Thomas' Calculus: Early Transcendentals

Mat104 Fall 2002, Improper Integrals From Old Exams
For the following integrals, state whether they are convergent or divergent, and give your reasons.
(1)
∞
0
dx
x
3
+ 2
converges. Break it up as
1
0
dx
x
3
+ 2
+
∞
1
dx
x
3
+ 2
. The first of these is proper
and finite. The second behaves like the integral of 1 /x 3 on [1 , ∞ ) and thus converges.
(2)
1
0
dx
x +
√
x
converges. As x → 0,
√
x goes to 0 much more slowly than x does. (Think
about the graphs.) Therefore when x is very close to 0, the denominator x +
√
x ≈
√
x .
So this integral will behave like the integral of 1 /
√
x on [0 , 1], and this integral converges.
(3)
∞
1
√
1 + x
x
3
converges. As x goes to ∞ , the integrand behaves like
√
x
x
3
=
1
x
5 / 2
.
(4)
∞
0
x
2
x
3
+ 1
dx diverges. Break it up into two integrals
1
0
x
2
x
3
+ 1
dx +
∞
1
x
2
x
3
+ 1
dx . The
first integral is proper and finite. The second can be compared to the integral of 1 /x on
[1 , ∞ ) which diverges.
(5)
1
0
ln x dx converges to - 1. Here we can compute directly since integration by parts tells
us that ln x dx = x ln x - x + C . Evaluating at the x = 1 endpoint gives ln 1 - 1 = - 1.
For the other endpoint we have to take the limit as x goes to 0. For this we need L’Hˆ opital’s
rule.
lim
x → 0
x ln x = lim
x → 0
ln x
1 /x
= lim
x → 0
1 /x
- 1 /x 2
= lim
x → 0
- x = 0 .
So evaluating at the x = 0 endpoint gives 0.
(6)
1
0
dx
e
x
- 1
diverges. The only difficulty is that the denominator is 0 when x = 0. There are
a couple of approaches we could take. The easiest is to use the Taylor series for e
x
. Then
we know that e
x
- 1 = x + higher powers of x and as x goes to zero, the higher powers of
x will vanish much more rapidly. So this function behaves essentially like 1 /x when x is
close to 0. Since dx/x diverges, this integral will also.
Alternatively, we could compute the integral, making the substitution u = e
x
and then
use partial fractions.
(7)
∞
0
dx
x
2
+ 2 x + 2
converges. The only difficulty is that we have an infinite endpoint. The
integrand is asymptotic to 1 /x 2 as x goes to infinity. Since
∞
1
dx/x
2
converges, this
integral will as well. (To compare these we should break up the integral. First integrate
from 0 to 1, which gives a finite value. Then integrate further from 1 out to ∞ . This gives
a finite value as well by comparison to 1 /x 2 .)
(8)
∞
1
x 3
ln x + x 4
dx diverges. Again the only problem is that we have an infinite endpoint. As
x goes to infinity, x
4
grows much faster than ln x . Thus the integrand will be asymptotic
to x
3
/x
4
= 1 /x as x goes to infinity.