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But by hypothesis the last integral exists. Thus b 1 f x dx exists, and hence f x dx converges lim
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12.4. Prove the quotient test (a) on Page 309.
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By hypothesis, lim x A N.
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   f x  f x A > 0. Then given any  > 0, we can nd N such that  A <  when  g x  x!1 g x Thus for x A N, we have A @ f x @A  g x b
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Then A  g x dx @ b
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f x dx @ A 
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There is no loss of generality in choosing A  > 0. 1 If g x dx converges, then by the inequality on the right of (1), a b 1 f x dx exists, and so f x dx converges lim b!1 N a 1 g x dx diverges, then by the inequality on the left of (1), If
IMPROPER INTEGRALS 1 f x dx 1 and so
[CHAP. 12
b!1 N
f x dx diverges
For the cases where A 0 and A 1, see Problem 12.41. As seen in this and the preceding problem, there is in general a marked similarity between proofs for in nite series and improper integrals.
1 12.5. Test for convergence:
(a) Method 1:
x dx ; 4 3x 5x2 1
1 b
x2 1 p dx. x6 16
For large x, the integrand is approximately x=3x4 1=3x3 . x 1 1 1 dx converges ( p integral with p 3), it follows by the @ 3 , and Since 4 5x2 1 3 1 x3 3x 3x 1 x dx comparison test that also converges. 4 2 1 3x 5x 1 Note that the purpose of examining the integrand for large x is to obtain a suitable comparison integral. 1 x 1 f x 1 Since lim g x dx converges, , and ; g x 3 . Method 2: Let f x 4 2 1 x!1 g x 3 3x 5x x 1
f x dx also converges by the quotient test.
Note that in the comparison function g x , we have discarded the factor 1. It could, however, just 3 as well have been included.   x 1 Hence, by Theorem 1, Page 309, the required integral . Method 3: lim x3 x!1 3 3x4 5x2 1 converges. p (b) Method 1: For large x, the integrand is approximately x2 = x6 1=x. 1 2 x2 1 1 1 1 1 dx x 1 p dx also diverges. diverges, For x A 2, p A . Since 2 x 2 2 x 2 x6 1 x6 16 1 2 x 1 1 f x Then since lim 1, and Method 2: Let f x p , g x . g x dx diverges, x!1 g x x x6 16 2 1 f x dx also diverges.
Method 3:
! x2 1 Since lim x p 1, the required integral diverges by Theorem 1, Page 309. x!1 x6 16
Note that Method 1 may (and often does) require one to obtain a suitable inequality factor (in this case 1, or any positive constant less than 1) before the comparison test can be applied. Methods 2 and 2 2 3, however, do not require this.
1 12.6. Prove that
e x dx converges.
lim x2 e x 0 (by L Hospital s rule or otherwise). Then by Theorem 1, with A 0; p 2 the given integral converges. Compare Problem 11.10(a), 11.
12.7. Examine for convergence: 1 ln x dx; where a is a positive constant; a x a 1
lim x ln x 1. x a
1 b
1 cos x dx: x2
Hence by Theorem 1, Page 309, with A 1; p 1, the given integral diverges.
CHAP. 12] 1 b
IMPROPER INTEGRALS 
1 cos x dx x2
1 cos x dx x2
1 cos x dx x2
The rst integral on the  right converges [see Problem 12.1(e)].  3=2 1 cos x Since lim x 0, the second integral on the right converges by Theorem 1, Page 309, x!1 x2 with A 0 and p 3=2. Thus, the given integral converges.
12.8. Test for convergence:
(a) Let x y.
ex dx; (a) 1 x
1
1 b
x3 x2 dx: 6 1 x 1
e y dy. y 1 1 1 y e y e Method 1: e y dy converges, @ e y for y @ 1. Then since dy converges; hence the y y 1 1 given integral converges.  y  e lim ye y 0. Then the given integral converges by Theorem 1, Page Method 2: lim y2 y!1 y!1 y 309, with A 0 and p 2. 0 1 3 x3 x2 x x2 (b) Write the given integral as dx dx. Letting x y in the rst integral, it 6 6 1 x 1 0 x 1 ! 1 3 y y2 y3 y2 1, this integral converges. dy. Since lim y3 becomes 6 y!1 y6 1 0 y 1 ! x3 x2 1, the second integral converges. Since lim x3 x!1 x6 1 Then the integral becomes Thus the given integral converges.
ABSOLUTE AND CONDITIONAL CONVERGENCE FOR IMPROPER INTEGRALS OF THE FIRST KIND 1 1 f x dx converges if j f x j dx converges, i.e., an absolutely convergent integral is 12.9. Prove that
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