Page 389 - Schaum's Outline of Theory and Problems of Advanced Calculus

380 GAMMA AND BETA FUNCTIONS [CHAP. 15

15.2. Evaluate each of the following.
5! 5 4 3 2
¼ 30
ð6Þ
2 2! 2 2
ðaÞ ¼ ¼
2 ð3Þ
5 3 3 3 1 1 3
2
2
2
2
ð Þ 2 ð Þ 2 ð Þ
1 1 1 4
ðbÞ ¼ ¼ ¼
2 2 2
ð Þ ð Þ ð Þ
16
ð3Þ ð2:5Þ 2!ð1:5Þð0:5Þ ð0:5Þ
315
ðcÞ ¼ ¼
ð5:5Þ ð4:5Þð3:5Þð2:5Þð1:5Þð0:5Þ ð0:5Þ
8 5 2 2
3 3 3 3
6 ð Þ 6ð Þð Þ ð Þ 4
2 2 3
ðdÞ ¼ ¼
3 3
5 ð Þ 5 ð Þ
15.3. Evaluate each integral.
ð
1 3 x
x e dx ¼ ð4Þ¼ 3! ¼ 6
ðaÞ
0
ð
1 6 2x
x e dx: Let 2x ¼ 7. Then the integral becomes
ðbÞ
0
y
1 6 y dy 1 1 6! 45
ð ð
6 y
e y e ð7Þ
2 2 ¼ 2 7 dy ¼ 2 7 ¼ 2 7 ¼ 8
0 0
1 p ffiffiffi
2
15.4. Prove that ð Þ¼ .
ð
1
2
1 x 1=2 x dx. Letting x ¼ u this integral becomes
e
2
ð Þ¼
0
ffiffiffi
ð p
1 2
2 e u du ¼ 2 ¼ p ffiffiffi
using Problem 12.31, Chapter 12
0 2
This result also is described in equation (11a,b)earlier in the chapter.
15.5. Evaluate each integral.
ð
1
ffiffiffi y 2 3
y e dy. Letting y ¼ x,the integral becomes
p
ðaÞ
0
1=3 x 1
ð ð p ffiffiffi
1 p ffiffiffiffiffiffiffiffi 1 1 1 1
e
x e x 2=3 dx ¼ x 1=2 x dx ¼ ¼
0 3 3 0 3 2 3
ð ð ð
1 4x 2 1 2 1 ð4ln3Þz 2 2
3 ðe ln 3 ð 4x Þ e dz. Let ð4ln 3Þz ¼ x and the integral becomes
ðbÞ dx ¼ Þ dz ¼
0 0 0
!
ð 1=2 ð p ffiffiffi
1 x x 1 1 1=2 x ð1=2Þ
e d p ffiffiffiffiffiffiffiffiffiffiffi ¼ p ffiffiffiffiffiffiffiffiffiffiffi x e dx ¼ p ffiffiffiffiffiffiffiffiffiffiffi ¼ p ffiffiffiffiffiffiffiffi
0 4ln 3 2 4ln 3 0 2 4ln 3 4 ln 3
ð 1 dx
u
(c) p ffiffiffiffiffiffiffiffiffiffiffiffi : Let ln x ¼ u. Then x ¼ e . When x ¼ 1; u ¼ 0; when x ¼ 0; u ¼1. The integral
0 ln x
becomes
ð 1 u ð 1
e

u 1=2 u p ffiffiffi
e
p ffiffiffi du ¼ du ¼ ð1=2Þ¼
0 u 0
ð
1
m ax n
15.6. Evaluate x e dx where m; n; a are positive constants.
0