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I am trying to prove that the improper integral $\int_{0}^{\infty} \cos(x^3)\mathrm{d}x$ is convergent (which seems counterintuitive since the integrand doesn't tend to $0$ as $x$ tends to $\infty$). I've tried integration by parts or looking for an appropriate trigonometric expression, but I don't get successful results. Any ideas?

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    $\begingroup$ Let $u = x^3$, with $dx = \frac{du}{3u^{2/3}}$. Then integrate by parts. $\endgroup$ – Joey Zou Sep 25 '16 at 8:07
  • $\begingroup$ ... you should obtain $\dfrac{\Gamma(\frac{1}{3})}{2\sqrt{3}}$. $\endgroup$ – Jean Marie Sep 25 '16 at 8:10
  • $\begingroup$ Thanks! Just realised the power of Euler's Gamma Function! However, the problem was posed in a part of the problem set where Euler's Gamma Function wasn't still introduced (note that it is only asking for convergence, not a concrete value), so I would appreciate any other solution regarding standard convergence criteria for improper integrals (direct comparison, Dirichlet, etc). $\endgroup$ – Mr Peanutbutter Sep 25 '16 at 9:16
  • $\begingroup$ Now, I have realized that in fact, I have fullfilled the mission (proving the convergence of the integral) by bringing back the issue to a known (converging) integral. But I understand that you would prefer a proof by majorations, bracketting between convergent series, etc... $\endgroup$ – Jean Marie Sep 25 '16 at 10:10
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I address the initial question by proving that the looked for integral is a recognized integral as a certain cosine transform. Working this way, we obtain the exact value of this integral.

Let us start from the expression given by the change of variables $X^3=x \Longleftrightarrow X=x^{1/3}$ suggested by @Joey Zou:

$$\tag{1}I:=\int_0^{\infty}cos(X^3)dX=\frac{1}{3}J \ \ \text{with} \ \ J:=\int_0^{\infty}cos(x)x^{-2/3}dx$$

Let us establish that:

$$\ \tag{2} I=\dfrac{\Gamma(\frac{1}{3})}{2\sqrt{3}}$$

by using the following result, deduced from a table of cosine transforms (classical tables of Gradshteyn and Ryzhik), valid for all $\nu$ such that $0 \leq \nu \leq 1$.

$$\ \sqrt{\frac{2}{\pi}}\int_0^{\infty}\cos(\xi x)x^{-\nu}dx=\sqrt{\frac{\pi}{2}}\dfrac{1}{\Gamma(\nu)}\dfrac{1}{\cos(\frac{\pi \nu}{2})}\xi^{\nu-1}$$

With $\xi=1$ and $\nu=\frac{2}{3}$, we get

$$\ \tag{3} J=\int_0^{\infty}\cos(x)x^{-\frac{2}{3}}dx=\dfrac{\pi}{2}\dfrac{1}{\Gamma(2/3)}\dfrac{1}{\cos(\frac{\pi}{3})}=\pi\dfrac{1}{\Gamma(2/3)}$$

Setting $x=1/3$ in the complement's formula

$$\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin{\pi x}}$$

we have:

$$\tag{4}\Gamma(2/3)=\dfrac{2\pi}{\sqrt{3} \Gamma({1/3})}$$

Taking account of (3) and (4) in the definition of $J$ (relationship (1)) gives (2).

As an additional information, here is a graphical representation of $f$ defined by $f(x)=\cos(x^3)$. The convergence of its integral over $(0,\infty)$ can be understood by a mutual (approximate) annihilation of the positive and negative areas of the increasingly smaller "spikes" of the curve.

For example $\ \int_0^{3}cos(x^3)dx\approx 0.808977$ is rather close to the exact value $\ \int_0^{\infty}cos(x^3)dx=\dfrac{\Gamma(\frac{1}{3})}{2\sqrt{3}}\approx 0.773343$.

enter image description here

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