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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 $n$, $f(x)=-f(1/x)$, and $f(x)$ vanishes at $x=\mu^2$ and $x=1$ and $f(x)=f(\mu^4x)$.

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

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# 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 $n$, $f(x)=-f(1/x)$, and $f(x)$ vanishes at $x=\mu^2$ and $f(x)=f(\mu^4x)$. I want to show that $f(x)>0$ for \$\mu^2