Improper integrals
Objective. Test convergence at infinite limits and singularities.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
An improper integral has an infinite limit of integration or an integrand that is unbounded at some point in the interval, and is evaluated as a limit, such as ∫[a,∞) f(x)dx = lim(b→∞) ∫[a,b] f(x)dx; the integral converges if this limit exists finitely and diverges otherwise. For ∫₁^∞ 1/x² dx, compute ∫₁^b 1/x² dx = [−1/x]₁^b = 1−1/b, which tends to 1 as b→∞, so the integral converges to 1. By contrast ∫₁^∞ 1/x dx = lim(b→∞)[ln b − ln 1] diverges to infinity. Singularities inside the interval, such as 1/√x near x=0, require the same limiting process applied at the problematic endpoint.
Worked example
Determine whether the improper integral ∫₁^∞ 1/x² dx converges, and if so find its value.
- 1Rewrite as a limit: ∫₁^∞ 1/x² dx = lim(b→∞) ∫₁^b x⁻² dx.
- 2Find the antiderivative: ∫x⁻²dx = −x⁻¹, so ∫₁^b x⁻²dx = [−1/x]₁^b = 1 − 1/b.
- 3Take the limit as b→∞: 1/b→0, so the expression tends to 1.
- 4The limit is finite, so the integral converges to 1.
Answer: Converges to 1
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Practice
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Start practisingCommon mistake. Students treat ∫₁^∞ 1/x dx as convergent by analogy with 1/x² without checking the limit, but ln(b)→∞, so this integral actually diverges; each improper integral must be tested individually via its limit.
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