Often you cannot find an antiderivative, yet you only need to know whether an improper integral converges. Compare the integrand with a simpler function whose integral you already know.
Why it works
Suppose $0 \le f(x) \le g(x)$ for $x \ge a$. The area under $f$ is part of the area under $g$. If $\int_a^{\infty} g$ is finite, then $F(t) = \int_a^{t} f$ grows as $t$ grows (because $f \ge 0$) but never passes $\int_a^{\infty} g$. A function that keeps increasing and stays below a ceiling must level off at a limit, so $\int_a^{\infty} f$ converges. In the other direction, if the smaller area $\int_a^{\infty} f$ is infinite, the larger one $\int_a^{\infty} g$ must be infinite too.
Key idea
If $0 \le f(x) \le g(x)$ on $[a, \infty)$: $\int_a^{\infty} g$ converges $\Rightarrow$ $\int_a^{\infty} f$ converges, and $\int_a^{\infty} f$ diverges $\Rightarrow$ $\int_a^{\infty} g$ diverges. Smaller than a finite area, or bigger than an infinite one: those are the only two conclusions. The same holds near a vertical asymptote.
| Benchmark | Behavior |
|---|---|
| $\int_1^{\infty}\frac{dx}{x^{p}}$ | converges for $p > 1$, diverges for $p \le 1$ |
| $\int_0^{1}\frac{dx}{x^{p}}$ | converges for $p < 1$, diverges for $p \ge 1$ |
| $\int_0^{\infty} e^{-cx}\,dx$, $c > 0$ | converges (to $\frac{1}{c}$) |
Worked example: Smaller than a convergent integral
Does $\int_1^{\infty}\frac{dx}{x^{2} + 3}$ converge?
- For $x \ge 1$, $x^{2} + 3 > x^{2}$, so $0 < \frac{1}{x^{2} + 3} < \frac{1}{x^{2}}$.
- $\int_1^{\infty}\frac{dx}{x^{2}}$ converges ($p = 2 > 1$), so the smaller integral converges too.
Answer: It converges (to some number less than 1).
Worked example: Bigger than a divergent integral
Does $\int_1^{\infty}\frac{2 + \cos x}{x}\,dx$ converge?
- Since $\cos x \ge -1$, the integrand is at least $\frac{1}{x} > 0$.
- $\int_1^{\infty}\frac{dx}{x}$ diverges, so the bigger integral diverges too.
Answer: It diverges.
Worked example: Beating a function with no antiderivative
Does $\int_1^{\infty} e^{-x^{2}}\,dx$ converge?
- $e^{-x^{2}}$ has no elementary antiderivative, so compare. For $x \ge 1$, $x^{2} \ge x$, so $e^{-x^{2}} \le e^{-x}$.
- $\int_1^{\infty} e^{-x}dx = \frac{1}{e}$ converges, so $\int_1^{\infty} e^{-x^{2}}dx$ converges (to less than $\frac{1}{e}$).
Answer: It converges.
Common mistake
Comparing in the wrong direction. "$\frac{1}{x^{2} + 3} < \frac{1}{x}$ and $\int\frac{dx}{x}$ diverges" proves nothing: being smaller than an infinite area allows any outcome. Likewise "bigger than something that converges" proves nothing. Check before concluding: convergence needs smaller than convergent; divergence needs bigger than divergent.
When an inequality is awkward to set up, the limit comparison test helps: if $\frac{f(x)}{g(x)} \to L$ with $0 < L < \infty$, then $\int^{\infty} f$ and $\int^{\infty} g$ both converge or both diverge. You will use the same idea for series in Comparison and Limit Comparison Tests.