Closed form of $\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}$?

Question

I considered the following function: $$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}.$$

It can be seen that $f(z)=0$ at every integer (division of a non-zero complex number by zero is assumed to be complex infinity, so its reciprocal is $0$). It also seems that the function is periodic on the real axis with period $2$, i.e. $$\forall x\in\mathbb{R}:\, f(x+2)=f(x),$$ though I was not able to prove that rigorously. I tried to use the $\sinh$ addition formula, but that's probably not useful.

On the whole real axis, $f$ is very close to $$g:\, \mathbb{C}\mapsto \mathbb{C},\, z\mapsto 0.31837572\sin \pi z,$$ but on the imaginary axis, there's a big difference between $f$ and $g$, for instance: $$f(2i)\approx 5.3796i,\,\text{but}\, g(2i)\approx 85.2499i.$$

Why is that? Can a closed form of $f$ in terms of Weierstrass/Jacobi elliptic functions be found?

Note: Curiously, any $$h:\,\mathbb{R}\mapsto\mathbb{R},\, x\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{a^{x+n}-a^{-x-n}}\right)^{-1}$$ where $a\gt 0$ and $a\ne 1$ seems to be $2$-periodic...

Edit: The function $$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}$$ is probably a doubly-periodic elliptic function with periods $2$ and $2\pi i$, so I added an appropriate tag.

Answer 1

Let $$g(z) = \sum_{n\in \mathbb{Z}} \frac{(-1)^n}{\sinh(z+n)}$$ $g$ has $2$ and $2\pi i$ as period, so an elliptic function. Poles of $g$, which are all simple, are exactly $0,1,\pi i,1+\pi i$ modulo periods, with residues $1,-1,-1,1$.

$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\ns}{ns}$ Since $g$ has only simple poles, it is easier to express it as Jacobi elliptic function. Recall that $\ns(z,k)$ has periods $4K,2iK'$ and residues $1$ at $0$. Hence $g(z)$ must be of form $$g(z) =a \ns(az,k) - a \ns(a(z-\pi i),k) + C$$ then it is easy to determine $K'/K = 2\pi, a = 2K, C=0$. This puts $g(z)$ in terms of Jacobi-elliptic function: $$g(z) = 2K[\ns(2Kz, k) - k \sn(2Kz,k)] \qquad K' = 2\pi K$$

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Mathematica code to calculate $g(z)$ numerically:

g[z_] := Module[{m = InverseEllipticNomeQ[Exp[-2 Pi^2]], K}, 
   K = EllipticK[m]; 
   2 K (JacobiNS[2 K*z, m] - Sqrt[m]*JacobiSN[2 K*z, m])];

This can give, within an eyeblink, value of any $g(z)$ to great precision. For example, try

N[g[1/3], 1000]

Answer 2

We can evaluate it in terms of Jacobi theta functions.

This identity is occasionally useful: $$ \sum _{n=-\infty }^{\infty }{\frac {b{u}^{n}}{1-a{q}^{2\,n}}}= \frac{b}{2i}\; {\frac { {\theta_1} \left( \frac{i}{2}\ln \left( au \right) ,q \right) { \theta_2} \left( 0,q \right) {\theta_3} \left( 0,q \right) {\theta_4} \left( 0,q \right) }{{\theta_1} \left( \frac{i}{2}\ln \left( u \right) ,q \right) {\theta_1} \left( \frac{i}{2}\ln \left( a \right) ,q \right) }} $$ provided $1 > |u| > |q^2|$ so that the series converges.

For this problem, write $\sinh$ in terms of $\exp$ to get the left side of this, with $$ a=e^{-2 z},\quad b=2e^{-z},\quad q=e^{-1},\quad u=-e^{-1} $$ to get the answer $$ \sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)} = {\frac {i{{\rm e}^{-z}}{\theta_2} \left( iz+i/2,{{\rm e}^{-1}} \right) {\theta_2} \left( 0,{{\rm e}^{-1}} \right) { \theta_3} \left( 0,{{\rm e}^{-1}} \right) {\theta_4} \left( 0,{{\rm e}^{-1}} \right) }{{\theta_2} \left( i/2,{ {\rm e}^{-1}} \right) {\theta_1} \left( iz,{{\rm e}^{-1}} \right) }} $$ Note, despite all those $i$s in there, this is real when $z$ is real.