Page 334 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 325

EVALUATION OF DEFINITE INTEGRALS
ð =2
12.23. Prove that ln sin xdx ¼ ln 2.
0 2
The given integral converges [Problem 12.42( f )]. Letting x ¼ =2 y,
ð =2 ð =2 ð =2
ln cos xdx
I ¼ ln sin xdx ¼ ln cos ydy ¼
0 0 0
Then
ð =2 ð =2 sin 2x
ln dx
2I ¼ ðln sin x þ ln cos xÞ dx ¼ 2
0 0

ð =2 ð =2 ð =2
ln 2
¼ ln sin 2xdx ln 2 dx ¼ ln sin 2xdx 2 ð1Þ
0 0 0
Letting 2x ¼ v,
ð =2 ð ð =2 ð
1 1
ln sin vdv
ln sin 2xdx ¼ ln sin vdv ¼ ln sin vdv þ
0 2 0 2 0 =2
1
¼ ðI þ IÞ¼ I (letting v ¼ u in the last integral)
2

ln 2.
2 2
Hence, (1)becomes 2I ¼ I ln 2 or I ¼
ð 2
12.24. Prove that x ln sin xdx ¼ ln 2.
0 2
Let x ¼ y. Then, using the results in the preceding problem,
ð ð ð
ð xÞ ln sin xdx
J ¼ x ln sin xdx ¼ ð uÞ ln sin udu ¼
0 0 0
ð ð
¼ ln sin xdx x ln sin xdx
0 0
2
¼ ln 2 J
2
ln 2:
or J ¼
2
dx
ð
1
is uniformly convergent for A 1.
12.25. (a) Prove that ð Þ¼ 2
0 x þ
ð 1 dx
ðbÞ Show that ð Þ¼ p . ðcÞ Evaluate :
2 2 2
ffiffiffi
0 ðx þ 1Þ
ð ð =2
dx 2n
1 1 3 5 ð2n 1Þ
ðdÞ Prove that :
2 nþ1 ¼ cos d ¼ 2
0 2 4 6 ð2nÞ
0 ðx þ 1Þ
1 1 ð 1 dx
(a) The result follows from the Weierestrass test, since @ for a A 1 and
2
2
2
x þ x þ 1 0 x þ 1
converges.
ð b dx 1 b 1
ð Þ¼ lim ¼ lim p tan 1 x ¼ lim p tan 1 b ffiffiffi ¼ p :
ðbÞ 2 ffiffiffi p ffiffiffi ffiffiffi p ffiffiffi
b!1 0 x þ b!1 0 b!1 2