6.4 · Improper integrals and convergence

An infinite bound is a limiting process

Define $\displaystyle \int_a^\infty f(x)dx$ as $\displaystyle \lim_{b\to\infty}\int_a^b f(x)dx$. The integral converges only if this limit is finite. Evaluate the finite-bound integral first, then take the limit.

For $\displaystyle \int_1^\infty x^{-2}dx$, the finite integral is $\displaystyle 1-1/b$, which tends to $\displaystyle 1$. An unbounded interval can enclose finite area.

$$\int_a^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$$

A singular endpoint also requires a limit

For an integrand unbounded at the left endpoint, replace that endpoint by a variable approaching from inside the interval. The integral $\displaystyle \int_0^1x^{-1/2}dx$ becomes a right-hand endpoint limit and converges to $\displaystyle 2$.

For an interior singularity, split the integral into two improper integrals. Both must converge separately. Opposite infinite contributions do not cancel in the ordinary improper-integral definition.

Distinguish the two power tests

The tail $\displaystyle \int_1^\infty x^{-p}dx$ converges exactly when $\displaystyle p>1$. The near-zero integral $\displaystyle \int_0^1x^{-p}dx$ converges exactly when $\displaystyle p<1$. At $\displaystyle p=1$, logarithmic divergence occurs in both cases.

Do not transfer the tail test unchanged to a finite singular endpoint. Locate the source of impropriety before deciding which limit and comparison apply.