## 17Calculus - Improper Integrals - Multiple Discontinuities

##### 17Calculus

On this page, we put together everything you learned on the previous 4 pages on improper integrals. We show you how to evaluate integrals with multiple discontinuities and possibly infinite intervals.

These are the 4 pages that you should know before studying the material on this page.

How To Apply The Techniques to Multiple Discontinuities

So, those are the only discontinuities that you had to deal with. However, it helps to have a set of logical steps to follow to make sure you have all possible cases covered. Here are our suggestions.
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, which include infinite limits of integration and discontinuities, points that are not in the domain of the integrand. 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. Note - - In this last step, you may get an indeterminate form $$\infty - \infty$$. This requires special handling discussed in the indeterminate differences section on the indeterminate forms page.

Here is a detailed example.

Set up the integrals and limits equivalent to $$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }$$ so that we can apply the Fundamental Theorem of Calculus, but do not evaluate the integrals.

Problem Statement

Set up the integrals and limits equivalent to $$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }$$ so that we can apply the Fundamental Theorem of Calculus, but do not evaluate the integrals.

Solution

 Problem Statement

Set up the integrals and limits equivalent to $$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{dx}{x(x+1)^9} } }$$ so that we can apply the Fundamental Theorem of Calculus, but do not evaluate the integrals.

 Solution

First we need to find the domain of the integrand. The problem points occur where the denominator is zero, i.e. at $$x=0$$ and $$x=-1$$. So the domain is all reals except for these two points.
The upper and lower limits of integration are infinite, so those are problems too. Here is a list of problem points, in numerical order.

 $$-\infty$$ $$x=-1$$ $$x=0$$ $$+\infty$$

Now we compare the discontinuities against the interval of integration, $$(-\infty,+\infty)$$. Both $$x=-1$$ and $$x=0$$ are inside this interval, so we need to keep both of them. Now we can start setting up the integrals.

First we handle case 3, discontinuities inside the interval. We break the integral at each discontinuity. To make these easier to write, we will write the integrand as $$f(x)$$, i.e. $$\displaystyle{ f(x)=\frac{1}{x(x+1)^9} }$$

$$\displaystyle{ \int_{-\infty}^{-1}{ f(x)~dx } + \int_{-1}^{0}{ f(x)~dx } + \int_{0}^{\infty}{ f(x)~dx } }$$

Okay, so at this point all the problem points are at the ends of integration limits and, within the limits of integration for each integral, the function is continuous. Now we will handle the end points of each integral separately.

 $$\displaystyle{ \int_{-\infty}^{-1}{ f(x)~dx } }$$ Since both endpoints are problem points, we need to break the integral into two integrals at any point between $$-\infty$$ and $$-1$$. We choose $$-5$$. $$\displaystyle{ \int_{-\infty}^{-5}{ f(x)~dx } + \int_{-5}^{-1}{ f(x)~dx } }$$ Now each integral has only one problem point.

Repeat the same procedure on the other two integrals.

 $$\displaystyle{ \int_{-1}^{0}{ f(x)~dx } = \int_{-1}^{-1/2}{ f(x)~dx } + \int_{-1/2}^{0}{ f(x)~dx } }$$
 $$\displaystyle{ \int_{0}^{\infty}{ f(x)~dx } = \int_{0}^{17}{ f(x)~dx } + \int_{17}^{\infty}{ f(x)~dx } }$$

Let\'s put our results together to see what we have at this point. $$\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \int_{-\infty}^{-5}{ f(x)~dx } + \int_{-5}^{-1}{ f(x)~dx } + \int_{-1}^{-1/2}{ f(x)~dx } + \int_{-1/2}^{0}{ f(x)~dx } + \int_{0}^{17}{ f(x)~dx } + \int_{17}^{\infty}{ f(x)~dx } }$$
We have six integrals with each integral having only one problem point and each problem point occurs at one of the limits of integration. We are now ready to set up limits as follows.

$$\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \lim_{a\to-\infty}{ \int_{a}^{-5}{ f(x)~dx } } + \lim_{b\to-1^-}{ \int_{-5}^{b}{ f(x)~dx } } + \lim_{c\to-1^+}{ \int_{c}^{-1/2}{ f(x)~dx } } + \lim_{d\to0^-}{ \int_{-1/2}^{d}{ f(x)~dx } } + \lim_{y\to0^+}{ \int_{y}^{17}{ f(x)~dx } } + \lim_{z\to\infty}{ \int_{17}^{z}{ f(x)~dx } } }$$
A few comments are in order.
1. Notice that we used different variables in each limit. This may or may not be required by your instructor but it is highly recommended by us since it will keep you from getting confused when evaluating the limits.
2. We have left $$f(x)$$ in the answer, but your instructor may require you to write out the integrand within each limit. However, as an instructor, I would be okay with this answer as long as you also write what $$f(x)$$ was equal to along with your answer as shown below.
3. As we mentioned in the solution, you may choose values other than $$-5$$, $$-1/2$$ and $$17$$. As long as you meet the requirement that the point is within each limit of integration, your answer is correct.
4. As always, check with your instructor to see what they require.

$$\displaystyle{ \int_{-\infty}^{\infty}{ f(x)~dx } = \lim_{a\to-\infty}{ \int_{a}^{-5}{ f(x)~dx } } + \lim_{b\to-1^-}{ \int_{-5}^{b}{ f(x)~dx } } + \lim_{c\to-1^+}{ \int_{c}^{-1/2}{ f(x)~dx } } + \lim_{d\to0^-}{ \int_{-1/2}^{d}{ f(x)~dx } } + \lim_{y\to0^+}{ \int_{y}^{17}{ f(x)~dx } } + \lim_{z\to\infty}{ \int_{17}^{z}{ f(x)~dx } } }$$

where $$\displaystyle{ f(x)=\frac{1}{x(x+1)^9} }$$

Okay, let's work some practice problems that may require multiple techniques. These are the types of problems that you will most likely see on your exam.

Practice

Unless otherwise instructed, evaluate these integrals giving your answers in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Basic

$$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$\displaystyle{ \frac{1}{2048} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

For this integral, there is only one problem point, the upper limit of integration. The discontinuity at $$x=-1$$ can be ignored since it is not within the limits of integration. So we have $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{(x+1)^9} } = \lim_{b\to\infty}{ \int_{1}^{b}{ \frac{dx}{(x+1)^9} } } }$$.
We will now integrate the indefinite integral $$\displaystyle{ \int{ \frac{dx}{(x+1)^9} } }$$ and plug the result back into the definite integral and limit.

 $$\displaystyle{ \int{ \frac{dx}{(x+1)^9} } }$$ Use integration by substitution. $$u = x+1 \to du = dx$$ $$\displaystyle{ \int{ \frac{du}{u^9} } }$$ $$\displaystyle{ \int{ u^{-9} ~du } }$$ $$\displaystyle{ \frac{u^{-8}}{-8} }$$ $$\displaystyle{ \frac{-1}{8(x+1)^8} }$$

Plugging this result back into the limit, we have

 $$\displaystyle{ \lim_{b\to\infty}{ \left[ \left. \frac{-1}{8(x+1)^8} \right|_1^b \right] } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{8(b+1)^8} + \frac{1}{8(2)^8} \right] } }$$ $$\displaystyle{ 0 + \frac{1}{2^{11}} = \frac{1}{2048} }$$

$$\displaystyle{ \frac{1}{2048} }$$

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

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x-1}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

The integral diverges.

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x-1}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

Looking at the limits of integration and comparing them with the integrand, it should be obvious that we have a problem at both ends. We have infinity at the upper limit and a discontinuity at the lower limit. However, the integrand is continuous in the interval $$(1,\infty)$$. So the first step is to break the integral into two integrals at any point inside the interval. This will give us two integrals with only one problem point each, allowing us to write limits for both. We choose to break the interval at 17.

 $$\displaystyle{ \int_{1}^{17}{ \frac{dx}{\sqrt{x-1}} } + \int_{17}^{\infty}{ \frac{dx}{\sqrt{x-1}} } }$$ $$\displaystyle{ \lim_{a\to1^+}{ \int_{a}^{17}{ \frac{dx}{\sqrt{x-1}} } } + \lim_{b\to\infty}{ \int_{17}^{b}{ \frac{dx}{\sqrt{x-1}} } } }$$

Now we will evaluate the indefinite integral.

 $$\displaystyle{ \int{ \frac{dx}{\sqrt{x-1}} } }$$ $$\int{ (x-1)^{-1/3}~dx }$$ $$\displaystyle{ \frac{(x-1)^{2/3}}{2/3} }$$ $$\displaystyle{ \frac{3}{2}(x-1)^{2/3} }$$

Next, we plug this result back into the both limits and evaluate.

 $$\displaystyle{ \lim_{a\to1^+}{ \left[ \frac{3}{2}(x-1)^{2/3} \right]_a^{17} } + \lim_{b\to\infty}{ \left[ \frac{3}{2}(x-1)^{2/3} \right]_{17}^{b} } }$$ $$\displaystyle{ \frac{3}{2} \lim_{a\to1^+}{ \left[ (17-1)^{2/3} - (a-1)^{2/3} \right] } + }$$ $$\displaystyle{ \frac{3}{2} \lim_{b\to\infty}{ \left[ (b-1)^{2/3} - (17-1)^{2/3} \right] } }$$ $$\displaystyle{ \frac{3}{2} \left[ (16)^{2/3} - 0 \right] + \frac{3}{2} \left[ \infty - (16)^{2/3} \right] = \infty }$$

The integral diverges.

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

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{x^2(x+1)} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$1 - \ln 2$$

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{x^2(x+1)} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

Looking at the limits of integration, we can see that the upper limit is a problem since it goes to infinity. Before we jump in, let\'s look at the domain of the integrand. The domain is all real number except for the values that make the denominator zero, in this case, $$x=0$$ and $$x=-1$$. So those points are potential problems as well. However, comparing these points to the limits of integration shows us that these points are outside the integration interval and, therefore, do not impact our problem. So even though it initially looked like there were several problem points, it turns out there is only one, the upper limit.
So now we will rewrite the integral with a limit.
$$\displaystyle{ \lim_{b\to\infty}{ \int_{1}^{b}{ \frac{dx}{x^2(x+1)} } } }$$
Now we will evaluate the indefinite integral.

 $$\displaystyle{ \int{ \frac{dx}{x^2(x+1)} } }$$ Use partial fractions to expand the integrand. $$\displaystyle{ \int{ \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{x+1} ~ dx } }$$ $$\displaystyle{ -\ln|x| + \frac{x^{-1}}{-1} + \ln|x+1| }$$

Plug this back into the limit.

 $$\displaystyle{ \lim_{b\to\infty}{ \left[ -\ln|x| - 1/x + \ln|x+1| \right]_1^b } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \left[ -\ln|b| - 1/b + \ln|b+1| \right] - \left[ -\ln|1| - 1/1 + \ln|1+1| \right] } }$$ $$\displaystyle{ \left[ -\infty - 0 + \infty \right] - \left[ -0 - 1 + \ln 2 \right] }$$

In the first set of brackets we have $$\infty-\infty$$, which is indeterminate. So we need to determine where these came from and evaluate the limit differently. We have
$$\displaystyle{ \lim_{b\to\infty}{ [ -\ln|b| + \ln|b+1| ] } }$$
Combining the natural log terms, we have
$$\displaystyle{ \lim_{b\to\infty}{ \ln\left| \frac{b+1}{b} \right| } }$$
$$\displaystyle{ \lim_{b\to\infty}{ \ln\left| 1 + \frac{1}{b} \right| } = 0 }$$
So that leaves us with $$-[-1+\ln 2] = 1 - \ln 2$$

$$1 - \ln 2$$

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$$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{1}{x^2+4} ~ dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{1}{x^2+4} ~ dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$\pi/2$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-\infty}^{\infty}{ \frac{1}{x^2+4} ~ dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

### Michael Penn - 3886 video solution

video by Michael Penn

$$\pi/2$$

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Intermediate

$$\displaystyle{ \int_{-\infty}^{\infty}{ \arctan \left[ \frac{1}{2x^2} \right] ~dx} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-\infty}^{\infty}{ \arctan \left[ \frac{1}{2x^2} \right] ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$\pi$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-\infty}^{\infty}{ \arctan \left[ \frac{1}{2x^2} \right] ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

He does not use correct notation when evaluating the integral. He needs to convert to limit notation BEFORE evaluating. Otherwise this is a good solution.

### Michael Penn - 3857 video solution

video by Michael Penn

$$\pi$$

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$$\displaystyle{ \int_{0}^{\infty}{ \arctan^2(1/x) ~dx} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\infty}{ \arctan^2(1/x) ~dx} }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$\pi \ln 2$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\infty}{ \arctan^2(1/x) ~dx} }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

### Michael Penn - 3858 video solution

video by Michael Penn

$$\pi \ln 2$$

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$$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\cos x) ~dx} }$$ and $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\sin x) ~dx} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\cos x) ~dx} }$$ and $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\sin x) ~dx} }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Hint

Both integrals are solved the same way with the same answer.
Use the initial substitution $$u = \pi/2 - x$$ and get an integrand with $$\ln(\sin x)$$ for the first integral or $$\ln(\cos x)$$ for the second integral. Then combine the integrals. You do not need to get a closed form for the indefinite integral to solve this problem. The idea is similar to the technique you might use for integration by parts involving sine and/or cosine.

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\cos x) ~dx} }$$ and $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\sin x) ~dx} }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$- ( \pi / 2 ) \ln 2$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\cos x) ~dx} }$$ and $$\displaystyle{ \int_{0}^{\pi/2}{ \ln(\sin x) ~dx} }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Hint

Both integrals are solved the same way with the same answer.
Use the initial substitution $$u = \pi/2 - x$$ and get an integrand with $$\ln(\sin x)$$ for the first integral or $$\ln(\cos x)$$ for the second integral. Then combine the integrals. You do not need to get a closed form for the indefinite integral to solve this problem. The idea is similar to the technique you might use for integration by parts involving sine and/or cosine.

Solution

Here is a written out solution of the first integral on twitter.
Here are three videos by three different instructors. The first two solve the first integral. The last video solves the second integral.

### Michael Penn - 3859 video solution

video by Michael Penn

### n choose k - 3859 video solution

video by n choose k

$$- ( \pi / 2 ) \ln 2$$

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$$\displaystyle{ \int_0^{\infty}{ \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)} ~ dx} }$$

Problem Statement

$$\displaystyle{ \int_0^{\infty}{ \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)} ~ dx} }$$

Hint

This solution requires several substitutions in the following order.
1. $$y=2^x-1$$
2. $$y=t^2$$
3. Now notice that there is an issue at $$t=1$$. So break the integral into two integrals. For the first integral just change the variable of integration to $$u$$ and for the second let $$t=1/u$$.
4. Now combine the integrals and use logarithm rules to simplify the fraction.
You should now have an integral that looks like $$\int{1/(u^2+1)~dx}$$ with a constant out front.

Problem Statement

$$\displaystyle{ \int_0^{\infty}{ \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)} ~ dx} }$$

$$\displaystyle{ \frac{\pi}{2(\ln 2)^2} }$$

Problem Statement

$$\displaystyle{ \int_0^{\infty}{ \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)} ~ dx} }$$

Hint

This solution requires several substitutions in the following order.
1. $$y=2^x-1$$
2. $$y=t^2$$
3. Now notice that there is an issue at $$t=1$$. So break the integral into two integrals. For the first integral just change the variable of integration to $$u$$ and for the second let $$t=1/u$$.
4. Now combine the integrals and use logarithm rules to simplify the fraction.
You should now have an integral that looks like $$\int{1/(u^2+1)~dx}$$ with a constant out front.

Solution

### Michael Penn - 4128 video solution

video by Michael Penn

$$\displaystyle{ \frac{\pi}{2(\ln 2)^2} }$$

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