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_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 \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 $(0,1)$.