In studying some physical propagator, I came across the following sum $$ \sum_{n = -\infty}^{+\infty} \frac{ a^n }{ \sin^2(z + n \pi \tau) }\ . $$ Obviously, my question is how to evaluate this sum.

To some extent, I understand the result when $a = 1$. Loosely speaking, without properly regularizing, we have $$ \sum_{n \in \mathbb{Z}} \frac{ 1 }{ \sin^2(z + n \pi \tau) } = - \sum_{n \in \mathbb{Z}} \partial_z \partial_z \ln \sin(z + n\pi\tau) = - \partial_z^2 \ln \prod_{n \in \mathbb{Z}} \sin(z+n\pi\tau) \ . $$ The final infinite product can be identified with $\theta_1(z/\pi|\tau)$, where $q = e^{2\pi i \tau}$, so up to regularization issue, we have $$ \sum_{n\in \mathbb{Z}} \frac{ 1 }{ sin^2(z + n\pi \tau) } = - \partial_z \partial_z \ln \theta_1(z/\pi|\tau) $$

However in the presence of $a^n$, I can't pull off this trick again (as far as I can see).

Suggestions on literature/references and more tricks are welcome!

  • $\begingroup$ The Fourier series in $\tau$ will have coefficients of the form $\sum_{k | m}a^{m/k} k e^{2kz} $, so it is close to the inverse Mellin transform (in $s$) of $Li_{s-1}(e^{2z})Li_s(a)$ $\endgroup$ – reuns Jan 26 at 9:43

The following is probably not mathematically rigorous, and is loosely based on the Ramanujan's identity \begin{align} \sum_{n = -\infty}^\infty \frac{ (A;q)_n }{ (B;q)_n }X^n = \frac{(q;q) (B/A;q) (AX;q) (q/(AX))}{(B;q)(q/A;q)(X;q)(B/(AX);q)} \ . \end{align}

  1. The original problem can be rephrased (assuming $\partial_z$ can be moved into the sum), \begin{align} F(z) \equiv & \ \sum_{n \in \mathbb{Z}} \frac{1}{\sin^2(\frac{z}{2} + n \pi \tau)} a^n = -2 \partial_z \sum_{n \in \mathbb{Z}}\frac{\cos(\frac{z}{2} + n\pi \tau)}{\sin(\frac{z}{2} + n\pi \tau)} a^n \equiv -2 \partial_z G(z). \end{align}

  2. To compute $G(z)$, we reorganize it \begin{align} G(z) = - i \sum_{n \in \mathbb{Z}} \frac{e^{i(z + 2 n\pi \tau)}}{1-e^{i(z + n2\pi \tau)}} a^n - i \sum_{n \in \mathbb{Z}} \frac{1}{1 - e^{i(z + 2 n\pi \tau)}} a^n \ . \end{align} Defining $x \equiv e^{iz}$, $q = e^{2 \pi i \tau}$, we have \begin{align} G(z) = -i \frac{x}{1-x} \sum_n \frac{(x;q)_n}{(qx;q)_n} (aq)^n - i \frac{1}{1-x} \sum_n \frac{(x;q)_n}{(qx;q)_n} a^n \ , \end{align} where we used \begin{align} \frac{1}{1-xq^n} = \frac{1}{1-x} \frac{(x;q)_n}{(qx;q)_n} \ . \end{align}

  3. Now Ramanujan's identity comes in, and using shift properties of the $q$-Pochhammer symbols, the two sums actually are equal and they add. The final result is \begin{align} G(z) = 2 \eta(\tau)^3 \frac{ \vartheta_1(\mathfrak{a} + \frac{ z}{ 2\pi }|\tau) }{ \vartheta_1(\frac{ z}{ 2\pi }|\tau)\vartheta_1(\mathfrak{a}|\tau )}\ . \end{align}

  4. The above computation is not rigorous because the Ramanujan's identities requires $|q| < 1$ ,$|B/A| < |X| < 1$ for convergence. However, these requirements applied to the two sums at the end of step 2 are not compatible: one sum requires $|a| < 1$, the other $|a| > 1$. Besides, we need to move the $\partial_z$ into a sum.

  5. However the final answer seems physically reasonable, since it does produce physical results that we expect, despite the above issue.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.