Comparison Test for Improper Integrals

The comparison test is simply a way to decide whether an improper integral is finite by checking it against a simpler one, without ever computing it. The name says it: you compare your function to one you already understand. It's good for the many integrals, like $\int_1^\infty e^{-x^2/2}\,dx$ from the bell curve, that have no antiderivative you can write down. All we're really saying with $0 \le g(x) \le f(x)$ is: $g$ (the smaller function) sits underneath $f$ (the bigger one), so $g$'s area can't be more than $f$'s. It looks like a formal test, but it's really just "less stuff fits under a lower curve"; bear with it.

In the last lesson, every improper integral had an antiderivative you could take a limit of. Try that with $\int_1^\infty \frac{dx}{x^3 + 1}$ and you'd be in for a long partial-fractions slog, and $\int_1^\infty e^{-x^2/2}\,dx$ can't be done at all. But often you don't need the number. You only need to know whether it's finite, and for that, a rough comparison is enough.

The comparison test. Suppose $0 \le g(x) \le f(x)$ for $x \ge a$.

• If $\int_a^\infty f(x)\,dx$ converges, then $\int_a^\infty g(x)\,dx$ converges.
• If $\int_a^\infty g(x)\,dx$ diverges, then $\int_a^\infty f(x)\,dx$ diverges.

Think about why that makes sense. A smaller positive function has less area under it. If the bigger area is finite, it caps everything below it. If the smaller area is infinite, everything above it is infinite too.

A common mistake. It's tempting to run the test the other way, like "$g \le f$ and $\int g$ converges, so $\int f$ converges." That tells you nothing. Being bigger than something finite doesn't stop you from being infinite, and being smaller than something infinite doesn't stop you from being finite. Only the two cases above give a conclusion, so before comparing, decide which way you expect the answer to go and pick a comparison that points that way.

Choosing what to compare with. For large $x$, keep only the terms that dominate and compare with a known integral, usually $\int_1^\infty \frac{dx}{x^p}$ or $\int_0^\infty e^{-x}dx$:

• $\dfrac{1}{x^3 + 1} \le \dfrac{1}{x^3}$, and $p = 3 > 1$, so $\int_1^\infty \frac{dx}{x^3 + 1}$ converges.
• $\dfrac{1 + \sin^2 x}{\sqrt{x}} \ge \dfrac{1}{\sqrt{x}}$, and $p = \tfrac12 \le 1$, so $\int_1^\infty \frac{1 + \sin^2 x}{\sqrt x}dx$ diverges.

The same test works for Type II integrals near a vertical asymptote: compare with $\int_0^1 \frac{dx}{x^p}$, which converges exactly when $p < 1$. Notice that's the opposite direction from the $p$-test at infinity. Near $0$, a smaller power blows up more gently, so it's the small powers that leave a finite area.

Physicists use comparison to show the total energy a system radiates, or the probability of rare events in a heavy-tailed distribution, is finite before anyone computes it. Statisticians use it to decide whether a distribution even has a mean.