Improper integrals and convergence tests

In one sentence

An improper integral must be split at every point where it misbehaves, and each piece compared with $1/x^p$ — where the threshold runs one way near zero and the other way near infinity.

Why the exam asks it

Because the two thresholds point in opposite directions, and a candidate who remembers only one of them will answer confidently and wrongly. Because conditional convergence gives a clean line to test. And because $\int \sin x / x$ is a single example that decides several options at once.

The idea

The two thresholds

$$\int_0^1 \frac{dx}{x^p} \text{ converges} \iff p < 1, \qquad \int_1^\infty \frac{dx}{x^p} \text{ converges} \iff p > 1.$$

The picture is worth more than the algebra here. Near zero, a mild singularity is survivable and a sharp one is not. Near infinity, you need decay fast enough that the tail has finite area, and $1/x$ is not fast enough.

The consequence, which the paper leans on: $\int_0^\infty dx/x^p$ never converges, for any $p$. Whatever value makes one end work breaks the other. Any option asserting convergence over the whole half-line without splitting is wrong before you compute anything.

The procedure

1. Find every bad point: where the integrand is unbounded, and each infinite endpoint.
2. Split so each piece has exactly one problem.
3. Compare each piece with $1/x^p$, choosing $p$ from the local behaviour.
4. The whole converges only if every piece does.

Step 2 is where the marks are. An integral with a singularity at $0$ and an infinite tail is two questions, and an option can be built so that one half converges and the other does not.

Conditional convergence

$$\int_1^\infty \frac{\sin x}{x}\,dx \text{ converges}, \qquad \int_1^\infty \left|\frac{\sin x}{x}\right| dx \text{ diverges}.$$

The first converges by Dirichlet's test: $\sin x$ has bounded partial integrals and $1/x$ decreases to zero. The second diverges because $|\sin x|$ has average value $2/\pi$, so the integrand is comparable to $1/x$ in the long run.

This is the integral analogue of $\sum (-1)^n/n$, and it plays the same role: the standard witness that convergence does not imply absolute convergence.

Where intuition breaks

"$\int_0^\infty f$ converges, so $f(x) \to 0$." False, and the counterexample is worth being able to build rather than merely recall.

Put a spike at each integer $n$: height $n$, width $2/n^3$. Each spike has area $n \cdot 2/n^3 = 2/n^2$, and $\sum 2/n^2$ converges, so the total area is finite. But the function reaches height $n$ at $x = n$, so it is unbounded and certainly does not tend to zero.

Why it is tempting: it is exactly true for series in the sense that $\sum a_n$ convergent forces $a_n \to 0$. Integrals have no such constraint, because area can be made small by making a spike thin rather than short. A converse assumed, imported from the wrong setting.

"$f \to 0$, so $\int f$ converges." $f = 1/x$. The same failure as $\sum 1/n$, and equally reliable as a distractor.

"$\int f$ converges, so $\int |f|$ converges." $\sin x / x$. Conditional convergence for integrals is as real as for series.

"$\int f$ and $\int g$ converge, so $\int fg$ converges." Take $f = g = 1/\sqrt x$ on $(0,1)$. Each converges, since $p = 1/2 < 1$. The product is $1/x$, which does not.

"It converges at infinity, so it converges." The question is whether it was split. $\int_0^\infty e^{-x}/\sqrt{x}\, dx$ converges at both ends and is fine; $\int_0^\infty dx/x$ fails at both. An option naming only one end is testing whether you checked the other.

The exam's angle

1. List the bad points before doing anything else. Zeros of the denominator, endpoints, infinity.
2. Near a singularity at $a$, ask what power of $(x-a)$ the integrand behaves like. Below $1$ converges.
3. Near infinity, ask what power of $x$ it decays like. Above $1$ converges.
4. If the integrand oscillates, separate convergence from absolute convergence. Dirichlet's test handles the first; comparison usually settles the second.
5. Check the option's interval. Half these questions are the same integrand over $(0,1)$, $(1,\infty)$ and $(0,\infty)$, with different answers.

The night before

• $\int_0^1 dx/x^p$ converges iff $p < 1$. $\int_1^\infty dx/x^p$ converges iff $p > 1$. Opposite directions.
• Therefore $\int_0^\infty dx/x^p$ never converges. Always split.
• $\int_1^\infty \sin x / x$ converges; the absolute version diverges. The standard conditional-convergence witness.
• Convergence of $\int f$ does not force $f \to 0$: thin tall spikes, height $n$, width $2/n^3$.
• $f \to 0$ does not force convergence: $1/x$.
• Products of convergent improper integrals can diverge: $1/\sqrt x$ against itself on $(0,1)$.