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# 7.6: Improper Integrals

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Compute by hand the integrals of a wide variety of functions by using the technique of Improper Integration.
• Combine this technique with other integration techniques to integrate.
• Distinguish between proper and improper integrals.

The concept of improper integrals is an extension to the concept of definite integrals. The reason for the term improper is because those integrals either

• include integration over infinite limits or
• the integrand may become infinite within the limits of integration.

We will take each case separately. Recall that in the definition of definite integral \begin{align*}\int_a^b f(x) dx\end{align*} we assume that the interval of integration \begin{align*}[a, b]\end{align*} is finite and the function \begin{align*}f\end{align*} is continuous on this interval.

## Integration Over Infinite Limits

If the integrand \begin{align*}f\end{align*} is continuous over the interval \begin{align*}[a, \infty),\end{align*} then the improper integral in this case is defined as

\begin{align*}\int_a^{\infty} f(x) dx = \lim_{l \to \infty} \int_a^l f(x) dx.\end{align*}

If the integration of the improper integral exists, then we say that it converges. But if the limit of integration fails to exist, then the improper integral is said to diverge. The integral above has an important geometric interpretation that you need to keep in mind. Recall that, geometrically, the definite integral \begin{align*}\int_a^b f(x) dx\end{align*} represents the area under the curve. Similarly, the integral \begin{align*}\int_a^l f(x) dx\end{align*} is a definite integral that represents the area under the curve \begin{align*}f(x)\end{align*} over the interval \begin{align*}[a, l],\end{align*} as the figure below shows. However, as \begin{align*}l\end{align*} approaches \begin{align*} \infty\end{align*}, this area will expand to the area under the curve of \begin{align*}f(x)\end{align*} and over the entire interval \begin{align*}[a, \infty).\end{align*} Therefore, the improper integral \begin{align*}\int_a^{\infty} f(x) dx\end{align*} can be thought of as the area under the function \begin{align*}f(x)\end{align*} over the interval \begin{align*}[a, \infty).\end{align*}

Example 1:

Evaluate \begin{align*}\int_1^{\infty} \frac{dx} {x}\end{align*} .

Solution:

We notice immediately that the integral is an improper integral because the upper limit of integration approaches infinity. First, replace the infinite upper limit by the finite limit \begin{align*}l\end{align*} and take the limit of \begin{align*}l\end{align*} to approach infinity:

\begin{align*}\int_1^{\infty} \frac{dx} {x} & = \lim_{l \to \infty} \int_1^l \frac{dx} {x}\\ & = \lim_{l \to \infty} [\ln x]_1^l\\ & = \lim_{l \to \infty} (\ln l - \ln 1)\\ & = \lim_{l \to \infty} \ln l\\ & = \infty.\end{align*}

Thus the integral diverges.

Example 2:

Evaluate \begin{align*}\int_2^{\infty} \frac{dx} {x^2}\end{align*} .

Solution:

\begin{align*}\int_2^{\infty} \frac{dx} {x^2} & = \lim_{l \to \infty} \int_2^l \frac{dx} {x^2}\\ & = \lim_{l \to \infty} \left [\frac{-1} {x} \right]_2^l\\ & = \lim_{l \to \infty} \left (\frac{-1} {l} + \frac{1} {2} \right) \\ & = \frac{1} {2}.\end{align*}

Thus the integration converges to \begin{align*}\frac{1} {2}.\end{align*}

Example 3:

Evaluate \begin{align*}\int_{+ \infty}^{- \infty} \frac{dx} {1 + x^2}\end{align*}.

Solution:

What we need to do first is to split the integral into two intervals \begin{align*}(-\infty, 0]\end{align*} and \begin{align*}[0, +\infty).\end{align*} So the integral becomes

\begin{align*}\int_{-\infty}^{+\infty} \frac{dx} {1 + x^2} = \int_{-\infty}^0 \frac{dx} {1 + x^2 } + \int_0^{+\infty} \frac{dx} {1 + x^2}.\end{align*}

Next, evaluate each improper integral separately. Evaluating the first integral on the right,

\begin{align*}\int_{-\infty}^0 \frac{dx} {1 + x^2} & = \lim_{l \to -\infty} \int_l^0 \frac{dx} {1 + x^2}\\ & = \lim_{l \to -\infty} \left [\tan^{-1} x \right]_l^0\\ & = \lim_{l \to -\infty} \left [\tan^{-1} 0 - \tan^{-1} l \right]\\ & = \lim_{l \to -\infty} \left [0 - \left (-\frac{\pi} {2}\right ) \right] = \frac{\pi} {2}.\end{align*}

Evaluating the second integral on the right,

\begin{align*}\int_0^{\infty} \frac{dx} {1 + x^2} & = \lim_{l \to \infty} \int_0^l \frac{dx} {1 + x^2}\\ & = \lim_{l \to \infty} \left [\tan^{-1} x \right]_0^l\\ & = \frac{\pi} {2} - 0 = \frac{\pi} {2}.\end{align*}

\begin{align*}\int_{-\infty}^{+\infty} \frac{dx} {1 + x^2} & = \frac{\pi} {2} + \frac{\pi} {2} = \pi.\end{align*}

Remark: In the previous example, we split the integral at \begin{align*}x = 0.\end{align*} However, we could have split the integral at any value of \begin{align*}x = c\end{align*} without affecting the convergence or divergence of the integral. The choice is completely arbitrary. This is a famous thoerem that we will not prove here. That is,

\begin{align*}\int_{-\infty}^{+\infty} f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^{+\infty} f(x) dx.\end{align*}

## Integrands with Infinite Discontinuities

This is another type of integral that arises when the integrand has a vertical asymptote (an infinite discontinuity) at the limit of integration or at some point in the interval of integration. Recall from Chapter 5 in the Lesson on Definite Integrals that in order for the function \begin{align*}f\end{align*} to be integrable, it must be bounded on the interval \begin{align*}[a, b].\end{align*} Otherwise, the function is not integrable and thus does not exist. For example, the integral

\begin{align*}\int_0^4 \frac{dx} {x - 1}\end{align*}

develops an infinite discontinuity at \begin{align*}x = 1\end{align*} because the integrand approaches infinity at this point. However, it is continuous on the two intervals \begin{align*}[0, 1)\end{align*} and \begin{align*}(1, 4].\end{align*} Looking at the integral more carefully, we may split the interval \begin{align*} [0,4] \rightarrow [0,1) \cup (1,4]\end{align*} and integrate between those two intervals to see if the integral converges.

\begin{align*}\int_0^4 \frac{dx} {x - 1} = \int_0^1 \frac{dx} {x - 1} + \int_1^4 \frac{dx} {x - 1}.\end{align*}

We next evaluate each improper integral. Integrating the first integral on the right hand side,

\begin{align*}\int_0^1 \frac{dx} {x - 1} &= \lim_{l \to 1^{-}} \int_0^l \frac{dx} {x - 1}\\ &= \lim_{l \to 1^{-}} [\ln |x - 1|]_0^l\\ &= \lim_{l \to 1^{-}} [\ln |l - 1| - \ln |- 1|]\\ &= -\infty.\end{align*}

The integral diverges because \begin{align*}\ln(0)\end{align*} is undefined, and thus there is no reason to evaluate the second integral. We conclude that the original integral diverges and has no finite value.

Example 4:

Evaluate \begin{align*}\int_1^3 \frac{dx} {\sqrt{x - 1}}\end{align*} .

Solution:

\begin{align*}\int_1^3 \frac{dx} {\sqrt{x - 1}} & = \lim_{l \to 1^{+}} \int_l^3 \frac{dx} {\sqrt{x - 1}}\\ & = \lim_{l \to 1^{+}} \left [2 \sqrt{x - 1} \right]_l^3\\ & = \lim_{l \to 1^{+}} \left [2 \sqrt{2} - 2\sqrt{l - 1} \right]\\ & = 2\sqrt{2}.\end{align*}

So the integral converges to \begin{align*}2\sqrt{2}\end{align*}.

Example 5:

In Chapter 5 you learned to find the volume of a solid by revolving a curve. Let the curve be \begin{align*}y = xe^{-x}, 0 \le x \le \infty\end{align*} and revolving about the \begin{align*}x-\end{align*}axis. What is the volume of revolution?

Solution:

From the figure above, the area of the region to be revolved is given by \begin{align*}A = \pi y^2 = \pi x^2 e^{-2x}\end{align*}. Thus the volume of the solid is

\begin{align*}V = \pi \int_0^{\infty} x^2 e^{-2x} dx = \pi \lim_{l \to \infty} \int_0^l x^2 e^{-2x} dx.\end{align*}

As you can see, we need to integrate by parts twice:

\begin{align*}\int x^2 e^{-2x} dx & = -\frac{x^2} {2} e^{-2x} + \int x e^{-2x} dx\\ & = - \frac{x^2} {2} e^{-2x} - \frac{x} {2} e^{-2x} - \frac{1} {4} e^{-2x} + C.\end{align*}

Thus

\begin{align*}V & = \pi \lim_{l \to \infty} \left [-\frac{x^2} {2} e^{-2x} - \frac{x} {2} e^{-2x} - \frac{1} {4} e^{-2x} \right]_0^l\\ & = \pi \lim_{l \to \infty} \left [\frac{2x^2 + 2x + 1} {-4e^{2x}} \right]_0^l\\ & = \pi \lim_{l \to \infty} \left [\frac{2l^2 + 2l + 1} {-4e^{2l}} - \frac{1} {-4e^0} \right]\\ & = \pi \lim_{l \to \infty} \left [\frac{2l^2 + 2l + 1} {4e^{2l}} + \frac{1} {4} \right].\end{align*}

At this stage, we take the limit as \begin{align*}l\end{align*} approaches infinity. Notice that the when you substitute infinity into the function, the denominator of the expression \begin{align*}\frac{2l^2 + 2l + 1} {-4e^{2l}},\end{align*} being an exponential function, will approach infinity at a much faster rate than will the numerator. Thus this expression will approach zero at infinity. Hence

\begin{align*}V = \pi \left [0 + \frac{1} {4} \right] = \frac{\pi} {4},\end{align*}

So the volume of the solid is \begin{align*} \pi/4.\end{align*}

Example 6:

Evaluate \begin{align*}\int_{- \infty}^{+ \infty} \frac {dx}{e^x+e^{-x}}\end{align*}.

Solution:

This can be a tough integral! To simplify, rewrite the integrand as

\begin{align*}\frac {1}{e^x+e^{-x}} = \frac {1}{e^{-x}(e^{2x}+1)} = \frac{e^x}{e^{2x}+1} = \frac {e^x}{1+(e^x)^2}.\end{align*}

Substitute into the integral:

\begin{align*}\int \frac {dx}{e^x+e^{-x}} = \int \frac {e^x}{1+(e^x)^2}dx.\end{align*}

Using \begin{align*}u-\end{align*}substitution, let \begin{align*}u = e^x, du = e^xdx.\end{align*}

\begin{align*}\int \frac {dx}{e^x+e^{-x}} & = \int \frac {du}{1+u^2} \\ & = \tan^{-1}u + C \\ & = \tan^{-1}e^x + C.\end{align*}

Returning to our integral with infinite limits, we split it into two regions. Choose as the split point the convenient \begin{align*}x = 0.\end{align*}

\begin{align*}\int_{- \infty}^{+ \infty} \frac {dx}{e^x+e^{-x}} = \int_{- \infty}^{0} \frac {dx}{e^x+e^{-x}} + \int_{0}^{+ \infty} \frac {dx}{e^x+e^{-x}}.\end{align*}

Taking each integral separately,

\begin{align*}\int_{- \infty}^{0} \frac {dx}{e^x+e^{-x}} & = \lim_{l \to -\infty} \int_{l}^{0} \frac {dx}{e^x+e^{-x}} \\ & = \lim_{l \to -\infty} \left [ {\tan^{-1} e^x} \right ]_{l}^{0} \\ & = \lim_{l \to -\infty} \left [ {\tan^{-1} e^0} - {\tan^{-1}e^l} \right ] \\ & = \frac {\pi}{4} - 0 \\ & = \frac {\pi}{4}.\end{align*}

Similarly,

\begin{align*}\int_{0}^{+ \infty} \frac {dx}{e^x+e^{-x}} & = \lim_{l \to \infty} \int_{0}^{1} \frac {dx}{e^x+e^{-x}} \\ & = \lim_{l \to \infty} \left [ {\tan^{-1} e^x} \right ]_{0}^{l}\\ & = \lim_{l \to \infty} \left [ {\tan^{-1} e^l} - {\tan^{-1} 1} \right ] \\ & = \frac {\pi}{2} - \frac {\pi}{4} = \frac {\pi}{4}.\end{align*}

Thus the integral converges to

\begin{align*}\int_{- \infty}^{+ \infty} \frac {dx}{e^x+e^{-x}} = \frac {\pi}{4} + \frac {\pi}{4} = \frac {\pi}{2}.\end{align*}

For a video presentation of Improper Integrals (22.0), see Improper Integrals, www.justmathtutoring.com (6:23).

For a video presentation of Improper Integrals with Infinity in the Upper and Lower Limits (22.0), see Improper Integrals, www.justmathtutoring.com (7:55).

## Review Questions

1. Determine whether the following integrals are improper. If so, explain why.
1. \begin{align*}\int_{1}^{7} \frac {x+2}{x-3} dx\end{align*}
2. \begin{align*}\int_{1}^{7} \frac {x+2}{x+3} dx\end{align*}
3. \begin{align*}\int_{0}^{1} {{ \ln}x} dx\end{align*}
4. \begin{align*}\int_{0}^{\infty} \frac {1}{\sqrt {x-2}} dx\end{align*}
5. \begin{align*}\int_{0}^{{\pi}/4} {\tan x} dx\end{align*}

Evaluate the integral or state that it diverges.

1. \begin{align*}\int_{1}^{\infty} \frac {1}{x^{2.001}} dx\end{align*}
2. \begin{align*}\int_{- \infty}^{-2} \left [ \frac{1}{x-1} - \frac {1}{x+1} \right ] dx\end{align*}
3. \begin{align*}\int_{-\infty}^{0} e^{5x} dx\end{align*}
4. \begin{align*}\int_{3}^{5} \frac {1}{(x-3)^4} dx\end{align*}
5. \begin{align*}\int_{-{\pi}/2}^{{\pi}/2} {\tan x} dx\end{align*}
6. \begin{align*}\int_{0}^{1} \frac {1}{\sqrt {1-x^2}} dx\end{align*}
7. The region between the \begin{align*}x-\end{align*}axis and the curve \begin{align*}y = e^{-x}\end{align*} for \begin{align*}x \ge 0\end{align*} is revolved about the \begin{align*}x-\end{align*}axis.
1. Find the volume of revolution, \begin{align*}V.\end{align*}
2. Find the surface area of the volume generated, \begin{align*}S.\end{align*}

1. Improper; infinite discontinuity at \begin{align*}x = 3.\end{align*}
2. Not improper.
3. Improper; infinite discontinuity at \begin{align*}x = 0.\end{align*}
4. Improper; infinite interval of integration.
5. Not improper.
1. \begin{align*}\frac {1}{1.001}\end{align*}
2. \begin{align*}\ln 3\end{align*}
3. \begin{align*}\frac {1}{5}\end{align*}
4. divergent
5. divergent
6. \begin{align*}\frac {\pi}{2}.\end{align*}
1. \begin{align*}V = \pi /2\end{align*}
2. \begin{align*}S = {\pi} \left [{\sqrt {2} + { \ln}(1+ \sqrt {2})} \right ].\end{align*}

## Homework

Evaluate the following integrals.

1. \begin{align*}\int {\sqrt {\sin x}} {\cos x} dx\end{align*}
2. \begin{align*}\int x \tan^2(x^2) \sec^2(x^2) dx\end{align*}
3. \begin{align*}\int_{0}^{{ \ln}3} {\sqrt {e^{2x} - 1}} dx\end{align*}
4. \begin{align*}\int_{0}^{\infty} \frac {1}{x^2} dx\end{align*}
5. \begin{align*}\int_{-1}^{8} \frac {1}{\sqrt[3]{x}} dx\end{align*}
6. \begin{align*}\int \frac {x^2+x-16}{(x+1)(x-3)^2} dx\end{align*}
7. Graph and find the volume of the region enclosed by the \begin{align*}x-\end{align*}axis, the \begin{align*}y-\end{align*}axis, \begin{align*}x = 2\end{align*} and \begin{align*}y = x^2/(9-x^2)\end{align*} when revolved about the \begin{align*}x-\end{align*}axis.
8. The Gamma Function, \begin{align*} \Gamma (x)\end{align*}, is an improper integral that appears frequently in quantum physics. It is defined as \begin{align*} \Gamma (x) = \int_{0}^{\infty} {t^{x-1}e^{-t}} dt.\end{align*} The integral converges for all \begin{align*}x \ge 0.\end{align*}
1. Find \begin{align*} \Gamma (1).\end{align*}
2. Prove that \begin{align*}\Gamma (x + 1) = x \Gamma(x)\end{align*}, for all \begin{align*}x \ge 0\end{align*}.
3. Prove that \begin{align*}\Gamma \left ( \frac{1}{2} \right ) = {\sqrt {\pi}}.\end{align*}
9. Refer to the Gamma Function defined in the previous exercise to prove that
1. \begin{align*}\int_{0}^{\infty} e^{-x^n} dx = \Gamma \left ( \frac{n+1}{n} \right ), {n \ge 0}\end{align*} [Hint: Let \begin{align*}t = x^n\end{align*}]
2. \begin{align*}\int_{0}^{1} ({ \ln}x)^n dx = {(-1)^n} \Gamma (n+1), {n \ge 0}\end{align*} [Hint: Let \begin{align*}t = - \ln x\end{align*}]
10. In wave mechanics, a sawtooth wave is described by the integral \begin{align*} \int_{- \pi / \omega}^{+ \pi / \omega} {t \sin(k \omega t)} dt,\end{align*} where \begin{align*}k\end{align*} is called the wave number, \begin{align*}\omega\end{align*} is the frequency, and \begin{align*}t\end{align*} is the time variable. Evaluate the integral.

1. \begin{align*}\frac {2}{3} \sin x^{3/2} + C\end{align*}
2. \begin{align*}\frac {1}{6} \tan^3 (x^2) + C\end{align*}
3. \begin{align*}\sqrt{8} - \sec^{-1}3\end{align*}
4. divergent
5. \begin{align*}\frac {9}{2}\end{align*}
6. \begin{align*}{ \ln}\frac {(x-3)^2}{\begin{vmatrix}{x+1} \end{vmatrix}} +\frac {1}{x-3} + C\end{align*}
7. \begin{align*}{\pi} \left ( \frac{19}{5} - \frac {9}{4} { \ln}5 \right)\end{align*}
1. \begin{align*} \Gamma (1) = 1\end{align*}
1. Hint: Let \begin{align*}t = x^n.\end{align*}
2. Hint: \begin{align*}t = -\ln x\end{align*}
8. \begin{align*}\frac {2}{(k \omega)^2} \sin (k \pi)\end{align*}

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