- Note that is continuous everywhere, so it is bounded on any finite interval.
- , so is unbounded on .
- , so even though is not continuous at , it is bounded on . Hence is proper.
- has a vertical asymptote at , so it is unbounded on .

We can use limits to integrate functions on unbounded domains or functions with unbounded range.

as a limit of Riemann sums. This limit need not always exist, as it depends on the
properties of the function on the given interval . When the function is *continuous*
on , this definite integral represents the net between the graph of and the -axis on

and the Fundamental Theorem of Calculus comes to the rescue and assures us that

where is an antiderivative of .

Note that while explicitly computing the limit of Riemann sums is an arduous task, the Fundamental Theorem of Calculus allows us to use antiderivatives to accomplish the same task. However, note that there are two requirements that the function must satisfy before we can apply the Fundamental Theorem.

- The interval over which we integrated, , must be a closed interval of finite length, and
- The function must be bounded on .

This says that the outputs of are trapped between the horizontal lines and ; the output values cannot become arbitrarily large in either the positive or negative sense on .

It turns out that there are many instances where these limitations are a problem. For instance, there are applications in statistics, physics, and engineering where we need to integrate over unbounded regions or we need to integrate unbounded functions. In this section, we will generalize the notion of the integral in such a way to overcome each of the restrictions above.

**improper integral**if one of, or both, of the conditions hold:

- The interval of integration is infinite.
- The function is unbounded on the interval of integration.

### Unbounded intervals

Consider the expression What does this expression mean? Let’s consider a particular example and see if we can make sense of it.

We interpret definite integrals as giving us the net area underneath the graph of the function over the given interval. If we used this interpretation here, this notation should mean that we want to find the area under from to . This idea needs to be more precisely defined. Let’s consider the definite integral

where is some fixed number.

This integral gives us the area underneath from to . If we wanted to make sense of the integral to , we could think of letting get larger and larger and seeing how this affects the area.

That is, we interpret the integral from 1 to as a limit of a definite integral:

We can interpret this result to mean that the area under from to is infinite.

It might seem like we should always get an answer of infinity or negative infinity if we integrate over an unbounded region, but the next example shows that is not always the case.

We calculate the definite integral

This gives the area under on the interval .

Now we take the limit of this area as .

In this case, when we take the limit, we get a finite number, namely . This suggests that we can consider the area under from to to be equal to despite the fact that the region is unbounded.

Thus in the two examples we have studied so far we have

and

Why do we get such different answers in these two cases? The difference lies in the speed with which each function approaches as . Although both functions approach as , the function becomes smaller much more rapidly than . We can see in the graphs above that shrinks down to zero much more quickly. The key idea is that the area that we are accumulating as is shrinking rapidly enough that the total area stays bounded. This is a theme we will see again when we study convergence of series.

We now give a precise definition for integrals over unbounded regions.

- Let be a continuous function on .
- Let be a continuous function on .
- Let be a continuous function on . Let be any real number.

An improper integral is said to **converge** if its corresponding limit exists
and is equal to a real number. Otherwise, the improper integral is said to
**diverge**.

Now we must look at each integral separately.

(your answer should be in terms of ) Similarly we obtain

(your answer should be in terms of )

Now we take the limits of each integral

Hence we have

In this example, both integrals converge and therefore the entire integral converges to .

**a single limit**? Since sine is an odd function on any bounded interval , we would have found this integral to be . However, We must compute

**two limits**to evaluate this integral correctly.

### Unbounded functions

We have just considered definite integrals where the interval of integration was unbounded. We now consider another type of improper integral, where the interval is finite but the function is unbounded on the interval.

Let’s begin with some examples.

How then do we work with integrals of unbounded functions? Can we define this integral in terms of integrals we know how to compute?

Suppose lies strictly between and . Then consider the definite integral

The function is now bounded on this smaller interval.

Now we can examine what will happen if we let begin to move towards from the right. What will happen to the area under the curve? Just as we did in the case of unbounded integrals, we can make sense of this integral by defining it as the limit of a definite integral

We can then evaluate our original improper integral:

Isn’t it quite surprising that the area bounded by a curve which has a vertical asymptote can be finite? Consider the graph of on the interval :

Essentially, the function approaches the vertical asymptote at “fast enough” so that only a finite amount of area is accumulated.

Now we look at another example:

earlier and saw that this integral converges. We remarked at the time this was due to how rapidly the function approached as .

However the integral

diverges. The issue is that the integral is approaching the vertical asymptote too slowly and thus the area builds up too quickly. This is due to the fact that the manner in which approaches the horizontal asymptote is different from the manner in which it approaches the vertical asymptote .

what would have happened if had we **not noticed the vertical asymptote** in the
integrand at ?

We probably would have blindly computed:

But the integrand is always positive, so this answer of is complete nonsense! The reason this error occurs is because this expression is not a definite integral. It only has meaning if we interpret it as a limit of definite integrals

**Be on the lookout for vertical asymptotes!**

We can now generalize the previous two examples to give a definition for such improper integrals of unbounded functions.

Let’s look at a couple more examples.

Since lies in the middle of the interval and the upper limit of integration is infinity, we need to break the integral into three pieces.

We chose for the upper and lower limit of the middle and last integral, respectively, but we could use any number greater than .

Since the integrand is a rational function, where the denominator is reducible, we could apply partial fractions decomposition to get

Now we find the antiderivative to get

Now we evaluate one integral at a time. Let’s start with

Thus this integral diverges, so the entire integral must diverge.