TITLE: Does $\int\limits_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge?
QUESTION [14 upvotes]: Does $\displaystyle\int_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge?

What is the standard method of checking convergence of this kind of improper integrals?

REPLY [8 votes]: An approach which goes directly to the heart of the matter instead of getting lost into irrelevant computational details, is to start with the basic fact that, after a while, say on $x\geqslant1$, the function $x\mapsto x^3-x$ is increasing, hence, by the change of variable $x\to u=x^3-x$, the convergence of the integral in the question is equivalent to the convergence when $u\to\infty$ of the integral
$$
\int^\infty_0\frac{\cos(u)}{\alpha(u)}\mathrm du,
$$
where the function $\alpha:[0,\infty)\to[0,\infty)$ is uniquely defined by the condition that, for every $x\geqslant1$,
$$
\alpha(x^3-x)=\frac{\mathrm d}{\mathrm dx}(x^3-x)=3x^2-1.
$$
A formula for $\alpha$ is not even useful here (this is a huge computational advantage of the approach), all that matters is that:

$\color{red}{\text{The function $\alpha$ is increasing and unbounded}}$. 

Thus, the integral above converges at infinity because the contributions of the successive $u$-intervals of length $\pi$ starting at every $\frac\pi2+n\pi$ are the terms of an alternating series (their signs alternate and their amplitudes decrease to zero).
More generally, the approach above shows that the integral
$$
\int_0^\infty\cos(\omega(x))\,\mathrm dx
$$
converges as soon as:

$\color{red}{\text{The function $\omega$ is differentiable for $x$ large enough and $\omega'(x)\to\infty$ when $x\to\infty$}}$.