Section 7.1 Improper Integrals

We’re going to think a bit about integration with a twist: what happens when our “definite” integrals can’t actually be evaluated? First, let’s try to sink ourselves back into the context we’ve been in for a while now: what kinds of problems have we encountered so far, and how do we use our calculus intuition to get around those problems?

Activity 7.1.1 Remembering a Theme so Far.

(a)

Let’s say that we want to find what the \(y\)-values of some function \(f(x)\) are when the \(x\)-values are “infinitely close to” some value, \(x=a\text{.}\) Since there is no single \(x\)-value that is “infinitely close to” \(a\) that we can evaluate \(f(x)\) at, we need to do something else. How do we do this?

(b)

Let’s say that we want to find the rate of change of some function instantaneously at a point with \(x=a\text{.}\) We can’t find a rate of change unless we have two points, since we need to find some differences in the outputs and inputs. How do we do this?

(c)

Suppose you want to find the total area, covered by an infinite number of infinitely thin rectangles. You have a formula for finding the dimensions and areas for some finite number of rectangles, but how do we get an infinite number of them?

(d)

Can you find the common calculus theme in each of these scenarios?

So moving forward, we want to remember how we typically have solved these problems. Now, let’s try to identify the types of problems with integrals that we need to figure our way around.

Activity 7.1.2 Remembering the Fundamental Theorem of Calculus.

We want to think about generalizing our notion of integrals a bit. So in this activity, section, we’re going to think about some of the requirements for the Fundamental Theorem of Calculus and try to loosen them up a bit to see what happens. We’ll try to construct meaningful approaches to these situations that fit our overall goals of calculating area under a curve.

This practice, in general, is a really good and common mathematical process: taking some result and playing with the requirements or assumptions to see what else can happen. So it might feel like we’re just fiddling with the “What if?” questions, but what we’re actually doing is good mathematics!

(a)

What does the Fundamental Theorem of Calculus say about evaluating the definite integral \(\displaystyle \int_{x=a}^{x=b} f(x)\, dx\text{?}\)

(b)

What do we need to be true about our setup, our function, etc. for us to be able to apply this technique to evaluate \(\displaystyle \int_{x=a}^{x=b} f(x)\,dx\text{?}\)

We are going to introduce the idea of “Improper Integrals” as kind-of-but-not-quite definite integrals that we can evaluate. They are going to violate the requirements for the Fundamental Theorem of Calculus, but we’ll work to salvage them in meaningful ways.

This should build a pretty good idea of a new “class” of integrals: ones that aren’t quite definite integrals that we can evaluate with the Fundamental Theorem of Calculus, but ones that we can use limits to get at.

Subsection Improper Integrals

Definition 7.1.3 Improper Integral.

An integral is an improper integral if it is an extension of a definite integral whose integrand or limits of integration violate a requirement in one of two ways:

The interval that we integrate the function over is unbounded in width, or infinitely wide.
The integrand is unbounded in height, or infinitely tall, somewhere on the interval that we integrate over.

With this definition, we can think about the strategies that we got from Activity 7.1.1: we’re going to identify the problems in our integral (infinite width of the interval or infinite height of the integrand function) and use a limit!

Before we formalize that, though, let’s try to think about how this works by being really explicit about what this limit actually is doing.

Activity 7.1.4 Approximating Improper Integrals.

In this activity, we’re going to look at two improper integrals:

\(\displaystyle \displaystyle \int_{x=2}^{\infty} \frac{1}{(x+1)^2}\;dx\)
\(\displaystyle \displaystyle \int_{x=-1}^{x=2} \frac{1}{(x+1)^2}\;dx\)

(a)

First, let’s just clarify to ourselves what it means for an integral to be improper. Why are each of these integrals improper? Be specific!

(b)

Let’s focus on \(\displaystyle \int_{x=2}^{\infty} \frac{1}{(x+1)^2}\;dx\) first. We’re going to look at the slightly different integral:

\begin{equation*} \int_{x=2}^{x=t} \frac{1}{(x+1)^2}\;dx\text{.} \end{equation*}

As long as \(t\) is some real number with \(t\gt 2\text{,}\) then our function is continuous and bounded on \([2,t]\text{,}\) and so we can evaluate this integral:

\begin{equation*} \int_{x=2}^{x=t} \frac{1}{(x+1)^2}\;dx = F(t)-F(2) \end{equation*}

where \(F(x)\) is an antiderivative of \(f(x)=\dfrac{1}{(x+1)^2}\text{.}\)

Find and antiderivative, \(F(x)\text{.}\)

Hint 1.

You’ll need to use \(u\)-substitution, with \(u=(x+1)\) and \(du=dx\text{.}\)

Hint 2.

You should be thinking about finding an antiderivative using an indefinite integral: \(\displaystyle\int \frac{1}{u^2}\;du = \int u^{-2}\;du\text{.}\)

(c)

Now we’re going to evaluate some areas for different values of \(t\text{.}\) Use your antiderivative \(F(x)\) from above!

Let’s start with making \(t=99\text{.}\) So we’re going to evaluate:

\begin{equation*} \int_{x=2}^{x=99} \frac{1}{(x+1)^2}\;dx = F(99)-F(2) \end{equation*}
Now let \(t=999\text{.}\) Evaluate:

\begin{equation*} \int_{x=2}^{x=999} \frac{1}{(x+1)^2}\;dx = F(999)-F(2) \end{equation*}
Now let \(t=9999\text{.}\) Evaluate:

\begin{equation*} \int_{x=2}^{x=9999} \frac{1}{(x+1)^2}\;dx = F(9999)-F(2) \end{equation*}

(d)

Based on what you found, what do you think is happening when \(t\to \infty\) to the definite integral

\begin{equation*} \int_{x=2}^{x=t} \frac{1}{(x+1)^2}\;dx = F(t)-F(2)\text{?} \end{equation*}

(e)

Ok, we’re going to switch our focus to the other improper integral,\(\displaystyle \int_{x=-1}^{x=2} \frac{1}{(x+1)^2}\;dx\text{.}\) again, we’ll look at a slightly different integral:

\begin{equation*} \int_{x=t}^{x=2} \frac{1}{(x+1)^2}\;dx\text{.} \end{equation*}

As long as \(t\) is some real number with \(-1\lt t \lt 2\text{,}\) then our function is continuous and bounded on \([t,2]\text{,}\) and so we can evaluate this integral:

\begin{equation*} \int_{x=t}^{x=2} \frac{1}{(x+1)^2}\;dx = F(2)-F(t) \end{equation*}

where \(F(x)\) is an antiderivative of \(f(x)=\dfrac{1}{(x+1)^2}\text{.}\) We can use the same antiderivative as before!

We’re going to evaluate this intergal for different values of \(t\) again, but this time we’ll use values that are close to \(-1\text{,}\) but slightly bigger, since we want to be in the interval \([-1,2]\text{.}\)

Let’s start with making \(t=-\frac{9}{10}\text{.}\) So we’re going to evaluate:

\begin{equation*} \int_{x=-9/10}^{x=2} \frac{1}{(x+1)^2}\;dx = F(2)-F\left(-\frac{9}{10}\right) \end{equation*}
Now let \(t=-\frac{99}{100}\text{.}\) Evaluate:

\begin{equation*} \int_{x=-99/100}^{x=2} \frac{1}{(x+1)^2}\;dx = F(2)-F\left(-\frac{99}{100}\right) \end{equation*}
Now let \(t=-\frac{999}{1000}\text{.}\) Evaluate:

\begin{equation*} \int_{x=-999/1000}^{x=2} \frac{1}{(x+1)^2}\;dx = F(2)-F\left(-\frac{999}{1000}\right) \end{equation*}

(f)

Based on what you found, what do you think is happening when \(t\to -1^+\) to the definite integral

\begin{equation*} \int_{x=t}^{x=2} \frac{1}{(x+1)^2}\;dx = F(2)-F(t)\text{?} \end{equation*}

We can think about putting this a bit more generally into limit notation, but we’ll get to this later.

Ok, let’s formalize these limits with some strategies for evaluating improper integrals!

Subsection Strategies for Evaluating Improper Integrals

Evaluating Improper Integrals (Infinite Width).

For a function \(f(x)\) that is continuous on \([a,\infty)\text{,}\) we can evaluate the improper integral \(\displaystyle\int_{x=a}^\infty f(x)\;dx\text{:}\)

\begin{equation*} \int_{x=a}^\infty f(x)\;dx = \lim_{t\to\infty} \int_{x=a}^{x=t} f(x)\;dx\text{.} \end{equation*}

If \(f(x)\) is continuous on \((-\infty,b]\text{,}\) we can evaluate the improper integral \(\displaystyle\int_{-\infty}^{x=b} f(x)\;dx\text{:}\)

\begin{equation*} \int_{-\infty}^{x=b} f(x)\;dx = \lim_{t\to -\infty} \int_{x=t}^{x=b} f(x)\;dx\text{.} \end{equation*}

Finally, if \(f(x)\) is continuous on \((-\infty,\infty)\) and \(m\) is some real number, then we can evaluate the improper integral \(\displaystyle\int_{-\infty}^{\infty}f(x)\;dx\text{:}\)

\begin{align*} \int_{-\infty}^{\infty}f(x)\;dx \amp = \int_{-\infty}^{x=m} f(x)\;dx + \int_{x=m}^{\infty} f(x)\;dx\\ \amp = \lim_{t\to -\infty} \int_{x=t}^{x=m} f(x)\;dx + \lim_{t\to\infty} \int_{x=m}^{x=t} f(x)\;dx \end{align*}

Example 7.1.5.

Evaluate the improper integral \(\displaystyle \int_{x=2}^{\infty}\frac{1}{(x+1)^2}dx\) by evaluating the limit:

\begin{equation*} \lim_{t\to\infty} \int_{x=2}^{x=t} \frac{1}{(x+1)^2}dx = \lim_{t\to\infty} \left(F(t)-F(2)\right)\text{.} \end{equation*}

Try to interpret this limit. What does it mean if this limit doesn’t exist? What does it mean if the limit does exist? What does the actual number represent?

Evaluating Improper Integrals (Infinite Height).

For a function \(f(x)\) that has an unbounded discontinuity (a vertical asymptote) at \(x=m\) with \(a\lt m\lt b\text{,}\) but is otherwise continuous on \([a,b]\text{,}\) then we can evaluate the improper integrals:

\begin{align*} \int_{x=m}^{x=b} f(x)\;dx \amp = \lim_{t\to m^+}\int_{x=t}^{x=b} f(x)\;dx \\ \int_{x=a}^{x=m} f(x)\;dx \amp = \lim_{t\to m^-} \int_{x=a}^{x=t} f(x)\;dx\\ \int_{x=a}^{x=b} f(x)\;dx \amp = \int_{x=a}^{x=m} f(x)\;dx + \int_{x=m}^{x=b} f(x)\;dx \\ \amp = \lim_{t\to m^-} \int_{x=a}^{x=t} f(x)\;dx + \lim_{t\to m^+}\int_{x=t}^{x=b} f(x)\;dx \end{align*}

Example 7.1.6.

Evaluate the improper integral \(\displaystyle \int_{x=-1}^{x=2}\frac{1}{(x+1)^2}dx\) by evaluating the limit:

\begin{equation*} \lim_{t\to -1^{+}} \int_{x=t}^{x=2} \frac{1}{(x+1)^2}dx = \lim_{t\to -1^+} \left(F(2)-F(t)\right)\text{.} \end{equation*}

Try to interpret this limit. What does it mean if this limit doesn’t exist? What does it mean if the limit does exist? What does the actual number represent?

Ok, let’s note that we can classify these improper integrals into two categories. We have already classified them based on the reason that they’re improper. But now we also can classify them based on the outcome of the limit:

Improper integrals (of any type) whose limit exists.
Improper integrals (of any type) where the limit doesn’t exist.

Let’s define a term for this, so that our classification isn’t so wordy.

Subsection Convergence and Divergence of an Improper Integral

Definition 7.1.7 Convergence of an Improper Integral.

We say that an improper integral converges if the limit of the appropriate definite integral exists. If the limit does not exist, then we say that the improper integral diverges.

All we’ve done here is added some language: we’ll say that an improper integral diverges if the limit doesn’t exist. And if the limit exists, we’ll say that the improper integral “converges to .”

Practice Problems Practice Problems

1.

Explain what it means for an integral to be improper. What kinds of issues are we looking at?

2.

Give an example of an integral that is improper due to an unbounded or infinite interval of integration (infinite width).

3.

Give an example of an integral that is improper due to an unbounded integrand (infinite height).

4.

What does it mean for an improper integral to converge? How does this connect with limits?

5.

What does it mean for an improper integral to diverge? How does this connect with limits?

6.

Why do we need to use limits to evaluate improper integrals?

7.

For each of the following improper integrals:

Explain why the integral is improper. Be specific, and point out the issues in detail.
Set up the integral using the correct limit notation.
Antidifferentiate and evaluate the limit.
Explain whether the integral converges or diverges.

(a)

\(\displaystyle \int_{x=0}^{\infty} \frac{1}{\sqrt{x+1}}\;dx\)

Solution.

This is an improper integral because of the infinite width of the integral: we cannot use the Fundamental Theorem of Calculus to evaluate this, since the function isn’t being considered on a closed interval (or even an open one with finite width).

\begin{align*} \int_{x=0}^{\infty} \frac{1}{\sqrt{x+1}}\;dx \amp = \lim_{t\to \infty} \int_{x=0}^{x=t} (x+1)^{-1/2}\;dx\\ \amp = \lim_{t\to\infty} 2(x+1)^{1/2}\bigg|_{x=0}^{x=t}\\ \amp = \lim_{t\to\infty} 2(t+1){1/2} - 2\\ \amp = \infty \end{align*}

The limit does not exist, and so this integral diverges.

(b)

\(\displaystyle \int_{x=0}^{\infty} e^{-2x}\;dx\)

Solution.

\begin{align*} \int_{x=0}^{\infty} e^{-2x} \;dx \amp = \lim_{t\to \infty} \int_{x=0}^{x=t} e^{-2x}\;dx\\ \amp = \lim_{t\to\infty} -\frac{1}{2} e^{-2x}\bigg|_{x=0}^{x=t}\\ \amp = \lim_{t\to\infty} -\frac{1}{2e^{2t}} + \frac{1}{2e^0}\\ \amp = \frac{1}{2} \end{align*}

The limit exists, and so the integral converges to \(\frac{1}{2}\text{.}\)

(c)

\(\displaystyle \int_{x=-1}^{x=3} \dfrac{1}{x+1}\;dx \)

Solution.

This integral is improper because at \(x=-1\) the integrand function has a vertical asymptote: this unbounded discontinuity violates the requirements of the Fundamental Theorem of Calculus.

\begin{align*} \int_{x=-1}^{x=3} \dfrac{1}{x+1}\;dx \amp = \lim_{t\to-1^+} \int_{x=t}^{x=3} \dfrac{1}{x+1}\;dx \\ \amp = \lim_{t\to-1^+} \ln|x+1|\bigg|_{x=t}^{x=3}\\ \amp = \lim_{t\to-1^+} \ln|4| - \ln|t+1|\\ \amp = \ln(4) - (-\infty)\\ \amp = \infty \end{align*}

The limit does not exist, and so this integral diverges.

(d)

\(\displaystyle \int_{-\infty}^{x=0} \sqrt{e^x}\;dx\)

Solution.

\begin{align*} \int_{-\infty}^{x=0} \sqrt{e^x}\;dx \amp = \lim_{t\to-\infty} \int_{x=t}^{x=0} e^{\frac{1}{2}x}\;dx\\ \amp = \lim_{t\to-\infty} \left(2e^{\frac{1}{2}x}\right)\bigg|_{x=t}^{x=0} \\ \amp = \lim_{t\to-\infty} 2e^{0} - 2e^{t/2} \\ \amp = 2 \end{align*}

The limit exists, and so the integral converges to \(2\text{.}\)

(e)

\(\displaystyle \int_{x=2}^{x=8} \dfrac{5}{(x-2)^3}\;dx\)

Solution.

This integral is improper because at \(x=2\) the integrand function has a vertical asymptote: this unbounded discontinuity violates the requirements of the Fundamental Theorem of Calculus.

\begin{align*} \int_{x=2}^{x=8} \dfrac{5}{(x-2)^3}\;dx \amp = \lim_{t\to 2^+} \int_{x=t}^{x=8} 5(x-2)^{-3}\;dx\\ \amp = \lim_{t\to 2^+} -\frac{5}{2(x-2)^2} \bigg|{x=t}^{x=8}\\ \amp = \lim_{t\to 2^+} -\frac{5}{2(6)^2} + \frac{5}{2(t-2)^2}\\ \amp = - \frac{5}{72} + \infty\\ \amp = \infty \end{align*}

The limit does not exist, and so this integral diverges.

(f)

\(\displaystyle \int_{x=1}^{x=12} \dfrac{1}{\sqrt[5]{12-x}}\;dx\)

Solution.

This integral is improper because at \(x=12\) the integrand function has a vertical asymptote: this unbounded discontinuity violates the requirements of the Fundamental Theorem of Calculus.

\begin{align*} \int_{x=1}^{x=12} \dfrac{1}{\sqrt[5]{12-x}}\;dx \amp = \lim_{t\to 12^-} \int_{x=1}^{x=t} (12-x)^{-1/5}\;dx\\ \amp = \lim_{t\to 12^-} -\frac{5(12-x)^{4/5}}{4}\bigg|_{x=1}^{x=t}\\ \amp = \lim_{t\to 12^-} -\frac{5(12-t)^{4/5}}{4} +\frac{5(11)^{4/5}}{4}\\ \amp = \frac{5(11)^{4/5}}{4} \end{align*}

The limit exists, and so the integral converges to \(\dfrac{5(11)^{4/5}}{4} \text{.}\)

8.

One of the big ideas in probability is that for a curve that defines a probability density function, the area under the curve needs to be 1. What value of \(k\) makes the function \(\dfrac{kx}{(x^2+3)^{5/4}}\) a valid probability distribution on the interval \([0,\infty)\text{?}\)

Solution.

We’ll set up the integral \(\displaystyle \int_{x=0}^\infty \frac{kx}{(x^2+3)^{5/4}}\;dx\) and find the value of \(k\) such that:

\begin{equation*} \int_{x=0}^\infty \frac{kx}{(x^2+3)^{5/4}}\;dx =1\text{.} \end{equation*}

\begin{align*} \int_{x=0}^\infty \frac{kx}{(x^2+3)^{5/4}}\;dx \amp = \lim_{t\to\infty} \int_{x=0}^{x=t} \frac{kx}{(x^2+3)^{5/4}}\;dx\\ u = x^2+3\;\;\; \amp \;\;\;dx = 2x\;du \\ \amp = \frac{k}{2}\lim_{t\to\infty} \int_{u=3}^{u=t^2+3} u^{-5/4}\;du\\ \amp =\frac{k}{2} \lim_{t\to\infty} -\frac{4}{u^{1/4}}\bigg|_{u=3}^{u=t^2+3}\\ \amp = \frac{k}{2} \lim_{t\to\infty} -\frac{4}{(t^2+3)^{5/4}} + \frac{4}{\sqrt[4]{3}}\\ \amp \frac{2k}{\sqrt[4]{3}} \end{align*}

Since we are hoping for this integral to be 1, we need \(k=\frac{\sqrt[4]{3}}{2}\text{.}\)

9.

Let’s consider the integral \(\displaystyle \int_{x=1}^{\infty} \dfrac{\sqrt{x^2+1}}{x^2}\;dx\text{.}\) This is a difficult integral to evaluate!

(a)

First, compare \(\sqrt{x^2+1}\) to \(\sqrt{x^2}\) using an inequality: which one is bigger?

Solution.

Since \(k^2+1 \gt k\text{,}\) we know that \(\sqrt{k^2+1} \gt \sqrt{k^2}\text{.}\)

(b)

Second, use this inequality to compare the function \(\dfrac{\sqrt{x^2+1}}{x^2}\) to \(\dfrac{1}{x}\) for \(x\gt 0\text{:}\) which one is bigger? Again, use your inequality from above to help!

Solution.

First, since \(k\gt 0\text{,}\) we know that \(\sqrt{k^2} = k\text{.}\) From our work above, we know that \(\sqrt{k^2+1} \gt k\text{.}\)

\begin{align*} \sqrt{k^2+1} \amp \gt k\\ \frac{\sqrt{k^2+1}}{k^2} \amp \gt \frac{k}{k^2}\\ \frac{\sqrt{k^2+1}}{k^2} \amp \gt \frac{1}{k} \end{align*}

(c)

Now compare \(\displaystyle \int_{x=1}^\infty \dfrac{\sqrt{x^2+1}}{x^2}\;dx\) to \(\displaystyle \int_{x=1}^\infty \dfrac{1}{x}\;dx\text{.}\) Which one is bigger?

Solution.

Since \(\dfrac{\sqrt{k^2+1}}{k^2} \gt \dfrac{1}{k}\) on the interval \([1,\infty)\text{,}\) we can say that:

\begin{equation*} \int_{x=1}^{\infty} frac{\sqrt{k^2+1}}{k^2}\;dx \gt \int_{x=1}^{\infty} \frac{1}{k}\;dx\text{.} \end{equation*}

(d)

Explain how we can use this result to make a conclusion about whether our integral, \(\displaystyle \int_{x=1}^\infty \dfrac{\sqrt{x^2+1}}{x^2}\;dx\) converges or diverges.

Solution.

\begin{align*} \int_{x=1}^{\infty} \frac{1}{x}\;dx \amp = \lim_{t\to\infty} \int_{x=1}^{x=t} \frac{1}{x}\;dx\\ \amp = \lim_{t\to\infty} \ln|x|\bigg|_{x=1}^{x=t}\\ \amp = \lim_{t\to\infty} \ln(t)\\ \amp = \infty \end{align*}

But since we know that

\begin{equation*} \int_{x=1}^{\infty} frac{\sqrt{k^2+1}}{k^2}\;dx \gt \int_{x=1}^{\infty} \frac{1}{k}\;dx \end{equation*}

we know that the integral \(\displaystyle \int_{x=1}^\infty \dfrac{\sqrt{x^2+1}}{x^2}\;dx\) must diverge as well: there is no real number that the limit of this integral could converge to that is greater than infinity.

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