Double Integral $\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x+y)\,dx\,dy=(\gamma+2\log 2)\pi^2$ 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. 
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
    {\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
    =\bracks{\gamma + 2\ln\pars{2}}\pi^{2}:\ {\large ?}}$

\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}

$$\color{#00f}{\large%
I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
=\bracks{\gamma + 2\ln\pars{2}}\pi^{2}}
$$
A: Using the identity 
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
The integral can be written 
$$
I=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\,dx\,dy $$
Now by splitting the integrals
$$\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x)\cos(y)\,dx\,dy-\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\sin(x)\sin(y)\,dx\,dy
$$
Notice by symmetry of the integrals we have
$$\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx \right)^2-\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx \right)^2
$$ 
Both inegrals are solvable by using the mellin transforms 
$$\int^\infty_0 x^{s-1}\sin(x)\,dx =  \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$
$$\int^\infty_0 x^{s-1}\cos(x)\,dx =  \Gamma (s) \cos\left( \frac{\pi s}{2} \right)$$
By differentiation under the integral sign and using $s=\frac{1}{2}$.
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx =-\frac{1}{2} \sqrt{\frac{π}{2}}  \left(2 \gamma +π+\log(16) \right) $$
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx=\frac{1}{2} \sqrt{\frac{π}{2}} (-2 \gamma +π- \log(16))
$$
Collecting the results together we have
$$I=(\gamma+2\log 2)\pi^2$$
A: Consider
$$\int_0^{\infty} dx \, x^{\alpha} e^{i x}$$
We know from Cauchy's theorem that this integral is equal to (when it converges)
$$i \, e^{i \pi \alpha/2} \int_0^{\infty} du \, u^{\alpha} \, e^{-u} = i \, e^{i \pi \alpha/2} \, \Gamma(\alpha+1)$$
Differentiating both sides with respect to $\alpha$, we get
$$\int_0^{\infty} dx \, x^{\alpha} e^{i x}\,  \log{x} = \Gamma(\alpha+1) e^{i \pi \alpha/2} \left [i \, \psi(\alpha+1)-\frac{\pi}{2} \right ] $$
Square both sides:
$$\begin{align}\int_0^{\infty} dx \, x^{\alpha} e^{i x}\,  \log{x} \int_0^{\infty} dy \, y^{\alpha} e^{i y}\,  \log{y} &= \Gamma(\alpha+1)^2 e^{i \pi \alpha} \left [\frac{\pi^2 }{4}-\psi(\alpha+1)^2-i \pi \psi(\alpha+1) \right ] \end{align}$$
Now plug in $\alpha=-1/2$ and consolidate; use the fact that $\Gamma(1/2)=\sqrt{\pi}$ and $\psi(1/2)=-\gamma-2 \log{2}$:
$$\int_0^{\infty} dx \, \int_0^{\infty} dy \frac{\log{x} \log{y}}{\sqrt{x y}} e^{i (x+y)} = -i \pi \left [\frac{\pi^2}{16} - (\gamma+2 \log{2})^2 + i \pi (\gamma+2 \log{2}) \right ]$$
Take the real part of both sides, and get
$$\int_0^{\infty} dx \, \int_0^{\infty} dy \frac{\log{x} \log{y}}{\sqrt{x y}} \cos{(x+y)} = \pi^2 (\gamma+2 \log{2}) $$
as was to be shown.
A: The integral is
$$I=\Re\left(\left(\int_0^{\infty} \frac{\ln x}{\sqrt{x}}e^{ix}\,dx\right)^2\right)$$
Evaluating the definite integral first:
$$J=\int_0^{\infty} \frac{\ln x}{\sqrt{x}}e^{ix}\,dx$$
Use the substitution $\sqrt{x}=e^{i\pi/4}t$ to obtain:
$$J=2e^{i\pi/4}\int_0^{\infty} e^{-t^2}\ln(t^2e^{i\pi/2})\,dt=2e^{i\pi/4}\int_0^{\infty} \left(2\ln te^{-t^2}+\frac{i\pi}{2}e^{-t^2}\right)\,dt$$
Using the following results:
$$\int_0^{\infty} e^{-t^2}\ln t\,dt=-\frac{\sqrt{\pi}}{4}(\gamma+2\ln2)$$
$$\int_0^{\infty} e^{-t^2}\,dt=\frac{\sqrt{\pi}}{2}$$
....we obtain:
$$J=2e^{i\pi/4}\left(-\frac{\sqrt{\pi}}{2}(\gamma+2\ln 2)+\frac{i\pi}{2}\frac{\sqrt{\pi}}{2}\right)$$
$$\Rightarrow e^{i\pi/4}\sqrt{\pi}\left(\frac{i\pi}{2}-(\gamma+2\ln 2)\right)$$
Squaring $J$ and taking the real part,
$$I=\pi^2(\gamma+2\ln 2)$$
