# How to Compute $\zeta (0)$?

Ultimately, I am interested in analytically continuing the function $$\eta _a(s):=\sum _{n=1}^\infty \frac{1}{(n^2+a^2)^s},$$ where $a$ is a non-negative real number, and calculating $\eta _a$ and its derivatives (at least the first derivative) at the origin: $\eta _a(0),\eta _a'(0),\ldots$.

It is well-known that $\zeta (0)=-\tfrac{1}{2}$ and that $\zeta '(0)=-\tfrac{1}{2}\ln (2\pi)$, but I do not actually know how to obtain these ($\zeta$ is of course the Riemann Zeta function). I figured that, perhaps if I knew how to calculate these values, I would be able to generalize the technique to be able to calculate the corresponding values of $\eta _a$.

So then, how does one calculate $\zeta (0)$, $\zeta '(0)$, etc.? If this technique does not obviously generalize to $\eta _a$, any ideas how I might go about calculating these values?

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Functional equation for $\zeta$ will be quite useful when calculating those values. Of course, this equation is a part of proof that $\zeta$ can be continuated to $\Bbb{C}\setminus \{1\}$. – Sangchul Lee Sep 14 '13 at 7:01
Concerning $\zeta^{(n)}(0)$ this thread could be of interest (at least numerically...). – Raymond Manzoni Sep 15 '13 at 17:27

By the functional equation of the zeta function:

$$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$

We now use the fact that the zeta function has a simple pole at $\,s=1\,$ with residue $\,1\,$ (this is, imo, one of the most beautiful elementary things that can be proved about this wonderful function), and this means that

$$\lim_{s\to 1}(s-1)\zeta(s)=1$$

Now, using the functional equation for the Gamma Function $\,s\Gamma(s)=\Gamma(s+1)\;$, we multiply the functional equation for zeta by $\,(1-s)\;$ and then pass to the limit when $\,s\to 1\;$:

$$(1-s)\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\left[(1-s)\Gamma(1-s)\right]\zeta(1-s)\implies$$

$$\lim_{s\to 1}(1-s)\zeta(s)=-1=\lim_{s\to 1}\;\Gamma(2-s)2^s\pi^{s-1}\zeta(1-s)=1\cdot 2\zeta(0)\implies$$

$$\zeta(0)=-\frac12$$

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And to do $\zeta'(0)$, etc., you use more terms in the Laurent series for $\zeta(1-s)$ and $\Gamma(1-s)$, and maybe also the Taylor series of $\sin(\pi s/2), 2^s, \pi^{s-1}$. – GEdgar Sep 14 '13 at 13:55
This is good, but how does one arrive at the functional equation itself? – Jonathan Gleason Sep 14 '13 at 19:42
Oh, my: that is, imo, way too long and involved to show it here. You need Mellin transforms and/or theta functions and/or some other heavy machinery from complex analysis lime Gamma Function and etc. to reach that. I suppose there's some "easier" way to get it, but as far as I know they all are lengthy and hard. – DonAntonio Sep 14 '13 at 21:57

There is a paper listed online which should answer your questions. If you let $P(x)=1$ and $Q(x)=x^2 + a^2$ then you are looking at $$\eta_a(s)= \sum P(k)/Q(k)^s.$$

This paper suggests that $\eta_a(s)$ continues to the $s$-plane and provides values of $\eta_a(0),$ $\eta_a'(0).$

http://www.math.nagoya-u.ac.jp/~kohjimat/weng.pdf

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