While going through a paper I stumbled on the following equality :
$\begin{equation} \rho(\alpha)=\frac{1}{\pi}\sum\limits_{k\in\mathbb{Z}}\frac{e^{2ik\alpha}}{\cosh(k\zeta)}=\frac{1}{\pi}\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)}\frac{\vartheta_3(\alpha,q)}{\vartheta_4(\alpha,q)}\end{equation}$
such that $\zeta \in \mathbb{R^+}$ and $\vartheta_i(\alpha,q)$ are the Jacobi theta functions with nome $q=e^{-\zeta}$ and are defined as in the Whittaker and Watson book :
$\begin{align}\begin{aligned}\vartheta_1(z)&=-i\sum\limits_{n=-\infty}^{+\infty}(-1)^nq^{(n+\frac{1}{2})^2}\exp((2n+1)iz)\\ \vartheta_2(z)&=\sum\limits_{n=-\infty}^{+\infty}q^{(n+\frac{1}{2})^2}\exp((2n+1)iz)\\ \vartheta_3(z)&=\sum\limits_{n=-\infty}^{+\infty}q^{n^2}\exp(2niz)\\ \vartheta_4(z)&=\sum\limits_{n=-\infty}^{+\infty}(-1)^nq^{n^2}\exp(2niz)\end{aligned}\end{align}$
Now, they proceed to say that $\rho(-iu-i\zeta/2)$ is an odd function of u. Such that u $\in [-\zeta/2,\zeta/2]$.
Thus one would expect to have:
$\begin{equation}\rho(-iu-i\zeta/2)=-\rho(iu-i\zeta/2)\end{equation}$
While trying to manipulate the infinite sum in the first equation I struggled to see why this function would be odd, in addition I put the infinite sum in Mathematica where I got a graph that also shows that the infinite sum is not odd in u. However, using some properties of the Jacobi theta functions I got to the answer I want.
- Inifinite Sum approach
$\begin{equation}\rho(-iu-i\zeta/2)=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\exp(2ik(-iu-i\zeta/2))}{\cosh(k\zeta)}=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\exp(2k(u+\zeta/2))}{\cosh(k\zeta)}=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\cosh(2k(u+\zeta/2))+\sinh(2k(u+\zeta/2))}{\cosh(k\zeta)}\end{equation}$
The part $\sum\limits_{k=-\infty}^{+\infty}\frac{\sinh(2k(u+\zeta/2))}{\cosh(k\zeta)}$ is zero since at k=0 it's zero as for the rest of the sum the positive and negative k's cancel out.
Thus we are left with:
$\begin{equation}\rho(-iu-i\zeta/2)=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\cosh(2k(u+\zeta/2))} {\cosh(k\zeta)}\end{equation}$
Similarly, we get :
$\begin{equation} \rho(iu-i\zeta/2)=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\cosh(2k(-u+\zeta/2))} {\cosh(k\zeta)}\end{equation}$
From this we can clearly see that (if I did everything correctly ):
$\begin{equation}\rho(-iu-i\zeta/2)+\rho(iu-i\zeta/2)=\frac{1}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\cosh(2k(u+\zeta/2))+\cosh(2k(-u+\zeta/2))} {\cosh(k\zeta)}=\frac{2}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\cosh(2ku)\cosh(k\zeta)} {\cosh(k\zeta)}\neq 0\end{equation}$
This normally implies that $\rho(-iu-i\zeta/2)$ is not odd.
(I also checked $\rho(-iu-i\zeta/2)-\rho(iu-i\zeta/2)=\frac{2}{\pi}\sum\limits_{k=-\infty}^{+\infty}\frac{\sinh(2ku)\sinh(k\zeta)} {\cosh(k\zeta)}$ which is also not zero)
- Theta functions approach
A couple properties of the theta functions that I will be using are :
$\begin{align}\begin{aligned}\vartheta_4(z+\tau \pi)&=-\frac{1}{q}e^{-2iz}\vartheta_4(z)\\ \vartheta_3(z+\tau \pi)&=\frac{1}{q}e^{-2iz}\vartheta_3(z)\end{aligned}\end{align}$
Such that $e^{i\tau\pi}=e^{ii\zeta}$ which implies that $i\zeta=\tau\pi$
Thus we can see that:
$\begin{equation}\rho(z+i\zeta)=\frac{1}{\pi}\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)}\frac{\vartheta_3(z+i\zeta,q)}{\vartheta_4(z+i\zeta,q)}=-\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)}\frac{\vartheta_3(z,q)}{\vartheta_4(z,q)}=-\rho(z)\end{equation}$
Now if we take $z=-iu-i\zeta/2$ in the last equation, we can see that:
$\begin{equation}\rho(-iu-i\zeta/2+i\zeta)=-\rho(-iu-i\zeta/2) \implies \rho(-iu+i\zeta/2)=-\rho(-iu-i\zeta/2)\end{equation}$
Finally on Mathematica
I also tried to examine the plot of this infinite series on Mathematica, although I did not obtain a very insightful as this sum blows up, the plot didn't show any sign of being odd in u.
Can you please help me figure out what I am doing incorrectly ? My main goal is to find the parity of $\rho(-iu-i\zeta/2)$ in u