Page 327 - Schaum's Outline of Theory and Problems of Advanced Calculus

318 IMPROPER INTEGRALS [CHAP. 12

ð b ð 1
lim f ðxÞ dx ¼1 and so f ðxÞ dx diverges
b!1 N a
For the cases where A ¼ 0 and A ¼1, see Problem 12.41.
As seen in this and the preceding problem, there is in general a marked similarity between proofs for
infinite series and improper integrals.

xdx x 1
ð ð 2
1 1
12.5. Test for convergence: (a) ; ðbÞ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx.
4
2
1 3x þ 5x þ 1 2 x þ 16
6
3
4
(a) Method 1: For large x,the integrand is approximately x=3x ¼ 1=3x .
x 1 1 ð 1 dx
Since @ , and converges ( p integral with p ¼ 3), it follows by the
4
2
3x þ 5x þ 1 3x 3 3 1 x 3
ð
1 xdx
comparison test that also converges.
2
4
1 3x þ 5x þ 1
Note that the purpose of examining the integrand for large x is to obtain a suitable comparison
integral.
x 1 1 ð 1
. Since lim f ðxÞ ¼ , and gðxÞ dx converges,
Method 2: Let f ðxÞ¼ 4 2 ; gðxÞ¼ 3
3x þ 5x þ 1 x 3
ð x!1 gðxÞ 1
1
f ðxÞ dx also converges by the quotient test.
1
1
Note that in the comparison function gðxÞ,we have discarded the factor .Itcould, however, just
3
as well have been included.
x 1
Method 3: lim x 3 ¼ . Hence, by Theorem 1, Page 309, the required integral
4
2
3x þ 5x þ 1 3
x!1
converges.
ffiffiffiffiffi
2
p
6
(b) Method 1: For large x,the integrand is approximately x = x ¼ 1=x.
2
2
x 1 1 1 1 ð 1 dx ð 1 x 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi A . Since diverges, ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx also diverges.
For x A 2, p p
6
6
x þ 1 2 x 2 2 x 2 x þ 16
2
x 1 1 ð 1
Method 2: Let f ðxÞ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, gðxÞ¼ . Then since lim f ðxÞ ¼ 1, and gðxÞ dx diverges,
6
x 16 x x!1 gðxÞ 2
ð
1
f ðxÞ dx also diverges.
2
!
2
x 1
Method 3: Since lim x p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1, the required integral diverges by Theorem 1, Page 309.
x þ 16
x!1 6
Note that Method 1 may (and often does) require one to obtain a suitable inequality factor (in this
1
1
case ,or any positive constant less than )before the comparison test can be applied. Methods 2 and
2 2
3, however, do not require this.
ð
1 2
12.6. Prove that e x dx converges.
0
2 x
lim x e 2 ¼ 0(by L’Hospital’s rule or otherwise). Then by Theorem 1, with A ¼ 0; p ¼ 2 the given
x!1
integral converges. Compare Problem 11.10(a), Chapter 11.
12.7. Examine for convergence:
ln x 1 cos x
ð ð
1 1
dx; where a is a positive constant; dx:
1 x þ a 0 x
ðaÞ ðbÞ 2
ln x
(a) lim x ¼1. Hence by Theorem 1, Page 309, with A ¼1; p ¼ 1, the given integral diverges.
x þ a
x!1