On series $\sum\limits_{n\in\mathbb{Z}}\frac{(-1)^n}{(n+x)^{\alpha}}$ In the course of some physics related work I met the following series,
\begin{align}
S_{\alpha}(x)=\sum\limits_{n\in\mathbb{Z}}\frac{(-1)^n}{(n+x)^{\alpha}}, && x\in[0,1], ~k\in\mathbb{Z}_{\geq0}
\end{align}
Which can be written in a compact form using a combination of Hurwitz zeta functions. However, with some help from Wolfram Alpha, I realised that computing by hand some particular values of the case I am interested, which are positive odd integers, it seems we can simply write the above as a combination of powers of $\sin^{-1}{\pi x}$. For example,
\begin{align}
S_1(x) &= \frac{\pi}{\sin{\pi x}}\\
S_3(x) &= \frac{\pi^3}{4}\frac{3+\cos{2\pi x}}{\sin^3{\pi x}} = \frac{\pi^2}{2}\left[\frac{2}{\sin^3{\pi x}}-\frac{1}{\sin{\pi x}}\right]\\
S_5(x)&=\pi^5\left[\frac{1}{\sin^5{\pi x}}-\frac{5}{6}\frac{1}{\sin^3{\pi x}}+\frac{1}{24}\frac{1}{\sin{\pi x}}\right]
\end{align}
$\textbf{Question}$: Can we write
\begin{align}
S_{2k+1}(x)=\sum\limits_{0<n\leq 2k+1}\frac{a_n}{\sin^n{\pi x}} 
\end{align}
for some explicit coefficients $a_n$?
I did some literature research related to the Hurwitz Zeta but did not find much. Suggestions in this direction are also be appreciated.
 A: We have
$$S_1(x)= \sum_{n\in\mathbb{Z}}\frac{(-1)^n}{x+n}=\frac{\pi}{\sin(\pi x)}\tag{1}$$
by Herglotz' trick. By applying $D^{\alpha}=\frac{d^\alpha}{dx^{\alpha}}$ to both sides of $(1)$ we get
$$ S_{\alpha+1}(x) = \frac{(-1)^{\alpha}}{\alpha!}\,D^{\alpha}\left(\frac{\pi}{\sin(\pi x)}\right) \tag{2}$$
and the claim follows from the fact that $f(x)=\frac{\pi}{\sin(\pi x)}$ fulfills the differential equation
$$ f''(x) = 2\,f(x)^3 - \pi^2 f(x) \tag{3}$$
leading to a recursion for the coefficients $a_n$.
A: First, we have:
$$S_1(x)=\sum_{n\in\mathbb Z}\frac{(-1)^n}{n+x}=\pi\csc(\pi x)$$
Now take the $\alpha$th derivative to find that
$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}S_1(x)=\sum_{n\in\mathbb Z}\frac{(-1)^{n+\alpha}\alpha!}{(n+x)^{\alpha+1}}=(-1)^\alpha\alpha!S_{\alpha+1}(x)$$
Thus,
$$S_\alpha(x)=\frac{(-1)^{\alpha+1}\pi}{(\alpha-1)!}\frac{\mathrm d^{\alpha-1}}{\mathrm dx^{\alpha-1}}\csc(\pi x)$$
where it may be seen that
$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}\csc(\pi x)=\pi^\alpha\sum_{k=1}^\alpha(-1)^kk!\csc^{\alpha+1}(\pi x)\cdot P_{n,k}(x)$$

where $P$ is a Bell Polynomial:
$$P_{n,k}(x)=B_{n,k}\left(\sin(\pi(x+0.5)),\sin(\pi(x+1)),\sin(\pi(x+1.5)),\dots,\sin\left(\pi\left(x+\frac{n-k+1}2\right)\right)\right)$$
