How to prove that $$\int_{-\infty}^{\infty} e^{-\delta r^2} \frac{\pi}{2 \cosh^2(\pi r)} \, dr = 1 + O(\delta),$$ as $\delta \to 0$?
2026-03-26 14:25:23.1774535123
How to prove the following asymptotic behavior?
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Almost trivial, with a CAS. Taylor expand the exponential and use $$ \int_{-\infty}^\infty \frac{dr}{\cosh^2{(\pi\,r)}}=\frac{2}{\pi} \, \text{ and } \int_{-\infty}^\infty \frac{r^2\,dr}{\cosh^2{(\pi\,r)}}=\frac{1}{6\pi}.$$