Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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? –  anon Jul 14 '11 at 21:25
add comment

1 Answer

up vote 4 down vote accepted

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,

share|improve this answer
    
This is exactly what I was looking for! Thanks! –  Daniel Parry Jul 14 '11 at 23:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.