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CHAP. 15]
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GAMMA AND BETA FUNCTIONS
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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 di culty of the proofs, they shall be omitted. In (2) introduce the logarithmic expansion  y y y2 y3 2 3 ln 1 x x 2x 3x and also let y Then p x 1 xx e x x For large values of x p x 1 % xx e x x p x v; dy 1
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When x is replaced by integer values n, then the Stirling relation p n! x 1 % 2x xx e x is obtained. It is of interest that from (4) we can also obtain the result (12) on Page 378. See Problem 15.72.
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1 dx dy dz where V is 15.21. Evaluate I
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the region in the rst octant bounded by the sphere x2 y2 z2 1 and the coordinate planes.
Let x2 u; y2 v; z2 w. Then du dv dw u 1 =2 v 1 =2 w
1 =2 p p p I 2 u2 v2 w r 1 u =2 1 v =2 1 w
=2 1 du dv dw 8
where r 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, 1 u 1 u v 1 1 I u =2 1 v =2 1 w
=2 1 du dv dw 2 8 u 0 v 0 w 0 1 u 1 1 u =2 1 v =2 1 1 u v
=2 du dv 4
u 0 v 0 & 1 u ' 1 1 u =2 1 v =2 1 1 u v
=2 dv du 4
u 0 v 0
Fig. 15-2
Letting v 1 u t, we have 1 1 u v =2 1 1 u v
=2 dv 1 u
=2 t =2 1 1 t
=2 dt
v 0 t 0
1 u
=2
=2 1
=2 1
GAMMA AND BETA FUNCTIONS
[CHAP. 15
so that (2) becomes I 1 4
1 4
=2
=2 1 1 u =2 1 1 u
=2 du
=2 1 u 0 =2
=2 1 =2
=2 1 =2 =2
=2
=2 1
=2 1 8 =2 1 3
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  x2 y2 z2 .
The required mass 8
x2 y2 z2 dx dy dz, where V is the region in the rst 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. required result is 8 a3 a3 a3 3=2 3=2 3=2 4s9 2 2 2 1 3=2 3=2 3=2 945 Then the
MISCELLANEOUS PROBLEMS 1 p f 1:4 g2 1 x4 dx p . 15.23. Show that 6 2 0
Let x4 y. Then the integral becomes p  f 1=4 g2 1 1 3=4 1 1=4 3=2 y 1 y 1=2 dy 4 0 4 7=4 6 1:4 3=4 p From Problem 15.17 with p 1=4; 1=4 3=4  2 so that the required result follows.
15.24. Prove the duplication formula 22p 1 p p 1 2
Let I =2
sin2p x dx; J
=2
p  2p .
sin2p 2x dx.
Then I 1 B p 1 ; 1 2 2 2 Letting 2x u, we nd
p p 1  2 2 p 1 
J 1 2 =2
sin2p u du
=2
sin2p u du I =2
2 sin x cos x 2p dx 22p
sin2p x cos2p x dx
22p 1 B p 1 ; p 1 2 2 Then since I J,
22p 1 f p 1 g2 2 2p 1
p p 1  22p 1 f p 1 g2 2 2 2p p 2p 2p
CHAP. 15]
GAMMA AND BETA FUNCTIONS
and the required result follows. simpler case of integers.)
(See Problem 15.74, where the duplication formula is developed for the
=2 15.25. Show that
d f 1=4 g2 p . q 4  1 1 sin2  2
=2
Consider I d p cos  =2
cos 1=2  d 1 B 1 ; 1 2 4 2
p 1  f 1 g2 4 p4 2 3 2 2 4
as in Problem 15.23. =2 =2 =2 d d d p p p : But I cos  0 0 0 cos2 =2 sin2 =2 1 2 sin2 =2 Letting follows. p p 2 sin =2 sin  in this last integral, it becomes 2 =2
d q , from which the result 1 1 sin2  2
1 15.26. Prove that
cos x  ; 0 < p < 1. dx xp 2 p cos p=2
We have
1 1 x p p
u p 1 e xu du. 1
Then
cos x 1 1 1 p 1 xu u e cos x du dx p dx p x 0 0 p 1 1 u du p 0 1 u2
where we have reversed the order of integration and used Problem 12.22, 12. Letting u2 v in the last integral, we have by Problem 15.17 1 up 1 1 v p 1 =2   du dv 2 2 0 1 v 2 sin p 1 =2 2 cos p=2 0 1 u Substitution of (2) in (1) yields the required result.
1 15.27. Evaluate
cos x2 dx.
1 Letting x y, the integral becomes 2
! p cos y 1  1 =2 by Problem 15.26. p dy 2 y 2 2 1 cos =4 2
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 Ans. (a) 7 ; 2 4 3 b 16=105; b c 3 3=2 ; 9=2
3 3=2 8
c 1=2 3=2 5=2 .
a 30;
1 15.29. Evaluate Ans. (a)
GAMMA AND BETA FUNCTIONS x4 e x dx; 1 b p 2 c 16 1 b
[CHAP. 15
x6 e 3x dx;
1 c
x2 e 2x dx:
a 24; 1
80 ; b 243 e x dx;
15.30. Find
0 1 3
p p 4 x e x dx; 4=5 c p 5 5 16
1 c
y3 e 2y dy.
Ans. a
1 ; 3
p 3  b ; 2 r  ; 8
1 15.31. Show that
e st p dt t
s > 0.
15.32. Prove that v 1 15.33. Evaluate Ans: (a)
 1  1 v 1 ln dx; x 0 b 1
v > 0. 1 p 3 ln 1=x dx.
0 1 3
ln x 4 dx;
x ln x 3 dx; 1 3 Ans:
a 24;
b 3=128;
15.34. Evaluate
(a) 7=2 ;
b 1=3 .
p a 16  =105;
b 3 2=3
15.35. Prove that lim x 1 where m 0; 1; 2; 3; . . .
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