Does $\int\limits_0^\infty \cos(x^3 -x) \, \mathrm dx$ converge? 
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? 
 A: 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 $x\to\infty$}}$.

A: In addition to @AndreNicolas' good answer, there is a somewhat faster or symbollically-lighter (I always have to think a little carefully to get integration-by-parts exactly right) heuristic of approximating (!?!) the $x^3-x$ by $x^3$ because the oscillation of cosine is dominated by that, then changing variables by replacing $x$ by $x^{1/3}$, gives $\int_0^\infty {\cos x\over 3x^{2/3}}\;dx$, which is "alternating decreasing" in a slightly generalized sense, therefore convergent although not absolutely so.
A: Integration by parts is useful here, and can be tweaked to give good estimates of the integral. 
We can start at $x=1$, since the function is well-behaved in the interval $[0,1]$. 
Let $u=\frac{1}{3x^2-1}$ and $dv=(3x^2-1)\cos(x^3-x)\,dx$. Then we have $du=-\frac{6x}{(3x^2-1)^2}\,dx$ and we can take $v=\sin(x^3-x)$. Now when we integrate from $1$ to $M$, everything turns out nicely. 
For note that $uv\to 0$ as $M\to\infty$, and the integral that is left to do converges absolutely because the integrand has absolute value $\le \frac{6x}{(3x^2-1)^2}$.
