Demonstrate the following trivial equation:
$$\int_0^{\infty} \int_0^{\infty}\frac{\log x \log y}{\sqrt{xy}} \cos(x + y)dxdy = (\gamma + 2 \log 2) \pi^2$$
where $\gamma$ is the Euler-Mascheroni constant.
Proposed by Wan Lang Zonneveld
This seems to me to be a matter of symmetry.
I don't know about solving it by symmetry, but here's one approach (provided you know $\psi\left(\frac12\right)=-\gamma-\ln 4$):$$\begin{align}\int_0^\infty x^{s-1}\exp-ixdx&=\Gamma(s)i^{-s}=\Gamma(s)\exp\frac{-\pi is}{2}\\\implies\int_0^\infty x^{s-1}\exp-ix\ln xdx&=\left(\psi(s)-\frac{\pi i}{2}\right)\Gamma(s)\exp\frac{-\pi is}{2}\\\implies\int_0^\infty\frac{1}{\sqrt{x}}\exp-ix\ln xdx&=\left(\gamma+\ln 4+\frac{\pi i}{2}\right)\sqrt{\pi}\exp\frac{-\pi i}{4}.\end{align}$$ Your integral is the real part of the square of this, i.e. $\pi^2(\gamma+\ln 4)$.