Page 393 - Schaum's Outline of Theory and Problems of Advanced Calculus

384 GAMMA AND BETA FUNCTIONS [CHAP. 15

=2 =2
ð ð ð 2
6
3
8
2
15.16. Evaluate (a) cos d ; ðbÞ sin cos d ; ðcÞ sin d .
0 0 0
1 3 5 5
(a)From Problem 15.15 the integral equals [compare Problem 15.14(a)].
2 4 6 ¼ 32
(b) The integral equals
ð =2 ð =2 ð =2 2 2 4 2
3 2 3 5
sin ð1 sin Þ d ¼ sin d sin d ¼ 1 3 1 3 5 ¼ 15
0 0 0
The method of Problem 15.14(b)can also be used.
ð =2 1 3 5 7 35
8
The given integral equals 4 sin d ¼ 4 :
ðcÞ 2 4 6 8 2 ¼ 64
0
x
ð p 1
1
15.17. Given dx ¼ , show that ð pÞ ð1 pÞ¼ , where 0 < p < 1.
0 1 þ x sin p sin p
x y
Letting ¼ y or x ¼ ,the given integral becomes
1 þ x 1 y
1
ð
y p 1 ð1 yÞ p dy ¼ Bð p; 1 pÞ¼ ð pÞ ð1 pÞ
0
and the result follows.
ð
dy
1
15.18. Evaluate .
0 1 þ y 4
p
1 ð 1 x 3=4 2 ffiffiffi
4
1
Let y ¼ x. Then the integral becomes dx ¼ ¼ by Problem 15.17 with p ¼ .
4 0 1 þ x 4 sinð =4Þ 4 4
2
The result can also be obtained by letting y ¼ tan .
ð
2 p ffiffiffiffiffiffiffiffiffiffiffiffiffi 16
3 3
15.19. Show that x 8 x dx ¼ p ffiffiffi.
0 9 3
3
Letting x 8y or x ¼ 2y 1=3 ,the integral comes
ð 1 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 1
2 2=3
8
2 4
2y 1=3 3 8ð1 yÞ y dy ¼ 8 y 1=3 ð1 yÞ 1=3 dy ¼ Bð ; Þ
0 3 3 0 3 3 3
2 4
3 3 1 2
8 ð ð Þ 8 8 16
3
3
¼ ¼ ð Þ ð Þ¼ ¼ p ffiffiffi
3 ð2Þ 9 9 sin =3 9 3
STIRLING’S FORMULA
p ffiffiffiffiffiffiffiffi n n
2 n n e approximately.
15.20. Show that for large positive integers n; n! ¼
ð
1 z 1 t
t e dt. Let lfz ¼ x þ 1then
By definition ðzÞ¼
0
ð ð ð
1 1 1
x t tþln t x tþx ln t
ðx þ 1Þ¼ t e dt ¼ e dt ¼ e dt ð1Þ
0 0 0
For a fixed value of x the function x ln t t has a relative maximum for t ¼ x (as is demonstrated by
elementary ideas of calculus). The substutition t ¼ x þ y yields
ð ð
1 1 y
x x
ðx þ 1Þ¼ e x e x lnðxþyÞ y dy ¼ x e e x lnð1þ Þ y dy ð2Þ
x
x x