

2− x+

1

2

)dx =

1

6.

Now

I1 =

∮ ∮

[OAB]

(0, 0, z) · −→n1dσ = 0,

because −→n1 = −−→k and z = 0 on [OAB].

I2 =

∮ ∮

[OCA]

(0, 0, z) · −→n2dσ =

∮ ∮

[OCA]

(0, 0, z) · (−−→j )dσ = 0.


I3 =

∮ ∮

[OBC]

(0, 0, z) · (−−→i )dσ = 0.

Hence the flux is equal to I1 + I2 + I3 + I4 = 16.

R 36. We shall now explain why the direct oriented areaof the projection prxOydσ of the oriented element of area dσ on thexOy-plane can be well approximated with cos γdσ, where γ is the anglebetween the Oz-axis and the normal versor in a fixed point M of thesmall surface which has the area dσ. We just approximated an elementdσ of a surface S with the area of a small parallelogram constructedwith one vertex at M and generated by the tangent vectors−→r u(M) and−→r v(M) (see Fig.7). The projection of this parallelogram on the xOy-plane is in general a quadrilateral in this plane. Let us divide ourparallelogram into two equal triangles. It is enough to prove that thearea of the projection of a triangle on the xOy-plane is equal to the areaof the initial triangle multiplied by cos γ, where γ is the angle between

the normal versor −→n (M) and the versor−→k . It is also sufficient to

consider the triangle ABC with a vertex, say A, in the xOy-plane (seeFig.15).

F 15

Let D = CB ∩ (xOy-plane). Moreover, it is sufficient to considerthe initial triangle to have a entire edge, say [AD], in the xOy-plane(area(∆ACB) = area(∆ADC) − area(∆ABD)).

Hence, let us take again a triangle [XY Z] with [XY ] ⊂ xOy-plane (see Fig.16). Let Z ′ be the projection of Z on the xOy-planeand let Z ′′ be the projection of Z ′ on [XY ]. Since [XY ]⊥[ZZ ′] and


F 16

[XY ]⊥[Z ′Z ′′], we see that [XY ] is perpendicular on the entire plane[ZZ ′Z ′′]. In particular it is perpendicular on [ZZ ′′]. Let −→n (Z ′′) be thenormal versor on the plane [XY Z] at the point Z ′′ (see Fig.16). Sinceγ + δ = 90 and γ′ + δ = 90 one has that γ = γ′ (see Fig.16). Thus,[Z ′Z ′′] = [ZZ ′′] cos γ and so,

area(∆XY Z) =1

2[XY ][ZZ ′′] =

1

cos γ

1

2[XY ] cos γ[ZZ ′′] =

=1

cos γ

1

2[XY ][Z ′Z ′′] =

1

cos γarea(∆XY Z ′].

Hence, area(prxOy∆[XY Z]) = area(∆XY Z) cos γ, what we wanted toprove.

We must be very careful with the notation dxdy, the area of thedirect oriented domain which is the projection of a small part of thedirect oriented surface S, of area dσ. Since dxdy = −dydx, one uses theexterior product notation dxdy = dx ∧ dy. To change dxdy with dydxmeans to take the other face of the oriented projection ω (see Fig.17).


1. Compute I =

∫∫

S

xydσ, where S is the part of the plane (P ) :

2x+ y + 2z = 1 which is inside the cylinder: x2 + y2 = 1.2. The cylindrical solid x2 + y2 ≤ R2 cut the semisphere x2 + y2 +

z2 = 4R2, z ≥ 0 into a surface S. Find its area and its centre of mass(the density function is considered to be 1).


F 17

3. Compute

∫∫

S

(x2 + y2)dσ is S is a surface with the following

parametrization: x = u, y = v, z = uv and u2 + v2 ≤ 1.4. Find the moment of inertia of the cone y2 +z2 = 4x2, x ∈ [−2, 2]

with respect to the Oz-axis.5. Find the centre of mass for the 3D-lamina (S, f), where S is the

surface x2 + z2 = 4y, y ≤ 4 and f(x, y, z) = y.

6. Compute I =

∫∫

S

(x+ y + z)dσ, where S is the union of all the

sides of the parallelepiped [0, 1] × [0, 2] × [0, 3].

7. Compute I =

∫∫

S

(x+ y+ z)dσ, where S is the surface x2 + y2 +

z2 = a2, z ≥ 0.8. Find the mass of the 3D-plate (S, f), where S is the surface

x2 + y2 + z2 = 1, x ≥ 0, y ≥ 0, z ≥ 0 and the density function is

f(x, y, z) = [x2 + y2 + (z − 1)2]− 12 .

9. Find the area of the part of the sphere x2 + y2 + z2 = a2, whichis inside the cylinder x2 + y2 = ax.

10. Compute I =

∮ ∮

(S)

xydydz + xzdzdx + 2x2ydxdy, where (S) is

the exterior side of the surface z = x2 + y2, x2 + y2 ≤ 4.

11. Compute the flux of the field−→F (x2, y2, z2) through the interior

side of the surface x2 + y2 = z, z ≤ 1.


12. Compute the flux of the field−→F (x3, y3, z3) through the total

exterior sides of the surface x2 + y2 = z2, 0 ≤ z ≤ 4.

13. Find the flux of the field−→F (x, y, z) = x

−→i + y

−→j + z

−→k through

the interior side of the sphere x2 + y2 + z2 = R2.

14. Find the flux of the field−→F (x2, y2, z2) through the total exterior

sides of tetrahedron [OABC], where A(1, 0, 2), B(0, 2, 0) and C(0, 0, 3).

15. Compute the flux of the field−→F

(1x, 1y, 1z

)through the exterior

side of the sphere x2 + y2 + z2 = R2.

16. Find the flux of the field−→F (yz, xz, xy) through the total exte-

rior face of the cylinder x2 + y2 = a2, 0 ≤ z ≤ h.

CHAPTER 9

Gauss and Stokes formulas. Applications

1. Gauss formula (divergence theorem)

If a surface S is a union of many sides with different parame-trizations, the computation of a surface integral of the second type∮ ∮

(S)

−→F · −→n dσ can be very long and not so easy. A famous formula of

Gauss can reduce this computation to a calculation of a triple integral.

D 32. We say that a surface S = −→g (D) ⊂ R3 is totallyclosed in R3 if it is bounded, closed as a topological subset of R3 (i.e.R3S is open) and if it is the boundary ∂Ω of a bounded, closed andconnected subset Ω of R3. Moreover, S must divide the space R3 intotwo disjoint (with their intersection empty!) subsets, Ω and R3Ωwhich each of them are connected.

For instance, any sphere is totally closed. The cylinder x2+y2 = R2,0 ≤ z ≤ h is closed but not totally closed. The cone x2 + y2 = z2,0 ≤ z ≤ h together the disc z = h, x2 + y2 ≤ h2 above it is totallyclosed but, without that disc is closed and no totally closed. The conicsurface x2 + y2 = z2, −3 ≤ z ≤ 3, z /∈ (−ε, ε) with 0 < ε < 3, togetherthe both discs z = 3, x2 + y2 ≤ 9 and z = −3, x2 + y2 ≤ 9, is closedbut it is not totally closed because the interior of the domain boundedby this surface is not connected (Why?).

T 95. (Gauss formula) Let Ω ⊂ R3 be a bounded, closedand connected subset with its boundary ∂Ω = S a piecewise smooth,oriented and totally closed surface, considered with its "exterior" side(see Fig.1). Let

−→F (x, y, z) = P (x, y, z)

−→i +Q(x, y, z)

−→j +R(x, y, z)

−→k

be a field defined on Ω, which is continuous and piecewise of class C1

(the set of points at which this property to be of class C1 fails is a finite

339

340 9. GAUSS AND STOKES FORMULAS. APPLICATIONS

union of points, smooth lines, or smooth surfaces of zero volume). Then

(1.1)

∮ ∮

(S)=∂Ω

−→F · −→n dσ =

∫∫∫

Ω

div−→F dxdydz,

where div−→F (x, y, z) = ∂P

∂x(x, y, z)+ ∂Q

∂y(x, y, z)+ ∂R

∂z(x, y, z) is called the

divergence of−→F .

F 1

Before to prove this basic result, let us make some commentarieson it.

The philosophical importance of Gauss formula (1.1) is the follow-ing. On the left side of this formula we have an "oriented" object, ingeneral very complicated, but only a 2D-object. On the right side onehas a triple integral, a simpler object (because it has no "orientation" atall), but a 3D-object. Hence this formula connects a 2D-computationwith a 3D-computation. On the left side of this formula we have in

fact the flux ΦS(−→F ) of the field

−→F through the surface S, "from inside

to outside", i.e. the normal versor at a point M ∈ S is "above" thetangent plane at M to S (see Fig.1). On the right side of the formula

one has "the global productivity" of the volume Ω w.r.t. the field−→F .

Let us explain in the following the word "productivity". Let us applythe mean formula (see theorem 87, e)) to the triple integral from theright side: ∫∫∫

Ω

div−→F dxdydz = div

−→F (P0) · vol(Ω),

1. GAUSS FORMULA (DIVERGENCE THEOREM) 341

where P0 is a point inside Ω. Thus, Gauss formula says that:

div−→F (P0) =

ΦS=∂Ω(−→F )

vol(Ω),

i.e. if diam(Ω) is very small, the "productivity" div−→F (P0) at P0 mea-

sures how much energy created by the field−→F inside Ω is transferred

from inside of Ω to outside. More exactly, it is equal to the quantity offlux produced per unity of volume. This is the true interpretation of

the divergence of−→F , defined above. For instance, if one has no "pro-

duction" (div−→F = 0) inside Ω, one has no flux of

−→F through S = ∂Ω.

Gauss formula explains many phenomena in Physics, Chemistry, Econ-omy, Engineering, etc.

Gauss formula or the divergence theorem was obtained by Gaussduring the examination of the electric fields, magnetic fields or gravita-tional fields. Let say some words on the Gauss law in Physics. Let q0 bea fixed point positive electric charge in the origin O and let M(x, y, z)

be an arbitrary point in space. Let −→r =−−→OM = x

−→i +y

−→j +z

−→k be the

position vector of M . We always assume that at M we have anotherfree point positive electric charge q = 1C (Coulomb). Then the famousCoulomb law says that the repulsive electric field at M(x, y, z) is:

−→E (x, y, z) =

q04πε0

·−→rr3,

where r = (x2+y2+z2)12 and ε0 is the electric constant, depending only

on the electric properties of the space itself. The Gauss law says thatif S is any closed continuous surface which contains the origin (i.e. the

electric charge q0), then the flux ΦS(−→E ) of the electric field through

S, from inside to outside (the force is a repulsive one!) is constant,namely it always is equal to q0

ε0. This means that:

(1.2)

∮ ∮

S

−→E · −→n dσ =

q0ε0.

Gauss succeeded to express this last law as a "local law" (at a pointP0 = O!):

(1.3) div−→E (P0) =

ρ(P0)

ε0,

where ρ(P0) is the charge density at P0 produced by q0. It appears thatfrom here we can deduce our Gauss formula (1.1) in this particular case.There is a mathematical problem! The electric field is not defined atthe origin O! Thus, our Gauss formula (1.1) cannot be applied in this


case. But, in higher mathematics one can extend Gauss formula tosuch more complicated situation and one can obtain the Gauss law ofPhysics as a particular case of it. In fact, Gauss law is equivalent to

Coulomb law (see [BPC]). Since

∫∫∫

Ω

ρ(P )dxdydz = q0, the formulas

(1.2) and (1.3) give rise to the equality:

(1.4)

∮ ∮

(S)=∂Ω

−→E · −→n dσ =

∫∫∫

Ω

div−→Edxdydz,

i.e. Gauss formula for the electric field−→E . But, what is the meaning

of the triple integral on the right, because the electric field−→E is not

defined at O? This is a big mathematical problem which forced scien-tists to create new concepts and notions like distributions, potentialtheory, etc. We shall only compute the surface integral of the second

type of formula (1.4), i.e. the flux of−→E through the surface of a sphere

of radius R > 0 :∮ ∮

(S): x2+y2+z2=R2

−→E · −→n dσ =

q04πε0

∮ ∮

(S): x2+y2+z2=R2

−→rr3

· −→n dσ =

=q0

4πε0

∮ ∮

(S): x2+y2+z2=R2

1

R2−→n · −→n dσ =

q0ε0,

because−→rr

= −→n and

∮ ∮

(S): x2+y2+z2=R2

dσ = 4πR2 (the area of the sphere

of radius R) in this particular case. Hence,∮ ∮

(S): x2+y2+z2=R2

−→E · −→n dσ =

q0ε0.

Gauss observed that the flux of−→E does not depend on the radius of

the sphere, so that he tried to prove that the sphere can be substitutedwith an arbitrary closed surface S, which contains O as an interiorpoint of it. The reader can find a proof of this last statement in anyadvanced course on Electricity, for instance in [BPC].

Now we prove the main theorem 95.

P. (for Gauss formula) We assume that our domain Ω canbe decomposed into a finite number of simple subdomains Ωi of thesame type like Ω such that for i = j, Ωi ∩ Ωj is at most a surface of


volume zero. A simple subdomain is a simple domain w.r.t. one of theaxes Ox, Oy, or Oz. Let us assume that we have proved the theorem forsimple domains and let us consider a nonsimple domain like in Fig.2.In this case Ω = Ω1 ∪ Ω2 ∪ Ω3 and let (S1) = Ω1 ∩ Ω2, (S2) = Ω1 ∩ Ω3.

Since the normal versors−→n′1,

−→n′2 on (S1) and

−→n′′1,

−→n′3 on (S2) respectively,

have opposite signs, the corresponding surface integrals annihilate oneto each other (see Fig.2).

F 2

The triple integral can be naturally decomposed into three tripleintegrals and so we can apply Gauss formula for the three simple do-mains:∫∫∫

Ω

div−→F dxdydz =

∫∫∫

Ω1

div−→F dxdydz

+

∫∫∫

Ω2

div−→F dxdydz

+

+

∫∫∫

Ω3

div−→F dxdydz

=

=

∮ ∮

(S)(∂Ω2)(∂Ω3)

−→F · −→n1dσ +

∮ ∮

(S1)

−→F ·

−→n′1dσ +

∮ ∮

(S2)

−→F ·

−→n′′1dσ

+

+

∮ ∮

(∂Ω2)(S1)

−→F · −→n2dσ +

∮ ∮

(S1)

−→F ·

−→n′2dσ

+


+

∮ ∮

(∂Ω3)(S2)

−→F · −→n3dσ +

∮ ∮

(S2)

−→F ·

−→n′3dσ

=

=

∮ ∮

(S)(∂Ω2)(∂Ω3)

−→F · −→n1dσ +

∮ ∮

(∂Ω2)(S1)

−→F · −→n2dσ +

∮ ∮

(∂Ω3)(S2)

−→F · −→n3dσ

+

+

∮ ∮

(S1)

−→F ·

−→n′1dσ −

∮ ∮

(S1)

−→F ·

−→n′1dσ

+

∮ ∮

(S2)

−→F ·

−→n′′1dσ −

∮ ∮

(S2)

−→F ·

−→n′′1dσ

=

=

∮ ∮

(S)

−→F · −→n dσ.

We see that it is sufficient to separately prove the following three equal-ities (then add them!):

(1.5) a)

∮ ∮

(S)

P (x, y, z)dydz =

∫∫∫

Ω

∂P

∂xdxdydz,

(1.6) b)

∮ ∮

(S)

Q(x, y, z)dzdx =

∫∫∫

Ω

∂Q

∂ydxdydz,

(1.7) c)

∮ ∮

(S)

R(x, y, z)dxdy =

∫∫∫

Ω

∂R

∂zdxdydz.

Since the proofs for all these formulae a), b) or c) are similar, we shallprove only formula c). For this, after dividing the domain Ω into a finiteunion of nonoverlaping subdomains (see what we did above!) which aresimple w.r.t. Oz-axis, we suppose that Ω is a simple domain w.r.t. Oz-axis like in Fig.3. This means that

Ω = (x, y, z) : ϕ(x, y) ≤ z ≤ ψ(x, y), (x, y) ∈ Ωxy = prxOyΩ ,where ϕ and ψ are continuous functions defined on Ωxy.

Let us look now at Fig.3 and write:

∫∫∫

Ω

∂R

∂zdxdydz =

∫ ∫

Ωxy

ψ(x,y)∫

ϕ(x,y)

∂R

∂zdz

dxdy =


F 3

=

∫ ∫

Ωxy

R(x, y, ψ(x, y))dxdy −∫ ∫

Ωxy

R(x, y, ϕ(x, y))dxdy =

=

∮ ∮

(S2)

R(x, y, z)dxdy +

∮ ∮

(S1)

R(x, y, z)dxdy+

+

∮ ∮

(S3)

R(x, y, z)dxdy =

∮ ∮

(S)

R(x, y, z)dxdy,

because

∮ ∮

(S3)

R(x, y, z)dxdy = 0, (γ = 90, so cos γ = 0 and dxdy =

cos γdσ = 0 for (S3)).For proving (b) for instance, we eventually decompose this last Ω

(which was considered to be simple w.r.t. Oz-axis) more, up to we getit as a finite union of simple subdomains w.r.t. Oy-axis.

E 102. Let us use Gauss formula to compute the flux ΦS(−→F )

of the field−→F = x2−→i +y2−→j +z2

−→k through the total exterior sides sur-

face S of the cube Ω = [0, 1] × [0, 1] × [0, 1] (see Fig.4).


F 4

Use then this result to compute the flux Φ1 of the same field−→F

through the exterior surface of the cube, except the hatched face [O′A′B′C ′].

ΦS(−→F ) =

∮ ∮

S

−→F · −→n dσ Gauss

=

∫∫∫

Ω

(2x+ 2y + 2z)dxdydz =

= 2

1∫

0

1∫

0

1∫

0

(x+ y + z)dx

dy

dz =

= 2

1∫

0

1∫

0

((x2

2+ xy + xz

)∣∣∣∣1

0

)dy

dz =

= 2

1∫

0

1∫

0

(1

2+ y + z

)dy

dz = 2

1∫

0

((y

2+y2

2+ yz

)∣∣∣∣1

0

)dz =

= 2

1∫

0

(1 + z) dz = 2

(z +

z2

2

)∣∣∣∣1

0

= 3.

Let (S1) be the hatched surface [O′A′B′C ′]. Since the normal versor−→n =

−→k = (0, 0, 1), one has that dydz = cosαdσ = 0 and dzdx =

cosβdσ = 0. Thus,

Φ(S1)(−→F ) =

∫ ∫

[0,1]×[0,1]

dxdy = 1.


Hence Φ1 = ΦS(−→F ) − Φ(S1)(

−→F ) = 2.

E 103. Compute the flux of the vector field−→F = (x2, y2, z)

through the sphere x2+y2+z2 = 1, from exterior to interior (see Fig.5).

F 5

We simply apply Gauss formula (95):

Φ(S)(−→F ) = −

∫∫ ∫

x2+y2+z2≤1

(2x+ 2y + 1)dxdydz.

Let us use spherical coordinates to compute the last triple integral:

−∫∫ ∫

x2+y2+z2≤1

(2x+ 2y + 1)dxdydz =

= −1∫

0

ρ2

2π∫

0

π∫

0

(2ρ cosϕ sin θ + 2ρ sinϕ sin θ + 1) sin θdθ

dϕ

dρ =

= −2

1∫

0

ρ3

2π∫

0

(cosϕ+ sinϕ)

π∫

0

sin2 θdθ

dϕ

dρ−

−1∫

0

ρ2

2π∫

0

π∫

0

sin θdθ

dϕ

dρ = 0 − 2π · 2 · ρ

3

3

∣∣∣∣1

0

= −4π

3.


2. A mathematical model of the heat flow in a solid

Let Ω be a bounded, closed and homogenous 3D-solid and let (S)be its boundary which is supposed to be a piecewise smooth, connectedand totally closed surface. The heat will flow in the direction of a lowtemperature. We assume that the rate of flow is proportional to thegradient of temperature. Let −→v be the velocity of the heat flow andlet u(x, y, z, t) be the temperature at the point M of coordinates x, y,and z at the moment t. Then

(2.1) −→v = −K−−−→grad u,

where K is the thermal conductivity (a constant) and the gradient isreferred to x, y and z. The amount of heat leaving Ω through (S), "frominterior to exterior", is given by the flux

Φ(S)(−→v ) =

∮ ∮

(S)

−→v · −→n dσ.

Let us use now Gauss formula to compute this flux:∮ ∮

(S)

−→v · −→n dσ = −K∫∫∫

Ω

div(−−−→gradu)dxdydz = −K

∫∫∫

Ω

∆u dxdydz,

where ∆u = ∂2u∂x2

+ ∂2u∂y2

+ ∂2u∂z2. The total amount of heat H in Ω is:

H =

∫∫∫

Ω

σρu dxdydz,

where σ is the specific heat of the material of the solid (a constant)and ρ is the density of Ω. Thus, the time rate of decrease of H is:

−∂H∂t

= −∫∫∫

Ω

σρ∂u

∂tdxdydz.

Since this last one must equal the amount of heat leaving Ω, one has:

−∫∫∫

Ω

σρ∂u

∂tdxdydz = −K

∫∫∫

Ω

∆u dxdydz,

or ∫∫∫

Ω

(σρ∂u

∂t−K∆u

)dxdydz = 0.

Since the integrand function is continuous w.r.t. x, y and z, since Ωwas arbitrarly chosen, by a local application of the mean formula, we

3. STOKES THEOREM 349

see that σρ∂u∂t−K∆u is identical to zero on Ω (prove this in general like

in the case of a simple integral!). Finally we get the famous parabolicPDE (partial differential equation) of the heat flow:

∂u

∂t= c2∆u,

where c2 = Kσρ. If for a small interval of time the heat flow does not

depend on time, we get ∆u = 0, which is called the Laplace equation.It is fundamental in many branches of Applied Mathematics. Morediscussions on this subject can be find in any course of PDE.

3. Stokes Theorem

This fundamental result is a generalization of the Green theorem(in plane), for the 3-dimensional case. This theorem says that in someconditions a surface integral of the second type can be transformed intoa line integral and conversely.

A surface (S) ⊂ R3 has the curve Γ as its boundary or borderif Γ ⊂ (S), Γ is a piecewise smooth, connected and "closed" curve(−→r (a) = −→r (b), −→r : [a, b] → Γ ⊂ R3) and, for any point M0 of Γ, thereexists a small δ > 0 such that the intersection B(M0; r)∩(S), 0 < r ≤ δ,between any ball (in R3) with centre at M0 and radius r, at mostδ, has the property that B(M0; r) [B(M0; r) ∩ (S)] is a connectedset. Moreover, if M0 ∈ (S) but M0 /∈ Γ, then there exists at leastone ball B(M0; r), r > 0 such that B(M0; r) [B(M0; r) ∩ (S)] is thedisjoint union of two connected subsets of B(M0; r) (their intersectionsis empty!).

A sphere S is not a surface with a border. But a semisphere is asurface with a border Γ, the unique "big" circle on the sphere. Otherexamples of surfaces with a border one can see in Fig.6, Fig.7 or Fig.8.We must be careful! For instance, the cylinder x2 +y2 = R2, 0 ≤ z ≤ hhas not a (unique) border in our acceptance because its candidate for a"border" is the nonconnected union between the following two circles:x2 + y2 = R2, z = 0 and x2 + y2 = R2, z = h,where h = 0. If we add tothis last cylinder the "above" disc x2 + y2 ≤ R2, z = h, the obtainedsurface has a unique (connected) border: x2 + y2 = R2, z = 0, thus itis a surface with a border in our previous acceptance.

Let −→g : D → R3, −→g (u, v) = x(u, v)−→i + y(u, v)

−→j + z(u, v)

−→k , be a

piecewise smooth deformation and let (S) = −→g (D) be its corresponding

(direct oriented) surface with a border like above.. Let−→F (x, y, z) =

P (x, y, z)−→i +Q(x, y, z)

−→j +R(x, y, z)

−→k be a field of class C1 defined on

a space domain which contains the surface (S). Let ∂D = γ be a direct


oriented closed piecewise smooth curve and let Γ = ∂(S) be the borderof (S) which is the image of γ through the deformation −→g . Notice thatΓ is "closed" and the orientation of Γ, which is that one inherited fromthe direct orientation of γ in the uov-plane, is compatible with thedirect orientation of (S). This means that if we walk on the border Γin the direction indicated by its direct orientation, the direct face ofthe surface (S) is always "on the left" (see Fig.6). The surface (S) ofFig.6 is not simple w.r.t. Oz-axis (i.e. it has no parametrization ofthe form z = z(x, y), (x, y) ∈ prxOy(S)), but we can divide it into twononoverlaping oriented surfaces (S1) and (S2) with the correspondingborders ∂(S1) = [AEBCA] and ∂(S2) = [FACBF ] respectively (seeFig.6). It will be easy to prove that if the bellow Stokes formula istrue for simple surfaces (S1) and (S2), then it is true for the entire(S) = (S1) ∪ (S2) (see this proof bellow!). This is why we shall provethis basic formula only for a simple surface (S) w.r.t. Oz-axis.

F 6

Let us recall that curl−→F is a new vector field associated to the field−→

F as follows:

(3.1) curl−→F = det

−→i

−→j

−→k

∂∂x

∂∂y

∂∂z

P Q R

=


=

(∂R

∂y− ∂Q

∂z

)−→i +

(∂P

∂z− ∂R

∂x

)−→j +

(∂Q

∂x− ∂P

∂y

)−→k .

Recall that the normal versor can be computed by the formula

−→n =

(A√

A2 +B2 + C2,

B√A2 +B2 + C2

,C√

A2 +B2 + C2

)

and dσ =√A2 +B2 + C2dudv. Let us put

nx =A√

A2 +B2 + C2, ny =

B√A2 +B2 + C2

, nz =C√

A2 +B2 + C2.

T 96. (Stokes formula) Let D, (S), −→g , Γ = ∂(S), etc. bethe above notation and hypotheses. Then

(3.2)

∮ ∮

(S)

(curl−→F ) · −→n dσ =

∮

Γ

Pdx+Qdy +Rdz.

The first surface integral is the flux of the curl−→F through the surface

(S). The last line integral is the circulation (or the work) of−→F on the

oriented border Γ of (S). Stokes formula says that this flux of curl−→F

and the circulation of−→F are equal.

P. Let us put the expression of curl−→F from (3.1) on the left

side of formula (3.2):∮ ∮

(S)

(curl−→F ) · −→n dσ =

∮ ∮

(S)

[

(∂R

∂y− ∂Q

∂z

)nx +

(∂P

∂z− ∂R

∂x

)ny+

+

(∂Q

∂x− ∂P

∂y

)nz]dσ =

∮ ∮

(S)

(∂P

∂zny −

∂P

∂ynz

)dσ+

+

∮ ∮

(S)

(∂Q

∂xnz −

∂Q

∂znx

)dσ +

∮ ∮

(S)

(∂R

∂ynx −

∂R

∂xny

)dσ.

To prove that this last sum is equal to

∮

Γ

Pdx+Qdy+Rdz it is enough

to prove the following three equalities:

a)

∮ ∮

(S)

(∂P∂zny − ∂P

∂ynz

)dσ =

∮

Γ

Pdx,


b)

∮ ∮

(S)

(∂Q∂xnz − ∂Q

∂znx

)dσ =

∮

Γ

Qdy,

c)

∮ ∮

(S)

(∂R∂ynx − ∂R

∂xny

)dσ =

∮

Γ

Rdz.

Since the proof of each of these equalities are similar, we shall proveonly a). Now we can suppose that (S) is simple w.r.t. Oz-axis (other-wise we decompose it into a finite simple surfaces like in Fig.6). Hence,one can find for (S) an explicit parametrization w.r.t. z :

(S) :

x = xy = y

z = z(x, y), (x, y) ∈ Ωxy = prxOy(S)

(see Fig.7). In this situation

−→n =

(− ∂z∂x,− ∂z

∂y, 1

)

√(∂z∂x

)2+

(∂z∂y

)2

+ 1

and the left side of a) becomes:

∮ ∮

(S)

(∂P

∂zny −

∂P

∂ynz

)dσ =

=

∫ ∫

Ωxy

[−∂P∂z

(x, y, z(x, y)) · ∂z∂y

− ∂P

∂y(x, y, z(x, y))

]dxdy =

=

∫ ∫

Ωxy

− ∂

∂y[P (x, y, z(x, y))] dxdy =

Green=

∮

γ

P (x, y, z(x, y))dx =

∮

Γ

P (x, y, z)dx,

i.e. the right side of a). Here we just applied Green formula for thedomain Ωxy with its boundary γ (see Fig.7).

If (S) is not simple with respect to Oz-axis we reduce everything tosimple components of (S). ‘Let us apply this last idea for the surface


F 7

of Fig.6.

∮ ∮

(S)

(curl−→F )·−→n dσ =

∮ ∮

(S1)

(curl−→F ) · −→n dσ

+

∮ ∮

(S2)

(curl−→F ) · −→n dσ

=

∮

[AEB]

Pdx+Qdy +Rdz +

∮

[BCA]

Pdx+Qdy +Rdz

+

+

∮

[BFA]

Pdx+Qdy +Rdz +

∮

[ACB]

Pdx+Qdy +Rdz

.

But ∮

[BCA]

Pdx+Qdy +Rdz = −∮

[ACB]

Pdx+Qdy +Rdz,

so that∮ ∮

(S)

(curl−→F ) · −→n dσ =

∮

[AEB]

Pdx+Qdy +Rdz+

+

∮

[BFA]

Pdx+Qdy +Rdz =

∮

Γ

Pdx+Qdy +Rdz.


E 104. Let us compute the line integral I =

∮

Γ

x2ydx +

y2dy + z2dz, where Γ is the oriented curve [O′A′B′C ′O′] from Fig.8.For this, let us consider the oriented surface

(S) :

x = xy = yz = 1

, (x, y) ∈ [0, 1] × [0, 1],

which has as its (unique) border the connected curve Γ. Since

curl−→F = det

−→i

−→j

−→k

∂∂x

∂∂y

∂∂z

x2y y2 z2

= −x2−→k ,

we can apply Stokes formula (3.2) and find:∮

Γ

x2ydx+ y2dy + z2dz =

∮ ∮

(S)

(curl−→F ) · −→n dσ = −

∫ ∫

[0,1]×[0,1]

x2dxdy =

= −1∫

0

x2

1∫

0

dy

dx = −

1∫

0

x2dx = −1

3.

The same curve Γ with the inverse orientation is also the border ofthe surface S′ obtained by the union of the five other sides of the cube,except the "above" side, situated on the plane z = 1. This surface isoriented by the exterior normal vectors like in Fig.8.

Hence,

∮ ∮

S′

(curl−→F ) · −→n dσ = 1

3. Let us now directly compute the

line integral I.

I =

∮

Γ

x2ydx+y2dy+z2dz =

∮

[O′A′]:x=x,y=0,z=1,x∈[0,1]

x2ydx+y2dy+z2dz+

+

∮

[A′B′]:x=1,y=y,z=1,y∈[0,1]

x2ydx+ y2dy + z2dz+

+

∮

[B′C′]:x=x,y=1,z=1,x∈[0,1]−

x2ydx+ y2dy + z2dz+


F 8

+

∮

[C′O′]:x=0,y=y,z=1,y∈[0,1]−

x2ydx+ y2dy + z2dz =

= 0 +

1∫

0

y2dy −1∫

0

x2dx−1∫

0

y2dy = −1

3.

We just obtained the same result like that one obtained by using Stokesformula. Hence, these last computation can also be viewed as a simpleverification.

We know that if−→F : D → R3 if a conservative field of class C1 on

a domain D, then curl−→F =

−→0 (see theorem 68) Conversely, if D is

a simple connected domain (any totally closed surface S ⊂ D can be

continuously shrunk to a point in D), then the field−→F of class C1 is

conservative. We easily see that in order to prove that−→F is conserva-

tive, it is enough to prove that the line integral

∮

γ

−→F · d−→r = 0 on any

"closed" piecewise smooth curve γ ⊂ D (construct a primitive for−→F of

the type: U(x, y, z) =

(x,y,z)∮

(a,b,c)

−→F · d−→r ). Now, for any "closed" piecewise

smooth curve γ ⊂ D, it is intuitively clear (at least for the usual balls,cylindrical domains, conical domains, paraboloidal domains, etc.) thatthere exists a bounded, connected piecewise smooth surface T ⊂ D


with the unique border γ (try to explain this existence at least for astarred domain D, i.e. if there exists a point P0 ∈ D such that if P isanother point in D, then the whole segment [P0P ] is included in D).Then we can apply Stokes formula for the surface T with its border γ :

∮ ∮

T

(curl−→F ) · −→n dσ =

∮

γ

Pdx+Qdy +Rdz.

But curl−→F =

−→0 , thus

∮

γ

Pdx+Qdy +Rdz = 0. Hence−→F is conserv-

ative.Sometimes Stokes formula cannot be easily applied, but this last

result can work easily.

E 105. Let us compute the line integral I =

∮

Γ

x2dx+y2dy+

z2dz, where Γ this time is the oriented curve [ABCD] of Fig.9.

F 9

Since−→F = (x2, y2, z2), curl

−→F =

−→0 and this implies that

−→F is

conservative. Hence, the integral

∮

Γ

x2dx+y2dy+z2dz does not depend

on the integration path which connects the points A and D. The simplestpath which connect A and D is the oriented segment [AD] : x = 0,


y = y, z = 0. Thus,

∮

Γ

x2dx+ y2dy + z2dz =

R∫

−R

y2dy =2

3R3.


1. Use Stokes formula to compute the circulation of the field−→F (y−

z, z−x, x− y) along the intersection between the cylinder x2 + y2 = a2

and the plane xa

+ zb− 1 = 0, a > 0, b > 0. The projection of the

orientation of this curve on the xOy-plane is "from x to y", i.e. thecounterclockwise orientation.

2. Use Stokes formula to compute I =

∮

C

ydx+ zdy+xdz, where C

is the intersection curve of the sphere x2 + y2 + z2 = a2 with the planex+ y+ z = a, a > 0. The projection of the orientation of this curve onthe xOy-plane is "from x to y", i.e. the counterclockwise orientation.

3. Use Gauss formula to compute I =

∮ ∮

S

xyz(xdydz + ydzdx +

zdxdy), where S is x2 + y2 + z2 = R2, x ≥ 0, y ≥ 0, z ≥ 0, the interiorside.


∮ ∮

S

xy2dydz + z3dzdx +

y2zdxdy, where S is the exterior side of the ellipsoid x2

a2+ y2

b2+ z2

c2= 1.


∮ ∮

S

x2dydz + y2dzdx +

z2dxdy, where S is the total interior side of the paraboloid: x2+y2 = z,0 ≤ z ≤ 1 together x2 + y2 ≤ 1, z = 1, the "above" part.


∮ ∮

S

x2dydz + y2dzdx +

z2dxdy, where S is the exterior side of the sphere (x− a)2 + (y− b)2 +(z − c)2 = R2.

7. Compute in two ways the flux of the field−→F = (y − z)

−→i + (z −

x)−→j + (x− y)

−→k through the "open" surface S : x2 +y2 = 4− z, z > 0,

on the exterior side.8. Compute the circulation of

−→F = xy

−→j + xz

−→k , along the circle

C, the intersection between the sphere x2 + y2 + z2 = 4 and the conex2 + y2 = 3z3, z > 0. The projection of the orientation of this curve onthe xOy-plane is "from x to y", i.e. the counterclockwise orientation.


9. Use Stokes formula to compute the circulation of−→F = x

−→j +y

−→k

along the "closed" curve S, the intersection between the sphere x2 +y2 + z2 = R2 and the plane x + y + z = 0, with the orientation givenby the direction of the vector (1, 1, 1).

10. Let−→F = yz

−→i + 2xz

−→j −x2−→k . Compute the flux of

−→F through

the ellipsoid 4x2 + y2 + 4z2 = 8, the exterior side and the circulation ofthe same field along the intersection curve γ, the intersection betweenthis last ellipsoid and the plane z = 1.The projection of the orientationof this curve on the xOy-plane is "from y to x", i.e. the clockwiseorientation.

CHAPTER 10

Some remarks on complex functions integration

1. Complex functions integration

The reader can find some theory on complex numbers, complexsequences, complex series, complex functions, complex differentiability,etc. in [Po], sections 1.2, 5.2 and 11.8. We shall review some of themhere.

To any point M of a plane (P ) with a fixed direct oriented (coun-

terclockwise) Cartesian coordinate system xOy or O;−→i ,

−→j one can

uniquely associate a pair (x, y) of two real numbers such that−−→OM =

x−→i + y

−→j . Conversely, to any pair (x, y) ∈ R2 we can uniquely as-

sociate a point M such that the position vector−−→OM = x

−→i + y

−→j .

The ordered pair x and y are called the coordinates of M w.r.t. theCartesian coordinate system xOy. We write this as M(x, y). This dou-ble association give rise to the famous Descartes’ bijection betweenthe points of the plane (P ) and the elements of the arithmetical spaceR2. To the arithmetical sum (x, y) + (x′, y′) = (x + x′, y + y′) from

R2 one associates the point N(x + x′, y + y′) and−−→ON =

−−→OM +

−−→OM ′,

where M(x, y) and M ′(x′, y′). If α ∈ R and M(x, y) then, to the prod-uct α(x, y) = (αx, αy) from R2 we associate the point T (αx, αy) and−→OT = α

−−→OM. We see that the vector space R2 is isomorphic with the

vector space V2 of all free 2D-vectors. But R2 is also isomorphic withthe vector space C of all complex numbers z = x+ iy, x, y ∈ R, wherei =

√−1 is the pure imaginary quantity, not real, which denote one

of the root of the equation x2 + 1 = 0. With an appropriate multipli-cation, C becomes a field, the complex number fields. The associationM(x, y) zM = x + iy Mz(x, y) gives rise to a bijection betweenthe points of the plane (P ) and the complex numbers set C. To the ad-dition of C corresponds the addition between the corresponding vectorsin V2. The plane (P ) becomes a representation of C. The distance be-tween two pointsM(x, y) andM ′(x′, y′) in plane is exactly the distance|zM − zM ′| in C. An ε-neighborhood of a fixed complex number z0 isan open ball of the form B(z0, r) = z ∈ C : |z − z0| < r, i.e. an opendisc in the plane (P ) with center at Mz0 and radius r > 0. An open

359

360 10. SOME REMARKS ON COMPLEX FUNCTIONS INTEGRATION

subset A of C is a subset of C with the property that if z ∈ A, thenthere exists a small ε > 0 such that the corresponding ε-neighborhoodof z is contained in A. A subset D of C is a domain if it is open andif it is connected, i.e. any two points M,N of D can be connected bya polygonal line (or a smooth curve) in D, with extremities at M andN respectively. D is bounded if there exists a real number R > 0 suchthat D ⊂ B(0, R). In C we have only "one" ∞, the circle with centreat 0 and radius equal to ∞ (see the Riemann sphere construction in[Po], 6.5). A neighborhood of ∞ is the exterior of any closed ball ofthe form B[0, R], i.e. z ∈ C : |z| > R. A sequence zn of complexnumbers is convergent to the complex number z if |zn − z| → 0, i.e.if and only if xn = Re zn → x = Re z and yn = Im zn → Im z. Herezn = xn+iyn and z = x+iy. If A is a subset of real numbers, a functionf : A→ C, f(t) = x(t)+ iy(t), where x : A→ R and y : A→ R are thereal functions components of f, is called a real function with complexvalues. The continuity, derivability and integrability of such functionsare componentwise operations. Any continuous and injective functiong : [a, b] → C, g(t) = x(t) + iy(t), is called a curve in C. In fact, theimage g([a, b]) is the support curve of g, or simple a continuous curvein C. If the real functions x(t) and y(t) are piecewise of class C1 wesay that the curve γ = g([a, b]) is a piecewise smooth curve. It has alength if and only if its parametric representation γ : x = x(t), y = y(t),

t ∈ [a, b] has a length and its length is

∫

γ

ds. Let A be a subset of C and

let f : A → C be a function defined on A with complex values. Sucha function is called a complex function. Let z = x + iy be a runningpoint in A and let f(z) = u(x, y) + iv(x, y). The real functions of twovariables (x, y) u(x, y) and (x, y) v(x, y) are called the real partRe f of f and the imaginary part Im f of f respectively. We say thatf : D → C is continuous at a point z0 of a fixed domain D, if for anysequence zn with elements of D and such that zn → z0, one has thatf(zn) → f(z0). We say that f is continuous on D if it is continuousat any point of D. Such a function is continuous on D if and only ifits component real functions u(x, y) and v(x, y) are continuous on D.Here we simultaneously look at D as being in C and in its correspond-ing representation plane (P ) as a subset of R2! The differentiabilityof a complex function f : D → C is defined like the differentiability(derivability) of a real function of one variable with real values. Wesaw in theorem 91, [Po] that f is differentiable (or analytic) on the do-main D if and only if its component functions u and v have continuous

1. COMPLEX FUNCTIONS INTEGRATION 361

partial derivatives on D and

(1.1)∂u

∂x=∂v

∂y,∂u

∂y= −∂v

∂x

(Cauchy-Riemann relations) and

(1.2) f ′(z) =∂u

∂x(x, y) + i

∂v

∂x(x, y) =

∂v

∂y(x, y) − i

∂u

∂y(x, y)

Any power series S(z) =∑∞n=0 cn(z − z0) has a set of convergence

M such that B(z0, R) ⊂ M ⊂ B[z0, R], where R is called a convergenceradius and it is computed exactly like in the real case:

(1.3) R =1

lim sup n√

|cn|=

1

lim sup∣∣∣ cn+1cn

∣∣∣.

All the basic results relative to sequences or series of functions fromthe real case preserve in the complex functions case. For instance, iffn(z) is a sequence of complex functions, such that fn is uniformlybounded by a number an, i.e. |fn(z)| ≤ an for any z ∈ D, and thenumerical series

∑an is convergent, then the series of functions

∑fn

is absolutely and uniformly convergent onD (WeierstrassM test). Theproofs follows exactly the same ideas like in the real case.

We shall now introduce the integral

∫

ab

f(z)dz of a complex function

f : D → C along a fixed arc of a curve ab ⊂ D. We assume that thearc ab is the image of g : [T0, T ] → C, g(t) = x(t) + iy(t), where g is acontinuous function of a real variable with complex values (a complexcurve!). Let ∆ : t0 = T0 < t1 < ... < tn = T be a division of the realinterval [T0, T ] and let ci, ci ∈ [ti−1, ti] be a set of marking points ofthis division. Let zj = g(tj) and ξj = g(cj), where j = 1, 2, ..., n, bethe corresponding points on the arc ab (see Fig.1). Let ∆∗ = zj bethe natural division of the arc ab obtained as the image of the division∆ of the real interval [T0, T ].

A sum of the type

(1.4) Sf(∆∗, ξj) =

n∑

j=1

f(ξj)(zj − zj−1)

is called a Riemann sum of f relative to the division ∆∗ and the setof marking points ξj.


F 1

D 33. We say that f : D → C is integrable on the arc ab

if there exists a complex number Idef=

∫

ab

f(z)dz such that for any small

ε > 0 there exists a small δ > 0 (depending on ε) with the property thatif ‖∆‖ < δ, then

∣∣I − Sf (∆∗, ξj)

∣∣ < ε for any set ξj of markingpoints of ∆∗. This means that I can well be approximated with Riemannsums of the above type.

It is easy to see that if f is piecewise continuous and g is smooth,then f is integrable on the arc ab and

(1.5)

∫

ab

f(z)dz =

T∫

T0

f(g(t))g′(t)dt,

where g′(t) = x′(t) + iy′(t). Indeed,

n∑

j=1

f(ξj)(zj−zj−1) =n∑

j=1

f(g(cj)) [(x(tj) − x(tj−1)) + i(y(tj) − y(tj−1))] =

(1.6)Lagrange

=n∑

j=1

f(g(cj)) [x′(dj) + iy′(ej)] (tj − tj−1),

where dj, ej , cj ∈ [tj−1, tj]. Since this interval is very small as ‖∆‖ → 0,and since g′ is continuous, we can well approximate the sum of (1.6)


withn∑

j=1

f(g(cj)) [x′(cj) + iy′(cj)] (tj − tj−1),

which is a Riemann sum of

T∫

T0

f(g(t))g′(t)dt. Hence this last integral can

be well approximated with Riemann sums of the formn∑

j=1

f(ξj)(zj −

zj−1). This means that f is integrable on the arc ab and

∫

ab

f(z)dz =

T∫

T0

f(g(t))g′(t)dt.

E 106. Let us compute In =

∫

|z−z0|=r

1(z−z0)ndz for any nat-

ural number n = 0, 1, ... . The equation |z − z0| = r describes a circlewith centre at z0 and radius r. It has the following parametrizationz = z0 + r exp(iθ), where θ ∈ [0, 2π]. It is equivalent to the real formx = x0 + r cos θy = y0 + r sin θ

, θ ∈ [0, 2π]. Applying formula (1.5) we get:

In =

2π∫

0

1

rneinθrieiθdθ = r−n+1i · e

iθ(1−n)

i(1 − n)

∣∣∣∣2π

0

= 0,

if n = 1 and I1 = i

2π∫

0

dθ = 2πi. Here we used the equality: e2πmi =

cos 2πm+ i sin 2πm = 1 for any integer number m. Hence

(1.7)

∫

|z−z0|=r

1

(z − z0)ndz =

0, if n = 12πi, n = 1.

We shall now express the complex integral

∫

ab

f(z)dz as a complex

number, with a real and an imaginary part. For this, let ξj = ηj + iθj,


with ηj , θj ∈ R and let us come back to the Riemann sum (1.4):

n∑

j=1

f(ξj)(zj − zj−1) =

=n∑

j=1

[u(ηj , θj) + iv(ηj , θj)

][(x(tj) − x(tj−1)) + i(y(tj) − y(tj−1))] =

=n∑

j=1

[u(ηj, θj)(x(tj) − x(tj−1)) − v(ηj, θj)(y(tj) − y(tj−1))

]+

+i

n∑

j=1

[u(ηj , θj)(y(tj) − y(tj−1)) + v(ηj, θj)(x(tj) − x(tj−1))

].

Since the real part of this last sum is a Riemann sum for the line

integral

∮

ab

udx − vdy and since the imaginary part is a Riemann sum

for the line integral

∮

ab

vdx+ udy, we find that:

(1.8)

∫

ab

f(z)dz =

∮

ab

udx− vdy + i

∮

ab

vdx+ udy.

Let us compute

∫

γ

(3xy + y2i) dz, where γ is the oriented segment

[AB] with A(0, 0) and B(1, 2). Since u(x, y) = 3xy and v(x, y) = y2

can be easily computed, we use formula (1.8). A parametrization for[AB] is x = x, y = 2x, x ∈ [0, 1]. Hence,

∫

ab

f(z)dz =

1∫

0

(6x2 − 8x2)dx+ i

1∫

0

(4x2 + 12x2)dx = −2

3+ i

16

3.

T 97. (Cauchy fundamental theorem) Let D be a simpleconnected domain of C and let f : D → C be a differentiable (ana-lytic) complex function defined on D. Then, for any piecewise smooth

"closed" oriented curve γ ⊂ D, one has that

∮

γ

f(z)dz = 0.


P. Since f is differentiable one has that ∂u∂x

= ∂v∂y, ∂u∂y

= − ∂v∂x.

We can apply Green formula (3.1), Ch.6, for the both line integrals onthe right side of the equality, which results from formula (1.8):

∮

γ

f(z)dz =

∮

γ

udx− vdy + i

∮

γ

vdx+ udy =

=

∫∫

Dγ

(−∂u∂y

− ∂v

∂x

)dxdy + i

∫∫

Dγ

(−∂v∂y

+∂u

∂x

)dxdy = 0,

where Dγ is the domain bounded by γ (see Fig.2)

F 2

The simple connected property of D is necessarily. Indeed, forinstance ∫

|z−z0|=r

1

z − z0dz = 2πi = 0

(see formula (1.7)) because D = Cz0 is not simple connected.A main consequence of this Cauchy fundamental theorem is the pos-

sibility to construct a primitive function F (z) for an analytic function

f(z) defined on a simple connected domainD. The formula

∮

γ

f(z)dz =

0 for any piecewise smooth "closed" oriented curve γ ⊂ D, implies

like usually that the complex integral

B∫

A

f(z)dz does not depend on


the complex path Γ which connects the points A and B. Hence, ifone fixes a point z0 in D and if z is an arbitrary point in D, then

F (z) =

z∫

z0

f(ζ)dζ is a primitive function for f(z), i.e. F (z) is differen-

tiable and F ′(z) = f(z) for any z ∈ D (see Fig.3). The proof of thisstatement is almost similar to the proof of the analogous result for lineintegrals of the second type (see also the remark 22). Indeed, let w0 be apoint in D and let a closed disc B[w0, r] = z ∈ C : |z − w0| ≤ r withcentre at w0 and radius r > 0, sufficiently small such that B[w0, r] ⊂ D.Since D is connected, there exists a polygonal line [z0w0] ⊂ D, whichconnect z0 and w0. For any w ∈ B[w0, r], let us evaluate the quotient:

F (w) − F (w0)

w − w0

=

∫

[z0w0]

f(ζ)dζ +

∫

segm[w0w]

f(ζ)dζ −∫

[z0w0]

f(ζ)dζ

w − w0

=

=

∫

segm[w0w]

f(ζ)dζ

w − w0,

where segm[w0w] is the segment of the straight line which connect w0

and w (see Fig.3).

F 3


Since our goal is to prove that

limw→w0

F (w) − F (w0)

w − w0= f(w0),

let us evaluate now the distance between F (w)−F (w0)w−w0 and f(z) :

∣∣∣∣F (w) − F (w0)

w − w0

− f(w0)

∣∣∣∣ =

∣∣∣∣∣∣∣

∫

segm[w0w]

f(ζ)dζ −∫

segm[w0w]

f(w0)dζ

∣∣∣∣∣∣∣|w − w0|

≤

≤

∫

segm[w0w]

|f(ζ) − f(w0)| |dζ|

|w − w0|≤ supζ∈segm[w0w]

|f(ζ) − f(w0)| .

Since f is continuous on D, this last supremum goes to zero when w →w0.Hence,

∣∣∣F (w)−F (w0)w−w0 − f(w0)

∣∣∣ → 0,whenw → w0, i.e. limw→w0

F (w)−F (w0)w−w0 =

f(w0), i.e. F is differentiable at w0 and F ′(w0) = f(w0).What happens when D is not simple connected, i.e. if it has some

"gaps" or "holes"? We shall describe in the following a very simplemodel of a n-connected domain Ω.We start with a complex domain D∗

and let us consider another simple connected bounded domain D ⊂ D∗

such that D = D ∪ ∂D ⊂ D∗. Let D1, D2, ..., Dn−1 be n − 1 nonover-laping (the intersection Di∩Dj, i = j, is empty or at most a piecewisesmooth curve) domains contained in D. We assume that the boundary∂D = Γ+

0 is a directly oriented piecewise smooth curve and the bound-aries ∂D1 = Γ−1 , ..., ∂Dn−1 = Γ−n−1 are included inD and they are piece-

wise smooth inverse oriented curves (see Fig.4). Then Ω = D∪n−1j=1Dj

is said to be a n-connected domain. For n = 1 we get Ω = D itself, i.e.a simple connected domain. For n = 2, Ω = DD1 and we obtain adouble connected domain (with a "hole"), etc. To obtain from Ω a newsimple connected domain Ω∗, it is enough to get out from it n−1 piece-wise smooth curves γ1, ..., γn−1 which connect ∂D with ∂D1, ..., ∂Dn−1

respectively, traced twice in the opposite directions (see Fig.4).Hence Ω∗, which has as its boundary the union between ∂D, ∂D1, ..., ∂Dn−1

and γ1, ..., γn−1 traced twice each of them, is a simple connected do-main.


F 4

T 98. (n-connected Cauchy fundamental theorem) Withthese notation and hypotheses, let f : D∗ ∪n−1

j=1 Dj → C be a con-tinuous function which is analytic on Ω. Then,

(1.9)

∫

Γ+0

f(z)dz =

∫

Γ+1

f(z)dz +

∫

Γ+2

f(z)dz + ...+

∫

Γ+n−1

f(z)dz.

In particular, if h : D → C is a continuous function which is analytic onD except the points z1, z2, ..., zm of D (see Fig.5), then one can uniquely

define the numbers res(f, zj)def= 1

2πi

∫

|z−zj |=rj

f(z)dz, j = 1, 2, ...,m for

any rj > 0 sufficiently small such that B(zj , rj) ⊂ D and B(zj , rj) ∩B(zk, rk) = ∅ for j = k. Here res(f, zj) is said to be the residue of f atzj. Moreover, formula (1.9) immediately implies the famous "residuesformula":

(1.10)

∫

Γ+0

f(z)dz = 2πim∑

j=1

res(f, zj).

P. It is sufficient to apply the Cauchy fundamental theoremfor the simple connected domain Ω∗ and taking count that the curvesγ1, ..., γn−1 are traced twice in the opposite directions:

∫

Γ+0

f(z)dz +n−1∑

j=1

∫

Γ−j

f(z)dz +n−1∑

j=1

∫

γ−j

f(z)dz +n−1∑

j=1

∫

γ+j

f(z)dz = 0,


F 5

or ∫

Γ+0

f(z)dz =n−1∑

j=1

∫

Γ+j

f(z)dz.

The residue res(f, zj)def= 1

2πi

∫

|z−zj |=rj

f(z)dz is well defined because if

we diminish the radius rj of the disc to another one 0 < r∗j < rjthen, on the annulus z ∈ C : r∗j < |z − zj| < rj the function f(z)is analytic and, applying the first statement of this theorem, we get

that 12πi

∫

|z−zj |=rj

f(z)dz = 12πi

∫

|z−zj |=r∗j

f(z)dz, i.e. the residue res(f, zj) is

well defined. The equality (1.10) is the particular case of the generalCauchy formula (1.9), where we take the discs |z − zj| < rj for thedomains Dj.

R 37. Let H be a complex domain, let D be a bounded do-main included in H such that ∂D = Γ is a direct oriented, continuousand piecewise smooth curve. Let f be an analytic function on H, excepta finite number of points z1, z2, ...zm inside D and a finite number ofpoints w1, w2, ..., ws which are on ∂D = Γ (see Fig.6). If in a smallneighborhood Vj of wj, which does not contain any other wt, the curveΓ is smooth, then we say that the angle in radians attached to wj isαj = π. If wh is an angular point on Γ and the angle in radians betweenthe two distinct tangent lines at wh to Γ is αh, we attach this αh to wh


(here αh measure the "openness" of the point wh relative to the domainD; more exactly, we take a small disk B(wh, r) and consider the sectorof it bounded by the two tangent lines and which contains a region ofD). Then, the above residues formula can be generalized as follows:

(1.11)

∫

Γ+

f(z)dz = 2πim∑

j=1

res(f, zj) + is∑

j=1

αj · res(f, wj)

F 6

We shall try in the following to compute such a new number res(f, z0) =

12πi

∫

|z−z0|=r

f(z)dz for a function f which is analytic on the disc |z − z0| <

r, except the point z0, its centre. Then, we shall apply the residues for-mula (1.10) in order to compute some complicated real or complexintegrals. But what about the ∞ point of C = C ∪ ∞? (see the Rie-mann sphere in [Po], 6.5). To study the behavior of a function f in aneighborhood V∞,R = z ∈ C : |z| > R of infinite means to study thebehavior of the functions g(w) = f(1/w) in the corresponding neigh-borhood V0,1/R = w ∈ C : |w| < 1/R and then to come back to ∞with the obtained informations. Let us assume that a complex functionf is analytic on C except a finite set of points z1, z2, ..., zk. Let R > 0large enough such that |zj| < R for j = 1, 2, ..., k. We define

(1.12) res(f,∞)def=

1

2πi

∫

γ−

f(z)dz,


where γ is the circle |z| = R. We take γ with its inverse directionbecause the neighborhood V∞,R = z ∈ C : |z| > R of ∞ has as itsboundary with direct orientation, from the point of view of ∞, exactlythe oriented curve γ−. Now, formula (1.10) says:

(1.13) res(f,∞) +k∑

j=1

res(f, zj) = 0.

Sometimes this formula is very useful because, if the number k is large,to compute separately each residue and to make their sum can be verydifficult. Instead of doing this, formula 1.13 says that it is sufficientto compute −res(f,∞). But, how to compute all this residues withoutcomplex integrals?

We saw in [Po] that many elementary functions are defined bycomplex power series. A general form of a complex power series at z0(related to this fixed point!) is the following:(1.14)

f(z) = ...+a−n

(z − z0)n+

a−n+1

(z − z0)n+1+ ...

a−1

(z − z0)+ a0 + a1(z − z0)+

(1.15) +a2(z − z0)2 + ...+ an(z − z0)

n + ... =∞∑

n=−∞an(z − z0)

n.

where an, n ∈ Z are the coefficients of this series. The series itselfis called a Laurent series. It can be proved (see [Kra], or [ST]) thatalways exist 0 ≤ r < R ≤ ∞ such that the series is convergent on themaximum open annulus z ∈ C : r < |z − z0| < R and uniformlyconvergent on any closed annulus of the form z ∈ C : r′ ≤ |z − z0| ≤R′ for any r′, R′ with r < r′ ≤ R′ < R. If all the negative indexedcoefficients a−1, a−2, ..., a−n, ... are zero, then r = 0 and the series isconvergent on a maximum open disc B(z0, R) with centre at z0 and ofradius R. For instance,

ez−1 = 1 +1

1!(z − 1) +

1

2!(z − 1)2 + ...+

1

n!(z − 1)n + ...

has r = 0, R = ∞ and the series is convergent on the entire plane, butnot in ∞, where it does not exist like an usual function ( lim

z→∞ez−1 does

not exist because limn→∞

e2nπi = 1 and limn→∞

en = ∞). For

e1z = 1 +

1

1!z+

1

2!z2+ ...+

1

n!zn+ ...

r = 0, R = ∞, but the series is not defined at z = 0. It is defined at∞ and its values is 1 at this point.


Let us now compute the residue of f(z) from formula (1.15) bytaking the integral on a circle |z − z0| = ρ, with r < ρ < R :

(1.16) res(f, z0) =1

2πi

∞∑

n=−∞an

∫

|z−z0|=ρ

(z − z0)ndz = a−1,

because of the formulas (1.7). We can express the residue of f at a pointz0 even in a more simple way if all the negative indexed coefficientsa−n are zero except a finite number of them, starting with a−k = 0, i.e.a−k−1 = a−k−2 = ... = a−n = ... = 0. Then we see that

(z−z0)kf(z) = a−k+a−k+1(z−z0)+...+a−1(z−z0)k−1+a0(z−z0)k+....

Hence

(1.17) res(f, z0) = a−1 =1

(k − 1)!limz→z0

[(z − z0)

kf(z)](k−1)

.

Here the exponent (k − 1) means the derivative of order (k − 1) ofthe function (z − z0)

kf(z). We say that z0 is a pole of order k for thefunction f.

E 107. Let us use the residues formula (1.11) and formulas

(1.17), (1.16) to compute the complex integrals a) I(r) =

∫

|z−i|=r

sin 1z

(z2+2)2dz

and b) J =

∫

[tuv]

1z2(z2−1)

dz, where [tuvt] is the perimeter line, direct ori-

ented, of the triangle with vertices t = 0, u = 1 and v = i.To compute I(r) we need to compute all the finite nonzero residues

of f(z) =sin 1

z

(z2+2)2, i.e. the residue in at most the singular points (the

points at which the function is not analytic) of f(z). We easily see thatz0 = 0 is an essential singular point (see the definition bellow) of fbecause sin 1

z= 1

z− 1

3!z3+ ... and 1

(z2+2)2is analytic (differentiable) in

a small neighborhood of z0 = 0. The points z1 = i√

2 and z2 = −i√

2are poles of order 2 because, for instance, the function g(z) = (z −i√

2)2f(z) is analytic in a small neighborhood of i√

2 which does notcontain the points z0 = 0 and z2 = −i

√2, etc. It is clear enough

that res(f, z0) = c−1 = 14. Now, to compute the residues res(f, z1) and

res(f, z2) we need formula (1.17):

res(f, z0) =1

(k − 1)!limz→z0

[(z − z0)

kf(z)](k−1)

.


Thus,

res(f, z1) = limz→i

√2

[sin 1

z

(z + i√

2)2

]′=

= limz→i

√2

− 1z2

(z + i√

2)2 cos 1z− 2(z + i

√2) sin 1

z

(z + i√

2)4=

=i√

2 sin i√2− cos i√

2

16.

res(f, z2) = limz→−i

√2

[sin 1

z

(z − i√

2)2

]′=

= limz→−i

√2

− 1z2

(z − i√

2)2 cos 1z− 2(z − i

√2) sin 1

z

(z − i√

2)4=

=−i

√2 sin i√

2− cos i√

2

16.

We can now start the discussion. If r <√

2 − 1, then inside the circle|z − i| = r there exists no singular point for f. Thus, in this caseI(r) = 0. If r =

√2−1, then z1 is on the circle |z − i| =

√2−1 and we

apply formula (1.11) (we say that at z1 we have a semiresidue. Hence,

I(√

2 − 1) = πii√

2 sin i√2−cos i√

2

16. If

√2 − 1 < r < 1, then z1 is inside the

circle√

2 − 1, so that I(r) = 2πii√


2

16. If r = 1, then, besides

z1, one has a semiresidue at z0 and I(1) = 2πii√


2

16+ πi.

If 1 < r < 1 +√

2, then we have residues at z0 and z1 inside the

circle, so I(r) = 2πii√


2

16+ 2πi. If r = 1 +

√2, then we

have residues at z0 and z1 inside the circle and a semiresidue at z2,

so that I(1 +√

2) = 2πii√


2

16+ 2πi + πi

−i√


2

16. If

r > 1 +√

2, we have all the three points as residues inside the circle,

so that I(r) = 2πii√


2

16+ 2πi + 2πi

−i√


2

16.

To compute J =

∫

[tuv]

1z2(z2−1)

dz, we can use formula (1.11). One has

a residue at t = 0, with the associated angle α1 = π2and a residue at

u = 1, with the associated angle α2 = π4. But, denoting h(z) = 1

z2(z2−1),

one has

res(h, 0) = limz→0

[1

z2 − 1

]′= −lim 2z

z→0(z2 − 1)−2 = 0


and

res(h, 1) = limz→1

1

z2(z + 1)=

1

2.

Hence, J = π2i · 0 + π

4i · 1

2= πi

8.

If an infinite number of negative indexed terms a−n are distinct ofzero, we say that z0 is an essential point. If all a−n are zero, i.e. if

(1.18) f(z) = a0 + a1(z − z0) + ...+ an(z − z0)n + ...

we say that f is analytic at z0 and z0 is a regular point of f . In this lastsituation we call the expansion on the right side, a Taylor expansion.It can be proved that any analytic function f in a disc B(z0, R) has aTaylor expansion of the type (1.18). Let us evaluate the coefficients anin this last case. For this, let us take a circle γr with centre at z0 andan arbitrary radius r > 0 and let us integrate terms by terms:

∫

γr

f(z)

(z − z0)n+1dz =

∞∑

k=0

ak

∫

γr

(z − z0)k−n−1dz = an · 2πi,

because of formulas (1.7) (

∫

γr

(z− z0)k−n−1dz = 0 for k < n and

∫

γr

(z−

z0)k−n−1dz = 2πi for k = n) and because of the Cauchy fundamental

theorem (97) for k > n (then (z − z0)k−n−1 is analytic and so

∫

γr

(z −

z0)k−n−1dz = 0). Hence,

an =1

2πi

∫

γr

f(z)

(z − z0)n+1dz.

If we substitute γr with any other closed piecewise smooth curve γwhich contains z0 inside the interior of the domain bounded by γ, it isimmediate from the general Cauchy fundamental theorem 98 that

(1.19) an =1

2πi

∫

γ

f(z)

(z − z0)n+1dz.

Since an = 1n!f (n)(z0) (compute this n-th derivate in the Taylor expan-

sion (1.18)), one obtains:

(1.20) f (n)(z0) =n!

2πi

∫

γ

f(z)

(z − z0)n+1dz.


For n = 0 we get the basic Cauchy integral formula:

(1.21) f(z0) =1

2πi

∫

γ

f(z)

z − z0dz.

This formula supply us the value of f at any point z0 of a boundeddomain D with its boundary ∂D = γ, a continuous piecewise smoothcurve. It can be proved independently and then we can recover theTaylor expansion of an analytic function f(z) inside a disc or the Lau-rent series expansion for an analytic function defined in an annulus.Indeed, let us take a small r > 0 such that the disc B(z0, r) ⊂ Dγ,the simple connected domain bounded by γ. The 2-connected Cauchyfundamental theorem (see theorem 98) says that

1

2πi

∫

γ

f(z)

z − z0dz =

1

2πi

∫

|z−z0|=r

f(z)

z − z0dz.

Let us use formula (1.7) to evaluate the difference:∣∣∣∣∣∣∣

1

2πi

∫

|z−z0|=r

f(z)

z − z0dz − f(z0)

∣∣∣∣∣∣∣=

1

2π

∣∣∣∣∣∣∣

∫

|z−z0|=r

f(z)

z − z0dz −

∫

|z−z0|=r

f(z0)

z − z0dz

∣∣∣∣∣∣∣=

=1

2π

∫

|z−z0|=r

|f(z) − f(z0)||z − z0|

|dz| ≤ 1

2πr· 2πr sup

|z−z0|=r|f(z) − f(z0)| → 0,

when r becomes smaller and smaller. Since the quantity∣∣∣∣∣∣∣

1

2πi

∫

|z−z0|=r

f(z)

z − z0dz − f(z0)

∣∣∣∣∣∣∣does not depend on r, we see that it must be zero, i.e. we just obtainedthe basic Cauchy integral formula. Looking at the equality

f(w) =1

2πi

∫

|z−z0|=r

f(z)

z − wdz

like at an integral with a parameter w and applying an analogous ofLeibniz formula (1.7), we successively get:

f ′(w) =1

2πi

∫

|z−z0|=r

f(z)

(z − w)2dz,


f ′′(w) =1 · 2

2πi

∫

|z−z0|=r

f(z)

(z − w)3dz, ..., f (n)(w) =

n!

2πi

∫

|z−z0|=r

f(z)

(z − w)n+1dz,

i.e. we just proved formula (1.20) and the amazing fact that if f(z)in differentiable "once" on a simple connected domain D, then it hasderivatives of an arbitrary order on D.

Now, let us come back to the Taylor expansion (1.18) of an analyticfunction f(z) in a disc B(z0, R) and let r < R. Let us evaluate the

coefficient an = 12πi

∫

γ: |z−z0|=r

f(z)(z−z0)n+1dz, n = 0, 1, 2, ... :

|an| ≤1

2πM

∫

γ: |z−z0|=r

1

rn+1|dz| =

1

2πM · 1

rn+1· 2πr =

M

rn,

where M is the greatest value of |f(z)| when z runs on the compactsubset z : |z − z0| = r. The inequalities

(1.22) |an| ≤M

rn

are called the Cauchy inequalities. They are surprisingly used to provethe following basic theorems.

T 99. (Liouville theorem) Let f(z) be an analytic functionin the entire complex plane such that its modulus is bounded. Thenf(z) must be a constant.

P. Since f(z) is analytic in the entire plane and bounded bya numberM, in formula 1.22 one can make r → ∞ for each n = 1, 2, .... Hence a1, a2, ..., an, ... are all zero, i.e. f(z) = a0.

An immediate and deep application is the fact that some real func-tions which are bounded and nonconstant on the real line (see sin x,cosx, etc.) do not continue to be bounded as functions of a complexvariable. For instance |sin z| > 1 for an infinite number of values ofz. Indeed, since sin z is analytic on the entire C, then it cannot bebounded on C (otherwise one can apply Liouville theorem and sin xwould be constant!). But another important consequence of the Liou-ville theorem is the following basic result in Mathematics.

T 100. (the fundamental theorem of algebra) Any noncon-stant polynomial P (z) = a0 + a1z + ...+ anz

n with complex coefficientshas at least one root in C. In particular it has all roots in C and it canbe written as a product of the form:

P (z) = an(z − ζ1)(z − ζ2)...(z − ζn),

2. APPLICATIONS OF RESIDUES FORMULA 377

where ζ1, ζ2,..., ζn are all the roots of the polynomial P (z).

P. Assume contrary, namely that P (z) has no root in allcomplex plane. Then f(z) = 1

P (z)is analytic and bounded, because

limz→∞

f(z) = 0. Indeed, since limz→∞

f(z) = 0, there exists a large number

T > 0 such that |f(z)| < 1 for any |z| > T. Let M be the greatestvalue of f(z) on the closed disc B[0, T ]. Then |f(z)| < 1 + M for anyz ∈ C, i.e. f(z) is bounded and we can apply the above Liouville the-orem and find that f(z) is a constant function, i.e. P (z) is constant, acontradiction, because we assumed that it is nonconstant. Hence P (z)must have at least one root in C. The other statements can be eas-ily derived by using successively the Euclidean division algorithm forpolynomials.

2. Applications of residues formula

a)How do we evaluate real integrals of the form I =

2π∫

0

R(cos θ, sin θ)dθ

Here R(cos θ, sin θ) is a rational function of cos θ and sin θ. This

means a quotient P (x,y)Q(x,y)

of two polynomials in two variables in which

we substituted the variables x and y with cos θ and sin θ respectively.To compute the integral I, we make a change of variables: z = eiθ.Thus,

(2.1) I =1

i

∫

|z|=1

R

(1

2

(z +

1

z

),

1

2i

(z − 1

z

))· 1

zdz.

This last integral is a complex integral of a rational function

P (z)

Q(z)= R

(1

2

(z +

1

z

),

1

2i

(z − 1

z

))· 1

z

in variable z. Since this last function has at most a finite number ofpoles (the zeros of Q(z)) inside the circle |z| = 1 (or on its circumfer-ence), we can apply the residue formula and get:

(2.2) I = 2πi · 1

i

∑

Q(zk)=0

res

(P (z)

Q(z), zk

).

For instance, if we want to compute the coefficient am =

2π∫

0

cosmx2+sinx

dx of

the Fourier expansion of the function f(x) = 12+sinx

, x ∈ [0, 2π] (see


"Fourier series" chapter in any course of Advanced Mathematics), thebest idea is to use the above change of variables: z = eix. Thus,

am =1

i

∫

|z|=1

1

2· 1 + z2m

zm· 2iz

z2 + 4iz − 1· 1

zdz =

=

∫

|z|=1

1

zm(z2 + 4iz − 1)dz +

∫

|z|=1

zm

z2 + 4iz − 1dz.

b) Integrals of the form

∞∫

−∞

f(x)dx

These improper integrals of the first type can be easily computedby using residues formula. First of all we need an auxiliary result.

T 101. Let f(z) be an analytic function everywhere in theupper half-plane z ∈ C : Im z > 0, except a finite number of pointsz1, z2, ..., zn. We assume that in a neighborhood |z| > R0 > 0 of ∞ thefunction f(z) is bounded as follows: |f(z)| < M

|z|1+δ , where M > 0 and

δ > 0. Then

(2.3) limR→∞

∫

C′R

f(z)dz = 0,

where C ′R is a semicircle |z| = R, with Im z > 0 (see Fig.7)

P. Let us take R > R0 and write:∣∣∣∣∣∣∣

∫

C′R

f(z)dz

∣∣∣∣∣∣∣≤

∫

C′R

|f(z)| |dz| <∫

C′R

M

|z|1+δ|dz| =

πM

Rδ,

which goes to zero when R → ∞. Hence

∫

C′R

f(z)dz → 0 when R → ∞

and the theorem is completely proved.

Now we can prove the basic result which will help us to computeimproper integrals just announced.

T 102. Let f(x) be a real function of class C∞, definedon the entire real axis and let f(z) be a unique (see [Po], section 11.8)analytic prolongation of it to the upper half-plane Im z > 0.We assume


F 7

that the function f(z) satisfies the conditions of the above theorem 101.

Then our improper integral

∞∫

−∞

f(x)dx is convergent and:

(2.4)

∞∫

−∞

f(x)dx = 2πin∑

k=1

res(f, zk),

where z1, z2, ..., zn are all the singular points (at which f is not analyti-

cal) of f in the upper half-plane Im z > 0. In particular, if f(z) = P (z)Q(z)

is a rational function which satisfies the supplementary conditions thatdegQ(z) ≥ degP (z) + 2 and that the equation Q(z) = 0 has no realroot, then

(2.5)

∞∫

−∞

P (x)

Q(x)dx = 2πi

∑

Re zk>0 and Q(zk)=0

res(f, zk)

P. Let us take R0 large enough such that |zk| < R0 for anyk = 1, 2, ..., n. Now we consider in the upper half-plane the closedcontour [−R,R] ∪ C ′R (see Fig.7). We apply the residue formula andfind:

R∫

−R

f(x)dx+

∫

C′R

f(z)dz = 2πin∑

k=1

res(f, zk).


Since the conditions of theorem 101 are satisfied, taking limit whenR→ ∞, we get exactly formula (2.4).

The last statement is true because, f(z) = P (x)Q(x)

together with the

above restrictions on the polynomials P (z), Q(z), satisfies the sameconditions of theorem 101.

For instance, we know that the integral I =∫∞−∞

1x4+1

dx is conver-

gent but its computation is not so easy. We see that∣∣ 1z4+1

∣∣ ≤ 1|z|3 , for

|z| large enough, i.e. R0 = 1, for instance, M = 1 and δ = 2 > 0 intheorem 101. Thus we can apply the last theorem. For this we need tofind all the zeros of z4 +1 = 0 in the half-plane Im z > 0. The solutions

of the equation z4 + 1 = 0 are z0,1,2,3 = eiπ+2kπ

4 , k = 0, 1, 2, 3. But only

z0 = eiπ4 and z1 = ei

3π4 are in the upper half-plane. Thus,

I = 2πi

[res

(1

z4 + 1, ei

π4

)+ res

(1

z4 + 1, ei

3π4

)].

But, there is a problem! Formula (1.17) is not practicable here! Why?Let us see:

res

(1

z4 + 1, ei

π4

)= limz→z0

[(z − z0)

1

z4 + 1

]=

1

(z0 − z1)(z0 − z2)(z0 − z3).

To compute this last number is not so easy. But, let us look carefullyto the general formula 1.17 for the case of a simple pole z0, i.e. fork = 1 and for f(z) = P (z)

Q(z), a rational function:

(2.6) res(f, z0) = limz→z0

(z − z0)P (z)

Q(z)=

P (z0)

limz→z0

Q(z)−Q(z0)z−z0

=P (z0)

Q′(z0).

Using this nice formula in our case, we get:

res

(1

z4 + 1, ei

π4

)=

1

4z3

∣∣∣∣z=ei

π4

=1

4eiπ4

.

Finally we obtain

I = 2πi

[res

(1

z4 + 1, ei

π4

)+ res

(1

z4 + 1, ei

3π4

)]=π√

2

2.

If the residues of f are in the lower half-plane and similar conditionsfor f are satisfied, a similar result is true (do it!).

c) Integrals of the form

∞∫

−∞

eiaxf(x)dx

Such integrals appear in the computation of a Fourier integral (ora Fourier transform).


We begin with an auxiliary result.

T 103. (Jordan’s lemma) Let f(z) be an analytic func-tion in the upper half-plane Im z > 0, except a finite number of pointsz1, z2, ..., zn. We also assume that there exists a real function µ(r),r ≥ 0, such that |f(z)| < µ(r) for any z on the semicircle |z| = r,Im z ≥ 0 and lim

r→∞µ(r) = 0. Then for any real number a > 0

(2.7) limR→∞

∫

C′R

eiazf(z)dz = 0,

where C ′R is a semicircular arc |z| = R in the upper half-plane Im z ≥ 0.

P. Let us make a change of variable, by putting z = R · eiθ,where 0 ≤ θ ≤ π. We then obtain∣∣∣∣∣∣∣

∫

C′R

eiazf(z)dz

∣∣∣∣∣∣∣≤

∫

C′R

∣∣eiaz∣∣ |f(z)| |dz| ≤ Rµ(R)

π∫

0

∣∣eiaz∣∣ dθ =

(2.8) = Rµ(R)

π∫

0

e−aR sin θdθ,

because∣∣eiaz

∣∣ =∣∣∣eiaR·eiθ

∣∣∣ =∣∣eiaR(cos θ+i sin θ)

∣∣ =∣∣eiaR cos θ

∣∣ ∣∣e−aR sin θ)∣∣ =

= |cos(aR cos θ) + i sin(aR cos θ)|∣∣e−aR sin θ)

∣∣ =∣∣e−aR sin θ)

∣∣ .Let us evaluate the real simple integral

I =

π∫

0

e−aR sin θdθ =

π2∫

0

e−aR sin θdθ +

π∫

π2

e−aR sin θdθ.

Making the change of variables η = π − θ in the last integral, we findthat

π∫

π2

e−aR sin θdθ = −0∫

π2

e−aR sin ηdη =

π2∫

0

e−aR sin ηdη =

π2∫

0

e−aR sin θdθ.

Thus, I = 2

π2∫

0

e−aR sin θdθ. Since [tan θ − θ]′ = 1cos2 θ

−1 ≥ 0, the function

θ tan θ−θ is increasing, so that tan θ ≥ θ if θ ∈ [0, π2]. Now

[sin θθ

]′=


θ cos θ−sin θθ2

= θ−tan θθ2 cos θ

≤ 0 if θ ∈(0, π

2

). Hence the function θ sin θ

θis

decreasing and its least value on (0, π2] is

sin π2

π2

= 2π, i.e. sin θ

θ≥ 2

πor

sin θ ≥ 2πθ. Therefore,

e−aR sin θ ≤ e−2aRπθ

and

I ≤ 2

π2∫

0

e−2aRπθdθ = − π

aRe−

2aRπθ∣∣∣π2

0= − π

aR

[e−aR − 1

].

Since ∣∣∣∣∣∣∣

∫

C′R

eiazf(z)dz

∣∣∣∣∣∣∣≤ Rµ(R)

π∫

0

e−aR sin θdθ ≤ µ(R)π

a

[1 − e−aR

]

and since µ(R) → 0, when R→ ∞, we obtain that∣∣∣∣∣∣∣

∫

C′R

eiazf(z)dz

∣∣∣∣∣∣∣→ 0,

when R → ∞.

R 38. a) If a < 0 and f(z) satisfies all the conditions whichappear in the statement of the theorem 103 for the lower half-planeIm z ≤ 0, then the formula 2.7 is again true for any semicircular arcC ′R in the lower half of the xOy-plane. b) Similar assertions hold fora = ±iλ, λ > 0, when we integrate on the right (Re z ≥ 0, see Fig.8)or on the left (Re z ≤ 0) half of the xOy-plane. We leave to the readerto state and prove such similar Jordan’s lemmas. Such variations ofJordan’s lemma are extensively used in operational calculus (Fourierand Laplace transforms).

T 104. Let f(x) be a continuous real valued function de-fined on the entire real axis R and let f(z) be an extension of it intothe entire upper half-plane Im z ≥ 0. We assume that f(z) satisfiesthe conditions of Jordan’s lemma and that it is analytic in this upperhalf-plane with the exception of a finite number of points z1, z2, ..., znin this half-plane, which are not on the real axis. Then the integral∫∞−∞ e

iaxf(x)dx, a > 0 exists and it can be computed as follows:

(2.9)

∫ ∞

−∞eiaxf(x)dx = 2πi

n∑

k=1

res[eiazf(z), zk

].


F 8

P. Let R0 be large enough such that |zk| < R0, k = 1, 2, ..., n,i.e. all the singularities (the points at which f is not analytic) of f beinside the circle |z| = R0. Take now R > R0 and consider in the upperhalf-plane the following "closed" contour Γ = [−R,R] ∪ C ′R, where C ′Ris the semicircle of the circle |z| = R, which are in the upper-half planeof the xOy-plane (draw it!). By applying the residues formula (1.10)we get:∫

Γ+

eiazf(z)dz =

∫ R

−Reiaxf(x)dx+

∫

C′R

eiazf(z)dz = 2πin∑

k=1

res[eiazf(z), zk

].

Making R → ∞ and applying Jordan’s lemma 103, we get formula(2.9).

E 108. In the operational calculus (Fourier transforms) weneed to compute integrals of the form:

I1 =

∫ ∞

−∞f(x) cosnx dx, I2 =

∫ ∞

−∞f(x) sinnx dx, n = 0, 1, ....

To do this we consider the following integral

I = I1 + iI2 =

∫ ∞

−∞einxf(x) dx

and try to apply formula (2.9) to it. For instance, let us computeI1 =

∫∞−∞

cosnxx2+h2

dx, where h > 0 is a real parameter and n = 0, 1, 2, ... .Thus,

I =

∫ ∞

−∞

einx

x2 + h2dx = 2πi · res

[einz

z2 + h2, ih

]=


2πi · limz→ih

[(z − ih)

einz

z2 + h2

]= 2πi · e

−nh

2ih=π

he−nh.

Thus I1 = πhe−nh and I2 =

∫∞−∞

sinnxx2+h2

dx = 0 (this was obvious from thebeginning because the function under the integration sign is odd and theinterval is symmetric w.r.t. the origin).

E 109. (Computing Fresnel’s integrals) In example 50 weproved that the Fresnel’s integrals I1 =

∫∞0

cosx2dx and I2 =∫∞0

sin x2dxare convergent. Let us compute them here. For this we naturally intro-

duce the auxiliary complex function f(z) = eiz2. Let A(r, 0), B

(r√2, r√

2

)

and let γr be the arc of the circle |z| = r, between A and B. LetΓ = [OA] ∪ γr ∪ [BO] with its direct orientation (draw it!). ApplyingCauchy fundamental theorem 97 to the analytic function f(z) = eiz

2,

we get:

(2.10) 0 =

∫

Γ+

eiz2

dz =

∫

[OA]

eix2

dx+

∫

γr

eiz2

dz +

∫

[BO]

eiz2

dz.

Now we show that

(2.11) limr→∞

∫

γr

eiz2

dz = 0.

Indeed, putting z2 = w, we obtain dz = dw2√w, where

√w is the first

branch (k = 0) of the square root multivalued function w = |w| ei argw √|w|ei argw+2kπ2 , k = 0, 1. Here θ = argw is the unique angle in [0, 2π)

which satisfies the relations: Rew = |w| cos θ and Imw = |w| sin θ.Hence, if we denote Γr2 the quarter of the circle |w| = r2 situated inthe first quadrant, one has:

∫

γr

eiz2

dz =

∫

Γr2

eiw

2√wdw = ir2

π2∫

0

eir2(cos θ+i sin θ)

2reiθ2

eiθdθ.

Since∣∣eiθ

∣∣ = 1 for any angle θ, we get:∣∣∣∣∣∣

∫

γr

eiz2

dz

∣∣∣∣∣∣≤ r

2

π2∫

0

e−r2 sin θdθ ≤ r

2

π2∫

0

e−r2 2πθdθ =

r

2

−2

r2πe−r

2 2πθ

∣∣∣∣π2

0

=1

rπ

(1 − e−r

2)→ 0,


when r → ∞. Here we used again the inequality sin θ ≥ 2πθ which was

deduced during the proof of Jordan’s lemma. Let us compute now theother two integrals of formula (2.10):

∫

[OA]

eix2

dx =

r∫

0

cosx2dx+ i

r∫

0

sinx2dx,

∫

[BO]

eiz2

dzz=ρei

π4

= −eiπ4r∫

0

e−ρ2

dρ.

Hence, formula (2.10) becomes:r∫

0

cosx2dx+ i

r∫

0

sinx2dx = eiπ4

r∫

0

e−ρ2

dρ−∫

γr

eiz2

dz.

Taking limit when r → ∞ and using formulas (2.11) and (2.21) weget:∞∫

0

cosx2dx+i

∞∫

0

sin x2dx = eiπ4

√π

2=

[1√2

+ i1√2

] √π

2=

√π

2√

2+i

√π

2√

2.

Hence ∞∫

0

cosx2dx =

∞∫

0

sin x2dx =

√π

2√

2.


1. Find the real and imaginary parts of the functions:a) f(z) = z − 3iez; b) f(z) = sin z; c) f(z) = tan z; d) f(z) = z

z+1.

2) Compute∫γ(1 + i−2z)dz, where γ is the segment of the straight

line which connect z1 = 0 and z2 = 2 − i.3) Compute

∫γ(1 + zz)dz, where γ is the semicircle |z| = 3, 0 ≤

arg z ≤ π.

4) Compute: a)∫ 2−ii

(3z − 1)dz; b)∫ i0z sin zdz; c)

∫ 2+i

−1zez

2dz;

5) Compute: a)∫|z|=1

ez

z3+3zdz; b)

∫|z−i|=1

eiz

z2+1dz; c)

∫|z|= 1

3

sin zz3dz;

d)∫|z|=2

ezdzz3(z+1)

; e)∫|z|=1

z3 sin 1zdz; f)

∫|z−1|=1

e2z

z3−1dz;

g)∫|z−2i|=a

1z2(z2+4)

dz (discussion on a > 0).

6) Compute: a)∫∞−∞

dx(x2+4)3

; b)∫∞−∞

132+x6

dx; c)∫∞−∞

dx(x2+4x+6)2

;

d)∫∞−∞

sin 5xx2+4

dx; e)∫∞0

x2 cos 2x(x2+2)2

dx; f)∫∞0

cos ax−cos bxx2

dx, a, b > 0.

APPENDIX A

Exam samples

1. June, 2010

1. Find the area of the plane domain bounded by the curves: y =3√x2, y = 1

4+x2, x = 2 and x = 2

√3.

2. Prove that the integral I(a) =∫∞0

e−at−1tet

dt is convergent for anya > 0 and compute it. What about its uniform convergence?

3. Compute

∫ ∫

x2+y2≤9

√x2 + y2dxdy.

4. Find the volume of the space domain

x2 + y2 ≤ 8z

0 ≤ z ≤ 2.

2. June, 2010

1. The graphic of y =√x 4√x2 + 1, x ∈ [0, 1] rotates around Ox-

axis. Find the volume of the resulting solid.2. Prove that the integral

∫∞0

1(x2+1)(x2+4)

dx is convergent and then

compute it.3. Find the area of the plane surface bounded by the curves: y =√

10x and y2 = 10x.

4. Compute

∫∫ ∫

x2+y2+z2≤9, z≥0

zdxdydz. Draw this domain and sup-

ply with a mechanical interpretation the obtained result.

3. June, 2010

1. Prove that I =∫∞1

tan−1 xx2

dx is convergent and compute it.

2. Compute the length of the curve Γ :

x = t2

y = t3, t ∈ [0, 1].

3. Let O(0, 0), A(1, 1), B(1,−1) and let the lamina OAB with thedensity function f(x) = x. Find its mass and the coordinates of itscentre of mass.

387

388 A. EXAM SAMPLES

4. Find the coordinates of the centre of mass for the total surface

of the cone:

x2 + y2 = 4z2

0 ≤ z ≤ 2.

4. June, 2010

1. The curve y = xe−x3, x ∈ [0,∞) rotates around the Ox-axis.

Find the volume of the obtained solid. First of all verify if this volumeis finite.

2. Starting with the formula∫ 1

0

tx−1(1 − t)y−1dt =Γ(x)Γ(y)

Γ(x+ y), x > 0, y > 0,

compute J =∫ π

2

0sin6 x cos6 xdx.

3. Let A(2, 1) and B(1, 0). The homogenous lamina OAB of density2, rotates around Ox-axis. Find its moment of inertia w.r.t. Ox-axis.

4. Let D : x2 + y2 + z2 ≤ 9, z ≥ 0. Compute I =

∫∫∫

D

z2dxdydz.

5. September, 2010

1. Let A(2,√

2), B(1, 0) and C(2,−√

2). Let γ be the arc of theparabola y2 = x which connect A and C. Let [ABC] the polygonal linewhich passes through A,B and C. Compute the area bounded by γand [ABC].

2. Find the work of the field−→F (x, y) = xy

−→j along the polygonal

arc [OABO], with this orientation, where A(1, 2) and B(3, 0).3. Compute the mass of the lamina D : x2 + y2 ≤ 9, x ≥ 0, y ≤ 0,

if the density function is f(x, y) = |y| .4. Find the coordinates of the centre of mass for the solid bounded

by the sphere x2 + y2 + z2 = 9 and inside the cylinder x2 + y2 = 4,z = 0, if the density function is f(x, y, z) = z.

6. September, 2010

1. Prove that I =∫∞0

1

(√x+1)

3dx is convergent and then compute

it.2. Draw the domain D : x2 + y2 ≤ 4, x2 + (y − 1)2 ≥ 1 and then

find the centre of mass of D (the density function is considered to bef = 1).

3. If a = Γ(23

), compute K =

∫∞0x10e−x

3dx as a function of a.

4. Find the volume of the domain D : x2+y2 ≤ 2z, x2+y2+z2 ≤ 3.

APPENDIX B

Basic antiderivatives

Prove all the following formulas (here∫f(x)dx means a primitive

of f(x)):

(0.1)

∫xαdx =

1α+1

xα+1, if α = −1.ln x, if α = −1 and x > 0

ln(−x), if α = −1 and x < 0.

(0.2)

∫axdx =

ax

ln a, if a > 0, a = 1.

(0.3)

∫1

x2 − a2=

1

2aln

∣∣∣∣x− a

x+ a

∣∣∣∣ , a = 0, x = ±a.

(0.4)

∫1

x2 + a2dx =

1

atan−1 x

a, a = 0.

(0.5)

∫1√

x2 + a2dx = ln

∣∣∣x+√x2 + a2

∣∣∣ , a = 0.

(0.6)

∫1√

a2 − x2dx = sin−1 x

a, a = 0, a2 − x2 > 0.

(0.7)

∫ √a2 − x2dx =

x

2

√a2 − x2 +

a2

2sin−1 x

a, a = 0, a2 − x2 > 0.

(0.8)

∫ √x2 ± a2dx =

x

2

√x2 ± a2 ± a2

2ln(x+

√x2 ± a2), a = 0.

(0.9)

∫sin xdx = − cosx.

(0.10)

∫cosxdx = sin x.

389

390 B. BASIC ANTIDERIVATIVES

(0.11)

∫1

cos2 xdx = tan x, x = (2k + 1)

π

2.

(0.12)

∫1

sin2 xdx = − cot x, x = kπ.

(0.13)

∫tan xdx = − ln |cosx| , x = (2k + 1)

π

2.

(0.14)

∫cot xdx = ln |sin x| , x = kπ.

(0.15)

∫1

sin xdx = ln

∣∣∣tanx

2

∣∣∣ , x = kπ.

(0.16)

∫1

cosxdx = ln

∣∣∣tan(π

4− x

2

)∣∣∣ , x = (2k + 1)π

2.

(0.17)

∫sinh xdx = cosh x.

(0.18)

∫cosh dx = sinh x.

(0.19)

∫1

cosh2 xdx = tanhx =

sinh x

cosh x.

(0.20)

∫1

sinh2 xdx = − coth x = −coshx

sinh x.

Index

"closed" curve, 206"closed" surface, 2091-dimensional deformation, 1833-D plate, 3153D-lamina, 315

divergence, 343general domain, 273

abelian integral, 27absolutely convergent, 119, 129abstract curve, 186algebraic function of one variable, 26almost everywhere, 61analytic, 362analytic function , 367analytic prolongation, 381annulus, 370approximation, 93area, 73area of a plane figure, 227automorphism, 14

Bezout’s theorem, 13binomial differential form, 32border, 351boundary, 351boundary of a subset, 70bounded domain, 362bounded function, 125bounded subset, 70

Cauchy criterion, 155Cauchy criterion for improper

integrals I, 118Cauchy criterion for improper

integrals II, 128Cauchy criterion for limits, 112

Cauchy criterion of uniform limit,149

Cauchy fundamental theorem, 366Cauchy inequalities, 378Cauchy integral formula, 377Cauchy-Riemann relations, 363central field, 204centre of mass, 317change of an elementary volume, 289change of variable formula, 9change of variables, 254change of variables formula, 257charge density, 343circulation, 198, 353closed differential form, 204, 218closure (topological), 70comparison test 1, 120, 130complementaries formula, 171complex convergence , 362complex curve, 362complex differentiability, 361complex function, 362complex functions, 361complex number field, 361complex path, 368conic surface, 327conical deformation, 313connected, 70connected domain, 362conservative field, 204continuity theorem, 151, 155continuity theorem for proper

integrals with parameters, 140continuous, 362coprime polynomials, 16Coulomb law, 343curl F, 209, 353

391

392 INDEX

cylindrical coordinates, 299cylindrical deformation, 312

Darboux criterion, 57, 229, 238, 276Darboux sums, 54decomposition into simple fractions,

20deformation, 183, 307density function, 51, 186, 227, 267,

315diameter, 274diffeomorphism, 254, 288differentiable, 362differential form, 4, 198differential form of order 2, 332differentiation theorem, 153Dirichlet’s integral, 124, 162Dirichlet’s test, 123, 133discontinuity points, 183distributions, 344divergence theorem, 341division, 51, 236, 274, 315, 328, 363domain, 70, 362double Darboux sums, 228double integral, 229double oriented, 328double Riemann sum, 228

electric constant, 343electric field, 343element of area, 229, 316element of length, 188elementary domain , 273elementary figure, 70elementary figures, 227elementary functions, 168elementary subsets, 114elliptic integrals, 46epsilon-delta criterion, 110equivalent paths, 186essential point, 376essential singular point, 374Euclid’s division algorithm, 11Euclidean division algorithm, 379Euler substitutions, 28Euler-Poisson integral, 164exact differential form, 204, 218Exam samples, 389explicitly described, 280

F is integrable on the orientedsurface (S), 329

factorial formula, 171factoriality theorem, 15figure, 70flux of the vector F(M) through the

surface (S), at the point M, 329flux-divergence formula, 341flux-divergence theorem, 341Fourier expansion, 379Fresnel’s, 124Fresnel’s integrals, 165, 386Froullani’s integrals, 159Fubini theorem, 155Fubini’s theorem, 146fundamental theorem of algebra, 378

Gabriel’s trumpet, 115gain principle, 202gamma function, 168Gauss formula, 341, 350Gauss law in Physics, 343Gauss-Laplace distribution, 181general change of variables in triple

integrals, 291general iterative formula I, 281general iterative formula II, 284generalized integrals of the first type,

115global flux, 329global productivity, 342gradient, 350gradient of temperature, 350greatest common divisor criterion, 16greatest common divisors for

polynomials, 16Green formula, 246, 367Green theorem, 351

heat, 350heat flow, 350

imaginary part, 362improper integral of the second type,

126improper integral with a parameter,

154improper integrals of the first type,

115

INDEX 393

independent on path, 203inferior Darboux sum, 238, 275integrable, 274integrable on the arc ab, 364integral by parts, 10integral cosine, 143integral with parameters, 139integration theorem, 153interior point, 70interpolating function, 95interpolation polynomial, 96interpolation with parabolas, 100irreducible polynomial, 11irrotational field, 208iterative formula, 229iterative formulas for parallelepipeds,

270iterative general formula, 241

Jordan’s lemma, 383

Koch’s curve flake of snow, 190

L-measure zero, 61Lagrange polynomial, 97lamina, 227Laplace equation, 351Laplace error function, 113Laurent series, 373Lebesgue criterion, 61Lebesgues measure zero, 61left limit, 107Leibniz formula, 155Leibniz’ formula for proper integrals

with parameters, 145Leibniz-Newton formula for line

integrals, 206length of a curve, 89limit comparison test, 121, 131limit of a function, 107limit point, 107line integral of the first type, 188line integral of the second type , 198linear bar, 51Liouville theorem, 378loaded curve, 186

Möbius surface, 325marking points, 51, 187, 236, 363mass center, 191

mass of a bar, 51mass of a wire, 188mean formula, 278mean formula for line integrals, 190mean theorem, 317mean theoremsmean formulas, 65

measurable, 71measure, 71

n-connected Cauchy fundamentaltheorem, 369

Newton area function, 3Newton-Leibniz formula, 67nodes, 96nonoverlaping, 215norm, 236, 274norm of a division, 51, 268, 315

open ball, 70, 361orientable surface, 324orientated smooth curve, 199orientation function, 324orientation of an arc, 196oriented smooth curve, 206

parabolic PDE, 351paraboloid, 319, 332parallelepiped, 70parallelepipedic domain, 268parameter, 183parametric equations of a surface,

307parametric path, 183parametric representation, 183parametric sheet, 307parametrisation, 183partial differential equation, 351partition, 274path, 183piecewise continuous, 227piecewise continuous function, 61,

267piecewise smooth, 183piecewise smooth curve, 362piecewise smooth surface, 267polar coordinates, 257pole, 374polygonal line, 70

394 INDEX

polynomial function, 10position vector, 183potential energy, 204potential field, 204potential function, 204potential theory, 344primitiveantiderivative, 3

primitive function, 204principal value, 114, 127proper integral with one parameter,

140

rate of flow, 350rational function of two variables, 26real function with complex values,

362real part, 362rectangles formula, 92rectifiable, 89reduction formula, 171reduction theorem, 301regular point, 376regular transformation, 288residue, 374residues formula, 370Riemann criterion, 57Riemann double sum, 148Riemann integrable, 52, 228, 237,

269, 274Riemann sphere, 372Riemann sum, 52, 188, 197, 237,

315, 329, 363right limit, 107

segment (oriented), 70series test, 118series test 1, 136side limits, 107simple connected, 209, 366simple convergent, 154simple fractions, 19simple limit along a set of

parameters, 149simple subdomain, 345simple w.r.t. Ox-axis, 213, 241simple w.r.t. Oy-axis, 214simple w.r.t. Oz-axis, 281Simpson formula, 101

singular point for a function, 126singular points, 183, 374smooth, 183smooth deformation, 288, 323solid, 267space domain, 267specific heat, 350spherical coordinates, 295spherical deformation, 313starred, 218starred domain, 357Stokes formula, 353substitution, 32superior Darboux sum, 238, 276support curve, 183support of a lamina, 227surface, 307surface integral, 316surface integral of the second type ,

329surface with two distinct faces, 324

temperature, 350test p, 122test-q, 132tetrahedron, 334thermal conductivity, 350thin plate, 227totally closed, 341transition matrix, 322trapezoids formula, 95

unbounded at a point, 126uniform limit along a set of

parameters, 149uniformly continuous, 51uniformly convergent, 155upper half-plane, 383

variation of a function, 51velocity of the heat flow, 350volume, 71

week limit, 114Weierstrass test, 156well approximate, 330well approximated, 65, 365wire, 186work, 198, 353work of a field, 91

INDEX 395

zero criterion, 116

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