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.
A: Thanks very much, you're right, it is not true for $\mu=1/4$, meaning that in general the real zeros of the function $f(x)$ are not just $\mu^{2n}$, $n$ an integer. I thought that this perhaps was the case, implying what I wanted to prove. I'll explain more:
Define $$\chi_q(t):=t^2\int_0^\infty \mu^{-4\langle\log_{\mu^4}(s)+q\rangle }s e^{-(\mu^{-1}-\mu) ts}ds ,\quad \Re(t)>0$$ where $\mu$ and $q$ are constants with $0<\mu<1$, $0\leq q < 1$, and $\langle x\rangle$ denotes the fractional part of $x$. 
The function I am looking at originally is
$$g_q(t)=\chi_q(t)-\chi_q(\mu^2/t),\quad \Re(t)>0.$$ This functions satisfies $g_q(\mu^4t)=g_q(t)$.
What I thought would be the case is that $g_q(t)$ has $\mu$ as its only zero in the region $\Re(t)>\mu^2R/2$, $|t-1/R|<1/R$. Here $R=\mu^{-1}+\mu$. This region is its own reflection about the circle of radius $\mu$, and the intersection of this region with the real axis is the interval $(\mu^2R/2, 2/R)$.
The function $s\mu^{-4\langle\log_{\mu^4}(s)+q\rangle }$ is a step function, and by integrating we can write $g_q(t)$ as the doubly infinite series (up to a multiplicative constant):$$t\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}t}-\frac{\mu^2}{t}\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}(\mu^2/t)}$$
For $q=0$, this is essentially the function in the original question since $f(x)=\mu^{-1} g_0(\mu x)$.
However, computational experiments with Mathematica indicate that for small values of $\mu$, g_0(t) has at least three real zeros in the interval $(\mu^2R/2, 2/R)$, and this seems to be the case as well for $\mu=1/5$ and $q=1/1000$, which disproves my conjecture. Below is the Mathematica code. Here m is $\mu$, and I truncate the series at $k=M$.
M = 300;
m = 1/5;
q = 0;
N[m]
R = m + 1/m;
g[x_] := x*(Sum[
       m^(4*(k - q))Exp[-(1/m - m) m^(4*(k - q))x], {k, 0, M}] + 
      Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))(m^2/x)], {k, 
        1, M}]) - (m^2/
      x)(Sum[m^(4*(k - q))*Exp[-(1/m - m)m^(4(k - q))(m^2/x)], {k,
         0, M}] + 
      Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))*(m^2/x)], {k, 
        1, M}]);
Plot[g[x], {x, m^2*R/2, 2/R}, PlotRange -> All] 
