1549901369-Elements_of_Real_Analysis__Denlinger_

7.8 *Improper Riemann Integrals 439

Definition 7.8.11 Suppose that 'ti a< b, f is integrable on [a, b]. Then we call

J~;: f an improper integral of type II. If for some c E IR, both J~= f and

J.c += 1 converge, we say that^1 - = += f converges, and wnt. e

Otherwise, we say that J~;: f diverges.

Examples 7.8.12 Determine the convergence or divergence of each of the fol-
lowing improper integrals.

(a) /,+=-;. dx (b) /,+= ~ dx (c) J+= exdx

1 X 1 yX -0<>

Solution. (a) The function f(x) = -;. is continuous on the closed interval

x

[1, b] for all b > 1, hence it is integrable there. Moreover,

lim f,b-;. d x = lim [-x-^1 ]~ = lim [1--b

1

] = 1.

b->O<> 1 x b->O<> b->O<>

/,
Therefore, +=^1 /,+=^1
1 2 x dx converges, and 1 2 x dx = 1.
1
(b) The function f(x) = Vx is continuous on the closed interval [1,b] for

all b > 1, hence it is integrable there. Moreover,

lim rb ~ dx = lim [ 2y'x ] ~ = lim [2v1b -1] = +oo.

b->O<> J 1 y x b->O<> b->O<>

Therefore, 1+= Jx dx diverges.

( c) To investigate 1-:= ex dx we must consider two separate integrals, say

J

o exd x and r+= exdx. First, lim {o exdx = lim [ex]~= lim [1 - ea] =

_ 00 } 0 a-+-oo l a a-+- oo a-+-oo

1 - 0 = 1. Thus, [

0

= ex dx converges (to 1).

Secondly, lim {b exdx = lim [ex]~= lim [eb - 1] = +oo. Thus,

b->+= } 0 b->+= b->+=

1+= ex d x diverges. Since one of these two improper integrals diverges, we

must say that 1-+: ex dx diverges. D

Theorem 7.8.13 (Comparison Test, Ila) Suppose that \ix ~ a, 0 :S

f(x) :S g(x). If J.a += g converges, then so does J.+a = f , and J.+a = f :S J.+a = g.