gamma and betta function adv calculus schaum

7/30/2019 Gamma and Betta Function Adv Calculus Schaum

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375

Gamma and Beta

Functions

THE GAMMA FUNCTION

The gamma function may be regarded as a generalization of n! (n-factorial), where n is any positive

integer to x!, where x is any real number. (With limited exceptions, the discussion that follows will be

restricted to positive real numbers.) Such an extension does not seem reasonable, yet, in certain ways,

the gamma function defined by the improper integral

x

10

tx1

et

dt 1

meets the challenge. This integral has proved valuable in applications. However, because it cannot be

represented through elementary functions, establishment of its properties take some effort. Some of the

important ones are outlined below.

The gamma function is convergent for x > 0. (See Problem 12.18, Chapter 12.)

The fundamental property

x 1 xx 2

may be obtained by employing the technique of integration by

parts to (1). The process is carried out in Problem 15.1.

From the form (2) the function x can be evaluated for all

x > 0 when its values in the interval 1 % x < 2 are known.

(Any other interval of unit length will suffice.) The table and

graph in Fig. 15-1 illustrates this idea.

TABLES OF VALUES AND GRAPH OF THE GAMMA

FUNCTION

n n

1.00 1.00001.10 0.9514

1.20 0.9182

1.30 0.8975

5

4

3

2

1

_1

_2

_3

_4

_5

_5 _4_3 _2 _1 1 2 3 4 5

n

(n)

Fig. 15-1

Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.


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1.40 0.8873

1.50 0.8862

1.60 0.8935

1.70 0.9086

1.80 0.93141.90 0.9618

2.00 1.0000

The equation (2) is a recurrence relationship that leads to the factorial concept. First observe that if

x 1, then (1) can be evaluated, and in particular,1 1:

From (2)

x 1 xx xx 1x 1 xx 1x 2 x kx k

Ifx n, where n is a positive integer, thenn 1 nn 1n 2 . . . 1 n! 3

Ifx is a real number, then x! x 1 is defined by x 1. The value of this identification is inintuitive guidance.

If the recurrence relation (2) is characterized as a differential equation, then the definition of xcan be extended to negative real numbers by a process called analytic continuation. The key idea is that

even though x is defined in (1) is not convergent for x < 0, the relation x 1x

x 1 allows themeaning to be extended to the interval 1 < x < 0, and from there to 2 < x < 1, and so on. Ageneral development of this concept is beyond the scope of this presentation; however, some information

is presented in Problem 15.7.The factorial notion guides us to information about x 1 in more than one way. In the

eighteenth century, Sterling introduced the formula (for positive integer values n)

limn!1

ffiffiffiffiffiffi

2p

nn1

en

n! 1 4

This is called Sterlings formula and it indicates that n! asymptotically approachesffiffiffiffiffiffi

2p

nn1

en for large

values ofn. This information has proved useful, since n! is difficult to calculate for large values ofn.

There is another consequence of Sterlings formula. It suggests the possibility that for sufficiently

large values ofx,

x!

x

1

%

ffiffiffiffiffiffi

2p

xx1

ex

5a

(An argument supporting this is made in Problem 15.20.)

It is known that x 1 satisfies the inequalityffiffiffiffiffiffi

2p

xx1

ex

< x 1 0;y > 0 and either or both x < 1 or y < 1, the

integral is improper but convergent.It is shown in Problem 15.11 that the beta function can be expressed through gamma functions in the

following way

Bx;y xy

xy17

Many integrals can be expressed through beta and gamma functions. Two of special interest are

=20

sin2x1 cos2y1 d1

2Bx;y

1

2

xy

xy18

1

0

xp1

1 xdx pp 1

sin p0 < p < 1 19

See Problem 15.17. Also see Page 377 where a classical reference is given. Finally, see Chapter 16,

Problem 16.38 where an elegant complex variable resolution of the integral is presented.

DIRICHLET INTEGRALS

IfV denotes the closed region in the first octant bounded by the surfacex

a

p

y

b

q

z

c

r 1 and

the coordinate planes, then if all the constants are positive,

V

x

1

y

1

z

1

dxdydz

abc

pqr

p

q

r

1

p

q

r

20

Integrals of this type are called Dirichlet integrals and are often useful in evaluating multiple

integrals (see Problem 15.21).

Solved Problems

THE GAMMA FUNCTION

15.1. Prove: (a) x 1 xx; x > 0; b n 1 n!; n 1; 2; 3; . . . .

a v 1

10

xv ex dx limM1

M0

xv ex dx

limM1

xvexjM0

M0

exvxv1 dx

& '

limM1

Mv eM v

M0

xv1 ex dx

& ' vv if v > 0

b 1 1

0

ex dx limM1

M

0

ex dx limM1

1 eM 1:

Put n 1; 2; 3; . . . in n 1 nn. Then

2 11 1; 3 22 2 1 2!; 4 33 3 2! 3!

In general, n 1 n! ifn is a positive integer.


6/17

15.2. Evaluate each of the following.

a 623

5!

2 2! 5 4 3 2

2 2 30

b 52

12

323

2

12

32 1

21

2

12

3

4

c 32:55:5

2!1:50:50:54:53:52:51:50:50:5

16

315

d 6 83

5 23

6532

32

3

5 23

4

3

15.3. Evaluate each integral.

a 10

x3 ex dx 4 3! 6

b1

0

x6 e2x dx: Let 2x 7. Then the integral becomes1

0

y

2

6ey

dy

2 1

27

10

y6 ey dy 727

6!27

458

15.4. Prove that 12 ffiffiffip .

12

1

0

x1=2 ex dx. Letting x u2 this integral becomes

2

10

eu2

du 2ffiffiffip

2

ffiffiffip using Problem 12.31, Chapter 12

This result also is described in equation (11a,b) earlier in the chapter.

15.5. Evaluate each integral.

a1

0

ffiffiffiy

pey

2

dy. Letting y3 x, the integral becomes1

0 ffiffiffiffiffiffiffiffix1=3

pex 1

3x2=3 dx 1

3

10

x1=2 ex dx 13

1

2

ffiffiffi

p

3

b1

0

34x2

dx 1

0

eln 34x2 dz 1

0

e4 l n 3z2

dz. Let 4 l n 3z2 x and the integral becomes1

0

ex dx1=2ffiffiffiffiffiffiffiffiffiffiffi4 l n 3

p !

12ffiffiffiffiffiffiffiffiffiffiffi

4 l n 3p

10

x1=2 ex dx 1=22ffiffiffiffiffiffiffiffiffiffiffi

4 l n 3p

ffiffiffi

p

4ffiffiffiffiffiffiffiffi

ln 3p

(c)

10

dxffiffiffiffiffiffiffiffiffiffiffiffi lnx

p : Let lnx u. Then x eu. When x 1; u 0; when x 0; u 1. The integralbecomes

1

0

euffiffiffiupdu

1

0

u1=2 eu du

1=2

ffiffiffip

15.6. Evaluate

10

xm eaxn

dx where m; n; a are positive constants.

380 GAMMA AND BETA FUNCTIONS [CHAP. 15


7/17

Letting axn y, the integral becomes1

0

y

a

1=n& 'mey d

y

a

1=n& ' 1

nam1=n

10

ym1=n1 ey dy 1nam1=n

m 1n

15.7. Evaluate (a) 1=2; b 5=2.

We use the generalization to negative values defined by x x 1x

.

a Letting x 12; 1=2 1=21=2 2

ffiffiffi

p:

b Letting x 3=2; 3=2 1=23=2 2 ffiffiffip3=2

4ffiffiffi

p

3; using a:

Then 5=2 3=25=2 8

15

ffiffiffi

p:

15.8. Prove that

10

xmln xn dx 1nn!

m 1n1, where n is a positive integer and m > 1.

Letting x ey, the integral becomes 1n1

0

yn em1y dy. Ifm 1y u, this last integral becomes

1

n

1

0

un

m 1n e

u du

m 1

1n

m 1n

1

1

0

un eu du

1n

m 1n

1

n

1

1nn!

m 1n

1

Compare with Problem 8.50, Chapter 8, page 203.

15.9. A particle is attracted toward a fixed point O with a force inversely proportional to its instanta-

neous distance from O. If the particle is released from rest, find the time for it to reach O.

At time t 0 let the particle be located on the x-axis at x a > 0 and let O be the origin. Then byNewtons law

md2x

dt2 k

x1

where m is the mass of the particle and k > 0 is a constant of proportionality.

Letdx

dt v, the velocity of the particle. Then d

2x

dt2 dv

dt dv

dx dxdt

v dvdx

and (1) becomes

mvdv

dx k

xor

mv2

2 k lnx c 2

upon integrating. Since v 0 at x a, we find c k ln a. Then

mv2

2 k ln a

xor v dx

dt

ffiffiffiffiffi2k

m

r ffiffiffiffiffiffiffiffiffiln

a

x

r3

where the negative sign is chosen since x is decreasing as t increases. We thus find that the time T taken for

the particle to go from x a to x 0 is given by

Tffiffiffiffiffim

2k

r a0

dxffiffiffiffiffiffiffiffiffiffi ffiffiffiln a=x

p 4

CHAP. 15] GAMMA AND BETA FUNCTIONS 381


8/17

Letting ln a=x u or x aeu, this becomes

T a

ffiffiffiffiffim

2k

r 10

u1=2 eu du a

ffiffiffiffiffim

2k

r1

2 a

ffiffiffiffiffiffiffim

2k

r

THE BETA FUNCTION

15.10. Prove that (a) Bu; v Bv; u; b Bu; v 2

=20

sin2u1 cos2v1 d.

(a) Using the transformation x 1 y, we have

Bu; v

10

xu11 xv1 dx

10

1 yu1yv1 dy

10

yv11yu1 dy Bv; u

(b) Using the transformation x sin2 , we have

Bu; v

10

xu11 xv1 dx

=20

sin2 u1cos2 v1 2sin cos d

2

=20

sin2u1 cos2v1 d

15.11. Prove that Bu; v uv

u vu; v > 0.

Letting z2 x2; we have u

10

zu1 ez dx 2

10

x2u1 ex2

dx:

Similarly, v 2

10

y2v1 ey2

dy: Then

uv 4

10

x2u1 ex2

dx

10

y2v1 ey2

dy

4

10

10

x2u1y2v1 ex2y2 dxdy

Transforming to polar coordiantes, x cos;y sin,

uv 4 =2

01

0

2uv1 e

2

cos2u1 sin2v1 d d

4

10

2uv1 e

2

d

=20

cos2u1 sin2v1 d

2u v

=20

cos2u1 sin2v1 d u vBv; u

u vBu; v

using the results of Problem 15.10. Hence, the required result follows.

The above argument can be made rigorous by using a limiting procedure as in Problem 12.31,

Chapter 12.

15.12. Evaluate each of the following integrals.

a

10

x41 x3 dx B5; 4 54

9

4!3!

8!

1

280



9/17

b2

0

x2 dxffiffiffiffiffiffiffiffiffiffiffi2 x

p : Letting x 2v; the integral becomes

4ffiffiffi

2p 1

0

v2ffiffiffiffiffiffiffiffiffiffiffi1 vp dv 4 ffiffiffi2p

1

0 v2

1 v1=2

dv 4 ffiffiffi2p B3; 12 4 ffiffiffi2p

3

1=2

7=2 64 ffiffiffi2

p

15

ca

0

y4ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffia2 y2

qdy: Letting y2 a2x or y ffiffiffixp ; the integral becomes

a61

0

x3=21 x1=2 dx a6B5=2; 3=2 a65=23=2

4 a6

16

15.13. Show that

=20

sin2u1 cos2v1 d uv2u v u; v > 0.

This follows at once from Problems 15.10 and 15.11.

15.14. Evaluate (a)

=20

sin6 d; b=2

0

sin4 cos5 d; c

0

cos4 d.

a Let 2u 1 6; 2v 1 0; i.e., u 7=2; v 1=2; in Problem 15.13:

Then the required integral has the value7=21=2

24 5

32:

b Letting 2u 1 4; 2v 1 5; the required integral has the value 5=23211=2 8315 :

c The given integral 2=2

0

cos4 d:

Thus letting 2u 1 0; 2v 1 4 in Problem 15.13, the value is 21=25=223

3

8.

15.15. Prove

=20

sinp d=2

0

cosp d a 1 3 5 p 12 4 6 p

2if p is an even positive integer,

(b)2 4 6 p 1

1

3

5

pis p is an odd positive integer.

From Problem 15.13 with 2u 1 p; 2v 1 0, we have=2

0

sinp d 12p 112

212p 2

(a) Ifp 2r, the integral equalsr 1

21

2

2r 1 r 1

2r 3

2 1

21

2 1

2

2rr 1 1 2r 12r 3 1

2r2r 2 2

2 1 3 5 2r 1

2 4 6 2r

2

(b) Ifp 2r 1, the integral equals

r

1

12

2r 32

r

r

1

1

ffiffiffi

p

2r 12r 1

2 1

2

ffiffiffip

2

4

6

2r

1 3 5 2r 1

In both cases

=20

sinp d=2

0

cosp d, as seen by letting =2 .



10/17

15.16. Evaluate (a)

=20

cos6 d; b=2

0

sin3 cos2 d; c2

0

sin8 d.

(a) From Problem 15.15 the integral equals1 3 52

4

6 5

32[compare Problem 15.14(a)].

(b) The integral equals=20

sin3 1 sin2 d=2

0

sin3 d=2

0

sin5 d 21 3

2 41 3 5

2

15

The method of Problem 15.14(b) can also be used.

c The given integral equals 4=2

0

sin8 d 4 1 3 5 72 4 6 8

2

35

64:

15.17. Given

10

xp1

1 x dx

sinp, show that p1 p

sinp, where 0 < p < 1.

Lettingx

1 x y or x y

1 y, the given integral becomes1

0

yp11 yp dy Bp; 1 p p1 p

and the result follows.

15.18. Evaluate

10

dy

1 y4.

Let y4 x. Then the integral becomes 14

10

x3=4

1

xdx

4sin=4

ffiffiffi2

p

4by Problem 15.17 with p 1

4.

The result can also be obtained by letting y2 tan .

15.19. Show that

20

xffiffiffiffiffiffiffiffiffiffiffiffiffi

8 x33p

dx 169ffiffiffi

3p .

Letting x3 8y or x 2y1=3, the integral comes1

0

2y1=3 ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffi

81 y3p

23y2=3 dy 8

3

10

y1=31 y1=3 dy 83B2

3; 4

3

83

234

3

2 8

91

32

3 8

9

sin =3 16

9 ffiffiffi3p

STIRLINGS FORMULA

15.20. Show that for large positive integers n; n! ffiffiffiffiffiffiffiffi

2np

nn en approximately.

By definition z 1

0

tz1 et dt. Let lfz x 1 then

x 1 1

0

tx et dt 1

0

etln tx

dt 1

0

etx ln t dt 1

For a fixed value ofx the function x ln t

t has a relative maximum for t

x (as is demonstrated by

elementary ideas of calculus). The substutition t xy yields

x 1 ex1x

ex lnxyy dy xx ex1x

ex ln1y

xy dy 2



11/17

To this point the analysis has been rigorous. The following formal steps can be made rigorous by

incorporating appropriate limiting procedures; however, because of the difficulty of the proofs, they shall be

omitted.

In (2) introduce the logarithmic expansion

ln 1 yx

y

x y

2

2x2 y

3

3x3 3

and also let

y ffiffiffixp v; dy ffiffiffixp dvThen

x 1 xx ex ffiffiffixp1x

ev2=2v3=3ffiffixp dv 4

For large values ofx

x 1 % xx ex ffiffiffixp 1

x ev

2

=2 dv xx ex ffiffiffiffiffiffiffiffi2xpWhen x is replaced by integer values n, then the Stirling relation

n! x 1 %ffiffiffiffiffiffiffiffi

2xp

xx ex 5is obtained.

It is of interest that from (4) we can also obtain the result (12) on Page 378. See Problem 15.72.

DIRICHLET INTEGRALS

15.21. Evaluate I

V

x1y1 z1 dxdydz where V is

the region in the first octant bounded by the sphere

x2 y2 z2 1 and the coordinate planes.Let x2 u;y2 v; z2 w. Then

I r

u1=2 v1=2 w1=2du

2ffiffiffiu

p dv2

ffiffiffiv

p dw2ffiffiffiffiw

p

18

r

u=21 v=21 w=21 dudvdw 1

wherer is the region in the uvw space bounded by the plane

u v w 1 and the uv; vw, and uw planes as in Fig. 15-2.Thus,

I 18

1u0

1uv0

1uvw0

u=21 v=21 w=21 dudvdw 2

14

1u0

1uv0

u=21 v=211 u v=2 dudv

14

1u0

u=211uv0

v=21 1 u v=2 dv& '

du

Letting v 1 ut, we have1uv0

v=21 1 u v=2 dv 1 u=2 1t0

t=21 1 t=2 dt

1 u=2 =2=2 1 =2 1


Fig. 15-2


12/17

so that (2) becomes

I 14

=2=2 1 =2 1

1u0

u=21 1 u=2 du 3

14

=2

=2 1 =2 1

=2

=2 1 =2 1

=2

=2

=28 =2 1

where we have used =2=2 =2 1.The integral evaluated here is a special case of the Dirichlet integral (20), Page 379. The general case

can be evaluated similarly.

15.22. Find the mass of the region bounded by x2 y2 z2 a2 if the density is x2y2z2.

The required mass 8

V

x2y2z2 dxdydz, where V is the region in the first octant bounded by the

sphere x2

y2

z2

a2 and the coordinate planes.

In the Dirichlet integral (20), Page 379, let b c a;p q r 2 and 3. Then therequired result is

8 a3 a3 a32 2 2

3=23=23=21 3=2 3=2 3=2

4s9

945

MISCELLANEOUS PROBLEMS

15.23. Show that

10

ffiffiffiffiffiffiffiffiffiffiffiffiffi1 x4

pdx f1:4g

2

6 ffiffiffiffiffiffi2p .

Let x4 y. Then the integral becomes1

4

10

y3=41 y1=2 dy 14

1=43=27=4

ffiffiffi

p

6

f1=4g21:43=4

From Problem 15.17 with p 1=4;1=43=4 ffiffiffi

2p

so that the required result follows.

15.24. Prove the duplication formula 22p1pp 1

2 ffiffiffip 2p.

Let I=2

0

sin2p xdx; J=2

0

sin2p 2xdx.

Then I 12Bp 1

2; 1

2 p

12 ffiffiffip

2p 1Letting 2x u, we find

J 12

0

sin2p u du =2

0

sin2p u du I

J=2

0

2sinx cosx2pdx 22p=2

0

sin2p x cos2p x dxBut

22p1 Bp 12;p 1

2 2

2p1fp 12g2

2p 1Then since I J,

p 12 ffiffiffip

2pp 22p1fp 1

2g2

2p2p



13/17

and the required result follows. (See Problem 15.74, where the duplication formula is developed for the

simpler case of integers.)

15.25. Show that =2

0

dffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi1 1

2sin2

q f1=4g2

4ffiffiffip

.

Consider

I=2

0

dffiffiffiffiffiffiffiffiffifficos

p =2

0

cos1=2 d 12B1

4; 1

2

14 ffiffiffip

234

f14g2

2ffiffiffiffiffiffi

2p

as in Problem 15.23.

But I=2

0

dffiffiffiffiffiffiffiffiffifficos

p =2

0

dffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficos2 =2 sin2 =2

p =20

dffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2sin2 =2

p :

Lettingffiffiffi

2p sin =2 sin in this last integral, it becomes ffiffiffi2p =2

0

dffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1

2sin2

q, from which the result

follows.

15.26. Prove that

10

cos x

xpdx

2p cosp=2 ; 0 < p < 1.

We have1

xp 1p

10

up1 exu du. Then

10

cosx

xpdx 1

p1

0

10

up1 exu cosxdudx

1

p 1

0

up

1 u2du

1

where we have reversed the order of integration and used Problem 12.22, Chapter 12.

Letting u2 v in the last integral, we have by Problem 15.1710

up

1 u2 du 1

2

10

vp1=2

1 v dv

2sinp 1=2

2cosp=22

Substitution of (2) in (1) yields the required result.

15.27. Evaluate

10

cos x2 dx.

Letting x2

y, the integral becomes

1

21

0

cosyffiffiffiyp

dy

1

2

212 cos=4 ! 12 ffiffiffiffiffiffiffiffi=2p

by Problem 15.26.

This integral and the corresponding one for the sine [see Problem 15.68(a)] are called Fresnel integrals.

Supplementary Problems

THE GAMMA FUNCTION

15.28. Evaluate (a)7

243 ; b33=29=2 ; c 1=23=25=2.

Ans. a 30; b 16=105; c 38

3=2



14/17

15.29. Evaluate (a)

10

x4 ex dx; b1

0

x6 e3x dx; c1

0

x2 e2x2

dx:

Ans. a 24; b 80243

; cffiffiffiffiffiffi

2p

16

15.30. Find (a)

10

ex2

dx; b1

0

ffiffiffix4

pe

ffiffix

pdx; c

10

y3 e2y5

dy.

Ans. a 13

13; b 3

ffiffiffi

p

2; c 4=5

5ffiffiffiffiffi

165p

15.31. Show that

10

estffiffit

p dt ffiffiffi

8

r; s > 0.

15.32. Prove that v 1

0

ln1

x

v1dx; v > 0.

15.33. Evaluate (a)

10

lnx4 dx; b1

0

x lnx3 dx; c1

0

ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiln1=x3

pdx.

Ans: a 24; b 3=128; c 13

13

15.34. Evaluate (a) 7=2; b 1=3. Ans: a 16 ffiffiffip =105; b 3 2=315.35. Prove that lim

x!mx 1 where m 0; 1; 2; 3; . . .

15.36. Prove that ifm is a positive interger, m 12 1

m2mffiffiffi

p1

3

5

2m

1

15.37. Prove that 01

10

ex lnx dx is a negative number (it is equal to , where 0:577215 . . . is calledEulers constant as in Problem 11.49, Page 296).

THE BETA FUNCTION

15.38. Evaluate (a) B3; 5; b B3=2; 2; c B1=3; 2=3: Ans: a 1=105; b 4=15; c 2=ffiffiffi

3p

15.39. Find (a)

10

x21 x3 dx; b1

0

ffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi1 x=x

pdx; c

20

4 x23=2 dx.

Ans: a 1=60; b =2; c 3

15.40. Evaluate (a)

40

u3=24 u5=2 du; b3

0

dxffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi3x x2

p : Ans: a 12; b

15.41. Prove that

a0

dyffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffia4 y4

p f1=4g24a

ffiffiffiffiffiffi2

p :

15.42. Evaluate (a)

=20

sin4 cos4 d; b2

0

cos6 d: Ans: a 3=256; b 5=8

15.43. Evaluate (a)

0sin

5

d; b =2

0cos

5

sin2

d: Ans: a 16=15; b 8=105

15.44. Prove that

=20

ffiffiffiffiffiffiffiffiffiffitan

pd =

ffiffiffi2

p.



15/17

15.45. Prove that (a)

10

x dx

1 x6

3ffiffiffi

3p ; b

10

y2dy

1 y4

2ffiffiffi

2p .

15.46. Prove that

11

e2x

ae3x

bdx

2

3ffiffiffi

3

pa

2=3

b

1=3where a; b > 0.

15.47. Prove that

11

e2x

e3x 1 dx 2

9ffiffiffi

3p

[Hint: Differentiate with respect to b in Problem 15.46.]

15.48. Use the method of Problem 12.31, Chapter 12, to justify the procedure used in Problem 15.11.

DIRICHLET INTEGRALS

15.49. Find the mass of the region in the xy plane bounded by x y 1;x 0;y 0 if the density is ffiffiffiffiffiffixyp .Ans: =24

15.50. Find the mass of the region bounded by the ellipsoidx2

a2 y2b2

z2c2

1 if the density varies as the square ofthe distance from its center.

Ans:abck

30a2 b2 c2;k constant of proportionality

15.51. Find the volume of the region bounded by x2=3 y2=3 z2=3 1.

Ans: 4=35

15.52. Find the centroid of the region in the first octant bounded by x2=3 y2=3 z2=3 1.

Ans: "xx "yy "zz 21=128

15.53. Show that the volume of the region bounded by xm ym zm am, where m > 0, is given by 8f1=mg3

3m2 3=m a3.

15.54. Show that the centroid of the region in the first octant bounded by xm ym zm am, where m > 0, is givenby

"xx "yy "zz 32=m3=m41=m4=m a

MISCELLANEOUS PROBLEMS

15.55. Prove that b

a x

a

p

b

x

q dx

b

a

pq1 B

p

1; q

1

where p >

1; q >

1 and b > a.

[Hint: Let x a b ay:]

15.56. Evaluate (a)

31

dxffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffix 13 x

p ; b7

3

ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffi7 xx 34

pdx.

Ans: a ; b 2 f1=4g2

3ffiffiffi

p

15.57. Show thatf1=3g21=6

ffiffiffi

p ffiffiffi2

3pffiffiffi

3p .

15.58. Prove that Bu; v 1

21

0

xu1

xv1

1 xuv dx where u; v > 0.[Hint: Let y x=1 x:



16/17

15.59. If 0 < p < 1 prove that

=20

tanp d 2

secp

2.

15.60. Prove that 1

0

xu11 xv1

x ru

v

Bu; v

ru

1 ru

v where u; v, and r are positive constants.

[Hint: Let x r 1y=ry.]

15.61. Prove that

=20

sin2u1 cos2v1 d

a sin2 b cos2 uv Bu; v2avbu

where u; v > 0.

[Hint: Let x sin2 in Problem 15.60 and choose r appropriately.]

15.62. Prove that

10

dx

xx 1

11 1

22 1

33

15.63. Prove that for m

2; 3; 4; . . .

sin

msin

2

msin

3

m sin m 1

m m

2m1

[Hint: Use the factored form xm 1 x 1x 1x 2 x n1, divide both sides by x 1, andconsider the limit as x! 1.]

15.64. Prove that

=20

ln sinx dx =2 ln 2 using Problem 15.63.[Hint: Take logarithms of the result in Problem 15.63 and write the limit as m!1 as a definite integral.]

15.65. Prove that 1

m

2

m

3

m

m 1

m

2m1=2ffiffiffiffimp

:

[Hint: Square the left hand side and use Problem 15.63 and equation (11a), Page 378.]

15.66. Prove that

10

ln x dx 12

ln2.

[Hint: Take logarithms of the result in Problem 15.65 and let m!1.]

15.67. (a) Prove that

10

sinx

xpdx

2 p sinp=2 ; 0 < p < 1.

(b) Discuss the cases p 0 and p 1.

15.68. Evaluate (a)

10

sinx2 dx; b1

0

x cosx3 dx.

Ans: a 12

ffiffiffiffiffiffiffiffi=2

p; b

3ffiffiffi

3p

1=3

15.69. Prove that

10

xp1 lnx1 x dx

2 cscp cotp; 0 < p < 1.

15.70. Show that

10

lnx

x4 1 dx 2

ffiffiffi2

p

16.

15.71. Ifa > 0; b > 0, and 4ac > b2, prove that11

11

eax2bxycy2 dxdy 2ffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi

4ac b2p



17/17

15.72. Obtain (12) on Page 378 from the result (4) of Problem 15.20.

[Hint: Expand ev

3=3ffiffinp in a power series and replace the lower limit of the integral by 1.]15.73. Obtain the result (15) on Page 378.

[Hint: Observe that x 1xx !, thus ln x lnx 1 lnx, and

0x

x 0x 1

x 1 1

x

Furthermore, according to (6) page 377.

x ! limk!1

k! kx

x 1 x kNow take the logarithm of this expression and then differentiate. Also recall the definition of the Euler

constant, .

15.74. The duplication formula (13a) Page 378 is proved in Problem 15.24. For further insight, develop it forpositive integers, i.e., show that

22n1n 12n 2n ffiffiffip

Hint: Recall that 12 , then show that

n 12 2n 1

2

2n 1 5 3 1

2nffiffiffi

p:

Observe that

2n 12nn 1

2n!2nn!

2n 1 5 3 1

Now substitute and refine.


gamma and betta function adv calculus schaum

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