Prove that
$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \log 2}{8}$$
Solution
$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \log 2}{8}$$
Solution
