Funky triple integral

Note: There was a sign error in the original solution, which I have since fixed.

Problem: Evaluate

$$ \int _0 ^{\infty}\int _0 ^{\infty}\int _0 ^{\infty} \log(x)\log(y)\log(z)\cos(x^2+y^2+z^2)dzdydx$$

Solution:

One way to attack this is to exploit the symmetry of the integral. Start by expanding the cosine term into individual pieces, i.e.,

$$\begin{align}\cos{(x^2+y^2+z^2)} &= \cos{x^2} \cos{(y^2+z^2)} – \sin{x^2} \sin{(y^2+z^2)}\\ &= \cos{x^2} \cos{y^2} \cos{z^2} – \cos{x^2} \sin{y^2} \sin{z^2} \\ &\quad -\sin{x^2} \sin{y^2} \cos{z^2} – \sin{x^2} \cos{y^2} \sin{z^2} \end{align}$$

Next define

$$C = \int_0^{\infty} dx \, \cos{x^2} \: \log{x} $$
$$S = \int_0^{\infty} dx \, \sin{x^2} \: \log{x} $$

Then the triple integral is equal to $C^3-3 C S^2$.

We obtain $C$ and $S$ using Cauchy’s theorem to find $C+i S$. We do this by considering the integral

$$\oint_{\eta} dz \, e^{i z^2} \log{z}$$

where $\eta$ is a circular wedge of radius $R$ and angle $\pi/4$ in the upper half plane with base along the positive real axis. One may show that the integral over the circular arc vanishes as $\pi \log{R}/(4 R)$ as $R \to \infty$. Thus, we have

$$C+i S = e^{i \pi/4} \int_0^{\infty} dt \, e^{-t^2} \left (\log{t} + i \frac{\pi}{4} \right ) $$

One may derive the value of this integral by noting that

$$\int_0^{\infty} dt \, e^{-t^2} \log{t} = -\frac{\sqrt{\pi}}{4} (\gamma + 2 \log{2})$$

This value is derived from using the value of

$$\frac12 \left [\frac{d}{d\alpha} \Gamma \left (\frac{\alpha+1}{2} \right ) \right ]_{\alpha=0} = \frac14 \Gamma \left (\frac12 \right ) \psi \left ( \frac12 \right )$$

We also use

$$\int_0^{\infty} dt \, e^{-t^2} = \frac{\sqrt{\pi}}{2}$$

to obtain

$$C = -\frac18 \sqrt{\frac{\pi}{2}} (\pi + 2 \gamma + 4 \log{2})$$
$$S = \frac18 \sqrt{\frac{\pi}{2}} (\pi – 2 \gamma – 4 \log{2})$$

and therefore the triple integral is

$$C^3-3 C S^2 = \frac{\pi ^{3/2} \left(4 \gamma ^2-8 \gamma \pi +\pi ^2+16 \log{2} \,(\gamma
-\pi +\log{2})\right) (2 \gamma +\pi +4 \log{2})}{512 \sqrt{2}}$$

claimtoken-52e5a4f889775

2 Comments

  • Shivam Patel wrote:

    Actually I first posted it in Maths SE……

  • rlgordonma@yahoo.com wrote:

    Shivam, just to clarify, you posted the problem asking whether there was a closed-form, not a solution. Please note that almost all of these problems are known to come from M.SE. I did notice, however, that I forgot to link to the problem in M.SE. I will do this.

Leave a Reply

Your email is never shared.Required fields are marked *