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Code Scholars Tutoring · AP Calculus BC

Improper integrals: limits before evaluation

Unit 6 · Topics 6.13 · Detailed notes, a worked example, and a self-check with reasoning.

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Understand the idea

An improper integral involves an infinite interval or an unbounded integrand. Its value is defined by a limit of ordinary definite integrals. A finite limit means the improper integral converges; an infinite or nonexistent limit means it diverges. An antiderivative alone does not settle convergence.

Replace an infinite bound with a variable, evaluate the finite integral, and only then take the limit. For an integrand unbounded at a finite endpoint, approach that endpoint from within the interval. The integral can converge even when the function becomes arbitrarily large near an endpoint.

If a singularity lies inside the interval, split the integral there. Both one-sided improper integrals must converge independently. For example, cancellation on symmetric intervals cannot rescue an integral whose two sides separately diverge; that would be a different limiting procedure.

∫1∞x−p dx converges iff p>1;∫01x−p dx converges iff p<1\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1;\qquad \int_0^1 x^{-p}\,dx\text{ converges iff }p<1

A method you can apply

  1. Identify every infinite bound or point where the integrand is unbounded.
  2. Write a separate limit for each improper part, then evaluate its finite integral.
  3. State explicitly whether every required limit is finite before combining the results.

Worked example

Determine whether ∫1∞4/x3 dx\displaystyle \int_1^\infty 4/x^3\,dx converges and find its value.

  1. Write lim⁡b→∞∫1b4x−3 dx\displaystyle \lim_{b\to\infty}\int_1^b4x^{-3}\,dx.
  2. An antiderivative is −2x−2\displaystyle -2x^{-2}, so the finite integral is 2−2/b2\displaystyle 2-2/b^2.
  3. The term 2/b2\displaystyle 2/b^2 approaches 0\displaystyle 0 as b\displaystyle b tends to infinity.
The integral converges to 2\displaystyle 2. This is a finite accumulated value over an infinite interval.

Try it yourself

Does ∫01x−1/2 dx\displaystyle \int_0^1x^{-1/2}\,dx converge? Find its value if it does.

Show solution and reasoning

Use lim⁡a→0+∫a1x−1/2 dx=lim⁡a→0+(2−2a)=2\displaystyle \lim_{a\to0^+}\int_a^1x^{-1/2}\,dx=\lim_{a\to0^+}(2-2\sqrt a)=2. The unbounded endpoint does not prevent convergence. In contrast, ∫01x−1 dx\displaystyle \int_0^1x^{-1}\,dx diverges.

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