Write the limit, then evaluate BC only

The definition of $\int_{a}^{b} f$ assumes a bounded interval and a bounded function. Break either assumption and the symbol needs new meaning.

1. Type 1: an infinite limit of integration. $\int_{1}^{\infty} \frac{dx}{x^{2}}$. The region has no right edge.
2. Type 2: an unbounded integrand. $\int_{0}^{1} \frac{dx}{\sqrt{x}}$. The region has no top.

Both are handled the same way: replace the offending endpoint with a variable, integrate normally, and take a limit.

$$\int_{1}^{\infty} f(x)\,dx = \lim_{b\to\infty}\int_{1}^{b} f(x)\,dx, \qquad \int_{0}^{1} f(x)\,dx = \lim_{a\to 0^{+}}\int_{a}^{1} f(x)\,dx.$$

If the limit exists as a finite number the integral converges to it; otherwise it diverges.

The limit notation is not a formality. Writing $F(\infty) - F(1)$ substitutes a symbol that is not a number into a function, which is not a step that exists. On a free-response question the limit notation carries its own mark, separate from the value.

Worked: $\int_{1}^{\infty}\frac{dx}{x^{2}} = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_{1}^{b} = \lim_{b\to\infty}\left(-\frac{1}{b} + 1\right) = 1$. Notice that the answer came from the lower limit; the upper one contributed nothing, and yet nothing could be concluded until it was shown to contribute nothing.

$\frac{1}{x}$ and $\frac{1}{x^{2}}$ both fall to zero as $x$ grows, and only one of them encloses a finite area.

$$\int_{1}^{\infty}\frac{dx}{x} = \lim_{b\to\infty}\Big[\ln x\Big]_{1}^{b} = \lim_{b\to\infty}\ln b = \infty,$$

which diverges, while $\int_{1}^{\infty}\frac{dx}{x^{2}} = 1$ converges. The integrand tending to zero is necessary for a Type 1 integral to converge and nowhere near sufficient. What matters is how fast.

The standard result worth memorising is the p-test:

$$\int_{1}^{\infty}\frac{dx}{x^{p}} \text{ converges} \iff p > 1, \qquad \int_{0}^{1}\frac{dx}{x^{p}} \text{ converges} \iff p < 1.$$

The two halves point opposite ways, and that is not a misprint. Out at infinity the danger is a tail that decays too slowly, so large $p$ is safe. Near zero the danger is a spike that rises too steeply, so small $p$ is safe. At $p = 1$ both diverge, which is why $\frac{1}{x}$ is the boundary case in both directions.

So $\int_{0}^{1}\frac{dx}{\sqrt{x}} = \lim_{a\to 0^{+}}\Big[2\sqrt{x}\Big]_{a}^{1} = 2$ converges with $p = \frac{1}{2}$, while $\int_{1}^{\infty}\frac{dx}{\sqrt{x}}$ diverges with the same $p$. Same integrand, different question.

The hardest case to notice is a discontinuity strictly inside the interval, because nothing in the notation announces it.

$\int_{-1}^{1}\frac{dx}{x^{2}}$ looks like an ordinary definite integral. Antidifferentiating and evaluating gives $\left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$, a perfectly ordinary number, and it is wrong twice over: the integrand is positive everywhere it is defined, so a negative area is already impossible, and the integrand is unbounded at $x = 0$, which is inside the interval.

The rule: split at the discontinuity and treat each piece as its own improper integral.

$$\int_{-1}^{1}\frac{dx}{x^{2}} = \lim_{c\to 0^{-}}\int_{-1}^{c}\frac{dx}{x^{2}} + \lim_{d\to 0^{+}}\int_{d}^{1}\frac{dx}{x^{2}},$$

and both pieces have to converge for the whole thing to. Here each one diverges, so the integral diverges.

Two habits follow.

1. Check the denominator's zeros against the interval before integrating anything. That is the same check Topic 6.12 required before evaluating a decomposed integral.
2. Never let the two pieces cancel. One piece going to $+\infty$ and the other to $-\infty$ does not produce a finite answer, because the limits are taken independently.

Once the setup is correct, the remaining work is Unit 1's.

1. Exponentials beat powers. $\int_{0}^{\infty} e^{-x}dx = \lim_{b\to\infty}\left(-e^{-b} + 1\right) = 1$, and $\lim_{b\to\infty} b\,e^{-b} = 0$, which is often needed after integrating by parts and may want L'Hospital's Rule from Topic 4.7.
2. Logarithms grow without bound, slowly. $\ln b \to \infty$, however slowly, so any answer that reduces to $\ln b$ diverges.
3. One-sided limits at a Type 2 endpoint. Approaching $0$ from inside $[0, 1]$ means $a \to 0^{+}$, and the sign on that superscript is part of the answer.

Two closing warnings that come up on almost every improper-integral question.

An integral with an infinite bound does not automatically diverge. Both verdicts occur often, so the question always has to be worked.

An antiderivative being finite at the endpoints proves nothing. The evaluation $\left[-\frac{1}{x}\right]_{-1}^{1}$ is finite and the integral diverges. The finiteness has to come out of a limit taken at the bad point, not from substituting into a formula that was never valid across it.