Page 392 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 15] GAMMA AND BETA FUNCTIONS 383

2 x dx
ð 2
ffiffiffiffiffiffiffiffiffiffiffi : Letting x ¼ 2v; the integral becomes
p
ðbÞ
0 2 x
1 1
ð 2 ð p ffiffiffi p ffiffiffi
p ffiffiffi v p ffiffiffi 2 1=2 p ffiffiffi 4 2 ð3Þ ð1=2Þ 64 2
1
4 2 p ffiffiffiffiffiffiffiffiffiffiffi dv ¼ 4 2 v ð1 vÞ dv ¼ 4 2 Bð3; Þ¼ ¼
2
0 1 v 0 ð7=2Þ 15
ð a q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi
2
2
y 4 a y dy: 2 2 p x; the integral becomes
ðcÞ Letting y ¼ a x or y ¼
0
1 6 a 6
ð
6
a 6 x 3=2 ð1 xÞ 1=2 dx ¼ a Bð5=2; 3=2Þ¼ a ð5=2Þ ð3=2Þ ¼
0 ð4Þ 16
ð =2
15.13. Show that sin 2u 1 cos 2v 1 d ¼ ðuÞ ðvÞ u; v > 0.
0 2 ðu þ vÞ
This follows at once from Problems 15.10 and 15.11.
ð =2 ð =2 ð
5
4
6
4
15.14. Evaluate (a) sin d ; ðbÞ sin cos d ; ðcÞ cos d .
0 0 0
ðaÞ Let 2u 1 ¼ 6; 2v 1 ¼ 0; i.e., u ¼ 7=2; v ¼ 1=2; in Problem 15.13:
5
Then the required integral has the value ð7=2Þ ð1=2Þ :
32
¼
2 ð4Þ
8
ðbÞ Letting 2u 1 ¼ 4; 2v 1 ¼ 5; the required integral has the value ð5=2Þ ð3Þ :
315
¼
2 ð11=2Þ
=2
ð
4
ðcÞ The given integral ¼ 2 cos d :
0
3
Thus letting 2u 1 ¼ 0; 2v 1 ¼ 4inProblem 15.13, the value is 2 ð1=2Þ ð5=2Þ ¼ .
8
2 ð3Þ
=2 =2
ð ð
p p 1 3 5 ð p 1Þ
15.15. Prove sin d ¼ cos d ¼ðaÞ if p is an even positive integer,
0 0 2 4 6 p 2
(b) 2 4 6 ð p 1Þ is p is an odd positive integer.
1 3 5 p
From Problem 15.13 with 2u 1 ¼ p; 2v 1 ¼ 0, we have
ð =2 1 1
p ½ ð p þ 1ފ ð Þ
2
2
1
sin d ¼
0 2 ½ ð p þ 2ފ
2
(a)If p ¼ 2r,the integral equals
1 1 1 3 1 1 1 ð2r 1Þð2r 3Þ 1 1 3 5 ð2r 1Þ
2 2 2 2 2 2 2
ðr þ Þ ð Þ ðr Þðr Þ ð Þ ð Þ
2rðr 1Þ 1 2rð2r 2Þ 2 2 2 4 6 2r 2
¼ ¼ ¼
2 ðr þ 1Þ
(b)If p ¼ 2r þ 1, the integral equals
1 p 2 4 6 2r
ffiffiffi
2
ðr þ 1Þ ð Þ rðr 1Þ 1
3 ¼ 1 1 1 p
ffiffiffi ¼
2 2 2 2
2 ðr þ Þ 2ðr þ Þðr Þ 1 3 5 ð2r þ 1Þ
ð =2 ð =2
p
p
In both cases sin d ¼ cos d ,as seen by letting ¼ =2 .
0 0