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
This observation can be incredibly useful
in determining whether or not an improper integral
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.
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.''
, the function
is a non-decreasing function.
converges to some value
Thus in this case
is a non-decreasing function bounded
above, hence the limit
This proves the first statement.
Likewise, the function
is also a non-decreasing function.
defined above is still non-decreasing and
does not exist, so
is not bounded. Since
is also unbounded, which proves the second statement.
converge? Answer: YES.
, we really do have
as illustrated in Figure 5.7.5
But why did we use
? It's a guess that turned out
to work. You could have used something else, e.g.,
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 anxious--instead, relax, take a deep breath and
For example, alternatively we could have done the following:
and this works just as well, since
. Does it converge
For large values of
, the term
very quickly goes to 0
so we expect this to diverge, since
, we have
(verify by cross multiplying)
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.
(Coming up with this comparison might take some work, imagination,
and trial and error.)
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.
Consider the integral
This is an improper integral since
has a pole at
Does it converge? NO.
On the interal