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You CAN Ace Calculus

17calculus > integrals > improper integrals

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Improper Integrals

When you initially learned integration, you were always given functions that were well-behaved in the sense that they were continuous over the interval you were integrating. Also, definite integrals were always between two finite values. Integration will work only under those conditions, i.e. you can't integrate over an interval that contains even one point where the function is discontinuous and, for definite integrals, the limits of integration must be finite.

However, under certain conditions we can use limits to be able to integrate over discontinuous functions or where the limits of integration are not finite. These types of integrals are called improper integrals. The way to handle these kinds of integrals is to convert the integral to a finite integral on a continuous finite interval, integrate and then take the limit of the result. We can handle only one type of discontinuity at a time, so we may end up with multiple integrals but, as long as each of them are continuous and finite, that shouldn't be a problem.

Before we get into the details, let's watch a quick overview video that will help get us started.

 Krista King Math - What makes an integral improper?! [4min-58secs]

The first thing we need to do is to determine the domain of the integrand and compare the domain with the limits of integration. Any points included in the limits of integration that are NOT in the domain are considered 'problem points'. Also, if one or either of the endpoints of is/are $$\pm$$infinity, they are also considered problems.

Let's start with the endpoints. The simplest case is where we have a continuous interval and one of the endpoints is infinite.

Special Improper Integral

$$\displaystyle{ \int_{1}^{\infty}{\frac{dx}{x^p}} = \left\{ \begin{array}{llr} \frac{1}{p-1} & & p \gt 1 \\ diverges & & p \leq 1 \end{array} \right. }$$

Case 1 - Infinite Interval

Case 1: Infinite Integral at One Endpoint - Problem and Example

$$\displaystyle{ \int_{a}^{\infty}{f(x) ~ dx} }$$

$$a$$ is finite

$$f(x)$$ is continuous on the interval $$[a,\infty)$$

$$\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2} ~ dx} }$$

$$\displaystyle{ \int_{-\infty}^{b}{g(x) ~ dx} }$$

$$b$$ is finite

$$g(x)$$ is continuous on the interval $$(-\infty,b]$$

$$\displaystyle{ \int_{-\infty}^{-1}{\frac{1}{x^2} ~ dx} }$$

Notes
1. We chose the function $$1/x^2$$ to demonstrate what we meant about being continuous on the interval of integration. Notice that as long as we stay away from $$x=0$$, the function $$f(x)=1/x^2$$ is continuous.
2. Also notice that we have a problem at only one endpoint in each integral. This is important, since we can handle only one problem at a time.
3. Using the techniques we know so far, we can't evaluate these integrals because it is required that the function be continuous and the definite integral can be evaluated only on a finite interval. So what do we do?

This is the case discussed in this video. He looks at the same function $$1/x^2$$ and shows how to evaluate it with limits. He uses a graph to explain really well what is going on with this type of improper integral.

 PatrickJMT - Improper Integral - Basic Idea and Example

So, now we can amend the table above with the equation showing how to evaluate the limit.

Case 1: Infinite Integral at One Endpoint - Solution

$$\displaystyle{ \int_{a}^{\infty}{f(x) ~ dx} }$$

$$a$$ is finite

$$f(x)$$ is continuous on the interval $$[a,\infty)$$

$$\displaystyle{\lim_{t \to \infty}{ \int_{a}^{t}{f(x) ~ dx} } }$$

$$t$$ is finite

$$\displaystyle{ \int_{-\infty}^{b}{g(x) ~ dx} }$$

$$b$$ is finite

$$g(x)$$ is continuous on the interval $$(-\infty,b]$$

$$\displaystyle{\lim_{t \to -\infty}{ \int_{t}^{b}{g(x) ~ dx} } }$$

$$t$$ is finite

This is an important starting point since the rest of the cases rely on your understanding of this first case. Take a few minutes to work these practice problems before you go on.

 Basic Problems

Practice 1

$$\displaystyle{\int_{1}^{\infty}{x^2~dx}}$$

solution

Practice 2

$$\displaystyle{\int_{1}^{\infty}{\frac{1}{x}dx}}$$

solution

Practice 3

$$\displaystyle{\int_{2}^{\infty}{\frac{dx}{x^5}}}$$

solution

Practice 4

$$\displaystyle{\int_{-\infty}^{0}{e^{0.1x}dx}}$$

solution

Practice 5

$$\displaystyle{\int_{1}^{\infty}{2x^{-2}~dx}}$$

solution

Practice 6

$$\displaystyle{\int_{0}^{\infty}{4e^{-2x}dx}}$$

solution

Practice 7

$$\displaystyle{\int_{0}^{\infty}{6e^{-3x}dx}}$$

solution

Practice 8

$$\displaystyle{\int_{0}^{\infty}{\frac{1}{2x+5}dx}}$$

solution

Practice 9

$$\displaystyle{\int_{-\infty}^{-3}{\frac{2}{x^2}dx}}$$

solution

Practice 10

$$\displaystyle{\int_{0}^{\infty}{0.5e^xdx}}$$

solution

Practice 11

$$\displaystyle{\int_{1}^{\infty}{\frac{8}{x^4}dx}}$$

solution

Practice 12

$$\displaystyle{\int_{-\infty}^{0}{\frac{1}{3-4x}dx}}$$

solution

 Intermediate Problems

Practice 13

$$\displaystyle{\int_{0}^{\infty}{x e^{-3x}dx}}$$

solution

Practice 14

$$\displaystyle{\int_{0}^{\infty}{se^{-5s}~ds}}$$

solution

Practice 15

$$\displaystyle{\int_{-\infty}^{2}{\frac{3}{x^2+1}dx}}$$

solution

Practice 16

$$\displaystyle{\int_{0}^{\infty}{\cos(x)~dx}}$$

solution

So, you should now be comfortable evaluating integrals where the integrand is continuous on the entire interval of integration and with one of the endpoints going to infinity or negative infinity. But what do you do if both limits of integration contain infinity, i.e. the lower limit is negative infinity AND the upper limit is infinity. The trick is to break the integral at a convenient point and then you have two integrals. For example, if we have $$\displaystyle{ \int_{-\infty}^{\infty}{ x ~ dx} }$$, we can break the integral at any point $$a$$ to get integrals.
$$\displaystyle{ \int_{-\infty}^{\infty}{ x ~ dx} = }$$ $$\displaystyle{ \int_{-\infty}^{0}{ x } + }$$ $$\displaystyle{ \int_{0}^{\infty}{ x ~ dx} }$$
We chose zero in the above example but you can choose any point or even use $$a$$. Then we use the technique we learned above to evaluate each integral, adding the results together to get our final answer.

Here are some practice problems to try.

Practice 17

$$\displaystyle{\int_{-\infty}^{\infty}{-3x^4~dx}}$$

solution

Practice 18

$$\displaystyle{\int_{-\infty}^{\infty}{x^2e^{-x^3}~dx}}$$

solution

Practice 19

$$\displaystyle{\int_{-\infty}^{\infty}{\frac{1}{1+x^2}dx}}$$

solution

Practice 20

$$\displaystyle{\ \int_{-\infty}^{\infty}{\frac{x^2}{9+x^6}dx} }$$

solution

Case 2 - Discontinuity at an Endpoint

This case occurs when we have a continuous function within a finite interval but the function is discontinuous at one of the endpoints. For example, we might have an integral like $$\displaystyle{ \int_{0}^{1}{ \frac{1}{x} ~ dx} }$$. In this case, $$f(x)=1/x$$ is discontinuous at one of the endpoints, $$x=0$$. So we can't integrate. However, we can use the same technique above that we used for an infinite interval and use limits. It looks something like this.

Case 2: Finite Integral with a Discontinuity at One Endpoint

$$\displaystyle{ \int_{a}^{b}{f(x) ~ dx} }$$

$$a$$ and $$b$$ are finite

$$f(x)$$ is continuous on the interval $$[a,b)$$ but discontinuous at $$x=b$$

$$\displaystyle{\lim_{t \to b}{ \int_{a}^{t}{f(x) ~ dx} } }$$

$$\displaystyle{ \int_{a}^{b}{g(x) ~ dx} }$$

$$a$$ and $$b$$ are finite

$$g(x)$$ is continuous on the interval $$(a,b]$$ but discontinuous at $$x=a$$

$$\displaystyle{\lim_{t \to a}{ \int_{t}^{b}{g(x) ~ dx} } }$$

Notice that we do the same thing as in Case 1, i.e. we replace the 'problem' point with a limit. Okay, try these practice problems before going on.

Practice 21

$$\displaystyle{\int_{-1}^{0}{\frac{1}{x^2}dx}}$$

solution

Practice 22

$$\displaystyle{\int_{0}^{6}{\frac{1}{x^{0.8}}dx}}$$

solution

Practice 23

$$\displaystyle{\int_{0}^{3}{\frac{1}{x^{1.2}}dx}}$$

solution

Practice 24

$$\displaystyle{\int_{5}^{8}{\frac{2}{\sqrt[3]{x-5}}dx}}$$

solution

Practice 25

$$\displaystyle{\int_{0}^{1}{\frac{\ln(x)}{x^{1/2}}dx}}$$

solution

So far, so good. Now, there is only one minor twist to these two cases that is easily handled. If you have a 'problem' (a discontinuity or infinity) at both ends, you just break the integral into two integrals at a convenient point. We know from basic integration that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = }$$ $$\displaystyle{ \int_{a}^{k}{f(x)~dx} + \int_{k}^{b}{f(x)~dx} }$$, right? So we just do the same thing with the improper integral and handle each case and limit separately. For example, we might want to evaluate $$\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} }$$, so we just split the integral into two and handle each end separately. We can break it anywhere in the interval, I will choose $$2$$. We end up with

$$\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} = }$$ $$\displaystyle{ \int_{0}^{2}{\frac{1}{x}~dx} + \int_{2}^{\infty}{\frac{1}{x}~dx} = }$$ $$\displaystyle{ \lim_{a \to 0}{\int_{a}^{2}{\frac{1}{x}~dx} } + \lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x}~dx} } }$$

Case 3 - Discontinuity Inside the Interval

Cases 1 and 2 above were very similar and dealing with them used the similar technique of taking the limit of one of ends of the interval. Now, Case 3 may seem very different but handling it is very straightforward and, if you understand Cases 1 and 2, you will be able to deal with Case 3 easily. Case 3 involves discontinuities INSIDE the interval, i.e. not at one of the endpoints or limits of integration. To handle this case, you just break the integral again (like you did in Case 2) but this time you break it AT THE POINT OF DISCONTINUITY (not at just any point). When you do, you end up with something like this.

Case 3: One Discontinuity Inside the Interval

$$\displaystyle{ \int_{a}^{b}{f(x) ~ dx} }$$

$$a$$ and $$b$$ are finite

$$f(x)$$ is continuous on the interval $$[a,b]$$ but discontinuous at $$x=c$$ which is somewhere inside the interval $$(a,b)$$

$$\displaystyle{ \int_{a}^{b}{f(x) ~ dx} = }$$ $$\displaystyle{ \int_{a}^{c}{f(x) ~ dx} + \int_{c}^{b}{f(x) ~ dx} }$$

Since $$f(x)$$ is discontinuous at $$x=c$$, you now have two integrals that fit Case 2. So you can use those techniques to evaluate the integrals. Pretty cool, eh? An example might be $$\displaystyle{ \int_{-1}^{1}{1/x~dx} = \int_{-1}^{0}{1/x~dx} + \int_{0}^{1}{1/x~dx} }$$.

Another minor twist might be that you have an infinite limit at one or both ends AND a discontinuity in the interval. In this case, break the integral at the discontinuity and drop back to Cases 1 and 2 until you have only one 'problem' in each integral. You probably also know what to do with multiple discontinuities, right? Just break the integral at each discontinuity.

Okay, try these practice problems before going on.

Practice 26

$$\displaystyle{\int_{-1}^{1}{\frac{e^x}{e^x-1}dx}}$$

solution

Practice 27

$$\displaystyle{\int_{-1}^{1}{x^{-2/3}dx}}$$

solution

Improper Integral Keys

There are several keys to handling improper integrals.
1. We can handle only one 'problem' per integral.
2. The 'problem' must occur at one of the limits of integration or we need to break the integral to get it at one of the limits of integration.
3. Use a limit to set up the integral so that it is finite and continuous and then deal with the discontinuity after the integral is evaluated.

In summary, in all three cases, we use limits to rewrite the integral. We can do this because of the Limit Key. We can also have a combination of cases, in which case we rewrite the definite integral by breaking it into pieces so that we have only one case to handle for each integral.

How To Apply All These Techniques

Okay, so you are given an integral. How do you know where to start with all these techniques? Here is a suggested list of steps. This may seem like a lot of steps but some may not be required depending on your integral. These steps cover the most involved integration possible.

 1. Make a list of all the 'problem' points including infinite limits of integration and discontinuities. 2. Discard any discontinuities that are outside the limits of integration. 3. List the 'problem' points left to right (or top to bottom) in increasing order with the endpoints of the integration at each end (the endpoints may or may not be 'problems'; so keep track of that information ). 4. Break the integral at each discontinuity. 5. Break the integral into pieces between discontinuities so that each integral has only one 'problem' point at one of the ends of the interval. 6. Set up each integral individually using limits. (It is best to use different limit variables for each integral but many instructors do not require it.) 7. Evaluate the integral once in a separate area on your paper. 8. Use your result from step 7 in each limit and evaluate them separately. 9. Add your answers together. In this step, you may get one or several indeterminate forms. If you do, drop back to step 8 and determine where they come from. Combine those results before taking the limit. This is something you will get a feel for as you get experience.

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