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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 |
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Let $0<\mu<1$ and $\alpha:=1-\mu^2$. Consider the function $f(x):=x\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}x}-\frac{1}{x}\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}/x}$, $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 |
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An interesting doubly infinite seriesLet $0<\mu<1$ 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
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