Compute explicitly $\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}(\frac{1}{2})^{x+y}$ It is possible to compute explicitly the following series?
$$\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}\Big(\frac{1}{2}\Big)^{x+y}$$
@Edit I tried to sum and subtract $2xy$ in the numerator.
In this way I get the following
\begin{align}
\sum_{x=1}^\infty\sum_{y=1}^\infty (x+y)\Big(\frac{1}{2}\Big)^{x+y}-2\sum_x x\Big(\frac{1}{2}\Big)^x\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
The first series should be easy to be computed. It remain the second one, in particular the following
\begin{align}
\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y=\sum_y\Big(\frac{1}{2}\Big)^y-x\sum_y\frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
Therefore the problem is reduced to computing:
\begin{align}
\sum_{y=1}^\infty \frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
But I don't know well how to do it.

$$ \sum_{y=1}^\infty \frac{1}{x+y} \left(\frac{1}{2}\right)^y = \Phi(1/2,1,x) - 1/x$$
where $\Phi$ is the Lerch Phi function.
