SalasSV_10_07 - 622 CHAPTER 10 SEQUENCES; INDETERMINATE...

10.7 IMPROPER INTEGRALS
In the de fi nition of the de fi nite integral
b
a
f ( x ) dx
it is assumed that [ a , b ] is a bounded interval and f is a bounded function. In this
section we use the limit process to investigate integrals in which either the interval
is unbounded or the integrand is an unbounded function. Such integrals are called
improper integrals .
Integrals over Unbounded Intervals
We begin with a function f continuous on an unbounded interval [ a , ∞ ). For each
number b > a we can form the de fi nite integral
b
a
f ( x ) dx .
(Figure 10.7.1)
If, as b tends to ∞ , this integral tends to a fi nite limit L ,
y
f
x
a b
Figure 10.7.1
lim
b →∞
b
a
f ( x ) dx = L ,
then we write
∞
a
f ( x ) dx = L

10.7 IMPROPER INTEGRALS
623
and say that
the improper integral
∞
a
f ( x ) dx converges to L .
Otherwise, we say that
the improper integral
∞
a
f ( x ) dx diverges .
In a similar manner,
improper integrals
b
−∞
f ( x ) dx arise as limits of the form lim
a →−∞
b
a
f ( x ) dx .
Example 1
(a)
∞
0
e
− 2 x
dx =
1
2
. (b)
∞
1
dx
x
diverges.
(c)
∞
1
dx
x 2
= 1. (d)
1
−∞
cos π x dx diverges.
VERIFICATION
(a)
∞
0
e
− 2 x
dx = lim
b →∞
b
0
e
− 2 x
dx = lim
b →∞
−
e
− 2 x
2
b
0
= lim
b →∞
1
2
−
1
2 e 2 b
=
1
2
.
(b)
∞
1
dx
x
= lim
b →∞
b
1
dx
x
= lim
b →∞
ln b = ∞ .
(c)
∞
1
dx
x 2
= lim
b →∞
b
1
dx
x 2
= lim
b →∞
−
1
x
b
1
= lim
b →∞
1 −
1
b
= 1.
(d) Note fi rst that
1
a
cos π x dx =
1
π
sin π x
1
a
= −
1
π
sin π a .
As a tends to −∞ , sin π a oscillates between − 1 and 1. Therefore the integral
oscillates between 1 /π and − 1 /π and does not converge.
The usual formulas for area and volume are extended to the unbounded case by
means of improper integrals.
Example 2 Let p be a positive number. If is the region between the x -axis and the
graph of
f ( x ) =
1
x p
, x ≥ 1,
(Figure 10.7.2)
y
x
Ω
1
f ( x ) =
x
p
Figure 10.7.2
then
area of =



1
p − 1
, if p > 1
∞ , if p ≤ 1.

624
CHAPTER 10 SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS
This comes about from setting
area of = lim
b →∞
b
1
dx
x p
=
∞
1
dx
x p
.
For p = 1,
∞
1
dx
x p
= lim
b →∞
b
1
dx
x p
= lim
b →∞
1
1 − p
( b
1 − p
− 1) =



1
p − 1
, if p > 1
∞ , if p < 1.
For p = 1,
∞
1
dx
x
p
=
∞
1
dx
x
= ∞ ,
as you have seen already.
For future reference we record the following: Let p > 0,
(10.7.1)
∞
1
dx
x p
converges for p > 1 and diverges for p ≤ 1.
Example 3 We know that the region below the graph of f ( x ) = 1 / x , x ≥ 1, has
in fi nite area. Suppose that this region with in fi nite area is revolved about the x -axis
(see Figure 10.7.3). What is the volume V of the resulting solid? It may surprise you
somewhat, but the volume is not in fi nite. In fact, it is π . Using the disc method to
calculate the volume (Section 6.2), we have
V =
∞
1
π [ f ( x )]
2
dx = π
∞
1
dx
x 2
= π lim
b →∞
b
1
dx
x 2
= π lim
b →∞
− 1
x
b
1
= π · 1 = π .