In this section we need to take a look at a couple of
different kinds of integrals. Both of
these are examples of integrals that are called Improper Integrals.
Let’s start with the first kind of improper integrals that
we’re going to take a look at.
Infinite Interval
In this kind of integral one or both of the limits of integration are
infinity. In these cases the interval of
integration is said to be over an infinite interval.
Let’s take a look at an example that will also show us how
we are going to deal with these integrals.
So, this is how we will deal with these kinds of integrals
in general. We will replace the infinity
with a variable (usually t), do the
integral and then take the limit of the result as t goes to infinity.
On a side note, notice that the area under a curve on an
infinite interval was not infinity as we might have suspected it to be. In fact, it was a surprisingly small number. Of course this won’t always be the case, but
it is important enough to point out that not all areas on an infinite interval
will yield infinite areas.
Let’s now get some definitions out of the way. We will call these integrals convergent if the associated limit
exists and is a finite number (i.e.
it’s not plus or minus infinity) and divergent
if the associated limit either doesn’t exist or is (plus or minus) infinity.
Let’s now formalize up the method for dealing with infinite
intervals. There are essentially three
cases that we’ll need to look at.
Let’s take a look at a couple more examples.
Example 2 Determine
if the following integral is convergent or divergent and if it’s convergent find
its value.
Solution
So, the first thing we do is convert the integral to a
limit.
Now, do the integral and the limit.
So, the limit is infinite and so the integral is
divergent.

If we go back to thinking in terms of area notice that the
area under on the interval is infinite.
This is in contrast to the area under which was quite small. There really isn’t all that much difference
between these two functions and yet there is a large difference in the area
under them. We can actually extend this
out to the following fact.
Fact
One thing to note about this fact is that it’s in essence
saying that if an integrand goes to zero fast enough then the integral will
converge. How fast is fast enough? If we use this fact as a guide it looks like
integrands that go to zero faster than goes to zero will probably converge.
Let’s take a look at a couple more examples.
Example 3 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
There really isn’t much to do with these problems once you
know how to do them. We’ll convert the
integral to a limit/integral pair, evaluate the integral and then the limit.
So, the limit is infinite and so this integral is divergent.

Example 4 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
In this case we’ve got infinities in both limits and so
we’ll need to split the integral up into two separate integrals. We can split the integral up at any point,
so let’s choose since this will be a convenient point for
the evaluation process. The integral is
then,
We’ve now got to look at each of the individual limits.
So, the first integral is convergent. Note that this does NOT mean that the
second integral will also be convergent.
So, let’s take a look at that one.
This integral is convergent and so since they are both
convergent the integral we were actually asked to deal with is also
convergent and its value is,

Example 5 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
First convert to a limit.
This limit doesn’t exist and so the integral is divergent.

In most examples in a Calculus II class that are worked over
infinite intervals the limit either exists or is infinite. However, there are limits that don’t exist,
as the previous example showed, so don’t forget about those.
Discontinuous
Integrand
We now need to look at the second type of improper integrals
that we’ll be looking at in this section.
These are integrals that have discontinuous integrands. The process here is basically the same with
one subtle difference. Here are the
general cases that we’ll look at for these integrals.
Note that the limits in these cases really do need to be
right or left handed limits. Since we
will be working inside the interval of integration we will need to make sure
that we stay inside that interval. This
means that we’ll use onesided limits to make sure we stay inside the
interval.
Let’s do a couple of examples of these kinds of integrals.
Example 6 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
The problem point is the upper limit so we are in the
first case above.
The limit exists and is finite and so the integral
converges and the integral's value is .

Example 7 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
This integrand is not continuous at and so we’ll need to split the integral up
at that point.
Now we need to look at each of these integrals and see if
they are convergent.
At this point we’re done.
One of the integrals is divergent that means the integral that we were
asked to look at is divergent. We
don’t even need to bother with the second integral.

Before leaving this section let's note that we can also have
integrals that involve both of these cases.
Consider the following integral.
Example 8 Determine
if the following integral is convergent or divergent. If it is convergent find its value.
Solution
This is an integral over an infinite interval that also
contains a discontinuous integrand. To
do this integral we’ll need to split it up into two integrals. We can split it up anywhere, but pick a
value that will be convenient for evaluation purposes.
In order for the integral in the example to be convergent
we will need BOTH of these to be convergent.
If one or both are divergent then the whole integral will also be
divergent.
We know that the second integral is convergent by the fact
given in the infinite interval portion above.
So, all we need to do is check the first integral.
So, the first integral is divergent and so the whole
integral is divergent.
