Recipe

Test an improper integral for convergence

Use when: the integral has an infinite limit of integration, or the integrand is undefined (blows up) somewhere on the interval

Goal: either the exact value the integral converges to, or a converges-or-diverges verdict backed by a valid comparison

Taught in Improper integrals, revisited

1. 1

   Decide whether the trouble is an unbounded interval or a point where the integrand blows up, then rewrite the integral with that trouble spot replaced by a variable t.

   Example

   \displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx = \lim_{t\to\infty}\int_1^t x^{-3}\,dx, an unbounded interval.

   • the interval itself is unbounded, such as a to \infty — rewrite as \lim_{t\to\infty}\int_a^t f(x)\,dx.
   • the integrand blows up at a point c in the interval, often an endpoint — rewrite as a limit of t toward c instead.

   Why: an improper integral is only ever defined as this limit; treating infinity as a number you can plug in skips the definition entirely.

2. 2

   Integrate the function normally up to t. Then take the limit as t moves toward the trouble spot; a finite limit means convergence, an infinite or non-existent one means divergence.

   Example

   \int_1^t x^{-3}\,dx = \frac12-\frac{1}{2t^2}, and as t\to\infty that goes to \frac12, so \int_1^{\infty}x^{-3}\,dx=\frac12, a genuine convergent value.

   • an antiderivative is available and the limit settles to a finite number — the integral converges to that number; done.
   • an antiderivative is available but the limit runs off to infinity or fails to exist — the integral diverges; done.
   • no elementary antiderivative can be found for the integrand — skip evaluating and continue to Step 3, the comparison test.

   Why: the limit's own behavior, finite versus infinite or non-existent, is literally what convergence and divergence mean.

3. 3

   Choose a comparison function g(x), matching the inequality's direction to what needs proving.

   Example

   Since -1\le\cos x\le1, 2\le 3+\cos x\le4, so 0\le\frac{3+\cos x}{x^3}\le\frac{4}{x^3} for every x\ge1.

   • the goal is proving convergence — find g already known convergent, with f(x)\le g(x) on the whole interval.
   • the goal is proving divergence — find g already known divergent, with f(x)\ge g(x) on the whole interval.

   Why: the direction of the inequality is what makes the argument valid; comparing against a smaller convergent integral, or a larger divergent one, proves nothing.

4. 4

   Once the comparison function's own status is settled, the original integral inherits the same converges-or-diverges verdict, with no exact value attached.

   Example

   4\int_1^{\infty}x^{-3}\,dx = 4\left(\frac12\right)=2, a finite number, so \int_1^{\infty}\frac{3+\cos x}{x^3}\,dx converges too, even though its own exact value stays unknown.

   Why: squeezing f underneath a finite g, or above an infinite one, only pins down converges-or-diverges, since an inequality is not an equality.