I was solving an integral I made up (not included because I would prefer if the answer was not just a different way of solving the initial integral) as a fun challenge (as one does) and my solution contained the following summation: $$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{n+k}}{\alpha n^{2}k+\beta nk^{2}} \text{ where }\alpha,\beta>0.$$ I found its symmetry very interesting and could find nothing about it or any similar double summations online. My only idea involves using the Lerch Transcendent to solve one summation to get: $$\sum_{k=1}^{\infty}\left(-1\right)^{k}\frac{\Phi\left(-1,1,\frac{k\beta}{\alpha}+1\right)-\ln2}{\beta k^{2}}$$
$$=\frac{\pi^{2}\ln2}{12\beta}+\sum_{k=1}^\infty\frac{\left(-1\right)^{k}\Phi\left(-1,1,\frac{k\beta}{\alpha}+1\right)}{\beta k^{2}}$$
$$=\frac{\pi^{2}\ln2}{12\beta}+-\frac{1}{\beta}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}\Phi\left(-1,1,\frac{k\beta}{\alpha}\right)}{k^{2}}+\frac{3}{4}\frac{\beta}{\alpha^{2}}\zeta\left(3\right)$$
$$=-\frac{1}{\beta}\sum_{k=1}^{30}\frac{\left(-1\right)^{k}}{k^{2}}\left(\psi\left(\frac{k\beta}{\alpha}\right)-\psi\left(\frac{k\beta}{2\alpha}\right)\right)+\frac{3}{4}\frac{\beta}{\alpha^{2}}\zeta\left(3\right)$$
None of these seem to be helpful. Does this sum have a closed form?
