Logarithmic Sum

Question

Is there a closed form for the following sum? $$\sum_{n=0}^{\infty}\sum_{m=1}^{\infty}(-1)^{m+n}\frac{\ln(m+n)}{(m+n)}$$ According to https://www.mathmash.org/contestprob.php?prob=227 it has a closed form. Using Mathematica we get $$\sum_{m=1}^{\infty}(-1)^{m+n}\frac{\ln(m+n)}{(m+n)}=\frac{1}{2} (-1)^n \left(\ln2 \left(-\text{PolyGamma}\left[0,1+\frac{n}{2}\right]+\text{PolyGamma}\left[0,\frac{1+n}{2}\right]\right)-\text{StieltjesGamma}\left[1,\frac{1+n}{2}\right]+\text{StieltjesGamma}\left[1,\frac{2+n}{2}\right]\right)$$

Answer 1

Hint:

Observe \begin{align} \sum^\infty_{n=0}\sum^\infty_{m=1}(-1)^{n+m}\frac{\log(n+m)}{n+m} =&\ \lim_{s\rightarrow 0^+} \sum^\infty_{n=0}\sum^\infty_{m=1}(-1)^{n+m}\frac{\log(n+m)}{(n+m)^{1+s}}\\ =&\ \lim_{s\rightarrow 0^+}\sum^\infty_{j=1}(-1)^j\frac{\log(j)}{j^s} = \frac{1}{2}\log\left(\frac{\pi}{2}\right) \end{align}

Edit: I used something called the Dirichlet regularization to sum up the series. For example,

\begin{align} ``\sum^\infty_{n=1} 1" = \lim_{s\rightarrow 0^+} ``\sum^\infty_{n=1}\frac{1}{n^s}" = \lim_{s\rightarrow 0^+} \zeta(s) = -\frac{1}{2}. \end{align} There is quite a bit of handwaviness to the above "solution/hint". The point is for the reader to rigorously justify each step.