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where x0 is a real number, and call the integral convergent or divergent according as the integrals on the right converge or not as in de nitions (1) and (2). (See the previous remarks in part (c) of the de nition of improper integrals of the rst kind.)
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SPECIAL IMPROPER INTEGRALS OF THE FIRST KIND 1 1. Geometric or exponential integral e tx dx, where t is a constant, converges if t > 0 and Note the analogy with the geometric series if r e t so that e tx rx . 1 dx The p integral of the rst kind p , where p is a constant and a > 0, converges if p > 1 and a x diverges if p @ 1. Compare with the p series. diverges if t @ 0.
CONVERGENCE TESTS FOR IMPROPER INTEGRALS OF THE FIRST KIND The following tests are given for cases where an integration limit is 1. Similar tests exist where an integration limit is 1 (a change of variable x y then makes the integration limit 1). Unless otherwise speci ed we shall assume that f x is continuous and thus integrable in every nite interval a @ x @ b. 1. Comparison test for integrals with non-negative integrands. (a) Convergence. Let g x A 0 for x A a, and suppose that all
1 g x dx converges. Then if
0 @ f x @ g x for all x A a,
f x dx also converges.
EXAMPLE.
Since
1 1 @ x e x and 1 e
e x dx converges,
dx also converges. ex 1
Divergence.
Let g x A 0 for all x A a, and suppose that f x dx also diverges.
g x dx diverges.
Then if
f x A g x for all x A a,
EXAMPLE. Since
1 1 > for x A 2 and ln x x
dx diverges ( p integral with p 1), x
dx also diverges. ln x
CHAP. 12]
IMPROPER INTEGRALS
Quotient test for integrals with non-negative integrands. 1 1 f x A 6 0 or 1, then f x dx and g x dx (a) If f x A 0 and g x A 0, and if lim x!1 g x a a either both converge or both diverge. 1 1 (b) If A 0 in (a) and g x dx converges, then f x dx converges. 1 (c) If A 1 in (a) and
1 g x dx diverges, then
f x dx diverges.
This test is related to the comparison test and is often a very useful alternative to it. In particular, taking g x 1=xp , we have from known facts about the p integral, the following theorem. Theorem 1. Let lim xp f x A. Then x!1 1 (i) f x dx converges if p > 1 and A is nite 1 (ii)
f x dx diverges if p @ 1 and A 6 0 (A may be in nite).
1 EXAMPLE 1.
x2 dx x2 1 converges since lim x2 4 . 4 x!1 4x 25 4 4x 25 x dx x p diverges since lim x p 1. x!1 4 x2 1 4 x2 1 x x
1 EXAMPLE 2.
Similar test can be devised using g x e tx . 1 f x dx converges or diverges accord3. Series test for integrals with non-negative integrands. a ing as un , where un f n , converges or diverges. 1 1 f x dx is called absolutely convergent if j f x j dx 4. Absolute and conditional convergence. a a 1 1 1 converges. If f x dx converges but j f x j dx diverges, then f x dx is called cona a a
ditionally convergent. 1 Theorem 2. If
1 j f x j dx converges, then
f x dx converges. In words, an absolutely convergent
integral converges.
1 cos x EXAMPLE 1. dx is absolutely convergent and thus convergent since x2 1 a  1 1 1  cos x  dx dx   dx @ and converges.  2  2 2 0 x 1 0 x 1 0 x 1 1 EXAMPLE 2. sin x dx converges (see Problem 12.11), but x 0 1 sin x Thus, dx is conditionally convergent. x 0  1 sin x    x  dx does not converge (see 0
Problem 12.12).
Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence.
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