interesting integral

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#1 h Apr 21, 2025, 3:12 PM

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$\int_0^\infty \frac{\sinh(t)}{t \cosh^3(t)} dt$

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#2 h Apr 22, 2025, 2:46 PM • 2 Y

Y by Svyatoslav, Martin.s

The answer is $\frac{7}{\pi^2}\zeta(3).$

Let $I$ be your integral. Applying IBP twice gives
$I=\int_0^{\infty}\frac{x-\tanh x}{x^3} \ dx,$ and so, using the identity
$\tanh x =\sum_{n \ge 1}\frac{8x}{(2n-1)^2\pi^2+4x^2},$ we get that
$I=\int_0^{\infty}\left[\frac{1}{x^2}-\frac{8}{\pi^2}\sum_{n \ge 1}\frac{1}{(2n-1)^2}\left(\frac{1}{x^2}-\frac{4}{(2n-1)^2\pi^2+4x^2}\right)\right]dx$ $=\int_0^{\infty}\left[\frac{1}{x^2}-\frac{8}{\pi^2x^2}\sum_{n \ge 1}\frac{1}{(2n-1)^2}+\frac{32}{\pi^2}\sum_{n \ge 1}\frac{1}{(2n-1)^2((2n-1)^2\pi^2+4x^2)}\right]dx$ $=\frac{32}{\pi^2}\sum_{n \ge 1}\frac{1}{(2n-1)^2} \int_0^{\infty} \frac{dx}{(2n-1)^2\pi^2+4x^2}=\frac{8}{\pi^2}\sum_{n \ge 1} \frac{1}{(2n-1)^3}=\frac{7}{\pi^2}\zeta(3).$

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