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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)$.

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    @Zander: Yes, Jose had a typo. (It comes from expanding $\exp(-tz^2)$ in the second line.) Also, $0^0=1$ in the contexts of polynomials and power series expansions, so in fact you get $\zeta(2s)=\zeta(2s)$, as expected.2012-06-26

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