6.5. IMPROPER INTEGRALS
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(b)
∞
e
1
x ln(x)
dx
(c)
∞
e
1
x(ln(x)) 2 dx
(d)
∞
e
1
x(ln(x)) p dx, where p is a positive real number
(e)
1
0
ln(x)
x
dx
(f)
1
0
ln(x) dx
2. Sometimes we may encounter an improper integral for which we cannot easily evaluate
the limit of the corresponding proper integrals. For instance, consider
∞
1
1
1+x 3 dx.
While it is hard (or perhaps impossible) to find an antiderivative for
1
1+x 3 , we can still
determine whether or not the improper integral converges or diverges by comparison
to a simpler one. Observe that for all x > 0, 1 + x 3 > x 3 , and therefore
1
1 + x 3 <
1
x 3 .
It therefore follows that
b
1
1
1 + x 3 dx <
b
1
1
x 3 dx
for every b > 1. If we let b → ∞ so as to consider the two improper integrals
∞
1
1
1+x 3 dx and
∞
1
1
x 3 dx, we know that the larger of the two improper integrals
converges. And thus, since the smaller one lies below a convergent integral, it follows
that the smaller one must converge, too. In particular,
∞
1
1
1+x 3 dx must converge, even
though we never explicitly evaluated the corresponding limit of proper integrals. We
use this idea and similar ones in the exercises that follow.
(a) Explain why x 2 + x + 1 > x 2 for all x ≥ 1, and hence show that
∞
1
1
x 2 +x+1
dx
converges by comparison to
∞
1
1
x 2 dx.
(b) Observe that for each x > 1, ln(x) < x. Explain why
b
2
1
x
dx <
b
2
1
ln(x)
dx
for each b > 2. Why must it be true that
b
2
1
ln(x) dx diverges?
(c) Explain why
x 4 +1
x 4 > 1 for all x > 1. Then, determine whether or not the