Now that we’ve seen how to actually compute improper
integrals we need to address one more topic about them. Often we aren’t concerned with the actual
value of these integrals. Instead we
might only be interested in whether the integral is convergent or divergent. Also, there will be some integrals that we
simply won’t be able to integrate and yet we would still like to know if they
converge or diverge.
To deal with this we’ve got a test for convergence or
divergence that we can use to help us answer the question of convergence for an
improper integral.
We will give this test only for a subcase of the infinite
interval integral, however versions of the test exist for the other subcases
of the infinite interval integrals as well as integrals with discontinuous
integrands.
Comparison Test
Note that if you think in terms of area the Comparison Test
makes a lot of sense. If is larger than then the area under must also be larger than the area under .
So, if the area under the larger function is finite (i.e. converges) then the area under the smaller
function must also be finite (i.e. converges).
Likewise, if the area under the smaller function is infinite (i.e. diverges) then
the area under the larger function must also be infinite (i.e. diverges).
Be careful not to misuse this test. If the smaller function converges there is no
reason to believe that the larger will also converge (after all infinity is
larger than a finite number…) and if the larger function diverges there is no
reason to believe that the smaller function will also diverge.
Let’s work a couple of examples using the comparison
test. Note that all we’ll be able to do
is determine the convergence of the integral.
We won’t be able to determine the value of the integrals and so won’t
even bother with that.
Example 1 Determine
if the following integral is convergent or divergent.
Solution
Let’s take a second and think about how the Comparison
Test works. If this integral is
convergent then we’ll need to find a larger function that also converges on
the same interval. Likewise, if this
integral is divergent then we’ll need to find a smaller function that also
diverges.
So, it seems like it would be nice to have some idea as to
whether the integral converges or diverges ahead of time so we will know
whether we will need to look for a larger (and convergent) function or a
smaller (and divergent) function.
To get the guess for this function let’s notice that the
numerator is nice and bounded and simply won’t get too large. Therefore, it seems likely that the
denominator will determine the convergence/divergence of this integral and we
know that
converges since by the fact in the previous section. So let’s guess that this integral will
converge.
So we now know that we need to find a function that is
larger than
and also converges.
Making a fraction larger is actually a fairly simple process. We can either make the numerator larger or
we can make the denominator smaller.
In this case we can’t do a lot about the denominator. However we can use the fact that to make the numerator larger (i.e. we’ll replace the cosine with
something we know to be larger, namely 1).
So,
Now, as we’ve already noted
converges and so by the Comparison Test we know that
must also converge.

Example 2 Determine
if the following integral is convergent or divergent.
Solution
Let’s first take a guess about the convergence of this
integral. As noted after the fact in
the last section about
if the integrand goes to zero faster than then the integral will probably converge. Now, we’ve got an exponential in the
denominator which is approaching infinity much faster than the x and so it looks like this integral
should probably converge.
So, we need a larger function that will also
converge. In this case we can’t really
make the numerator larger and so we’ll need to make the denominator smaller
in order to make the function larger as a whole. We will need to be careful however. There are two ways to do this and only one,
in this case only one, of them will work for us.
First, notice that since the lower limit of integration is
3 we can say that and we know that exponentials are always
positive. So, the denominator is the
sum of two positive terms and if we were to drop one of them the denominator
would get smaller. This would in turn
make the function larger.
The question then is which one to drop? Let’s first drop the exponential. Doing this gives,
This is a problem however, since
diverges by the fact. We’ve got a larger function that is
divergent. This doesn’t say anything
about the smaller function. Therefore,
we chose the wrong one to drop.
Let’s try it again and this time let’s drop the x.
Also,
So, is convergent. Therefore, by the Comparison test
is also convergent.

Example 3 Determine
if the following integral is convergent or divergent.
Solution
This is very similar to the previous example with a couple
of very important differences. First,
notice that the exponential now goes to zero as x increases instead of growing larger as it did in the previous
example (because of the negative in the exponent). Also note that the exponential is now
subtracted off the x instead of
added onto it.
The fact that the exponential goes to zero means that this
time the x in the denominator will
probably dominate the term and that means that the integral probably
diverges. We will therefore need to
find a smaller function that also diverges.
Making fractions smaller is pretty much the same as making
fractions larger. In this case we’ll
need to either make the numerator smaller or the denominator larger.
This is where the second change will come into play. As before we know that both x and the exponential are
positive. However, this time since we
are subtracting the exponential from the x
if we were to drop the exponential the denominator will become larger and so
the fraction will become smaller. In
other words,
and we know that
diverges and so by the Comparison Test we know that
must also diverge.

Example 4 Determine
if the following integral is convergent or divergent.
Solution
First notice that as with the first example, the numerator
in this function is going to be bounded since the sine is never larger than
1. Therefore, since the exponent on
the denominator is less than 1 we can guess that the integral will probably
diverge. We will need a smaller
function that also diverges.
We know that . In particular, this term is positive and so
if we drop it from the numerator the numerator will get smaller. This gives,
and
diverges so by the Comparison Test
also diverges.

Okay, we’ve seen a few examples of the Comparison Test
now. However, most of them worked pretty
much the same way. All the functions
were rational and all we did for most of them was add or subtract something
from the numerator or denominator to get what we want.
Let’s take a look at an example that works a little
differently so we don’t get too locked into these ideas.
Example 5 Determine
if the following integral is convergent or divergent.
Solution
Normally, the presence of just an x in the denominator would lead us to guess divergent for this
integral. However, the exponential in
the numerator will approach zero so fast that instead we’ll need to guess
that this integral converges.
To get a larger function we’ll use the fact that we know
from the limits of integration that . This means that if we just replace the x in the denominator with 1 (which is
always smaller than x) we will make
the denominator smaller and so the function will get larger.
and we can show that
converges. In fact,
we’ve already done this for a lower limit of 3 and changing that to a 1 won’t
change the convergence of the integral.
Therefore, by the Comparison Test
also converges.

We should also really work an example that doesn’t involve a
rational function since there is no reason to assume that we’ll always be
working with rational functions.
Example 6 Determine
if the following integral is convergent or divergent.
Solution
We know that exponentials with negative exponents die down
to zero very fast so it makes sense to guess that this integral will be
convergent. We need a larger function,
but this time we don’t have a fraction to work with so we’ll need to do
something different.
We’ll take advantage of the fact that is a decreasing function. This means that
In other words, plug in a larger number and the function
gets smaller.
From the limits of integration we know that and this means that if we square x it will get larger. Or,
Note that we can only say this since . This won’t be true if ! We can now use the fact that is a decreasing function to get,
So, is a larger function than and we know that
converges so by the Comparison Test we also know that
is convergent.

The last two examples made use of the fact that . Let’s take a look at an example to see how do
we would have to go about these if the lower limit had been smaller than 1.
Example 7 Determine
if the following integral is convergent or divergent.
Solution
First, we need to note that is only true on the interval as is illustrated in the graph below.
So, we can’t just proceed as we did in the previous
example with the Comparison Test on the interval . However, this isn’t the problem it might at
first appear to be. We can always
write the integral as follows,
We used Mathematica
to get the value of the first integral.
Now, if the second integral converges it will have a finite value and
so the sum of two finite values will also be finite and so the original
integral will converge. Likewise, if
the second integral diverges it will either be infinite or not have a value
at all and adding a finite number onto this will not all of a sudden make it
finite or exist and so the original integral will diverge. Therefore, this integral will converge or
diverge depending only on the convergence of the second integral.
As we saw in Example 6 the second integral does converge
and so the whole integral must also converge.

As we saw in this example, if we need to, we can split the
integral up into one that doesn’t involve any problems and can be computed and
one that may contain a problem that we can use the Comparison Test on to
determine its convergence.