Page 396 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 15] GAMMA AND BETA FUNCTIONS 387

and the required result follows. (See Problem 15.74, where the duplication formula is developed for the
simpler case of integers.)
=2 d
ð 2
15.25. Show that q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ f ð1=4Þg
ffiffiffi .
p
2
1
0 1 sin 4
2
Consider
d 1=2 ð Þ f ð Þg
ð =2 ð =2 1 p ffiffiffi 1 2
cos 1 1 1 4 4
2
4 2
I ¼ p ffiffiffiffiffiffiffiffiffiffi ¼ d ¼ Bð ; Þ¼ 3 ¼ p ffiffiffiffiffiffi
0 cos 0 2 ð Þ 2 2
4
as in Problem 15.23.
ð =2 d ð =2 d ð =2 d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :
p p p
2
But I ¼ ffiffiffiffiffiffiffiffiffiffi ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2
2
0 cos 0 cos =2 sin =2 0 1 2 sin =2
ð =2
p ffiffiffi p ffiffiffi d
Letting 2 sin =2 ¼ sin in this last integral, it becomes 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, from which the result
1
2
0 1 sin
follows. 2
cos x
ð
1
15.26. Prove that p dx ¼ ; 0 < p < 1.
0 x 2 ð pÞ cosð p =2Þ
1 1 ð 1 p 1 xu
We have ¼ u e du. Then
x p ð pÞ 0
ð ð ð
1 cos x 1 1 1 p 1 xu
x p dx ¼ u e cos xdu dx
0 ð pÞ 0 0
1 ð 1 u p
du
ð pÞ 0 1 þ u
¼ 2 ð1Þ
where we have reversed the order of integration and used Problem 12.22, Chapter 12.
2
Letting u ¼ v in the last integral, we have by Problem 15.17
ð p ð ð p 1Þ=2
1 u 1 1 v
0 1 þ u 2 du ¼ 2 0 1 þ v dv ¼ 2 sinð p þ 1Þ =2 ¼ 2cos p =2 ð2Þ
Substitution of (2)in (1)yields the required result.
ð
1
2
15.27. Evaluate cos x dx.
0
!
1 ð 1 cos y 1 p ffiffiffiffiffiffiffiffi
2
Letting x ¼ y,the integral becomes p ffiffiffi dy ¼ 1 ¼ 1 2 =2 by Problem 15.26.
2 0 y 2 2 ð Þ 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 (a) ð7Þ ; ðbÞ ð3Þ ð3=2Þ ; ðcÞ ð1=2Þ ð3=2Þ ð5=2Þ.
2 ð4Þ ð3Þ ð9=2Þ
Ans. ðaÞ 30; ðbÞ 16=105; ðcÞ 3 8 3=2