Recipe

Evaluate an improper integral with an infinite discontinuity

Use when: the integrand has a vertical asymptote at an endpoint or interior point of the interval of integration

Goal: a convergence verdict for each piece touching the discontinuity, and the integral's exact value when every piece converges

Taught in Improper integrals with infinite discontinuities · extends Evaluate an improper integral with an infinite limit

1. 1

   Find where the integrand is undefined or unbounded inside the interval.

   Example

   \displaystyle\int_0^3\frac{dx}{(x-1)^2} has its asymptote at x=1, inside [0,3], so split it: \displaystyle\int_0^1\frac{dx}{(x-1)^2}+\int_1^3\frac{dx}{(x-1)^2}.

   • the discontinuity sits at an endpoint of the interval — no split needed; go to Step 2 with that endpoint as the one-sided limit.
   • the discontinuity sits at an interior point c — split the integral into two pieces at c and check each piece separately.

   Why: treating an interior discontinuity as if FTC Part 2 applied straight across it gives a finite-looking but completely wrong answer.

2. 2

   Rewrite each piece touching the discontinuity as a one-sided limit, approaching the bad point from inside the interval, never at it.

   Example

   \int_0^1\frac{dx}{(x-1)^2} = \lim_{t\to1^-}\int_0^t (x-1)^{-2}\,dx

   Why: the function is never actually defined at the discontinuity, so the only meaningful approach is from the side that stays inside the domain.

3. 3

   Antidifferentiate, evaluate at the finite bounds, then let the variable bound approach the discontinuity.

   Example

   \int_0^t(x-1)^{-2}\,dx = \left[\frac{-1}{x-1}\right]_0^t = \frac{-1}{t-1}-1

   As t\to1^-, t-1\to0^-, so \dfrac{-1}{t-1}\to+\infty. This piece already diverges.

   Why: this is exactly the same limit-taking machinery as an infinite-limit improper integral, just aimed at a finite point instead of at infinity.

4. 4

   If every piece converges, add their values for the total. If any single piece diverges, the whole integral diverges immediately, with no need to check the rest.

   Example

   Since \displaystyle\int_0^1\frac{dx}{(x-1)^2} already diverges, \displaystyle\int_0^3\frac{dx}{(x-1)^2} diverges, without ever evaluating the second piece.

   Why: one divergent piece makes the sum meaningless, no matter how well-behaved the other piece is.