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The only definition that I have ever seen of this Riemann-Hurtwitz zeta-function is this,

For $0 < a \leq 1$ we have the identity $$ \zeta(z, a) = \frac{2 \Gamma(1 - z)}{(2 \pi)^{1-z}} \left[\sin \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^{1-z}} + \cos \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\sin 2 \pi a n}{n^{1-z}} \right] \,$$

  • Now from this (or otherwise!) how do I show that, $\xi(s,\frac{1}{2}) = \sum_{n=0}^{\infty} \frac{1}{ (n + \frac{1}{2} )^s }$

  • It seems that for any $k$ even and $d$ a positive integer there is such a sum over inverse powers representation of $\xi(s-k,\frac{d}{2}-1 )$ - it would be great to know if there is a general expression here!

  • Any insight/motivation about this $\xi(z,a)$ function would be a great help.

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Is $\zeta$ same as $\xi$? – user11000 Dec 8 '13 at 4:59
A functional equation might help. – Mustafa Said Dec 8 '13 at 5:59
@user17762 Yes! Sorry for the typo! – user6818 Dec 8 '13 at 7:38
@MustafaSaid May be you can elaborate... – user6818 Dec 8 '13 at 7:38
@user6818, the Hurwitz zeta function satisfies the functional equation, $\zeta(1-s, \frac{m}{n}) = \frac{2 \Gamma (s)}{(2 \pi n)^s} \sum_{k=1}^n [\cos(\frac{\pi s}{2} - \frac{2\pi km}{n}) \zeta(s, k/n)]$ that holds for all values of $s$ and integers, $1 \leq m \leq n$. Now let $m/n = 1/2$ and compare both sides of the functional equation. I suspect that this will give you the desired expression for $\zeta(s, 1/2)$. – Mustafa Said Dec 8 '13 at 8:23

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