In this section we discuss using comparison to determine if an
improper integrals converges or diverges.
Recall that if and are continuous functions on an
interval and
, then
This observation can be incredibly useful
in determining whether or not an improper integral
converges.
Not only does this
technique help in determing whether integrals converge, but it also
gives you some information about their values, which is often much
easier to obtain than computing the exact integral.
Proof.
Since
for all
, the function
is a nondecreasing function.
If
converges to some value
, then
for any
we have
Thus in this case
is a nondecreasing function bounded
above, hence the limit
exists.
This proves the first statement.
Likewise, the function
is also a nondecreasing function.
If
diverges then
the function
defined above is still nondecreasing and
does not exist, so
is not bounded. Since
we
have
for all
, hence
is also unbounded, which proves the second statement.
The theorem is very intuitive if you think about areas under a graph.
``If the bigger integral converges then so does the smaller one, and
if the smaller one diverges so does the bigger ones.''
Example 5.7.9
Does
converge? Answer: YES.
Figure:
Graph of
and

Since
, we really do have
as illustrated in Figure
5.7.5.
Thus
so
converges.
But why did we use
? It's a guess that turned out
to work. You could have used something else, e.g.,
for
some constant . This is an illustration of how in mathematics
sometimes you have to use your imagination or guess and see what
happens. Don't get anxiousinstead, relax, take a deep breath and
explore.
For example, alternatively we could have done the following:
and this works just as well, since
converges (as
is continuous).
Example 5.7.10
Consider
. Does it converge
or diverge?
For large values of
, the term
very quickly goes to
0,
so we expect this to diverge, since
diverges.
For
, we have
, so
for all
we have
(verify by cross multiplying)
But
Thus
must also diverge.
Note that there is a natural analogue of Theorem 5.7.8
for integrals of functions that ``blow up'' at a point, but
we will not state it formally.
Example 5.7.11
Consider
We have
(Coming up with this comparison might take some work, imagination,
and trial and error.)
We have
thus
converges, even though
we haven't figured out its value. We just know that it is
.
(In fact, it is
.)
What if we found a function that is bigger than
and its integral diverges?? So what! This does
nothing for you. Bzzzt. Try again.
Example 5.7.12
Consider the integral
This is an improper integral since
has a pole at
.
Does it converge? NO.
On the interal
we have
.
Thus
Thus
diverges.
William Stein
20060315