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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?

Jonathan Gleason
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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 07:01
  • Concerning $\zeta^{(n)}(0)$ this thread could be of interest (at least numerically...). – Raymond Manzoni Sep 15 '13 at 17:27

2 Answers2

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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 $\,s\to 1\;$:

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

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

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

Maximilian Janisch
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DonAntonio
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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
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    @SamanthaWyler That follows from Cauchy's Formulae for residue of isolated poles of a meromorphic complex function. – DonAntonio Mar 07 '21 at 17:51
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    @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. – DonAntonio Mar 07 '21 at 17:52
  • Sorry took complex last year and now I remember doin this. going to delete my previous comments. Also This is 11.7.8 https://math.byu.edu/~bakker/Math346/BeamerSlides/Lec30annotated.pdf – Samantha Wyler Mar 07 '21 at 17:59
  • I'm trying to understand why $\lim_{s\to 1}(1-s)\zeta(s)=-1=\lim_{s\to 1};\Gamma(2-s)2^s\pi^{s-1}\zeta(1-s)=1\cdot 2\zeta(0)$. I think to get $\lim_{s\to 1}(1-s)\zeta(s)=-1$, you essentially multiply $\lim_{s\to 1}(s - 1)\zeta(s)= 1$ by $-1$, but then if you multiplied the equation above by $-1$ and take the limit, you get $ \lim_{s\to 1}(1-s)\zeta(s)= \lim_{s\to 1}; -sin \frac{\pi s}{2} [\Gamma(1-s)2^s\pi^{s-1}]\zeta(1-s)$. – Samantha Wyler Mar 07 '21 at 18:19
  • Dose $\lim_{s\to 1};\Gamma(2-s)2^s\pi^{s-1}\zeta(1-s) \lim_{s\to 1}(1-s)\zeta(s)= \lim_{s\to 1}; -sin \frac{\pi s}{2} \Gamma(1-s)2^s\pi^{s-1}\zeta(1-s)$? – Samantha Wyler Mar 07 '21 at 18:22
  • @SamanthaWyler I'm not sure I completely understand what you're asking. I propose you to open up a new question about whatever is bugging you. You can also pin me in case you're interested in my input. – DonAntonio Mar 07 '21 at 20:49
  • Let us continue this discussion in chat. – Samantha Wyler Mar 08 '21 at 15:56
  • If you multiply the right hand side of the functional equation by $(1 - s)$, don't you get $ \lim_{s\to 1}; -sin \frac{\pi s}{2} \Gamma(1-s)2^s\pi^{s-1}\zeta(1-s)$ instead of $\lim_{s\to 1};\Gamma(2-s)2^s\pi^{s-1}\zeta(1-s) \lim_{s\to 1}(1-s)\zeta(s)$? – Samantha Wyler Mar 08 '21 at 16:00
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    @SamanthaWyler No. I get what is in the third line of equations before the bottom. – DonAntonio Mar 08 '21 at 16:13
  • Yes but then how are you getting $-1 = \lim_{s\to 1};\Gamma(2-s)2^s\pi^{s-1}\zeta(1-s)$? – Samantha Wyler Mar 08 '21 at 18:13
  • Also how do I pin you on another question? – Samantha Wyler Mar 08 '21 at 18:16
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    @SamanthaWyler Write @ and then my name...or any other user's name. – DonAntonio Mar 08 '21 at 19:23
  • @SamanthaWyler Because $$;(s-1)\zeta(s)\xrightarrow[s\to1^+]{}1\implies(1-s)\zeta(s)\xrightarrow[s\to1^+]{}-1$$ – DonAntonio Mar 08 '21 at 19:24
  • Ya I understand why $(1 - s)\zeta(s) = -1$, but I don't understand why the equality afterwards holds? – Samantha Wyler Mar 08 '21 at 19:57
  • Actually I think I figured it out. The two things I was asking if they were equal before are equal, because $(1 - s)\gamma(1 - s) = \gamma(2 - s)$, and lim$_{s \rightarrow 1}$ sin$\frac{\pi s}{2} =$ sin$\frac{\pi}{2} = 1$ – Samantha Wyler Mar 08 '21 at 20:05
  • @DonAntonio tried to pin you on another post I made but I don't think it worked cause your name didn't turn to a blue link? – Samantha Wyler Mar 08 '21 at 20:11
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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

Daniel Parry
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