We begin this section by considering the following definite integrals:

• $\ds \int_0^{100}\frac1{1+x^2}\ dx \approx 1.5608\text{,}$

• $\ds \int_0^{1000}\frac1{1+x^2}\ dx \approx 1.5698\text{,}$

• $\ds \int_0^{10,000}\frac1{1+x^2}\ dx \approx 1.5707\text{.}$

Notice how the integrand is $1/(1+x^2)$ in each integral (which is sketched in Figure 6.8.1). As the upper bound gets larger, one would expect the “area under the curve” would also grow. While the definite integrals do increase in value as the upper bound grows, they are not increasing by much. In fact, consider: \begin{equation*} \int_0^b \frac{1}{1+x^2}\ dx = \tan^{-1}(x) \Big|_0^b = \tan^{-1}(b) -\tan^{-1}(0) = \tan^{-1}(b) . \end{equation*}

As $b\rightarrow \infty\text{,}$ $\tan^{-1}(b) \rightarrow \pi/2\text{.}$ Therefore it seems that as the upper bound $b$ grows, the value of the definite integral $\ds \int_0^b\frac{1}{1+x^2}\ dx$ approaches $\pi/2\approx 1.5708\text{.}$ This should strike the reader as being a bit amazing: even though the curve extends “to infinity,” it has a finite amount of area underneath it.

When we defined the definite integral $\ds\int_a^b f(x)\ dx\text{,}$ we made two stipulations:

1. The interval over which we integrated, $[a,b]\text{,}$ was a finite interval, and

2. The function $f(x)$ was continuous on $[a,b]$ (ensuring that the range of $f$ was finite).

In this section we consider integrals where one or both of the above conditions do not hold. Such integrals are called improper integrals.

# Subsection6.8.1Improper Integrals with Infinite Bounds

##### Definition6.8.2Improper Integrals with Infinite Bounds; Converge, Diverge
1. Let $f$ be a continuous function on $[a,\infty)\text{.}$ Define \begin{equation*} \int_a^\infty f(x)\ dx \text{ to be } \lim_{b\to\infty}\int_a^b f(x)\ dx. \end{equation*}

2. Let $f$ be a continuous function on $(-\infty,b]\text{.}$ Define \begin{equation*} \int_{-\infty}^b f(x)\ dx \text{ to be } \lim_{a\to-\infty}\int_a^b f(x)\ dx. \end{equation*}

3. Let $f$ be a continuous function on $(-\infty,\infty)\text{.}$ Let $c$ be any real number; define \begin{equation*} \int_{-\infty}^\infty f(x)\ dx \text{ to be } \lim_{a\to-\infty}\int_a^c f(x)\ dx\ +\ \lim_{b\to\infty}\int_c^b f(x)\ dx. \end{equation*}

An improper integral is said to converge if its corresponding limit exists; otherwise, it diverges. The improper integral in part 3 converges if and only if both of its limits exist.

##### Example6.8.3Evaluating improper integrals

Evaluate the following improper integrals.

1. $\ds\int_1^\infty \frac1{x^2}\ dx$

2. $\ds\int_1^\infty \frac1x\ dx$

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

4. $\ds\int_{-\infty}^\infty \frac1{1+x^2}\ dx$

Solution

The previous section introduced l'Hôpital's Rule, a method of evaluating limits that return indeterminate forms. It is not uncommon for the limits resulting from improper integrals to need this rule as demonstrated next.

##### Example6.8.8Improper integration and l'Hôpital's Rule

Evaluate the improper integral $\ds \int_1^\infty \frac{\ln(x) }{x^2}\ dx\text{.}$

Solution

Improper Integrals with Infinite Range We have just considered definite integrals where the interval of integration was infinite. We now consider another type of improper integration, where the range of the integrand is infinite.

##### Definition6.8.10Improper Integration with Infinite Range

Let $f(x)$ be a continuous function on $[a,b]$ except at $c\text{,}$ $a\leq c\leq b\text{,}$ where $x=c$ is a vertical asymptote of $f\text{.}$ Define \begin{equation*} \int_a^b f(x)\ dx = \lim_{t\to c^-}\int_a^t f(x)\ dx + \lim_{t\to c^+}\int_t^b f(x)\ dx. \end{equation*}

##### Example6.8.11Improper integration of functions with infinite range

Evaluate the following improper integrals: $\ds 1.\ \int_0^1\frac1{\sqrt{x}}\ dx 2. \ \int_{-1}^1\frac{1}{x^2}\ dx\text{.}$

Solution

# Subsection6.8.2Understanding Convergence and Divergence

Oftentimes we are interested in knowing simply whether or not an improper integral converges, and not necessarily the value of a convergent integral. We provide here several tools that help determine the convergence or divergence of improper integrals without integrating.

Our first tool is to understand the behavior of functions of the form $\ds \frac1{x^p}\text{.}$

##### Example6.8.14Improper integration of $1/x^p$

Determine the values of $p$ for which $\ds \int_1^\infty \frac1{x^p}\ dx$ converges.

Solution

The result of Example 6.8.14 provides an important tool in determining the convergence of other integrals. A similar result is proved in the exercises about improper integrals of the form $\ds \int_0^1\frac1{x^p}\ dx\text{.}$ These results are summarized in the following Key Idea.

##### Key Idea6.8.16Convergence of Improper Integrals $\ds \int_1^\infty\frac1{x^p}dx$ and $\ds \int_0^1\frac1{x^p}dx\text{.}$
1. The improper integral $\ds \int_1^\infty\frac1{x^p}\ dx$ converges when $p>1$ and diverges when $p\leq 1\text{.}$

2. The improper integral $\ds \int_0^1\frac1{x^p}\ dx$ converges when $p\lt 1$ and diverges when $p\geq 1\text{.}$

A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. We often use integrands of the form $1/x^p$ to compare to as their convergence on certain intervals is known. This is described in the following theorem.

We used the upper and lower bound of “1” in Key Idea 6.8.16 for convenience. It can be replaced by any $a$ where $a>0\text{.}$

##### Example6.8.18Determining convergence of improper integrals

Determine the convergence of the following improper integrals.

1. $\ds \int_1^\infty e^{-x^2}\ dx$ 2. $\ds \int_3^\infty \frac{1}{\sqrt{x^2-x}}\ dx$

Solution

Being able to compare “unknown” integrals to “known” integrals is very useful in determining convergence. However, some of our examples were a little “too nice.” For instance, it was convenient that $\ds \frac{1}x \lt \frac{1}{\sqrt{x^2-x}}\text{,}$ but what if the “$-x$” were replaced with a “$+2x+5$”? That is, what can we say about the convergence of $\ds \int_3^\infty\frac{1}{\sqrt{x^2+2x+5}}\ dx\text{?}$ We have $\ds \frac{1}{x} > \frac1{\sqrt{x^2+2x+5}}\text{,}$ so we cannot use Theorem 6.8.17.

In cases like this (and many more) it is useful to employ the following theorem.

##### Example6.8.22Determining convergence of improper integrals

Determine the convergence of $\ds \int_3^{\infty} \frac{1}{\sqrt{x^2+2x+5}}\ dx\text{.}$

Solution

Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a little more difficult to employ, they are omitted from this text.

This chapter has explored many integration techniques. We learned Substitution, which “undoes” the Chain Rule of differentiation, as well as Integration by Parts, which “undoes” the Product Rule. We learned specialized techniques for handling trigonometric functions and introduced the hyperbolic functions, which are closely related to the trigonometric functions. All techniques effectively have this goal in common: rewrite the integrand in a new way so that the integration step is easier to see and implement.

As stated before, integration is, in general, hard. It is easy to write a function whose antiderivative is impossible to write in terms of elementary functions, and even when a function does have an antiderivative expressible by elementary functions, it may be really hard to discover what it is. The powerful computer algebra system Mathematica™ has approximately 1,000 pages of code dedicated to integration.

Do not let this difficulty discourage you. There is great value in learning integration techniques, as they allow one to manipulate an integral in ways that can illuminate a concept for greater understanding. There is also great value in understanding the need for good numerical techniques: the Trapezoidal and Simpson's Rules are just the beginning of powerful techniques for approximating the value of integration.

The next chapter stresses the uses of integration. We generally do not find antiderivatives for antiderivative's sake, but rather because they provide the solution to some type of problem. The following chapter introduces us to a number of different problems whose solution is provided by integration.

# Subsection6.8.3Exercises

Terms and Concepts

In the following exercises, evaluate the given improper integral.