Recipe

Evaluate an improper integral with an infinite limit

Use when: at least one limit of integration is infinity or negative infinity

Goal: a convergence verdict, and the integral's exact value when it converges

Taught in Improper integrals with infinite limits · a later section extends this recipe with more cases

1. 1

   Rewrite the improper integral as a limit of a proper (ordinary) definite integral, replacing infinity with a variable t.

   Example

   \int_2^\infty \frac{1}{x^3}\,dx = \lim_{t\to\infty}\int_2^t x^{-3}\,dx

   Why: FTC Part 2 only applies to a definite integral over a closed, finite interval; infinity is never a legal bound to plug directly into it.

2. 2

   Antidifferentiate and evaluate at the finite bounds 2 and t, exactly as with any ordinary definite integral.

   Example

   \int_2^t x^{-3}\,dx = \left[-\frac{1}{2x^2}\right]_2^t = -\frac{1}{2t^2}+\frac18

   Why: this step uses no new technique at all; every earlier integration method in this chapter still applies here.

3. 3

   Let t\to\infty and see whether the resulting expression settles on a finite number.

   Example

   As t\to\infty, -\dfrac{1}{2t^2}\to0, so the limit is \dfrac18.

   • the limit exists and is finite — the improper integral converges to that value.
   • the limit is infinite, or doesn't exist at all — the improper integral diverges.

   Why: convergence or divergence of the improper integral is, by definition, exactly whether this limit exists and is finite.

4. 4

   State whether the integral converges (and to what value) or diverges.

   Example

   \int_2^\infty\frac{1}{x^3}\,dx = \frac18 \quad\text{(convergent)}
   • Not yet: the improper behavior comes from an interior discontinuity in the integrand instead of an infinite limit — covered in Improper integrals with infinite discontinuities

   Why: a numeric value alone, without stating convergent or divergent, doesn't answer the question an improper integral asks.