Improper Integrals

An improper integral is a type of definite integral in which the integrand is undefined at one or both of the endpoints. Strictly speaking, it is the limit of the definite integral as the interval approaches its desired size.

Improper integrals may be evaluated by finding a limit of the indefinite integral of the integrand. However, such a value is meaningful only if the improper integral converges in the first place.

Its graph

Improper integrals appear frequently in the study of probability distributions, asymptotic behavior, and calculus in general. In order to evaluate them properly, it is crucial to understand precisely what is meant by these integrals.

One type of improper integral is an integral where one of the endpoints is (approaching) infinity. For instance,

$$\int_0^\infty \frac{1}{1 + x^2} \, dx = \lim_{b \to \infty} \int_0^b \frac{1}{1 + x^2} \, dx.$$

Another type of improper integral is an integral where both of the endpoints are (approaching) infinity. For instance,

$$\int_{-\infty}^\infty e^{-x^2} \, dx = \int_{-\infty}^0 e^{-x^2} \, dx + \int_0^\infty e^{-x^2} \, dx$$

is an integral used in Gaussian distributions.

A third type of improper integral is an integral for which an asymptote appears at one (or both) of the endpoints. For instance,

$$\int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx.$$

A fourth type of improper integral is one that is not properly defined: an integral whose integrand is not defined at all the values within the interval of integration. For instance,

$$\int_0^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx$$

is undefined because the function $f(x) = \tfrac{1}{\sqrt[3]{(x-1)^2}}$ is not defined throughout the interval of integration, even though its antiderivative $F(x) = 3\sqrt[3]{x-1}$ is defined throughout the interval of integration. A concept known as the Cauchy principal value permits a reinterpretation of the integral as

$$\int_0^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx = \lim_{b \to 1^-} \int_0^b \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx + \lim_{a \to 1^+} \int_a^2 \frac{1}{\sqrt[3]{(x - 1)^2}} \, dx$$

in order to assign a value to determine a value for it, provided each of the integrals converges. In general, the Cauchy principal value splits the interval of integration of an improper integral into closed intervals, on whose interiors the integrand is defined.

An improper integral may be interpreted unambiguously if its integrand is defined at all points in the interior of the interval of integration (as opposed to the fourth possibility above) and one of the endpoints. For such improper integrals, the value may either exist and be finite, diverge to $\pm \infty$, or diverge to no particular value.

For instance, $\int_{-\infty}^\infty \cos x \, dx$ diverges to no particular value, since the cosine function oscillates.

Note that it may be impossible to define clearly an improper integral like $\int_{-\infty}^\infty x \, dx$ where the integrand is undefined at both endpoints. In such a case, the Cauchy principal value would be $\lim_{a \to \infty} \int_{-a}^a x \, dx = 0$, but the expression itself is undefined.

An improper integral converges if its limit exists (and is finite). This may be determined by considering the limiting behavior of an antiderivative of the integrand, but it is not necessary to consider the antiderivative. It is possible to prove existence by showing that, taking the integral as a geometric statement, the area in question is bounded and the limiting behavior is monotonic (strictly increasing or strictly decreasing).

Determine whether or not the improper integral

$$\int_0^\infty \frac{1}{1 + x^2} \, dx$$

exists.

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The function in question is defined for all real numbers. Observe that an antiderivative $\int \tfrac{1}{1 + x^2} \, dx = \arctan x$ is finite:

$$\lim_{b \to \infty} \arctan b = \frac{\pi}{2},$$

so the limit of the antiderivative exists, thus making the improper integral exist.

On the other hand, the integrand is positive for all real numbers, so the limiting behavior is strictly increasing. Furthermore, comparison to a left Riemann sum yields the inequality

$$\begin{aligned} \int_0^\infty \frac{1}{1 + x^2} \, dx &< 1 + \int_1^\infty \frac{1}{1 + x^2} \, dx \\ &< 1 + \int_1^\infty \frac{1}{x^2} \, dx \\ &< 1 + \sum_{n=1}^\infty \frac{1}{n^2} \\ &= 1 + \frac{\pi^2}{6}. \end{aligned}$$

Since the limit is bounded and increasing, it must exist, and the improper integral must also exist. $_\square$

If an improper integral is clearly defined, its value may be determined by evaluating the limit.

Determine the value of

$$\int_0^\infty \frac{1}{1 + x^2} \, dx.$$

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As proved above, this value exists. Then

$$\int_0^\infty \frac{1}{1 + x^2} \, dx = \arctan(\infty) - \arctan(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \ _\square$$

Evaluate

$$\int_0^1 \frac{1}{\sqrt{x}} \, dx.$$

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We have

$$\begin{aligned} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx \\ &= \lim_{a \to 0^+} \left[ 2\sqrt{x} \right]_a^1 \\ &= \lim_{a \to 0^+} 2\sqrt{1} - 2\sqrt{a} \\ &= 2. \ _\square \end{aligned}$$

• Antiderivatives
• Definite Integrals