Evaluation of a particular type of integral involving logs and trigonometric function

Question

Is there any closed form for $$ \int _0 ^{\infty}\int _0 ^{\infty}\int _0 ^{\infty} \log(x)\log(y)\log(z)\cos(x^2+y^2+z^2)dzdydx$$ if yes then how to prove it?

Answer 1

Since $\log x \rightarrow \infty$ as $x\rightarrow\infty$, try proving that this integral does not exist (due to the oscilation the $\cos$ factor brings along.

Answer 2

Actually, solving $$J(a):=\int_0^\infty \log(x) \cos(x^2 + a) dx$$ with Maple yields $$J(a) = -\frac{\sqrt{2\pi}}{16} (4\log 2 \cos(a) + 2\cos(a) \gamma + \cos(a) \pi - 4\log 2 \sin(a) - 2\sin(a) \gamma + \sin(a) \pi)$$ Wich gives hope toward the existence of the integral, since $$I = \int_0^\infty \int_0^\infty \log x \log y J(x^2 + y^2)\ dy\ dx$$ And $J$ can be expressed as a linear combination of $\cos(a)$ and $\cos(a + \frac{\pi}2) = \sin(a)$, so $$\int \log x J(x^2+y^2) dx = \sum_i \lambda_i J(?_i)$$

Answer 3

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

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