Page 391 - Schaum's Outline of Theory and Problems of Advanced Calculus

382 GAMMA AND BETA FUNCTIONS [CHAP. 15

u
Letting ln a=x ¼ u or x ¼ ae ,this becomes
r ffiffiffiffiffi ð r ffiffiffiffiffi r ffiffiffiffiffiffiffi
m 1 m m
1
T ¼ a u 1=2 u du ¼ a ð Þ¼ a
e
2
2k 0 2k 2k
THE BETA FUNCTION
ð =2
15.10. Prove that (a) Bðu; vÞ¼ Bðv; uÞ; ðbÞ Bðu; vÞ¼ 2 sin 2u 1 cos 2v 1 d .
0
(a)Using the transformation x ¼ 1 y,we have
ð 1 ð 1 ð 1
x u 1 v 1 u 1 y v 1 y v 1 u 1
Bðu; vÞ¼ ð1 xÞ dx ¼ ð1 yÞ dy ¼ ð1 yÞ dy ¼ Bðv; uÞ
0 0 0
2
(b)Using the transformation x ¼ sin ,we have
ð 1 ð =2
x u 1 v 1 2 u 1 2 v 1 2 sin cos d
Bðu; vÞ¼ ð1 xÞ dx ¼ ðsin Þ ðcos Þ
0 0
ð =2
¼ 2 sin 2u 1 cos 2v 1 d
0

u; v > 0.
ðuÞ ðvÞ
15.11. Prove that Bðu; vÞ¼
ðu þ vÞ
ð ð
1 1
2 2 u 1 z 2u 1 x 2
Letting z ¼ x ; we have ðuÞ¼ z e dx ¼ 2 x e dx:
0 0
ð
1 2
e
Similarly, ðvÞ¼ 2 y 2v 1 y dy: Then
0
ð ð

1 2 1 2
ðuÞ ðvÞ¼ 4 x 2u 1 x dx y 2v 1 y dy
e
e
0 0
ð ð
1 1 2 2
¼ 4 x 2u 1 y 2v 1 ðx þy Þ dx dy
e
0 0
Transforming to polar coordiantes, x ¼ cos ; y ¼ sin ,
=2 1
ð ð
e
ðuÞ ðvÞ¼ 4 2ðuþvÞ 1 2 cos 2u 1 sin 2v 1 d d
¼0 ¼0
ð ð =2

1 2
e
¼ 4 2ðuþvÞ 1 d cos 2u 1 sin 2v 1 d
¼0 ¼0
=2
ð
cos 2u 1 sin 2v 1
¼ 2 ðu þ vÞ d ¼ ðu þ vÞ Bðv; uÞ
0
¼ ðu þ vÞ Bðu; vÞ
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.
ð 1 4!3! 1
4 3 ð5Þ ð4Þ
ðaÞ x ð1 xÞ dx ¼ Bð5; 4Þ¼ ¼ 8! ¼ 280
0 ð9Þ