Calculate an approximation of $\int_0^1\int_0^1\frac{\log(xy)xy}{-1+\log(xy)}dxdy$ Motivation. I was thinking in the more simple case of (6), I am saying the case $r=s=1$, that tell us this MathWorld. Then in this case one can check that the identity holds using integration methods. When I did it I've considered from Hata's integral the following different integral in this 

Question. What's about $$\int_0^1\int_0^1\frac{\log(xy)xy}{-1+\log(xy)}dxdy?$$

Remark. The context was that I was trying with my imagination different expressions for an hypothetical similar (6) but now for $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$, where $\mu(n)$ is the Möbius function. For example I tried also consider $\int_0^1\left(\int_1^\infty\left(\int_1^\infty\frac{dx}{xy(1-(1-xy)z)}\right)dy\right)dz$, but I understand that it is science fiction to find an expression for $\sum_{k=r+1}^\infty\frac{\mu(n)}{n^3}$ from this random method.
My attempt. I can deduce the following statement  $$\int_0^1\int_0^1\frac{\log(xy)xy}{-1+\log(xy)}dxdy=\frac{1}{4}+\int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)},$$ and I know that $$\int\frac{dx}{-1+\log(xy)}=e\frac{\operatorname{Ei}(\log(xy)-1)}{y}+\operatorname{ constant}.$$
Then can we finish the example to get the integral in the Question in terms of particular values of special functions? Thanks in advance.
References:
Hata, A New Irrationality Measure for $\zeta(3)$, Acta Arith. 92, 47-57 (2000).
 A: Change variables $xy/e=t$ to get
$$
\int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)}=\int_0^1 dy \frac{e}{y}\int_0^{y/e}\frac{dt}{\log t}=\int_0^1 dy \frac{e}{y}\mathrm{li}(y/e)\ ,
$$
where $\mathrm{li}(z)$ is the logarithmic integral. The second integral admits an antiderivative
$$
\int dy \frac{1}{y}\mathrm{li}(y/e)=\text{li}\left(\frac{y}{e}\right) \ln \left(\frac{y}{e}\right)-\frac{y}{e}+C\ ,
$$
therefore your double integral reads
$$
\int_0^1\int_0^1\frac{dxdy}{-1+\log(xy)}=-1-e \text{li}\left(\frac{1}{e}\right)\ .
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{0}^{1}\int_{0}^{1}{\ln\pars{xy}xy \over -1 + \ln\pars{xy}}\,\dd x\,\dd y =
\int_{0}^{1}\int_{0}^{1}\ln\pars{xy}xy
\bracks{-\int_{0}^{\infty}\pars{xy \over \expo{}}^{t}\,\dd t}\,\dd x\,\dd y
\\[5mm] = &\
-\int_{0}^{\infty}\expo{-t}\int_{0}^{1}\int_{0}^{1}\ln\pars{xy}x^{t + 1}
y^{t + 1}\,\dd x\,\dd y\,\dd t
\\[5mm] = &\
-2\int_{0}^{\infty}\expo{-t}
\bracks{\int_{0}^{1}\ln\pars{x}x^{t + 1}
\,\dd x}\pars{\int_{0}^{1}y^{t + 1}\,\dd y}\,\dd t =
2\int_{0}^{\infty}{\expo{-t} \over \pars{t + 2}^{3}}\,\dd t
\\[5mm] = &\
2\expo{}^2\int_{2}^{\infty}{\expo{-t} \over t^{3}}\,\dd t =
2\expo{}^{2}\bracks{\mrm{E}_{3}\pars{2} \over 2^{3 - 1}} =
\bbx{\ds{{1 \over 2}\,e^{2}\,\mrm{E}_{3}\pars{2}}} \approx 0.1113
\end{align}

$\ds{\mrm{E}_{p}\pars{z}}$ is the
  Generalized Exponenential Integral Function.

A: \begin{align}
I&=\int_0^1 \int_0^1 \frac{xy\ln(xy)}{xy-1}\ dx \ dy\\
&=\int_0^1 \int_0^1 \frac{xy\ln(x)}{xy-1}\ dx \ dy+\int_0^1 \int_0^1 \frac{xy\ln(y)}{xy-1}\ dx \ dy\\
&=2\int_0^1 \int_0^1 \frac{xy\ln(x)}{xy-1}\ dx \ dy\\
&=2\int_0^1x\ln(x)\left(\int_0^1\frac{y}{xy-1}\ dy\right)\ dx\\
&=2\int_0^1x\ln(x)\left(\frac1x+\frac{\ln(1-x)}{x^2}\right)\ dx\\
&=2\int_0^1\ln x\ dx+2\int_0^1\frac{\ln x\ln(1-x)}{x}\ dx\\
&=2(-1)-2\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\ln x\ dx\\
&=-2+2\sum_{n=1}^\infty\frac1{n^3}\\
&=-2+2\zeta(3)
\end{align}
A: Different approach
\begin{align}
I&=\int_0^1\left(\int_0^1\frac{xy\ln(xy)}{xy-1}\ dx\right)\ dy\overset{xy=z}{=}\int_0^1\left(\int_0^y\frac{z\ln(z)}{y(z-1)}\ dz\right)\ dy\\
&\int_0^1\left(\int_z^1\frac{z\ln(z)}{y(z-1)}\ dy\right)\ dz=\int_0^1\frac{z\ln^2(z)}{1-z}\ dz\\
&=\sum_{n=1}^\infty\int_0^1 z^n \ln^2(z)\ dx=2\sum_{n=1}^\infty\frac1{(n+1)^3}=2(\zeta(3)-1)
\end{align}
