Page 395 - Schaum's Outline of Theory and Problems of Advanced Calculus

386 GAMMA AND BETA FUNCTIONS [CHAP. 15

so that (2)becomes
ð 1
1 ð =2Þ ð
=2 þ 1Þ ð =2Þ 1 ð þ
Þ=2
u du
I ¼ ð1 uÞ ð3Þ
4
½ð þ
Þ=2 þ 1Š u¼0
1 ð =2Þ ð
=2 þ 1Þ ð =2Þ ½ð þ
Þ=2 þ 1Š ð =2Þ ð =2Þ ð
=2Þ
¼ ¼
4
½ð þ
Þ=2 þ 1Š ½ð þ þ
Þ=2 þ 1Š 8 ½ð þ þÞ=2 þ 1Š
where we have used ð
=2Þ ð
=2Þ¼ ð
=2 þ 1Þ.
The integral evaluated here is a special case of the Dirichlet integral (20), Page 379. The general case
can be evaluated similarly.
2
2 2 2
2
2
2
15.22. Find the mass of the region bounded by x þ y þ z ¼ a if the density is ¼ x y z .
ðð ð
2 2 2
The required mass ¼ 8 x y z dx dy dz, where V is the region in the first octant bounded by the
V
2
2
2
2
sphere x þ y þ z ¼ a and the coordinate planes.
In the Dirichlet integral (20), Page 379, let b ¼ c ¼ a; p ¼ q ¼ r ¼ 2 and ¼ ¼
¼ 3. Then the
required result is
3
3
a a a 3 ð3=2Þ ð3=2Þ ð3=2Þ 4 s 9
945
8 ¼
2 2 2 ð1 þ 3=2 þ 3=2 þ 3=2Þ
MISCELLANEOUS PROBLEMS
ð 2
1 p ffiffiffiffiffiffiffiffiffiffiffiffiffi
4
ffiffiffiffiffiffi .
15.23. Show that 1 x dx ¼ f ð1:4Þg
p
0 6 2
4
Let x ¼ y. Then the integral becomes
ffiffiffi
1 ð 1 3=4 1=2 1 ð1=4Þ ð3=2Þ p f ð1=4Þg 2
y ð1 yÞ dy ¼ ¼
4 0 4 ð7=4Þ 6 ð1:4Þ ð3=4Þ
p ffiffiffi
From Problem 15.17 with p ¼ 1=4; ð1=4Þ ð3=4Þ¼ 2 so that the required result follows.
15.24. Prove the duplication formula 2 2p 1 1 p ð2pÞ.
ffiffiffi
2
ð pÞ ð p þ Þ¼
ð =2 ð =2
2p 2p
Let I ¼ sin xdx; J ¼ sin 2xdx.
0 0
ð p þ Þ
1
ffiffiffi
p
1 1 1 2
2 2 2
Then I ¼ Bð p þ ; Þ¼
2 ð p þ 1Þ
Letting 2x ¼ u,we find
ð ð =2
2p
1 2p sin udu ¼ I
2
J ¼ sin udu ¼
0 0
ð =2 ð =2
2p
2p
2p
But J ¼ ð2 sin x cos xÞ dx ¼ 2 2p sin x cos xdx
0 0
2 2p 1 1 2
2
¼ 2 2p 1 1 1 f ð p þ Þg
2 2
Bð p þ ; p þ Þ¼
ð2p þ 1Þ
Then since I ¼ J,
1
ffiffiffi
ð p þ Þ 2 2p 1 1 2
p
2
2 f ð p þ Þg
¼
2p ð pÞ 2p ð2pÞ