Page 330 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 321

ð ðnþ1Þ
1 1 ð 1 ð
X X sin v
sin x sin x
dv
dx ¼ dx ¼ ð1Þ
x x 0 v þ n
0 n¼0 n n¼0
1 1
Now A for 0 @ v @ : Hence,
v þ n ðn þ 1Þ
ð ð
sin v 1 2
dv A
0 v þ n ðn þ 1Þ 9 sin vdv ¼ ðn þ 1Þ ð2Þ
2
1
X
Since diverges, the series on the right of (1)diverges by the comparison test. Hence,
ðn þ 1Þ
n¼0
ð
1

sin x
dx diverges and the required result follows.
0 x
IMPROPER INTEGRALS OF THE SECOND KIND, CAUCHY PRINCIPAL VALUE
ð 7 dx
12.13. (a) Prove that ffiffiffiffiffiffiffiffiffiffiffi converges and (b) find its value.
p
3
1 x þ 1
The integrand is unbounded at x ¼ 1. Then we define the integral as

ð 7 dx 2=3 7
3 2=3
lim ffiffiffiffiffiffiffiffiffiffiffi ¼ lim ðx þ 1Þ ¼ lim 6 ¼ 6
3 p 2=3 2
1þ x þ 1
!0þ !0þ 1þ !0þ
This shows that the integral converges to 6.
ð 5 dx
12.14. Determine whether converges (a)in the usual sense, (b)in the Cauchy principal
3
1 ðx 1Þ
value sense.
(a)By definition,
dx dx dx
ð 5 ð 1 1 ð 5
3 ¼ lim 3 þ lim 3
1 !0þ 2 !0þ
1 ðx 1Þ 1 ðx 1Þ 1þ 2 ðx 1Þ

1 1 1 1
¼ lim þ lim
1 !0þ 8 2 2 2 !0þ 2 2 32
1 2
and since the limits do not exist, the integral does not converge in the usual sense.
(b)Since
ð 1 dx ð 5 dx 1 1 1 1 3
lim ¼ lim
3 þ 3 !0þ 8 2 þ 2 32 ¼ 32
!0þ 2 2
1 ðx 1Þ 1þ ðx 1Þ
the integral exists in the Cauchy principal value sense. The principal value is 3/32.
12.15. Investigate the convergence of:
ð 3 dx ð 5 dx ð =2 dx
(a) (c) (e) ; n > 1:
2 3 2=3 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1=n
1 0
2 x ðx 8Þ ð5 xÞðx 1Þ ðcos xÞ
1
ð sin x ð 1 2 sin x
(b dx (d) dx
0 x 3 1 1 x
2=3
1 1 1 1

(a) 2=3 ¼ lim ¼ p ffiffiffiffiffi. Hence, the integral converges by
2 3 2=3 x!2þ x 2 2 3
lim ðx 2Þ

x!2þ x þ 2x þ 4 8 18
x ðx 8Þ
Theorem 3(i), Page 312.