A while ago I computed pretty easily the series $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{k+n} \frac{\log(k+n)}{k+ n}$ and then I thought of tackling the case where we have the product instead of sum in denominator, but this one seems far harder than the previous one. What would you suggest me to do?
$$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{k+n} \frac{\log(k+n)}{k n}$$
By Frullani's integral, $$\log(k+n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-(k+n)x}}{x}\,dx \tag{1}$$ and since $\sum_{k=1}^{+\infty}\frac{z^k}{k}=-\log(1-z)$ we have: $$ \sum_{k,n\geq 1}\frac{(-1)^{k+n}}{kn}=\log^2 2\tag{2}$$ and the whole sum equals: $$\int_{0}^{+\infty}\left(e^{-x}\log^2 2 -\log^2(1+e^{-x})\right)\frac{dx}{x}=\int_{0}^{1}\frac{\log^2(1+t)-t \log^2 2}{t \log t}\,dt\tag{3}$$ that is not so horrible.