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Let the function $$ F(x)=\frac{\pi\sinh(x)}{x}\sum_{n=-\infty}^{\infty}\frac{1}{\cosh\left[\frac{(2n+1)\pi^2}{2x}\right]} $$ where $\cosh(x)=\lambda\geq1$. For $\lambda\to1$, i.e., $x\to0$, what's the asymptotic expansion of $F$? The expansion can be expressed via $\lambda$, e.g., $F\simeq 3\sqrt{\lambda-1}+\,\mathcal{O}((\lambda-1)^2)$ (NOTE: This is NOT the right answer, however).

My naive solution

To begin with, we have $\lim_{x\to0}\frac{\sinh(x)}{x}=1$ and $x\simeq\sqrt{2(\lambda-1)}$, and $$ F(x)=\frac{2\pi\sinh(x)}{x}\sum_{n=0}^{\infty}\frac{1}{\cosh\left[\frac{(2n+1)\pi^2}{2x}\right]}\\ \simeq4\pi\sum_{n=0}^{\infty}\exp\left[-\frac{(2n+1)\pi^2}{2x}\right]\\ =4\pi\exp\left[-\frac{\pi^2}{2\sqrt{2(\lambda-1)}}\right]+\textrm{HIGH-ORDER} $$

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  • $\begingroup$ You mentioned the Poisson summation formula. What do you get using that $\frac{1}{\cosh}= \text{sech}$ is its own Fourier transform ? $\endgroup$ – reuns Aug 25 '17 at 3:34
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I don't know to what order you want the asymptotic expansion, but finding it up to leading order is simple:

$$F(x)=\pi \frac{\sinh(x)}{x} = \pi \frac{x+\frac{x^3}{6}+o(x^3)}{x} = \pi\left(1+\frac{x^2}{6}+o(x^2)\right) $$ Further, as you mentioned correctly $$ \cosh(x)=1+\frac{x^2}{2}+o(x^2) =\lambda \Rightarrow x = \sqrt{2(\lambda-1)}+ o(\sqrt{\lambda-1}).$$

Simply combining these two expressions gives the leading order term of the asymptotic expansion of $F$: $$ F(x) = \frac{\pi}{3}\left(2+\lambda+o(\lambda-1)\right).$$

The relative error for $x=0.5$ is about $\frac{|F(x)-\frac{\pi}{3}(2+\lambda)|}{F(x)}\approx 4\%$, for $x=0.1$ it is $0.1\%$.

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