# Analytic Continuation of Zeta-like function

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?

• Write $k=(k+a)-a$ and use Hurwitz zeta. – EFinat-S Sep 1 '18 at 11:04
• There appears a $\zeta(1-s,a)$ which is not finite at $s=0$. :-/ – YoungMath Sep 1 '18 at 11:23
• @YoungMath: It might help us if you gave a link to the paper, if it is in electronic form. – Alex M. Sep 1 '18 at 14:26
• Of course. Linked it. It's on page 34. – YoungMath Sep 1 '18 at 14:40
• No, it appears a $\zeta(s-1,a)$ at $s=0$. – EFinat-S Sep 1 '18 at 15:25

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).
• 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? – YoungMath Sep 1 '18 at 21:32