TITLE: A double sum for the square of the natural logarithm of $2$.
QUESTION [8 upvotes]: I am trying to show

\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^2 2^{n}} =(\ln(2))^2. 
\end{eqnarray*}

Motivation : I want to use this to calculate $ \operatorname{Li}_2(1/2)$. So I want a solution to the above that does not use any reference to dilogarithms and please avoid rational multiples of $\pi^2$ (if possible).
Right, lets turn it into a double integral. ( I know lots of you folks prefer integrals to plums.)
Show

\begin{eqnarray*}
\int_0^1 \int_0^1 \frac{xy \ dx  \ dy}{(1-xy)(2-xy)} =(\ln(2))^2. 
\end{eqnarray*}

Reassuringly Wolfy agrees
My try: Let $u=xy$, & the double integral becomes
\begin{eqnarray*}
\int_0^1 \frac{dy}{y}  \int_0^y \frac{u \ du }{(1-u)(2-u)} . 
\end{eqnarray*}
Partial fractions
\begin{eqnarray*}
 \frac{u}{(1-u)(2-u)} =\frac{1}{1-u} - \frac{2}{2-u}.
\end{eqnarray*}
Do the $u$ integrations to leave the $y$ integrals
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  +2 \int_0^1 \frac{\ln(2-y) dy}{y}.
\end{eqnarray*}
The first integral is
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  = \frac{ \pi^2}{6}.
\end{eqnarray*}
which I was hoping to avoid and even worse Wolfy says the second integral is divergent
So you have a choice of questions, where did I go wrong in the above ? OR how can we show the initially stated result?

REPLY [3 votes]: A solution without dilogarithm.
\begin{align}J&=\int_0^1 \int_0^1 \frac{xy}{(1-xy)(2-xy)}\,dx\,dy\\
&\overset{x=1-uv,y=\frac{1-u}{1-uv}}=\int_0^1 \int_0^1 \frac{1-u}{(1+u)(1-uv)}\,du\,dv\\
&=\int_0^1\int_0^1\left(\frac{2}{(1+u)(1+v)}-\frac{1-v}{(1+v)(1-uv)}\right)\,du\,dv\\
&=\int_0^1\int_0^1\frac{2}{(1+u)(1+v)}\,du\,dv-J\\
&=2\left(\int_0^1\frac{1}{1+u}\,du\right)^2-J\\
&=2\ln^2 2 -J\\
J&=\boxed{\ln^2 2}
\end{align}
NB:
$\displaystyle \,dx\,dy=\frac{u\,du\,dv}{1-uv}$