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? 



An interesting doubly infinite series
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](https://mathoverflow.net/questions/11255/80730#80730)
where such a sum (and its Fourier expansion with complex-Gamma coefficients) arises naturally.


