Evaluate the double integral $\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)} \,dx\,dy$ We know that
$$\int _0^1\int _0^1\frac{x-1}{(1+x y) \ln  (x y)} \, dx\,dy=\gamma$$
$$\int _0^1\int _0^1\frac{x+1}{(1-x y) \ln  (x y)}\,dx\,dy=\ln \frac4\pi$$
I wonder what would be 
$$\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln  (x y)}\,dx\,dy$$
Mathematica fails.
 A: It is quite easy to prove that:
$$\iint_{(0,1)^2}\frac{(xy)^n}{\log(xy)}\,dx\,dy = -\frac{1}{n+1},\tag{1}$$
$$\iint_{(0,1)^2}\frac{x^{n+1}y^n}{\log(xy)}\,dx\,dy = -\log\frac{n+2}{n+1},\tag{2}$$
hence:
$$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{(x+i)}{(1-i xy)\log(xy)}\,dx\,dy = \sum_{n\geq 0}\iint_{(0,1)^2}\frac{i^n(x+i)(xy)^n}{\log(xy)}\,dx\,dy\\&=&-\sum_{n\geq 0}i^n \left(\frac{i}{n+1}+\log\frac{n+2}{n+1}\right)=\frac{1}{2}\log 2-\frac{\pi}{4}i+\sum_{n\geq 0}i^n\left(\log(n+1)-\log(n+2)\right)\\&=&\frac{1}{2}\log 2-\frac{\pi}{4}i+\color{red}{S_1}+i\color{blue}{S_2}\tag{3}\end{eqnarray*} $$
where:
$$\color{red}{S_1} = \sum_{n\geq 0}(-1)^n\left(\log(2n+1)-\log(2n+2)\right),$$
$$\color{blue}{S_2} = \sum_{n\geq 0}(-1)^n\left(\log(2n+2)-\log(2n+3)\right),\tag{4}$$
from which:
$$ \color{red}{S_1} = \log\prod_{n=0}^{+\infty}\frac{4n+1}{4n+2}\cdot\frac{4n+4}{4n+3},$$
$$ \color{blue}{S_2} = \log\prod_{n=0}^{+\infty}\frac{4n+2}{4n+3}\cdot\frac{4n+5}{4n+4}.\tag{5}$$
Using now the Euler product for the $\Gamma$ function it is not difficult to check that:
$$ \color{red}{S_1}=\log\frac{\sqrt{2\pi^3}}{\Gamma\left(\frac{1}{4}\right)^2},\qquad \color{blue}{S_2}=\log\frac{4\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}.\tag{6}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}\int_{0}^{1}{x + \ic \over \pars{1 - \ic xy}\ln\pars{xy}}
     \,\dd x\,\dd y:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large
\int_{0}^{1}\int_{0}^{1}{x + \ic \over \pars{1 - \ic xy}\ln\pars{xy}}
\,\dd x\,\dd y}
=\int_{0}^{1}\pars{1 + {\ic \over x}}
\int_{0}^{x}{\dd y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd x
\\[5mm]&=\int_{0}^{1}{\dd y \over \pars{1 - \ic y}\ln\pars{y}}
\int_{y}^{1}\pars{1 + {\ic \over x}}\,\dd x
=\int_{0}^{1}{1 - y -\ic\ln\pars{y} \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y
\\[5mm]&=\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y
-\ic\int_{0}^{1}{\dd y \over 1 - \ic y}\,\dd y
\\[5mm]&=\dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y}
+\half\,\ln\pars{2} - {1 \over 4}\,\pi\ic\tag{1}
\end{align}

Lets evaluate the '$\dsc{\mbox{red integral}}$':
\begin{align}&\overbrace{
\dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y}}
^{\ds{\dsc{y}=\dsc{\expo{-t}}\ \imp\ \dsc{t}=\dsc{-\ln\pars{y}}}}
=\int_{\infty}^{0}
{1 - \expo{-t} \over \pars{1 - \ic\expo{-t}}\pars{-t}}\,\pars{-\expo{-t}\,\dd t}
=\int_{0}^{\infty}
{\expo{-2t} - \expo{-t} \over 1 - \ic\expo{-t}}\,{\dd t \over t}
\\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \ic^{n}
\int_{0}^{\infty}\bracks{\expo{-\pars{n + 2}t} - \expo{-\pars{n + 1}t}}
\,{\dd t \over t}
\\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \ic^{n}
\bracks{\pars{n + 2}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{n + 2}t}\,\dd t
-\pars{n + 1}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{n + 1}t}\,\dd t}
\end{align}

Since
$\ds{\left.\int_{0}^{\infty}\ln\pars{t}\expo{-a t}\,\dd t\,
     \right\vert_{\Re\pars{a}\ >\ 0} = -\,{\gamma + \ln\pars{a} \over a}}$:
\begin{align}&\dsc{\int_{0}^{1}{1 - y \over \pars{1 - \ic y}\ln\pars{y}}\,\dd y}
=-\sum_{n\ =\ 0}^{\infty}\ \ic^{n}\ln\pars{n + 2 \over n + 1}
\\[5mm]&=-\sum_{n\ =\ 0}^{\infty}\ \pars{-1}^{n}\ln\pars{n + 1 \over n + 1/2}
-\ic\sum_{n\ =\ 0}^{\infty}\ \pars{-1}^{n}\ln\pars{n + 3/2 \over n + 1}
\\[5mm]&=\sum_{n\ =\ 0}^{\infty}\ \bracks{
\ln\pars{n + 1 \over n + 3/4} -\ln\pars{n + 1/2 \over n + 1/4}}
+\ic\sum_{n\ =\ 0}^{\infty}\ \bracks{
\ln\pars{n + 5/4 \over n + 1} -\ln\pars{n + 3/4 \over n + 1/2}}
\\[5mm]&=\fermi\pars{1 \over 4} + \fermi\pars{\half}\ic\quad\mbox{where}\quad
\fermi\pars{a}\equiv
\sum_{n\ =\ 0}^{\infty}\ \bracks{
\ln\pars{n + a + 3/4 \over n + a + 1/2} -\ln\pars{n + a + 1/4 \over n + a}}
\end{align}
$\ds{\fermi\pars{a}}$ is easily evaluated by expressing the $\ds{\ln}$'s functions in integral form:
$\ds{\ln\pars{n + \beta \over n + \alpha}
     =\pars{\beta - \alpha}\int_{0}^{1}{\dd t \over n + \alpha + \pars{\beta - \alpha}t}}$. Sum over $\ds{n}$ becomes somehow Digamma Functions which are integrated to yield $\ds{\ln\Gamma}$'s functions. The final result is:
$$
\fermi\pars{a}=\ln\pars{\Gamma\pars{a + 1/4}\Gamma\pars{a + 1/2}\over
\Gamma\pars{a}\Gamma\pars{a + 3/4}}
$$
which yields:
\begin{align}
\fermi\pars{1 \over 4}&=
\ln\pars{\root{\pi}\Gamma\pars{3/4} \over \Gamma\pars{1/4}}
\\[5mm]
\fermi\pars{\half}&=
\ln\pars{\Gamma\pars{3/4} \over \root{\pi}\Gamma\pars{5/4}}
\end{align}


The evaluation is completed.

