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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?

Did
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Arpan Dutta
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    The Mathematica command $$ Integrate[Cos[x^3 - x], {x, 0, Infinity}] $$ produces $$\pi AiryAi[-(1/3^{1/3})])/3^{1/3}.$$ – user64494 Aug 03 '13 at 05:03

3 Answers3

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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 $\le \frac{6x}{(3x^2-1)^2}$.

André Nicolas
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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 $x\to\infty$}}$.

Did
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  • I do not think $\omega$ being increasing and convex is sufficient for the terms in the alternating series you refer to to converge to zero. The identity function is a counter-example, as is $\omega(x) = \sqrt{x^2+1}$ (or more generally any function with an oblique asymptote). – user88377 Aug 03 '13 at 01:32
  • @user88377 Indeed, I tried to compress the conditions on $\omega$, and got misled. Revised to the previous version. Thanks. – Did Aug 03 '13 at 06:50
  • In the last, you assert $\int_0^\infty\cos(\omega(x)),\mathrm dx$ converges under the red condition, it is an nice result. I have a question: Under the same condition, can we prove that $\int_0^\infty\cos(\omega(x)),\mathrm dx$ is absolutely convergent! – Riemann Apr 19 '23 at 03:02
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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.

paul garrett
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