The Fundamental Theorem (FTC Part 2) needs a continuous function on a closed, bounded interval. Two natural questions break those rules: what is the area under $\frac{1}{x^{2}}$ all the way to infinity, and what is the area under $\frac{1}{\sqrt{x}}$ near $x = 0$, where the graph shoots up? Both answers are limits, like the ones in limits at infinity.
Why it works
An ordinary integral over $[1, t]$ is a finite number for every $t$. Letting $t \to \infty$ asks whether those areas settle down. If the limit exists, the infinite region has a finite area, and the integral converges; if not, it diverges. The same idea handles a vertical asymptote: stop a little short of it, then let the stopping point slide up to it.
Key idea
$$\int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx,$$ $$\int_a^{b} f(x)\,dx = \lim_{s\to a^{+}}\int_s^{b} f(x)\,dx$$ when $f$ blows up at $a$.
If the trouble is inside the interval, split there; the integral converges only if every piece does. Benchmarks: $\int_1^{\infty}\frac{dx}{x^{p}}$ converges exactly when $p > 1$ (to $\frac{1}{p - 1}$); $\int_0^{1}\frac{dx}{x^{p}}$ converges exactly when $p < 1$ (to $\frac{1}{1 - p}$).
Worked example: An infinite interval
Evaluate $\int_1^{\infty}\frac{dx}{x^{2}}$.
- $\int_1^{t}\frac{dx}{x^{2}} = \left[-\frac{1}{x}\right]_1^{t} = 1 - \frac{1}{t}$.
- As $t \to \infty$, $\frac{1}{t} \to 0$, so the integral converges to 1.
Answer: $1$
Worked example: A divergent one
Does $\int_1^{\infty}\frac{dx}{x}$ converge?
- $\int_1^{t}\frac{dx}{x} = \ln t$.
- $\ln t \to \infty$ as $t \to \infty$, so the integral diverges, even though $\frac{1}{x} \to 0$.
Answer: It diverges.
Worked example: A vertical asymptote at an end
Evaluate $\int_0^{1}\frac{dx}{\sqrt{x}}$.
- The integrand blows up at 0, so use $$\begin{aligned} &\lim_{s\to 0^{+}}\int_s^{1} x^{-1/2}\,dx \\ &= \lim_{s\to 0^{+}}\left(2 - 2\sqrt{s}\right). \end{aligned}$$
- The limit is 2: an unbounded function with a finite area.
Answer: $2$
Common mistake
Integrating straight across a vertical asymptote. For $\int_{-1}^{1}\frac{dx}{x^{2}}$, plugging into $\left[-\frac{1}{x}\right]_{-1}^{1}$ gives $-2$, a negative area for a positive function, which is impossible. The FTC does not apply, because $\frac{1}{x^{2}}$ is not continuous at 0. Split at 0: $\int_0^{1}\frac{dx}{x^{2}} = \lim_{s\to 0^{+}}\left(\frac{1}{s} - 1\right) = \infty$, so the whole integral diverges. Check every integrand for points where it is undefined inside the interval before using the FTC.
Worked example: Exponential decay
Evaluate $\int_0^{\infty} e^{-2x}\,dx$.
- $$\begin{aligned} &\int_0^{t} e^{-2x}dx \\ &= \left[-\frac{1}{2}e^{-2x}\right]_0^{t} \\ &= \frac{1}{2} - \frac{1}{2}e^{-2t}. \end{aligned}$$
- $e^{-2t} \to 0$, so the integral converges to $\frac{1}{2}$.
Answer: $\frac{1}{2}$