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6.13 Evaluating Improper Integrals

Syllabus
2020
Topic
6.13
Level
—

Learning objectives

—LIM-6.A—Evaluate an improper integral or determine that the integral diverges• LIM-6.A Evaluate an improper integral or determine that the integral diverges.• LIM-6.A.1 An improper integral is an integral that has one or both limits infinite or has an integrand that is unbounded in the interval of integration.• LIM-6.A.2 Improper integrals can be determined using limits of definite integrals.• Enduring understanding LIM-6: The use of limits allows us to show that the areas of unbounded regions may be finite.

Improper Integrals Are Defined by Limits

An integral is improper when an endpoint is infinite or the integrand becomes unbounded on the interval. Replace each improper feature by a limit of proper definite integrals; the improper integral converges only when every required limit exists and is finite.

Improper feature Required rewrite
upper endpoint ∞\infty ∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx
unbounded at endpoint aa ∫acf(x) dx=lim⁡t→a+∫tcf(x) dx\int_a^c f(x)\,dx=\lim_{t\to a^+}\int_t^c f(x)\,dx
unbounded at interior point cc split at cc and evaluate two one-sided limits separately

For an infinite interval, ∫1∞x−2 dx=lim⁡b→∞[−x−1]1b=lim⁡b→∞(1−1/b)=1\int_1^{\infty}x^{-2}\,dx=\lim_{b\to\infty}[-x^{-1}]_1^b=\lim_{b\to\infty}(1-1/b)=1. The finite limit means the integral converges to 11.

The integrand 1/x21/x^2 is unbounded at 00, so ∫−11x−2 dx\int_{-1}^{1}x^{-2}\,dx must be split. On the right, lim⁡t→0+∫t1x−2 dx=lim⁡t→0+(1/t−1)=∞\lim_{t\to0^+}\int_t^1x^{-2}\,dx=\lim_{t\to0^+}(1/t-1)=\infty. Because one required one-sided integral diverges, the original integral diverges.

Never substitute ∞\infty into an antiderivative or integrate straight across an interior singularity. Opposite infinite contributions cannot be canceled: if any required one-sided limit fails to be finite, the improper integral diverges.