# Laurent expansion of $\log\zeta(s)$

Is it possible to expand the logarithm of the zeta function

$$\log\zeta (s)= a_{0}+a_{1}s^{-1}+a_{2}s^{-2}+.... ,$$

with coefficients $a_{n} = \frac{1}{2\pi i}\oint dz \frac{\log\zeta(z)}{z^{n+1}}$ ?

My idea is to improve the Gram series based on the solution of the integral equation

$$\log\zeta (s)=s\int_{0}^{\infty}dt \frac{\pi(e^{t})}{e^{st}-1}.$$

Of course I am almost sure that the coefficients $a_{n}$ must depend on the nontrivial zeros of Riemann zeta $\zeta (\rho)=0$.

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Do you really mean $a_1^{-s}$ ... or $a_1 s^{-1}$, which would be a Laurent series. –  GEdgar Aug 19 '13 at 14:21
Your expansion should be valid where? And your integral $\oint$ is around what contour? –  GEdgar Aug 19 '13 at 14:24
mistake corrected, the idea is to expand the logarithm of riemann zeta $log\zeta (s)$ as a Z-transform , see en.wikipedia.org/wiki/Z_transform –  Jose Garcia Aug 19 '13 at 17:43
This question and its answers might be relevant: math.stackexchange.com/questions/71095/…. –  marty cohen Aug 19 '13 at 19:32
My guess: $\log \zeta(s)$ is not regular at infinity, and thus does not have an expansion of this form. –  GEdgar Aug 19 '13 at 21:33