Recently there were some discussions [1], [2], [3] of sums of the type
$$S(k) = \sum_{n=1}^{\infty} \frac{n^{k-1}}{e^{2 \pi n}-1}\tag{1}$$
Here $k$ was assumed to be a positive integer.
For a positive integer $k$ I have given here and here a closed expression with a slightly different definition of $S(m)$.
Now I would like to drop this restriction an allow $k$ to assume real values.
Questions
(1) as a typical example: is there a closed expression for $S(\frac{3}{2})$?
(2) what are the analytic properties of $S(z)$ in the complex $z$-plane?
My attempts regarding question (1)
It is more convenient to study the "generating" sum
$$S(z,t) = \sum_{n=1}^{\infty} \frac{n^{z-1} e^{-n t}}{e^{2 \pi n}-1}\tag{2}$$
so that $S(z) = S(z,t=0)$ and also
$$S(z+1,t) = - \frac{\partial S(z,t)}{\partial t}\tag{3}$$
I now replace the power of $n$ by an integral
$$n^{z-1} = \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} s^{-z} e^{-n s} \tag{4}$$
which leads to an integral representation as follows
$$S(z,t)=\sum _{n=1}^{\infty } \frac{e^{-n t} n^{z-1}}{e^{2 \pi n}-1}=\frac{1}{\Gamma (1-z)} \int_0^{\infty } s^{-z} f(s,t) \, ds, \Re(z)<1 \tag{5}$$
where
$$\begin{align} f(s,t) & = \sum _{n=1}^{\infty } \frac{e^{ - n (s+t)}}{e^{2 \pi n}-1}\\ & =\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } e ^{ -n (s+t)-2 \pi n m}\\ & = \sum _{m=1}^{\infty } \sum _{n=1}^{\infty } e^{ -n(2 \pi m + s+t)}\\ & =\sum _{m=1}^{\infty } \frac{1}{e^{2 \pi m+s+t}-1}\\ & =-\frac{\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)+\log \left(1-e^{-2 \pi }\right)}{2 \pi } \end{align}\tag{6}$$
Inserting this into $(5)$ we obtain
$$\begin{align} S(z,t) & =\sum _{n=1}^{\infty } \frac{e^{-n t} n^{z-1}}{e^{2 \pi n}-1}\\ & =\frac{-1}{2 \pi\Gamma (1-z)} \int_0^{\infty } s^{-z} \left(\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)+\log \left(1-e^{-2 \pi }\right)\right) \, ds, \Re(z)<1 \end{align}\tag{7}$$
where $\psi _{q }^{(0)}\left(x\right)$ is the q-polygamma function.
Now we can calculate the sum in question as
$$\sum _{n=1}^{\infty } \frac{\sqrt{n}}{e^{2 \pi n}-1} =\frac{1}{\sqrt{\pi}}\left(\frac{1}{2 \pi }\right)^2 \int_0^{\infty } \frac{1}{\sqrt{s}} \psi _{e^{-2 \pi }}^{(1)}\left(\frac{s}{2 \pi }+1\right) \, ds\tag{8}$$
here we have used
$$S\left(\frac{3}{2}\right)=-\frac{\partial }{\partial t}S\left(\frac{1}{2},t\right)|_{t\to 0}\tag{9}$$
and
$$\frac{\partial }{\partial t}\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)=\frac{1}{2 \pi } \psi _{e^{-2 \pi }}^{(1)}\left(\frac{s+t}{2 \pi }+1\right)\tag{10}$$
Here I am stuck, as I'm not sure if $(8)$ can be further simplified. The analytic properties should be deducible from $(7)$.
