Page 317 - Schaum's Outline of Theory and Problems of Advanced Calculus

308 IMPROPER INTEGRALS [CHAP. 12

Similarly, we define
b b
ð ð
f ðxÞ dx ¼ lim f ðxÞ dx ð2Þ
a! 1 a
1
where a is a variable on the negative real numbers. And we call the integral on the left convergent or
divergent according as the limit on the right does or does not exist.
ð ð b ð
1 dx dx 1 1 dx
EXAMPLE 1. 2 ¼ lim 2 ¼ lim 1 ¼ 1sothat 2 converges to 1.
1 x b!1 1 x b!1 b 1 x
u u u
ð ð ð
EXAMPLE 2. cos xdx ¼ lim cos xdx ¼ lim ðsin u sin aÞ.Since this limit does not exist, cos xdx
a! 1 a a! 1
1 1
is divergent.
In like manner, we define
ð ð ð
1 x 0 1
f ðxÞ dx
f ðxÞ dx ¼ f ðxÞ dx þ ð3Þ
x 0
1 1
where x 0 is a real number, and call the integral convergent or divergent according as the integrals on the
right converge or not as in definitions (1)and (2). (See the previous remarks in part (c)of the definition
of improper integrals of the first kind.)

SPECIAL IMPROPER INTEGRALS OF THE FIRST KIND
ð
1
tx
1. Geometric or exponential integral e dx, where t is a constant, converges if t > 0 and
a
x
diverges if t @ 0. Note the analogy with the geometric series if r ¼ e t so that e tx ¼ r .
ð
dx
1
2. The p integral of the first kind , where p is a constant and a > 0, converges if p > 1 and
a x p
diverges if p @ 1. Compare with the p series.
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 ¼ y then makes the integration limit 1). Unless
otherwise specified we shall assume that f ðxÞ is continuous and thus integrable in every finite interval
a @ x @ b.

1. Comparison test for integrals with non-negative integrands.
ð
1
(a) Convergence. Let gðxÞ A 0 for all x A a, and suppose that gðxÞ dx converges. Then if
a
ð
1
0 @ f ðxÞ @ gðxÞ for all x A a, f ðxÞ dx also converges.
a
1 1 ð 1 ð 1 dx
EXAMPLE. Since @ ¼ e x and e x dx converges, also converges.
x
x
e þ 1 e x 0 0 e þ 1
ð
1
(b) Divergence. Let gðxÞ A 0 for all x A a,and suppose that gðxÞ dx diverges. Then if
ð
a
1
f ðxÞ A gðxÞ for all x A a, f ðxÞ dx also diverges.
a
1 1 ð 1 dx ð 1 dx
EXAMPLE. Since > for x A 2 and diverges ( p integral with p ¼ 1), also diverges.
ln x x 2 x 2 ln x