1549901369-Elements_of_Real_Analysis__Denlinger_

7.8 *Improper Riemann Integrals 441

Solution. We consider two separate improper integrals:

/

+ oo 1

r;;; dx.

1 yxex

1

1 1

r;;; dx and

o yxex

(a) Consider f

1

; dx. For 0 < x < 1, ex > 1 so fiex > ft, so

Jo yxex

1 1 f

1

1

Vx < vx· In Example 7.8.4 we proved that Jn Vx dx converges, so by the

co::arison xtest, f

1

; dx converges.

0

x

Jo yxex

(b) Consider /+oo ; dx. For x > 1, ft > 1, so fiex > ex, so

1 yxex

r;;;^1 < -.^1 Now, hm. lb -x

v xex ex b-->oo^1 e dx = b-->oo lim [-e-x]~ = b-->oo lim [~ e - b lb] = l/e.

/

+oo 1
Thus, r;;; dx converges.
1 yxex

1

+00 1

(c) By (a) and (b) together, r;;; dx converges. D

o yxex

FINAL CAUTION ON IMPROPER INTEGRALS

Care must be taken not to conclude that improper integrals behave alge-
braically like ordinary integrals. They do not. For example, we know that if f
and g are integrable over [a, b], then so is their product f g. But it is not true

that if the improper integrals l: f and l: g converge, then so does l: f g. For

fl 1 fl ( 1 ) 2
example, Jo Vx dx converges, but Jo Vx dx does not. For further exam-

ples showing that improper integrals do not obey a ll the algebraic properties
of ordinary integrals, see Exercises 4 and 5 below.

EXERCISE SET 7.8-B

l. Determine the convergence or divergence of each of the following improper

integrals. Find the values of those that converge.