I am investigating closed forms of the function $$f(s;q)=\sum_{k\ge1}\frac{(-1)^k}{k^s+q^s}.$$ I of course don't expect a closed form for general $s$ and $q$ (such that the series converges), but I am rather interested in special cases. There is of course the easily-shown relation $$f(s;0)=(2^{1-s}-1)\zeta(s),$$ from which it is immediately evident that $f$ is some sort of analog of the classic zeta function.

My efforts have mostly been focused on special cases for fixed values of $s$, such as the easily shown $$f(1;q)=\int_0^1\frac{x^qdx}{1+x}.$$ Note that since this integral has no closed form for general $q$, I am willing to count the integral itself as a closed form, as it is certainly easy to evaluate for $q\in\Bbb Q_{\ge0}$. If something similar is the best we can do in other cases, so be it.

My own work has been primarily regarding the cases $f(2^n;q)$, and I have succeeded in finding a recurrence in $n$. First, we evaluate $f(2;q)$. To do so we define $\zeta_m=\exp\frac{i\pi}{m}$, as we will use it a lot later.

Recall the formula $$\frac{\pi}{\sin\pi z}=\sum_{k\in\Bbb Z}\frac{(-1)^k}{z+k},$$ so that $$\sum_{k\ge 1}\frac{(-1)^k}{k^2-z^2}=-\frac\pi{2z\sin\pi z}+\frac1{2z^2}.$$ Setting $z=iq$ and simplifying gives $$f(2;q)=\sum_{k\ge1}\frac{(-1)^k}{k^2+q^2}=\frac{\pi e^{\pi q}/q}{e^{2\pi q}-1}-\frac1{2q^2}.$$ This, interestingly enough, gives the limit $$\lim_{q\to 0}\left(\frac{\pi e^{\pi q}/q}{e^{2\pi q}-1}-\frac1{2q^2}\right)=-\frac{\pi^2}{12}.$$ I would like to discuss this limit further, but that may not be on topic on this post.

Back to the investigation at hand, we see that $$\begin{align} f(2M;q)&=\sum_{k\ge1}\frac{(-1)^k}{(k^{M}-iq^{M})(k^{M}+iq^{M})}\qquad [M=2^n]\\ &=\frac{1}{2iq^M}\sum_{k\ge1}(-1)^k\left(\frac{1}{k^M-iq^M}-\frac{1}{k^M+iq^M}\right)\\ &=\tfrac{1}{2iq^M}f(M;\zeta_{2M}^3q)-\tfrac{1}{2iq^M}f(M;\zeta_{2M}q). \end{align}$$ Thus, since we know $f(2;q)$, we also know $f(M;q)$.

So, my question:

For what other values of $s$ can $f(s;q)$ be evaluated in closed form?


The main thing to see is that it has some kind of closed-form for $s\ge 2$ even and $q\ne 0$.

This is because you want to evaluate $\frac{g(0)-q^{-s}}{2}$ where $$g(z)=\sum_{k\in \Bbb{Z}} \frac{(-1)^k}{(z+k)^s+q^s}=\sum_{k\in \Bbb{Z}} (-1)^k\sum_{m=1}^s \frac{1}{c_m (z+k-a_m)}=\sum_{m=1}^s \frac{\pi}{c_m \sin(\pi (z-a_m))}$$ $$ a_m=qe^{i\pi (2m+1)/s}, c_m= sa_m^{s-1}$$

It works the same way for $s$ odd except that $\sum_{k\ge 1} \frac{(-1)^k}{(z+k)^s+q^s}$ isn't periodic anymore, instead of $\pi/\sin(\pi z)$ you'll have $\Gamma'/\Gamma(z-1)-\Gamma'/\Gamma(z/2-1)$ which does have only a few closed-forms (for $z$ rational where it can be expressed as a linear combination of $\log e^{2i\pi l /r}$).

It is supposedly obvious that there is no closed-form for $s\not\in \Bbb{Z}$.

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  • $\begingroup$ Sorry, but where did the index $k\in\Bbb Z$ go in the inner sum (over $m$) of $$\sum_{k\in \Bbb{Z}} (-1)^k\sum_{m=1}^s \frac{1}{c_m (z-a_m)}?$$ Shouldn't it be $$\sum_{k\in \Bbb{Z}} (-1)^k\sum_{m=1}^s \frac{1}{c_m (z\color{red}{+k}-a_m)}?$$ Otherwise great answer (+1) $\endgroup$ – clathratus Jan 3 at 2:02
  • $\begingroup$ sure ${}{}{}{}$ $\endgroup$ – reuns Jan 3 at 18:01

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