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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.

kurtachovo
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2 Answers2

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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 $a_n$.

Jack D'Aurizio
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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:

$$P_{n,k}(x)=B_{n,k}\left(\sin(\pi(x+0.5)),\sin(\pi(x+1)),\sin(\pi(x+1.5)),\dots,\sin\left(\pi\left(x+\frac{n-k+1}2\right)\right)\right)$$

Simply Beautiful Art
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