How to show that $\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \frac{e^{-2\pi(m+n)}}{m+n+1}=\frac{1}{2} (1+\coth \pi ).$ How to show $$\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \frac{e^{-2\pi(m+n)}}{m+n+1}=\frac{1}{2} (1+\coth \pi ).$$
I know that sum of IGP should work here, but how to disentangle m and n. Please help.
 A: Denote the rquired sum by $S$  and use $\frac{1}{x}=\int_{0}^{1} t^{x-1} dt.$ Then  $$S=\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \frac{e^{-2\pi(m+n)}}{m+n+1} =\int_{0}^{1}  \sum_{m=0}^{\infty} \sum _{n=0}^{\infty}~ (t e^{-2\pi})^{m+n} ~dt =\int_{0}^{1} \left(\sum_{k=0}^{\infty} (t e^{-2\pi})^k \right)^2 dt. $$
$$ \Rightarrow S=\int_{0}^{1} (1-te^{-2\pi})^{-2} dt .= \frac{1}{1-e^{-2\pi}}.$$
A: Notice for any non-negative sequence $(f_{k})_{k\ge 0}$, we have
$$\sum_{m=0}^\infty \sum_{n=0}^\infty f_{m+n} = 
\sum_{\ell=0}^\infty\sum_{n=0}^\ell f_{\ell} = 
\sum_{\ell=0}^\infty f_{\ell}\left(\sum_{n=0}^\ell 1\right) =
\sum_{\ell=0}^\infty (\ell+1)f_{\ell}$$
Since the sequence is non-negative, it is legal to perform the sum in any order.
For this particular problem, substitute $f_k$ by $\frac{e^{-2\pi k}}{k+1}$ will turn your sum into a geometric series.
$$\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{e^{-2\pi(m+n)}}{m+n+1}
= \sum_{\ell=0}^\infty e^{-2\pi \ell}
= \frac{1}{1 - e^{-2\pi}}
= \frac{e^{\pi}}{2\sinh(\pi)}\\
= \frac{\cosh(\pi)+\sinh(\pi)}{2\sinh(\pi)} =
\frac12(\coth(\pi) +1)
$$
