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I'm looking for literature – in particular, methods of evaluation – for zeta functions defined by the sum

$$\zeta(s,a) = \sum_{k=-\infty}^\infty \frac{1}{(k^2+a)^s}\,.$$

I am interested in the continuation to $s=-3/2$. It looks like the Hurwitz zeta function, except that in the denominator, we have $k^2$ instead of $k$. Is there any work on this kind of sum? Does this zeta function have a name?

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  • $\begingroup$ Ok, I think this is related to the Epstein zeta function $\endgroup$ – QuantumDot Jan 21 '18 at 18:30
  • $\begingroup$ Is my answer helpful to you? If not, please leave me some feedback so that I know how to improve it. $\endgroup$ – Franklin Pezzuti Dyer Jan 26 '18 at 0:30
  • $\begingroup$ @Nilknarf Yes, it was helpful to some extent (upvoted); but I really need to be evaluate this thing accurately for a greater domain in $a$ and $s$. $\endgroup$ – QuantumDot Jan 26 '18 at 0:37
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I don't have any literature for you, and I have no insight about the case of $s=-3/2$, but I can provide some methods for evaluation. First of all, I'd like you to recall the formula $$\zeta(1,a)=\sum_{k=-\infty}^\infty \frac{1}{k^2+a}=\frac{\pi \coth(\pi \sqrt a)}{\sqrt a}$$ This formula can be easily derived using the residue theorem, which I am just assuming that you already know how to use. If not, correct me and I will write you a short proof as soon as I can.

Now observe the following interesting and useful property of the function $\zeta$ as you have defined it. If $s\gt 0$, then $$\begin{align} \frac{d}{da}\zeta(s,a) &= \frac{d}{da} \sum_{k=-\infty}^\infty \frac{1}{(k^2+a)^s}\\ &= \sum_{k=-\infty}^\infty \frac{d}{da} \frac{1}{(k^2+a)^s}\\ &= \sum_{k=-\infty}^\infty \frac{-s}{(k^2+a)^{s+1}}\\ \end{align}$$ and so we have the following recurrence: $$\zeta(s+1,a)=-\frac{1}{s}\frac{d}{da}\zeta(s,a),\space\space\space s\gt 0$$ From this recurrence, we can derive the following formulas: $$\zeta(2,a)=\frac{\pi\coth(\pi\sqrt a)}{2a^{3/2}}+\frac{\pi^2\text{csch}^2(\pi\sqrt a)}{2a}$$ $$\zeta(3,a)=\frac{3\pi\coth(\pi \sqrt a)}{8a^{5/2}}+\frac{\pi^3\coth(\pi\sqrt a)\text{csch}^2(\pi\sqrt a)}{4a^{3/2}}+\frac{3\pi^2\text{csch}^2(\pi\sqrt a)}{8a^2}$$ I could keep going, but the algebra just keeps getting messier. So instead, I will generalize this by writing $$\color{green}{\zeta(n,a)=\frac{(-1)^{n+1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\frac{\pi \coth(\pi \sqrt x)}{\sqrt x}_{x=a}}$$ which holds for all $n\in\mathbb N$ and $a$ not a negative perfect square.

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  • $\begingroup$ Could you explain why upon differentiation over $a$ you got $s$ instead of $-s$ in the numerator of the expression under sum? $\endgroup$ – user Jan 27 '18 at 13:54
  • $\begingroup$ @user355705 Sorry, that was a mistake on my part. Thanks for catching it! It was supposed to be a $-s$, and I had incorporated that into my final (green) formula, but for some reason I didn't write it in the recurrence. Thanks again! $\endgroup$ – Franklin Pezzuti Dyer Jan 27 '18 at 14:06

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