5
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $s\not\in \Bbb{Z}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.