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Reading a paper about eta invariants I came across a zeta-like function.

I'm looking for the analytic continuation of $$\sum_{k=1}^\infty k(k+a)^{-s}$$ at $s=0$, where $a$ is positive.

In the paper he just says "The [...] term causes no problem and at $s=0$ has the value $\left[\frac{(4a^2-1)}{12}\right]$." Unfortunately, I really don't see the that.

My first approach so far; I tried a Taylor series at $s=3$: \begin{align} \sum_{k=1}^\infty k(k+a)^{-s} &= \sum_{k=1}^\infty k \sum_{l=0}^\infty (-1)^l \begin{pmatrix}2+l \\ l\end{pmatrix} (k+a)^{-3-l} (s-3)^{l} \end{align} and inserted $s=0$ $$\sum_{k=1}^\infty \frac{k}{(k+a)^3} \sum_{l=0}^\infty \begin{pmatrix}2+l \\ l\end{pmatrix} \left( \frac{3}{k+a} \right)^l$$ which is for $a\geq 3$ $$ \sum_{k=1}^\infty \frac{k}{(k+a-3)^3}~. $$ That seems to converge if I didn't miscalculate. But what are the further steps in order to get the result above? Or is there a more skilful approach?

Thanks in advance!

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  • $\begingroup$ Write $k=(k+a)-a$ and use Hurwitz zeta. $\endgroup$
    – efs
    Sep 1, 2018 at 11:04
  • $\begingroup$ There appears a $\zeta(1-s,a)$ which is not finite at $s=0$. :-/ $\endgroup$
    – YoungMath
    Sep 1, 2018 at 11:23
  • $\begingroup$ @YoungMath: It might help us if you gave a link to the paper, if it is in electronic form. $\endgroup$
    – Alex M.
    Sep 1, 2018 at 14:26
  • $\begingroup$ Of course. Linked it. It's on page 34. $\endgroup$
    – YoungMath
    Sep 1, 2018 at 14:40
  • $\begingroup$ No, it appears a $\zeta(s-1,a)$ at $s=0$. $\endgroup$
    – efs
    Sep 1, 2018 at 15:25

1 Answer 1

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Let $a>0$. We can write $$f(s,a):=\sum_{k=1}^\infty k(k+a)^{-s}=\sum_{k=0}^\infty (k+a)^{-s+1}-a\sum_{k=0}^\infty (k+a)^{-s}=\zeta(s-1,a)-a\zeta(s,a).$$ Hence, $f$ has a meromorphic continuation to $\mathbb{C}$ with simple poles at $s=1$ and $s=2$.

Now, if $s=-n$ is a non-positive integer, it is known that $$\zeta(-n,a)=-\frac{B_{n+1}(a)}{n+1}$$ where $B_n(X)$ is the $n$-th Bernoulli polynomial. Hence, $$f(0,a)=\zeta(-1,a)-a\zeta(0,a)=-\frac{B_{2}(a)}{2}+aB_1(a).$$ Since $B_1(X)=X-1/2$ and $B_2(X)=X^2-X+1/6$, we obtain that $$f(0,a)=-\frac{6a^2-6a+1}{12}+a\frac{2a-1}{2}=\frac{6a^2-1}{12},$$ (unless I made some stupid mistake which I don't have the time to fix now).

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  • $\begingroup$ Thank you for that concise and precise answer. Unfortunately that is not the result as in the paper. Putting $a=\frac{\lambda^2}{2}$ and multiply by a factor 2 should give the value noted in the paper but it doesn't. Although, I came to the same computation and so does WolframAlpha. Did I miss something silly? $\endgroup$
    – YoungMath
    Sep 1, 2018 at 21:32
  • $\begingroup$ That is why I wrote that I may have done a mistake, but I do not see where. $\endgroup$
    – efs
    Sep 1, 2018 at 21:46
  • $\begingroup$ Me neither. Seems perfectly right. I will calculate the second term and maybe in the end they fit together properly. Must be so since the final result equals half the A-genus. $\endgroup$
    – YoungMath
    Sep 1, 2018 at 22:06
  • $\begingroup$ Ok. Please tell me then the reason of why the different results. Maybe a misprint? $\endgroup$
    – efs
    Sep 1, 2018 at 22:18
  • $\begingroup$ The final result actually fits. Apparently it's a misprint. $\endgroup$
    – YoungMath
    Sep 12, 2018 at 14:54

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