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Recall that the periodic zeta function has the Dirichlet series

$$F(\lambda,s)= \sum_{n=1}^\infty \frac{e^{2\pi i n\lambda}}{n^s}.$$

This defines an analytic function for $\Re s>0$ and has a functional equation

$$F(\lambda,s) = \frac{\Gamma(1-s)}{(2\pi)^s}\left(i^{1-s}\zeta(1-s,\lambda)+i^{s-1}\zeta(1-s,1-\lambda)\right)$$

which is supposed to analytically continue $F(\lambda,s)$ to the entire complex plane.

My problem is that the right hand side does not appear to be entire but meromorphic with poles at $s=0,1,2,....$

Can anyone help me reconcile this?

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  • $\begingroup$ I assume the periodic zeta function is already able to be evaluated for $\lambda \in \mathbb{Q} / \mathbb{Z}$, $\mathrm{Re}(s)>0$. By plugging $-s$ into the functional equation you've written you may then evaluate it for $\mathrm{Re}(s)\le0$ without any issues with the poles of $\Gamma$ - right? $\endgroup$ – anon Jul 14 '11 at 21:25
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First a comment:

In most scenarios, it is more helpful to view the periodic zeta function as the polylogarithm, $\mathrm{Li}_s(z)$, evaluated at $z=e^{2i\pi x}$. The functional equation you have above comes from the functional equation for the polylogarithm on this Wikipedia page:

$$\mathrm{Li}_s(z) = \frac{\Gamma(1 - s)}{(2\pi)^{1-s}}\left(i^{1-s}~\zeta\!\left(1-s,~\frac12+{\frac{\ln(-z)}{2\pi i}}\right)+i^{s-1}~\zeta\!\left(1-s,~\frac12-{\frac{\ln(-z)}{2\pi i}}\right)\right) .$$

Solution to your problem:

Recall that when $n\in\mathbb N$ we can relate the Hurwitz zeta function to the Bernoulli polynomials by $$\zeta(1-n,x)=-\frac{B_n(x)}{n}.$$ Then for $s=n\in\mathbb N$ the part in parentheses in your question becomes $$\left(i^{1-s}\zeta(1-s,\lambda)+i^{s-1}\zeta(1-s,1-\lambda)\right)$$

$$=i^{1-s}\left(\zeta(1-n,\lambda)+(-1)^{n-1}\zeta(1-n,1-\lambda)\right)=-\frac{i^{s-1}}{n}\left(B_n(\lambda)+(-1)^{n-1}B_n(1-\lambda)\right).$$

But this is always zero since the Bernoulli polynomials have the following symmetry $$B_n(1-x)=(-1)^n B_n(x).$$ The fact that this factor is zero cancels the pole coming from $\Gamma(1-s)$ when $s$ is an integer.

Hope that helps,

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  • $\begingroup$ This is exactly what I was looking for! Thanks! $\endgroup$ – Daniel Parry Jul 14 '11 at 23:34

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