I would like to know if this formula is true:
$\sum_{n=1}^{\infty}\frac{1}{(z^{2}+n^{2})^s}=\frac{1}{\Gamma(s)} \sum_{n=0}^{\infty}\Gamma(s+n)\zeta(2s+2n)\frac{ (-z^2)^n}{n!}.$
I have used the fact that
$\Gamma(s)\sum_{n=1}^{\infty}\frac{1}{(z^{2}+n^{2})^s}=\int_{0}^{\infty}t^{s-1}\exp(-tz^{2})\sum_{n=1}^{\infty}\exp(-tn^{2})~dt $
and that the Mellin transform of $\displaystyle \sum_{n=1}^{\infty}\exp(-tn^{2}) $ is just $ \Gamma(s) \zeta (2s)$.