10
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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 $\Bbb{C}\setminus \{1\}$. $\endgroup$ – Sangchul Lee Sep 14 '13 at 7:01
  • $\begingroup$ Concerning $\zeta^{(n)}(0)$ this thread could be of interest (at least numerically...). $\endgroup$ – Raymond Manzoni Sep 15 '13 at 17:27
18
$\begingroup$

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, in my opinion, 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 $\,s\to 1\;$:

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

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

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

$\endgroup$
19
  • $\begingroup$ 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 $\sin(\pi s/2), 2^s, \pi^{s-1}$. $\endgroup$ – GEdgar Sep 14 '13 at 13:55
  • 1
    $\begingroup$ @SamanthaWyler That follows from Cauchy's Formulae for residue of isolated poles of a meromorphic complex function. $\endgroup$ – DonAntonio Mar 7 at 17:51
  • 1
    $\begingroup$ @SamanthaWyler Yes. That's basic and very important stuff in complex analysis. You can either try to evaluate the residue by means of Cauchy's Formulae, or else develop a Laurent series in some annulus around some pole and check the coeffficient $\;a_{-1}\;$ in that series. $\endgroup$ – DonAntonio Mar 7 at 17:52
  • 1
    $\begingroup$ @SamanthaWyler No. I get what is in the third line of equations before the bottom. $\endgroup$ – DonAntonio Mar 8 at 16:13
  • 1
    $\begingroup$ @SamanthaWyler Write @ and then my name...or any other user's name. $\endgroup$ – DonAntonio Mar 8 at 19:23
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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