Evaluating Improper Integrals

Learning Outcomes

  • Evaluate an integral over an infinite interval
  • Evaluate an integral over a closed interval with an infinite discontinuity within the interval

Integrating over an Infinite Interval

How should we go about defining an integral of the type [latex]{\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx?[/latex] We can integrate [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] for any value of [latex]t[/latex], so it is reasonable to look at the behavior of this integral as we substitute larger values of [latex]t[/latex]. Figure 1 shows that [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] may be interpreted as area for various values of [latex]t[/latex]. In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.

Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.

Definition


  1. Let [latex]f\left(x\right)[/latex] be continuous over an interval of the form [latex]\left[a,\text{+}\infty \right)[/latex]. Then

    [latex]{\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex],



    provided this limit exists.

  2. Let [latex]f\left(x\right)[/latex] be continuous over an interval of the form [latex]\left(\text{-}\infty ,b\right][/latex]. Then

    [latex]{\displaystyle\int }_{\text{-}\infty }^{b}f\left(x\right)dx=\underset{t\to \text{-}\infty }{\text{lim}}{\displaystyle\int }_{t}^{b}f\left(x\right)dx[/latex],



    provided this limit exists.

    In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

  3. Let [latex]f\left(x\right)[/latex] be continuous over [latex]\left(\text{-}\infty ,\text{+}\infty \right)[/latex]. Then

    [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx={\displaystyle\int }_{\text{-}\infty }^{0}f\left(x\right)dx+{\displaystyle\int }_{0}^{+\infty }f\left(x\right)dx[/latex],



    provided that [latex]{\displaystyle\int }_{\text{-}\infty }^{0}f\left(x\right)dx[/latex] and [latex]{\displaystyle\int }_{0}^{+\infty }f\left(x\right)dx[/latex] both converge. If either of these two integrals diverge, then [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx[/latex] diverges. (It can be shown that, in fact, [latex]{\displaystyle\int }_{\text{-}\infty }^{+\infty }f\left(x\right)dx={\displaystyle\int }_{\text{-}\infty }^{a}f\left(x\right)dx+{\displaystyle\int }_{a}^{+\infty }f\left(x\right)dx[/latex] for any value of [latex]a.[/latex])

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] and the [latex]x[/latex] -axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] finite or infinite?

Example: Finding an Area

Determine whether the area between the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] and the x-axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] is finite or infinite.

Example: Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] and the x-axis over the interval [latex]\left[1,\text{+}\infty \right)[/latex] about the [latex]x[/latex] -axis.

In conclusion, although the area of the region between the x-axis and the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex] over the interval [latex]\left[1,\text{+}\infty \right)[/latex] is infinite, the volume of the solid generated by revolving this region about the x-axis is finite. The solid generated is known as Gabriel’s Horn.

Example: Evaluating an Improper Integral over an Infinite Interval

Evaluate [latex]{\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx[/latex]. State whether the improper integral converges or diverges.

Because improper integrals require evaluating limits at infinity, at times we may be required to use L’Hôpital’s Rule to evaluate a limit.

Recall: L’Hôpital’s Rule

Suppose [latex]f[/latex] and [latex]g[/latex] are differentiable functions over an open interval [latex]\left(a, \infty \right) [/latex] for some value of [latex] a [/latex]. If either:

  1. [latex]\underset{x\to \infty}{\lim}f(x)=0[/latex] and [latex]\underset{x\to \infty}{\lim}g(x)=0[/latex]
  2. [latex]\underset{x\to \infty}{\lim}f(x)= \infty [/latex] (or [latex] -\infty [/latex]) and [latex]\underset{x\to \infty}{\lim}g(x)= \infty [/latex] (or [latex] -\infty [/latex]), then
[latex]\underset{x\to \infty}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to \infty}{\lim}\dfrac{f^{\prime}(x)}{g^{\prime}(x)}[/latex]

 

assuming the limit on the right exists or is [latex]\infty [/latex] or [latex]−\infty[/latex].

Example: Evaluating an Improper Integral on [latex]\left(-\infty ,+\infty \right)[/latex]

Evaluate [latex]{\displaystyle\int }_{-\infty }^{+\infty }x{e}^{x}dx[/latex]. State whether the improper integral converges or diverges.

try it

Evaluate [latex]{\displaystyle\int }_{-3}^{+\infty }{e}^{\text{-}x}dx[/latex]. State whether the improper integral converges or diverges.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).

Try It

Integrating a Discontinuous Integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex], where [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,b\right)[/latex] and discontinuous at [latex]b[/latex]. Since the function [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,t\right][/latex] for all values of [latex]t[/latex] satisfying [latex]a<t<b[/latex], the integral [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] is defined for all such values of [latex]t[/latex]. Thus, it makes sense to consider the values of [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] as [latex]t[/latex] approaches [latex]b[/latex] for [latex]a<t<b[/latex]. That is, we define [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex], provided this limit exists. Figure 4 illustrates [latex]{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex] as areas of regions for values of [latex]t[/latex] approaching [latex]b[/latex].

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.

Figure 4. As [latex]t[/latex] approaches b from the left, the value of the area from a to [latex]t[/latex] approaches the area from a to b.

We use a similar approach to define [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex], where [latex]f\left(x\right)[/latex] is continuous over [latex]\left(a,b\right][/latex] and discontinuous at [latex]a[/latex]. We now proceed with a formal definition.

Definition


  1. Let [latex]f\left(x\right)[/latex] be continuous over [latex]\left[a,b\right)[/latex]. Then,

    [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\displaystyle\int }_{a}^{t}f\left(x\right)dx[/latex].

     

  2. Let [latex]f\left(x\right)[/latex] be continuous over [latex]\left(a,b\right][/latex]. Then,

    [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {a}^{+}}{\text{lim}}{\displaystyle\int }_{t}^{b}f\left(x\right)dx[/latex].

    In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

  3. If [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,b\right][/latex] except at a point [latex]c[/latex] in [latex]\left(a,b\right)[/latex], then

    [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx={\displaystyle\int }_{a}^{c}f\left(x\right)dx+{\displaystyle\int }_{c}^{b}f\left(x\right)dx[/latex],

    provided both [latex]{\displaystyle\int }_{a}^{c}f\left(x\right)dx[/latex] and [latex]{\displaystyle\int }_{c}^{b}f\left(x\right)dx[/latex] converge. If either of these integrals diverges, then [latex]{\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex] diverges.

The following examples demonstrate the application of this definition.

Example: Integrating a Discontinuous Integrand

Evaluate [latex]{\displaystyle\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx[/latex], if possible. State whether the integral converges or diverges.

Example: Integrating a Discontinuous Integrand

Evaluate [latex]{\displaystyle\int }_{0}^{2}x\text{ln}xdx[/latex]. State whether the integral converges or diverges.

Example: Integrating a Discontinuous Integrand

Evaluate [latex]{\displaystyle\int }_{-1}^{1}\frac{1}{{x}^{3}}dx[/latex]. State whether the improper integral converges or diverges.

try it

Evaluate [latex]{\displaystyle\int }_{0}^{2}\frac{1}{x}dx[/latex]. State whether the integral converges or diverges.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).