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b g x dx converges. Then if
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1 1 EXAMPLE. p < p for x > 1. x 1 x4 1 5 dx p 1), p also converges. 2 x4 1 1
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dx p converges ( p integral with a 1, x 1
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b g x dx diverges.
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(b) Divergence.
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Let g x A 0 for a < x @ b, and suppose that b f x A g x for a < x @ b, f x dx also diverges.
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Then if
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ln x 1 EXAMPLE. > for x > 3. Then since x 3 4 x 3 4 6 ln x dx also diverges. 4 3 x 3
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dx diverges ( p integral with a 3, p 4), x 3 4
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2. Quotient test for integrals with non-negative integrands. b f x A 6 0 or 1, then f x dx and (a) If f x A 0 and g x A 0 for a < x @ b, and if lim x!a g x b a g x dx either both converge or both diverge.
312 b
IMPROPER INTEGRALS
[CHAP. 12
(b) (c)
If A 0 in (a), then If A 1 in (a), and
b g x dx converges, then b g x dx diverges, then
f x dx converges.
f x dx diverges.
This test is related to the comparison test and is a very useful alternative to it. In particular taking g x 1= x a p we have from known facts about the p integral the following theorems. Theorem 3. Let lim x a p f x A. Then x!a b (i) f x dx converges if p < 1 and A is nite a b (ii) f x dx diverges if p A 1 and A 6 0 (A may be in nite).
If f x becomes unbounded only at the upper limit these conditions are replaced by those in Theorem 4. Let lim b x p f x B. Then x!b b f x dx converges if p < 1 and B is nite (i) a b f x dx diverges if p A 1 and B 6 0 (B may be in nite). (ii)
EXAMPLE 1.
r dx 1 x 1 1 p converges, since lim x 1 1=2 4 lim . x!1 x 1 1=2 x!1 x4 1 2 x4 1 1 dx 1 1 p diverges, since lim 3 x p p . 2 1 2 1 x!3 10 3 x x 3 x x
EXAMPLE 2.
b b Absolute and conditional convergence. f x dx is called absolute convergent if j f x j dx a a b b b converges. If f x dx converges but j f x j dx diverges, then f x dx is called conditiona a a
ally convergent. Theorem 5. If b
j f x j dx converges, then
f x dx converges.
In words, an absolutely convergent
integral converges.
  4  sin x  1 dx p converges ( p integral with a ; p 1), it follows that EXAMPLE. Since p  @ p and  3 x  3 3 3 x  x    4 4   sin x  sin x p  dx converges and thus p dx converges (absolutely).  3 x  3 x   
Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence.
CHAP. 12]
IMPROPER INTEGRALS
IMPROPER INTEGRALS OF THE THIRD KIND Improper integrals of the third kind can be expressed in terms of improper integrals of the rst and second kinds, and hence the question of their convergence or divergence is answered by using results already established.
IMPROPER INTEGRALS CONTAINING A PARAMETER, UNIFORM CONVERGENCE Let 
1 f x; dx 8
This integral is analogous to an in nite series of functions. In seeking conditions under which we may di erentiate or integrate  with respect to , it is convenient to introduce the concept of uniform convergence for integrals by analogy with in nite series. We shall suppose that the integral (8) converges for 1 @ @ 2 , or brie y 1 ; 2 . De nition. The integral (8) is said to be uniformly convergent in 1 ; 2 if for each  > 0, we can nd a number N depending on  but not on , such that   u    f x; dx <  for all u > N and all in 1 ; 2  
  1   u     f x; dx, which is analogous in This can be restated by nothing that  f x; dx      a u an in nite series to the absolute value of the remainder after N terms. The above de nition and the properties of uniform convergence to be developed are formulated in terms of improper integrals of the rst kind. However, analogous results can be given for improper integrals of the second and third kinds.
SPECIAL TESTS FOR UNIFORM CONVERGENCE OF INTEGRALS 1. Weierstrass M test. If we can nd a function M x A 0 such that (a) j f x; j @ M x 1 @ @ 2 ; x > a 1 M x dx converges, (b) 1 then
f x; dx is uniformly and absolutely convergent in 1 @ @ 2 .
  1 1 cos x 1 dx cos x @ Since  2 and converges, it follows that dx is uniformly x 1 2 2 x2 1 0 x 1 0 x 1 and absolutely convergent for all real values of . EXAMPLE.
As in the case of in nite series, it is possible for integrals to be uniformly convergent without being absolutely convergent, and conversely.
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