residue theoryfacstaff.cbu.edu/~wschrein/media/m413 notes/m413c6.pdf · chapter 6 residue theory we...

CHAPTER 6

Residue Theory

We developed three methods of evaluating contour integrals around closedcurves:

(1) The Cauchy-Goursat Theorem - but this requires the integrand to be ana-lytic in and on the curve;

(2) Cauchy’s integral formulas - but only applies to integrands that have a veryspecial form;

(3) Theorem 4.8 - but replaces a contour integral along a curve containinga number of singularities with contour integrals along curves each of whichencloses only one singularity.

Cauchy’s residue theorem, developed in this chapter, encompasses all of these.

6.1. Cauchy’s Residue Theorem

Cauchy’s residue theorem is a simple consequence of the equations for coe�-cients of a Laurent series.

To illustrate, suppose f(z) =1X

n=�1an(z�z0)

n is the Laurent series of a function

f valid in some annulus 0 < |z�z0| < R about an isolated singularily z0. WhenC is a simple, closed, piecewise smooth curve in this annulus that contains z0

in its interior, coe�cient a�1 is defined by

146

6.1. CAUCHY’S RESIDUE THEOREM 147

a�1 =1

2⇡i

I

Cf(z) dz =)

I

cf(z)dz = 2⇡ia�1.

Cauchy’s residue theorem is a simple extension of this result to the case whenC encloses an arbitrary number of isolated singularities.

Definition (6.1). The residue of a function f at an isolated singularityz0 is denoted by Res[f, z0]. It is the coe�cient a�1 in the Laurent expansion off about z0.

From the examples on pages 141-142 of these notes,

Reshsin z

z, 0i

= 0, Res[e1/z, 0] = 1

Reshz � sin z

z6, 0i

= � 1

120

Theorem (6.1 – Cauchy’s Residue Theorem). Suppose a function f isanalytic inside and on a simple, closed, piecewise smooth curve C, exceptat singularies z1, . . . , zn in its interior. Then

I

Cf(z) dz = 2⇡i

nX

j=1

Res[f, zj].

Proof. Construct around the zj disjoint circles Cj : |z � zj| = rj suchthat each Cj is interior to C (see the diagram at the top of the next page).By Theorem 4.8, the contour integral around C can be replaced by contourintegrals around the Cj,

148 6. RESIDUE THEORY

I

Cf(z) dz =

nX

j=1

I

Cj

f(z) dz.

If f is expanded in a Laurent series about zj,

f(z) =1X

n=�1an(z � zj)

n where a�1 =1

2⇡i

I

Cj

f(z) dz =)

I

Cj

f(z) dz = 2⇡ia�1 = 2⇡i Res[f, zj] =)

I

Cf(z) dz = 2⇡i

nX

j=1

Res[f, zj]. ⇤

Problem (Page 268 #13). Evaluate

I

C

1

z2 + 4z + 16dz where C : |z| = 1.

Solution.

z2 + 4z + 16 = 0 () z =�4 ±

p16� 64

2= �2 ± 2

p3 i.

Both of these values are exterior to C, as seen in the diagram at the top of thenext page. Thus I

C

1

z2 + 4z + 16dz = 0.

⇤


Example. Evaluate

I

Csin⇣1

z

⌘dz where C : |z| = 2.

Solution. Since the Laurent series for sin⇣1

z

⌘around z = 0 is

sin⇣1

z

⌘=

1

z� 1

3!z3+

1

5!z5� · · · ,

the residue of sin⇣1

z

⌘is Res

sin⇣1

z

⌘, 0

�= 1. Thus

I

Csin⇣1

z

⌘dz = 2⇡i(1) = 2⇡i. ⇤

The following theorem gives a formula for calculating residues at poles.

Theorem (6.2). If f has a pole of order m at z0, then

Res[f, z0] = limz!z0

1

(m� 1)!

dm�1

dzm�1

⇥(z � z0)

mf(z)⇤

Proof. f has a pole of order m at z0 =) (Laurent expansion)

f(z) =a�m

(z � z0)m+

a�m+1

(z � z0)m�1+ · · · +

a�1

(z � z0)+ a0 + · · · ,

valid in some annulus 0 < |z � z0| < R. Thus

(z � z0)mf(z) = a�m + a�m+1(z � z0) + · · · + a�1(z � z0)

m�1 + · · · =)dm�1

dzm�1

⇥(z � z0)

mf(z)⇤

= (m� 1)!a�1 + m!a0(z � z0) + · · · =)


limz!z0

1

(m� 1)!

dm�1

dzm�1

⇥(z � z0)

mf(z)⇤

= a�1 = Res[f, z0] ⇤

Problem (Page 268 # 1). Calculate the residue for the isolated singularity

of(z + 3)6

(z � 4i)4at z = 4i.

Solution. The isolated singularity at z = 4i is a pole of order 4 =)

Res

(z + 3)6

(z � 4i)4, 4i

�=

1

3!limz!4i

⇢d3

dz3

(z � 4i)4(z + 3)6

(z � 4i)4

��

=1

6limz!4i

⇥6(5)(4)(z + 3)3

⇤= �2340 + 880i. ⇤

Problem (Page 268 # 21). Evaluate

I

C

1

z2 � z + 2dz where C : |z| = 3.

Solution.1

z2 � z + 2=

1

(z + 2)(z � 1)has simple poles at z = �2 and

z = 1, both interior to C.

Res

1

z2 � z + 2, 1

�= lim

z!1

z � 1

(z + 2)(z � 1)=

1

3, and

Res

1

z2 � z + 2,�2

�= lim

z!�2

z + 2

(z + 2)(z � 1)= �1

3=)

I

C

1

z2 � z + 2dz = 2⇡i

✓1

3� 1

3

◆= 0. ⇤


An important conequence of the residue theorem is the principle of the ar-gument.

Theorem (6.3 – Principle of the Argument). Let f be analytic in a do-main D except at a finite number of poles. If C is a simple, closed, piece-wise smooth curve in D which does not pass through any poles or zeros off , then

(⇤)I

C

f 0(z)

f(z)dz = 2⇡i(N � P )

where N and P are the sums of the orders of the zeroes and poles of fwithin C.

To see why this is called the principle of the argument,we transform the contourintegral (⇤) from the z-plane to the w-plane with w = f(z). It follows that

N � P =1

2⇡i

I

C

f 0(z)

f(z)dz =

1

2⇡i

I

f(C)

1

wdw,

where f(C) is the image of C in the w-plane. But according to text Example4.22 (page 182), the contour integral on the right is the number of times thatf(C) encircles w = 0. In other words, N � P measures how many times thecurve f(C) encircles w = 0 as z traverses C in the counterclockwise direction;that is N �P measures how many times the argument of f(z) increases by 2⇡as z traverses C.


Problem (Page 269 # 39). Use the principle of the argument to evaluateI

C

z sinh z + 3 cosh z

z cosh zdz where C : |z � i| = 7.

Solution. Letting f(z) = z cosh z, we haveI

C

z sinh z + 3 cosh z

z cosh zdz =

I

C

✓z sinh z + cosh z

z cosh z+

2

z

◆dz

=

I

C

f 0(z)

f(z)dz + 2

I

C

1

zdz = 2⇡i(N � P ) + 2(2⇡i),

where N and P are the sums of the orders of the zeroes and poles of f(z) insideC.

There are 6 zeroes of order 1 and no poles within this circle, so N � P = 6.Then I

C

z sinh z + 3 cosh z

z cosh zdz = 2⇡i(6) + 4⇡i = 16⇡i. ⇤


Theorem (6.4 – Rouche’s Theorem). Let fand g be analytic in a domainD. If C us a simple, closed, piecewise smooth curve in D which containsin its interior only points of D, and if |g(z)| < |f(z)| on C, then sums ofthe orders of the zeros of f + g and f inside C are the same.

Proof. |g| < |f | on C =) f(z) 6= 0 and |f(z)+g(z)| � |f(z)|�|g(z)| > 0

on C. The function F (z) =f(z) + g(z)

f(z), then, does not have any zeros or poles

on C. Applying the principle of the argument to F and C,

1

2⇡i

I

C

F 0(z)

F (z)dz = N � P,

and each side may be interpreted as the number of times w = F (z) encirclesw = 0 as z traverses C. But on C,

|F (z)� 1| =

��f(z) + g(z)

f(z)� 1

�� =

��g(z)

f(z)

�� < 1 =)

the curve F (C) cannot encircle w = 0, and thus N = P .

Since the zeros of F occur at the zeros of f + g and the poles of F occur atthe zeros of f , it follows that the sum of the orders of the zeros of f and f + ginside C must be the same. ⇤


Problem (Page 269 # 45). Show that if g is analytic inside and on C :|z| = 1 and if |g(z)| < 1, then there is exactly one point z inside C for whichg(z) = z.

Solution. Let f(z) = �z. Then, on C, |f(z)| = 1. Since |g(z)| < |f(z)|on C, it follows from Rouche’s theorem that f(z) + g(z) = g(z) � z has thesame number of zeros inside C as f(z) = �z, namely one. Thus, g(z) � zhas a single zero inside C =) there is exactly one solution for g(z) = z insideC. ⇤

6.2. Evaluation of Definite Intrgrals

Contour integrals and residues can be used to evaluate certain classes of definiteintegrals that are otherwise di�cult or impossible.

Definite Integrals Involving Trigonometric Functions

Contour integrals and residues can be useful in the evaluation of definite inte-grals that are rational functions of sines and cosines,

Z b

a

P (cos ✓, sin ✓)

Q(cos ✓, sin ✓)d✓

where P (cos ✓, sin ✓) and Q(cos ✓, sin ✓) are polynomials in cos ✓ and sin ✓ pro-vided Q is never equal to 0 in the interval [a, b].

The technique is to set z = ei✓. This transforms the definite integral to a contourintegral along an arc of the unit circle in the complex z-plane. When a = 0 andb = 2⇡, it is the complete circle, and residues are used to evaluate the contourintegral. For other intervals of integration, residues may not be applicable, butother techniques for evaluating contour integrals may be advantageous. Theintegrand for the contour integral is obtained by substituting

cos ✓ =e✓i + e�✓i

2=

1

2

✓z +

1

z

◆, sin ✓ =

e✓ih � e�✓i

2i=

1

2i

✓z � 1

z

◆.

6.2. EVALUATION OF DEFINITE INTRGRALS 155


Z 2⇡

0

1

6 + 5 sin ✓d✓.

Solution. Set z = e✓i =) dz = ie✓id✓ =)Z 2⇡

0

1

6 + 5 sin ✓d✓ =

I

C

1

6 + 5⇣

z�1/z2i

⌘dz

iz=

I

C

2

5z2 + 12iz � 5dz,

where C is the circle |z| = 1. Since

5z2 + 12iz � 5 = 0 =) z =�12i ±

p�144 + 100

10=

(�6 ±p

11)i

5

with only z =(�6 +

p11)i

5interior to C. Then

Res

2

5z2 + 12iz � 5,

✓�6 +

p11

5

◆i

�

= limz!(�6+

p11)i/5

2⇥z � (�6 +

p11)i/5

⇤

5⇥z � (�6 +

p11)i/5

⇤⇥z � (�6�

p11)i/5

⇤

=2

5⇥(�6 +

p11)i/5� (�6�

p11)i/5

⇤ = � ip11

.

ThusZ 2⇡

0

1

6 + 5 sin ✓d✓ =

I

C

2

5z2 + 12iz � 5dz = 2⇡i

✓� ip

11

◆=

2⇡p11

. ⇤



Z 0

�⇡

1

4 + cos ✓d✓.

Solution. Set z = e✓i =) dz = ie✓id✓ =)Z 0

�⇡

1

4 + cos ✓d✓ =

Z

C

1

4 + z+1/z2

dz

iz= �2i

Z

C

1

z2 + 8z + 1

where C is the semicircle shown below.

z2 + 8z + 1 = 0 =) z =�8 ±

p64� 4

2= �4 ±

p15.

Using partial fractions,Z 0

�⇡

1

4 + cos ✓d✓ = �2i

Z

C

✓ p15/30

z + 4�p

15�

p15/30

z + 4 +p

15

◆dz

=ip15

nlog⇡/2(z + 4 +

p15)� log⇡/2(z + 4�

p15)o1

�1

=ip15

hlog⇡/2(5+

p15)�log⇡/2(5�

p15)�log⇡/2(3+

p15)+log⇡/2(3�

p15)i

=i

15

hln(5+

p15)+2⇡i�ln(5�

p15)�2⇡i�ln(3+

p15)�2⇡i+[ln(

p15�3)+⇡i

i

=⇡p15

+ip15

ln

✓5 +

p15

5�p

15·p

15� 3p15 + 3

◆=

⇡p15

+ip15

ln2p

15

2p

15=

⇡p15

.

⇤


Real Improper Integrals

Contour integrals are also e↵ective in evaluating improper integrals which haveinfinite upper and/or lower limits.

Imprper Integrals of rational Functions

The first type of improper integral we consider involve integrands that arerational functions, integrals of the form

Z b

a

P (x)

Q(x)dx,

where P (x) and Q(x) are polynomials, and at least one of a and b is infi-nite. They are simplest when Q(x) 6= 0 for any value of x in the interval ofintegration.

Example (6.8). Evaluate

Z 1

�1

1

1 + x4dx

Solution. We want to evaluate the contour integral

I

C

1

1 + z4dz where C

is the contour shown below:

If we let R !1, then the part of the contour integral along the real axis wouldgive us the required improper integral. We first find the contour integral overC. The integrand (1 + z4)�1 has simple poles at the four fourth roots of �1:

e⇡i/4, e3⇡i/4, e5⇡i/4, e7⇡i/4.


As shown in the preceding diagram, only the first two are within C. UsingL’Hopital’s rule (Theorem 5.24),

Res

1

1 + z4, e⇡i/4

�= lim

z!e⇡i/4

z � e⇡i/4

1 + z4= lim

z!e⇡i/4

1

4z3=

1

4e3⇡i/4= �

p2

8(1 + i).

Similarly, Res

1

1 + z4, e3⇡i/4

�=

p2

8(1� i). By Cauchy’s residue theorem,

I

C

1

1 + z4dz = 2⇡i

�p

2

8(1 + i) +

p2

8(1� i)

�=

⇡p2

.

Supposxe we now divide C into a semicircular part � and a straight line partas in the diagram below:

Then⇡p2

=

Z R

�R

1

1 + x4dx +

Z

�

1

1 + z4dz.

Set z = Re✓i, 0 ✓ ⇡ on � =) (by the triangle inequality)��

1

1 + z4

�� 1

|z4|� 1=

1

R4 � 1.

By the inequality at the top of page 96 of these notes,��

Z

�

1

1 + z4dz

�� 1

R4 � 1(⇡R) ! 0 as R !1 =)

Z 1

�1

1

1 + x4dx =

⇡p2. ⇤

The following theorem, based on the idea of the previous example, makes theevaluation of many examples easier.


Theorem (6.5). Suppose that P (x) and Q(x) are polynomials of degreesm and n, where n � m + 2, and Q(x) 6= 0 for all x. Then

Z 1

�1

P (x)

Q(x)dx = 2⇡i

(sum of the residues of P (z)/Q(z) at

its poles in the half-plane Im z > 0


Z 1

�1

x2 + 3

(x2 + 1)(x2 � x + 1)dx.

Solution. The functionz2 + 3

(z2 + 1)(z2 � z + 1)has simple poles at z = ±i

and z =1 ±

p3 i

2, only two of which have positive imaginary parts.

Res

z2 + 3

(z2 + 1)(z2 � z + 1), i

�= lim

z!i

(z � i)(z2 + 3)

(z � i)(z + i)(z2 � z + 1)=

2

(2i)(�i)= 1.

Res

z2 + 3

(z2 + 1)(z2 � z + 1),1

2+

p3i

2

�

= limz!1/2+

p3i/2

⇣z � 1

2 �p

3i2

⌘(z2 + 3)

(z2 + 1)⇣z � 1

2 +p

3i2

⌘⇣z � 1

2 �p

3i2

⌘

=

⇣12 +

p3i2

⌘2+ 3

⇣12 +

p3i2

⌘2+ 1

�(p

3 i)

=5 +

p3 i

�3 +p

3 i· �3�

p3 i

�3�p

3 i

=�12� 8

p3 i

12=�3� 2

p3 i

3.

Then by Theorem 6.5,Z 1

�1

x2 + 3

(x2 + 1)(x2 � x + 1)dx = 2⇡i

✓1� 1� 2

p3 i

3

◆=

4⇡p

3

3.

⇤



Z 1

0

1

1 + x3dx.

Solution. We consider

I

C

1

1 + z3dz where C is some appropriate contour.

The integrand has simple poles at e⇡i/3,�1,and e5⇡i/3. Only the first is in theupper half-plane, but we cannot use Theorem 6.5 since z = �1 is on thenegative real axis. We choose the line z = re�i, 0 < � < ⇡.

Let’s choose � such that⇡

3< � < ⇡, but, for the moment, leave it arbitrary.

Cauchy’s residue theorem and L’Hopital’s rule giveI

C

1

1 + z3dz = 2⇡i Res

1

1 + z3, e⇡i/3

�= 2⇡i lim

z!e⇡i/3

z � e⇡i/3

1 + z3

= 2⇡i limz!e⇡i/3

1

3z2=

2⇡i

3e2⇡i/3=

(p

3� i)⇡

3.

Thus

(⇤) (p

3� i)⇡

3=

Z

�1

1

1 + z3dz +

Z

�2

1

1 + z3dz +

Z

�3

1

1 + z3dz.

Set z = Re✓i on �1. On this arc,��1

1 + z3

�� 1

|z|3 � 1=

1

R3 � 1=)

��

Z

�1

1

1 + z3dz

�� 1

R3 � 1(R�) ! 0

as R !1. Thus, letting R !1,


(p

3� i)⇡

3=

Z

�2

1

1 + z3dz +

Z 1

0

1

1 + x3dx

where �2 : z = re�i, 0 r <1,⇡

3< � < ⇡. Then

Z

�2

1

1 + z3dz =

Z 0

1

1

1 + r3e3�ie�i dr = �e�i

Z 1

0

1

1 + r3e3�idr

Now we choose � =2⇡

3=)

Z

�2

1

1 + z3dz = �e2⇡i/3

Z 1

0

1

1 + r3dr =)

(p

3� i)⇡

3=

�✓� 1

2+

p3 i

2

◆+ 1

� Z 1

0

1

1 + x3dx =)

Z 1

0

1

1 + x3dx =

(p

3� i)⇡

3· 2

3�p

3 i=

2⇡

3p

3⇤

Improper Integrals of Rational Functions Multiplied by Sines and/or Cosines

We now consider improper integrals of the formsZ 1

�1

P (x)

Q(x)cos ax dx and

Z 1

�1

P (x)

Q(x)sin ax dx,

where a > 0 is a constant. The following theorem accomplishes the same thinghere that Theorem 6.5 did earlier for rational functions.


Theorem (6.6). Suppose that P (x) and Q(x) are polynomials of degreesm and n, where n � m + 1, and Q(x) 6= 0 for all real x. When a > 0 is aconstant,Z 1

�1

P (x) cos ax

Q(x)dx = �2⇡ Im

(sum of the residues of P (z)eazi/Q(z)

at its poles in the half-plane Im z > 0

andZ 1

�1

P (x) sin ax

Q(x)dx = �2⇡ Re

(sum of the residues of P (z)eazi/Q(z)

at its poles in the half-plane Im z > 0.


Z 1

�1

x2 cos x

(x2 + 9)2dx.

Solution.z2 cos z

(z2 + 9)2has double poles at ±3i, but only 3i is in the upper

half-plane.

Res

z2ezi

(z2 + 9)2, 3i

�= lim

z!3i

d

dz

z2ezi(z � 3i)2

(z + 9)2

�= lim

z!3i

d

dz

z2ezi

(z + 3i)2

�

= limz!3i

(z + 3i)2(2zezi + iz2ezi)� z2ezi(2)(z + 3i)

(z + 3i)4

�

=6ih6ie�3 + i(�9)e�3

i� 2(�9)e�3

(6i)3=

i

6e3.

Then, by Theorem 6.6,Z 1

�1

x2 cos x

(x2 + 9)2dx = �2⇡ Im

✓i

6e3

◆= � ⇡

3e3.

⇤

6.3. MORE IMPROPER INTEGRALS BY RESIDUES 163

6.3. More Improper Integrals by Residues

The improper integral of a continuous function f(x) over the real line is definedby

(⇤)Z 1

�1f(x) dx = lim

a!�1

Z 0

af(x) dx + lim

b!1

Z b

0f(x) dx

provided both limits exist.

The Cauchy principal value of the integral s defined by

(⇤⇤) PV

Z 1

�1f(x) dx = lim

b!1

Z b

�bf(x) dx

provided the limit exists.

(⇤) =) (⇤⇤) with the same value, but the converse is not universally true.Z 1

�1

x

x4 + 1dx = 0 in both senses, but PV

Z 1

�1

x

x2 + 1dx = 0, but does not

converge in the sense of (⇤).Cauchy principal values are also defined for improper integrals due to discon-tinuities in integrands. When the only discontinuity of a function f(x) in theinterval a < x < b is x = d, its improper integral over the interval a x bis defined by

(#)

Z b

af(x) dx = lim

✏!0+

Z d�✏

af(x) dx + lim

✏!0+

Z b

d+✏f(x) dx

provided both limits exist.

The integral is said to have a Cauchy principal value when the limit

(##) PV

Z b

af(x) dx = lim

✏!0+

Z d�✏

af(x) dx +

Z b

d+✏f(x) dx

�

exists. Again, (#) =) (##), but the converse is not uiversally true.

PV

Z 1

�1

1

xdx = 0, but has no value in the sense of (#).


Improper Integrals of Rational Functions with Simple Poles.

The first type of integral we consider is

Z 1

�1

P (x)

Q(x)dx where P (x) and Q(x) are

polynomials, and Q(x) has real simple zeros.

Theorem (6.7). When P (x) and Q(x) are polynomials of degrees m andn where n � m + 2, and the only zeros of Q(x) on the real axis are simplezeros,

PV

Z 1

�1

P (x)

Q(x)dx = 2⇡i

8<

:sum of the residues of

P (z)

Q(z)at


+ ⇡i

8<


P (z)

Q(z)at

its simple poles on the real axis.

Problem (Page 295 # 3). Evaluate PV

Z 1

�1

x2 + 2x

(x� 1)(x3 + 1)dx.

Solution.z2 + 2z

(z � 1)(z3 + 1)=

z(z + 2)

(z � 1)(z + 1)(z2 � z + 1)has poles at z =

±1 on the real axis and z =1 ±

p3

2. Residues at ±1 and at

1 +p

3

2are

Res

z2 + 2z

(z � 1)(z3 + 1), 1

�= lim

z!1

(z � 1)(z2 + 2z)

(z � 1)(z3 + 1)=

3

2,

Res

z2 + 2z

(z � 1)(z3 + 1),�1

�= lim

z!�1

(z + 1)(z2 + 2z)

(z � 1)(z + 1)(z2 � z + 1)=

1

6,


Res

z2 + 2z

(z � 1)(z3 + 1),1 +

p3

2

�= lim

z!(1+p

3 i)/2

⇣z � 1

2 �p

3 i2

⌘(z2 + 2z)

(z � 1)(z3 + 1)

= limz!(1+

p3 i)/2

⇣z � 1

2 �p

3 i2

⌘(2z + 2) + (z2 + 2z)

(z3 + 1) + (z � 1)(3z2)

�

= limz!(1+

p3 i)/2

z2 + 2z

3z2(z � 1)= lim

z!(1+p

3 i)/2

z + 2

3z(z � 1)

=12 +

p3 i2 + 2

3⇣

12 +

p3 i2

⌘⇣� 1

2 +p

3 i2

⌘ =5 +

p3 i

�6.

Then, by Theorem 6.7,

PV

Z 1

�1

x2 + 2x

(x� 1)(x3 + 1)dx = 2⇡i

✓5 +

p3 i

�6

◆+ ⇡i

✓3

2+

1

6

◆=

⇡p3. ⇤

Improper Intergrals of Rational Functions with Simple Poles Multiplied by Sinesand/or Cosines

Theorem (6.8). When P (x)and Q(x) are polynomials of degrees m andn where n � m + 1, and the only zeros of Q(x) are simple zeros, then fora > 0,

PV

Z 1

�1

P (x) cos ax

Q(x)dx = �2⇡ Im

8<


P (z)eazi

Q(z)at


� ⇡ Im

8<


P (z)eazi

Q(z)at


and


PV

Z 1

�1

P (x) sin ax

Q(x)dx = 2⇡ Re

8<


P (z)eazi

Q(z)at


+ ⇡ Re

8<


P (z)eazi

Q(z)at


Problem (Page 295 # 7). Evaluate PV

Z 1

�1

sin x cos x

x3 � 8dx.

Solution. First,

PV

Z 1

�1

sin x cos x

x3 � 8dx =

1

2PV

Z 1

�1

sin 2x

x3 � 8dx.

z3 � 8 = (z � 2)(z2 + 2z + 4) has simple poles at z = 2 and z = �1 ±p

3 i.

Res

e2zi

z3 � 8, 2

�= lim

z!2

(z � 2)e2zi

(z � 2)(z2 + 2z + 4)=

e4i

12,

Res

e2zi

z3 � 8,�1 +

p3 i

�= lim

z!�1+p

3 i

(z + 1�p

3 i)e2zi

z3 � 8

= limz!�1+

p3 i

e2zi + 2i(z + 1�p

3 i)e2zi

3z2=

e2(�1+p

3 i)i

3(1 +p

3 i)2

= � e�2(p

3+1)

6(1 +p

3 i)· 1�

p3 i

1�p

3 i= �(1�

p3 i)e�2(

p3+i)

24.

By Theorem 6.8,


PV

Z 1

�1

sin x cos x

x3 � 8dx =

1

2PV

Z 1

�1

sin 2x

x3 � 8dx

= ⇡ Re

� 1

24(1�

p3 i)e�2(

p3+i)

�+

⇡

2Re

e4i

12

◆

= �⇡e�2p

3

24(cos 2�

p3 sin 2) +

⇡

24cos 4. ⇤

Improper Integrals Involving Logarithms


Z 1

0

ln x

x2 + a2dx where a > 0.

Solution. This integral exists in the ordinary sense since both integrals onthe right side of the following converge:

Z 1

0

ln x

x2 + a2dx =

Z 1

0

ln x

x2 + a2dx +

Z 1

1

ln x

x2 + a2dx.

Because Z 1

0

ln x

x2 + a2dx =

Z 0

�1

ln(�x)

x2 + a2dx

we consider

I

C

log�⇡/2 z

z2 + a2dz where C is the contour below.


The integrand has simple poles at ±ai, so thatI

C

log�⇡/2 z

z2 + a2dz = 2⇡i Res

log�⇡/2 z

z2 + a2, ai

�= 2⇡i lim

z!ai

(z � ai) log�⇡/2 z

(z � ai)(z + ai)

=⇡

alog�⇡/2(ai) =

⇡

a

ln a +

⇡i

2

�.

On the semicircle �, set z = Re✓i, 0 ✓ ⇡ =)��log�⇡/2 z

z2 + a2

�� |ln R + ✓i|R2 � a2

ln R + ⇡

R2 � a2provided R > a and R > 1 =)

��

I

�

log�⇡/2 z

z2 + a2dz

�� ln R + ⇡

R2 � a2(⇡R) ! 0 as R !1.

On the semicircle �0, set z = re✓i,⇡ � ✓ � 0 =)��log�⇡/2 z

z2 + a2

�� |ln r + ✓i|a2 � r2

|ln r| + ⇡

a2 � r2=)

��

I

�0

log�⇡/2 z

z2 + a2dz

�� |ln r| + ⇡

a2 � r2(⇡ r) ! 0 as r ! 0.

Referring to the equation at the top of this page,

⇡

a

ln a +

⇡i

2

�= lim

r!0R!1

I

C

log�⇡/2 z

z2 + a2dz =

Z 1

�1

log�⇡/2 x

x2 + a2dx

=

Z 0

�1

ln(�x) + ⇡i

x2 + a2dx +

Z 1

0

ln x

x2 + a2dx =

= 2

Z 1

0

ln x

x2 + a2dx +

Z 0

�1

⇡i

x2 + a2dx.

Then real parts give Z 1

0

ln x

x2 + a2dx =

⇡

2aln a.

⇤


Improper Integrals Involving Fractional Powers


Z 1

0

1

xa(x + 1)dx, where 0 < a < 1.

Solution. To evaluate this improper integral by contour integration, wereplace the real integrand with the conplex

⇥za(z + 1)

⇤�1. Because za, for

fractional a, is multiple-valued, we must choose some branch of the function.This is dictated by the choice of contour. To obtain the required real improperintegral, part of the contour should include a part of the positive real axis. Toavoid the branch point, we include a small circle (of radius r) around z = 0.We also need a circle of large radius R centered at z = 0. This gives us thecurves on the left below, but they do not form a closed contour.

Our final choice of contour C is that shown on the right. It consists of aportion of the circle of radius r < 1 centered at the origin, and two horizontalline segmennts joining this circle to a portion of the large circle with radiusR > 1.We choose the branch za = ea log0 z with branch cut along the positivereal axis. To get the required real integral, we eventually take limits as r ! 0and R !1. Since the integrand has a simple pole at z = �1 interior to C,


I

C

1

za(z + 1)dz = 2⇡i Res

1

za(z + 1),�1

�

=2⇡i

(�1)a=

2⇡i

ea log0(�1)=

2⇡i

ea⇡i= 2⇡ie�a⇡i.

Breaking C into its four component parts,

2⇡ie�a⇡i =

Z

�1

1

za(z + 1)dz +

Z

�2

1

za(z + 1)dz

+

Z

�3

1

za(z + 1)dz +

Z

�4

1

za(z + 1)dz.

On �1,��1

za(z + 1)

�� =

��1

(z + 1)ea log0 z

�� =

��1

(z + 1)ea(lnR+arg0 z i)

�� 1

Ra(R� 1)=)

��

Z

�1

1

za(z + 1)dz

�� 1

Ra(R� 1)(2⇡R) ! 0 as R !1.

On �2,��1

za(z + 1)

�� =

��1

(z + 1)ea log0 z

�� =

��1

(z + 1)ea(ln r+arg0 z i)

�� 1

ra(r � 1)=)

��

Z

�1

1

za(z + 1)dz

�� 1

ra(r � 1)(2⇡r) ! 0 as r !1.

When limits are taken on the result of Cauchy’s residue theorem as r ! 0 andR !1,

2⇡ie�a⇡i = limr!0

R!1

Z

�3

1

za(z + 1)dz + lim

r!0R!1

Z

�4

1

za(z + 1)dz

= limr!0

R!1

Z

�3

1

(z + 1)ea log0 zdz + lim

r!0R!1

Z

�4

1

(z + 1)ea log0 zdz.


Each of these last two limits are along the positive real axis. Since the integralalong �3 is in the first quadrant, the limit of the argument in log0 z is 0. Since�4 is in the fourth quadrant, the limit of the argument in log0 z is 2⇡. Thus,when limits are taken,

2⇡ie�a⇡i =

Z 1

0

1

(x + 1)ea lnxdx +

Z 0

1

1

(x + 1)ea lnx+2⇡idx

=

Z 1

0

1

xa(x + 1)dx�

Z 1

0

1

xa(x + 1)e2a⇡idx

= (1� e�2a⇡i)

Z 1

0

1

xa(x + 1)dx.

Solving this equation for the required integral,Z 1

0

1

xa(x + 1)dx =

2⇡ie�a⇡i

1� e�2a⇡i=

2⇡i

ea⇡i � e�a⇡i=

2⇡i

2i sin a⇡=

⇡

sin a⇡. ⇤

residue theoryfacstaff.cbu.edu/~wschrein/media/m413 notes/m413c6.pdf · chapter 6 residue theory we...

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