Q:

An interesting doubly infinite series

Let $0<\mu<1$ and $\alpha:=1-\mu^2$. Consider the function 
$$f(x):=x\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}x}-\frac{1}{x}\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}/x},$$
defined for all $x>0$. Three properties are easy to check: $f(\mu^{2n})=0$ for every integer $n$, $f(x)=-f(1/x)$, $f(x)$ vanishes at $x=\mu^2$ and $x=1$ and $f(x)=f(\mu^4x)$.
I want to show that $f(x)<0$ for $\mu^2<x<1$, but I have not been able to prove it. Has anybody seen anything like this? 

A:

I've seen sums like this, and they can get quite amusing, e.g. the Fourier coefficients of $f(x)$ as a periodic function of $\log(x)$ involve values of the Gamma function at complex arguments (see below); but it seems that this is overkill for the question at hand: there are several ranges of $\mu$ for which $f(\mu) > 0$, e.g. $\mu = 1/4$ works, giving $f(1/4) = 0.0892157+ > 0$.  Are you sure this is what you meant?
If I computed everything correctly (and gp corroborates numerically), the following sine-Fourier expansion holds: write $\mu = \exp(-\lambda)$ and $x = \mu^t = \exp(-\lambda t)$; then
$$
f(x) = \sum_{n=1}^\infty \phantom. c_n \sin \frac{\pi n t}{2}
$$
where
$$
c_n = \frac1\lambda \mathop{\rm Im} \left(
\Gamma\bigl(1 + \frac{\pi i n}{2\lambda}\bigr) \Bigl/ \alpha^{1 + \frac{\pi i n}{2\lambda}}
\right).
$$
This does not depend on the choice $\alpha = 1 - \mu^2$.
P.S. See
this Mathoverflow answer
where such a sum (and its Fourier expansion with complex-Gamma coefficients) arises naturally.

