Improper Integrals

You’ll be able to

Define improper integrals over infinite intervals and at discontinuities
Evaluate improper integrals using limits
Determine whether an improper integral converges or diverges

When an integral becomes improper

A definite integral is improper if either the interval is infinite (a limit is ±∞) or the integrand has an infinite discontinuity (a vertical asymptote) somewhere on the interval. You cannot just plug endpoints into an antiderivative in these cases. Instead you replace the troublesome limit with a variable, integrate normally, and then take a limit. If that limit is a finite number, the integral converges; if it is infinite or does not exist, it diverges.

Improper integral over an infinite interval

∫ₐ^∞ f(x) dx = lim(b→∞) ∫ₐᵇ f(x) dx

Convert the infinite bound to a finite b, integrate, then let b → ∞. A finite result means convergence.

The p-integral benchmark

The most important convergence fact is the behavior of ∫₁^∞ 1/xᵖ dx: it converges when p > 1 and diverges when p ≤ 1. This single result — often called the p-test — anchors most convergence questions and directly parallels the p-series test in Unit 8. Notably ∫₁^∞ 1/x dx diverges (p = 1) even though 1/x → 0, a reminder that a function shrinking to zero is not enough to guarantee finite area.

Worked example

Evaluate ∫₁^∞ 1/x² dx.

1.Rewrite with a limit: ∫₁^∞ x⁻² dx = lim(b→∞) ∫₁ᵇ x⁻² dx.
2.Find the antiderivative: ∫x⁻² dx = −x⁻¹ = −1/x.
3.Evaluate from 1 to b: [−1/x]₁ᵇ = (−1/b) − (−1/1) = 1 − 1/b.
4.Take the limit as b → ∞: 1/b → 0, so the value is 1 − 0 = 1.

Answer: The integral converges to 1. Since p = 2 > 1, convergence is guaranteed by the p-test, and the exact value is 1.

Watch out

Never evaluate an improper integral by directly substituting ∞ into an antiderivative — you must write it as a limit first. Skipping the limit notation loses credit on free response even when the numerical answer is right.

Checkpoint

Does ∫₁^∞ (1/x) dx converge or diverge, and to what?

Checkpoint

For which values of p does ∫₁^∞ (1/xᵖ) dx converge?

On the exam

Improper integrals also arise from a vertical asymptote inside or at an endpoint of the interval, such as ∫₀¹ 1/√x dx. Split at the discontinuity and take a one-sided limit there — do not treat it as an ordinary definite integral.

Answer the 2 checkpoints as you read.