TITLE: Double Integral $\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x+y)\,dx\,dy=(\gamma+2\log 2)\pi^2$
QUESTION [23 upvotes]: Hi I am trying to solve this double integral
$$
I:=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x+y)\,dx\,dy=(\gamma+2\log 2)\pi^2.
$$
Thank you.
The constant in the result is given by $\gamma\approx .577$, and is known as the Euler-Mascheroni constant.  I was thinking to write
$$
I=\Re \bigg[\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt{xy}}\, e^{i(x+y)}\, dx\, dy\bigg]
$$
and using Leibniz's rule for differentiation under the integral sign to write
$$
I(\eta, \xi)=\Re\bigg[ \int_0^\infty \int_0^\infty \ \frac{\log (\eta x)\log(\xi y)}{\sqrt{xy}} e^{i(x+y)}dx\,dy.   \bigg]\\
$$
After taking the derivatives it became obvious that I need to try another method since the x,y constants cancel out.  How can we solve this integral I?  Thanks.

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
    {\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
    =\bracks{\gamma + 2\ln\pars{2}}\pi^{2}:\ {\large ?}}$

\begin{align}
I&=\Re\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\expo{\ic\pars{x + y}}\,\dd x\,\dd y
=\Re\braces{\bracks{\color{#c00000}{\int_{0}^{\infty}
{\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}}^{2}}
\end{align}

\begin{align}
&\color{#c00000}{\int_{0}^{\infty}
{\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\lim_{\mu \to -1/2}\partiald{}{\mu}\
\overbrace{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x}
^{\ds{t\ \equiv\ -\ic x\ \imp\ x\ =\ \ic t}}\
\\[3mm]&=\lim_{\mu \to -1/2}\partiald{}{\mu}
\int_{0}^{-\ic\infty}\expo{\ic\pi\mu/2}t^{\mu}\expo{-t}\,\ic\,\dd t
\\[3mm]&=\ic\lim_{\mu \to -1/2}\partiald{}{\mu}\braces{\expo{\ic\pi\mu/2}\bracks{%
\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd t
-\overbrace{\left.\lim_{R \to \infty}\int_{-\pi/2}^{0}z^{\mu}\expo{-z}\,\dd z\,
\right\vert_{z\ \equiv\ R\expo{\ic\theta}}}^{\ds{=\ 0}}}}
\\[3mm]&=\ic\lim_{\mu \to -1/2}\partiald{}{\mu}
\bracks{\expo{\ic\pi\mu/2}\Gamma\pars{\mu + 1}}
\end{align}
where $\ds{\Gamma\pars{z}}$ is the
Gamma Function
${\bf\mbox{6.1.1}}$.

\begin{align}
I&=\color{#c00000}{\int_{0}^{\infty}{%
\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\ic\lim_{\mu \to -1/2}
\bracks{\expo{\ic\pi\mu/2}\,{\ic\pi \over 2}\,\Gamma\pars{\mu + 1}
+\expo{\ic\pi\mu/2}\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1}}
\end{align}
  where $\ds{\Psi\pars{z}}$ is the
  Digamma Function ${\bf\mbox{6.3.1}}$.

\begin{align}
I&=\color{#c00000}{\int_{0}^{\infty}{%
\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\ic\expo{-\ic\pi/4}\Gamma\pars{\half}
\bracks{{\ic\pi \over 2} + \Psi\pars{\half}}
\\[3mm]&=\root{\pi \over 2}\pars{1 + \ic}\bracks{{\ic\pi \over 2} - \gamma - 2\ln\pars{2}}
\end{align}
$\ds{\gamma}$ is the
Euler-Mascheroni Constant ${\bf\mbox{6.1.3}}$ and we used the identities
$\ds{\Gamma\pars{\half} = \root{\pi}}$ and
$\ds{\Psi\pars{\half}=-\gamma - 2\ln\pars{2}}$.

\begin{align}
I&=\Re\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\expo{\ic\pars{x + y}}\,\dd x\,\dd y
\\[3mm]&=\Re\pars{\braces{\root{\pi \over 2}\pars{1 + \ic}\bracks{{\ic\pi \over 2} - \gamma - 2\ln\pars{2}}}^{2}}
\\[3mm]&=\Re\pars{{\pi \over 2}\,2\ic\braces{\bracks{\gamma + 2\ln\pars{2}}^{2} - {\pi^{2} \over 4} - \ic\pi\bracks{\gamma + 2\ln\pars{2}}}}
\end{align}

$$\color{#00f}{\large%
I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
=\bracks{\gamma + 2\ln\pars{2}}\pi^{2}}
$$