ITLR - Improper Intervals

Improper Intervals

An integral having at least one nonfinite limit or an integrand that becomes infinite between the limits of integration is an improper integral. The adjective improper calls attention to integrals that differ from the typical definite integral $LaTeX: \int_a^bf\left(x\right)dx$ $\int_a^bf\left(x\right)dx$ with the function f defined on a finite interval [a, b] and continuous, i.e., f does not have an infinite discontinuity. There are two types of improper integrals and a description and their geometric interpretation in terms of area under a curve follow:

types of improper integrals: one or both limits of integration are infinity (producing an infinity large area). f(x) becomes infinite at some value x between a and b (producing an infinitely large area).

Improper Integrals with Infinite Integration Limits

There are three possible situations for which improper integrals have one or more infinite limits of integration. Given definitions governing these situations, it is possible to determine whether or not a limit exists, and subsequently, whether or not the improper integral converges or diverges. Convergence of an integral refers to the existence of a finite limit defining an improper integral. Divergence of an integral occurs when no finite limiting value of an improper integral exists.

Improper Integrals with Infinite Discontinuities

The second type of improper integral contains an infinite discontinuity in [a, b]. Definitions governing these types of improper integrals are described below.

Evaluating an Improper Integral

When an integral is improper, it must be calculated in terms of limits. The presentation below reviews types of improper integrals and how to evaluate them, illustrates how to determine whether or not improper integrals converge or diverge, and explores an application of improper integrals in a real-world context.

Watch the video on Improper Integrals Links to an external site.

Consider a common type of improper integral, $LaTeX: \int_1^{\infty}\frac{1}{x^P}dx$ $\int_1^{\infty}\frac{1}{x^P}dx$ equation image indicator . The presentation below investigates the behavior in terms of convergence and divergence of this improper integral for various values of p.

It may be helpful to explore the examples and options for self-entered problems available through the improper integral applet by CLICKING HERE Links to an external site..

Comparison Tests for Convergence and Divergence of Improper Integrals

It is often impossible to find the exact value of an improper integral, but it is sufficient to know whether or not it converges or diverges. The Direct Comparison Test for Convergence and Divergence of Improper Integrals is helpful in these situations. Suppose f and g are continuous functions over equation image indicator $LaTeX: \left[a,\infty\right)$ $\left[a,\infty\right)$ with f(x) > g(x) > 0 for x > a. Then the following is true:

1. If equation image indicator $LaTeX: \int_a^{\infty }f\left(x\right)dx$ $\int_a^{\infty }f\left(x\right)dx$ converges, then $LaTeX: \int_a^{\infty }g\left(x\right)dx$ $\int_a^{\infty }g\left(x\right)dx$ converges.

2. If equation image indicator $LaTeX: \int_a^{\infty }g\left(x\right)dx$ $\int_a^{\infty }g\left(x\right)dx$ diverges, then $LaTeX: \int_a^{\infty }f\left(x\right)dx$ $\int_a^{\infty }f\left(x\right)dx$ diverges.

Determining the convergence or divergence of integrals such as equation image indicator $LaTeX: \int_1^{\infty}\frac{\sin^2x}{x^2}dx$ $\int_1^{\infty}\frac{\sin^2x}{x^2}dx$ is difficult. Comparing it to a simpler integral such as $LaTeX: \int_1^{\infty}\frac{1}{x^2}dx$ $\int_1^{\infty}\frac{1}{x^2}dx$ makes determining whether or not $LaTeX: \int_1^{\infty}\frac{\sin^2x}{x^2}dx$ $\int_1^{\infty}\frac{\sin^2x}{x^2}dx$ converges or diverges much easier. Both functions are continuous equation image indicator $LaTeX: \left[1,\infty\right)$ $\left[1,\infty\right)$ and $LaTeX: \int_1^{\infty}\frac{1}{x^2}dx\ge \int_1^{\infty}\frac{\sin^2x}{x^2}dx\:for\:x\ge 1$ $\int_1^{\infty}\frac{1}{x^2}dx\ge \int_1^{\infty}\frac{\sin^2x}{x^2}dx\:for\:x\ge 1$ .

graph of direct comparison test

Since equation image indicator $LaTeX: \int_1^{\infty}\frac{1}{x^2}dx$ $\int_1^{\infty}\frac{1}{x^2}dx$ converges based on the convergence of $LaTeX: \int_1^{\infty}\frac{1}{x^P}dx$ $\int_1^{\infty}\frac{1}{x^P}dx$ with p = 2, then $LaTeX: \int_1^{\infty}\frac{\sin^2x}{x^2}dx$ $\int_1^{\infty}\frac{\sin^2x}{x^2}dx$ also converges.

Another test is also used to determine the convergence or divergence of an improper integral. The Limit Comparison Test for Convergence and Divergence of Improper Integrals states that if f and g are positive continuous functions on equation image indicator $LaTeX: \left[a,\infty\right)$ $\left[a,\infty\right)$ and if $LaTeX: \lim_{x \to \infty }\frac{f(x)}{g(x)}=L$ $\lim_{x \to \infty }\frac{f(x)}{g(x)}=L$ where 0 < L < ∞, then $LaTeX: \int_a^{b}f\left(x\right)dx$ $\int_a^{b}f\left(x\right)dx$ and $LaTeX: \int_a^{b}g\left(x\right)dx$ $\int_a^{b}g\left(x\right)dx$ both converge or both diverge. Note that even though both of these integrals may converge, their individual limiting values are not necessarily the same value.

Improper Integrals Practice

Improper Integrals: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of improper integrals.

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