Double Euler sum
Posted: Mon May 29, 2017 11:46 am
Here is something that came up today while evaluating something else.
Prove that
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )} = \frac{13 \pi^8}{113400}$$
Prove that
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )} = \frac{13 \pi^8}{113400}$$