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.

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:
$$ \color{red}{S_1}=\log\frac{\sqrt{2\pi^3}}{\Gamma\left(\frac{1}{4}\right)^2},\qquad \color{blue}{S_2}=\log\frac{4\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}.\tag{6}$$
