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Given $f(x)=\frac{x^2}{\sinh ^2 x}$

I computed the Laplace transform $\mathscr{L}\left(f(x)\right)(s)=\frac{1}{4} \left(4 \psi ^{(1)}\left(\frac{s}{2}\right)+s \psi ^{(2)}\left(\frac{s}{2}\right)\right)$

Then $$\int_0^{\infty } \frac{x^2}{\sinh ^2 x} \, dx=\underset{s\to 0}{\text{lim}}\mathscr{L}\left(f(x)\right)(s)=\frac{\pi ^2}{6}=\zeta(2)$$ which is $$\sum _{n=1}^{\infty } \frac{1}{n^2}=\zeta(2)$$

Do you think there is a deeper link between these two results or is it just a coincidence?

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In fact,

$$ \begin{align} \int_0^{\infty } \frac{x^2}{\sinh ^2 x} \, dx&=\int_0^\infty\frac{4x^2}{(e^x-e^{-x})^2}\,dx\\[1ex] &=\int_0^\infty\frac{4e^{-2x}x^2}{(1-e^{-2x})^2}\,dx\\[1ex] &=\int_0^\infty\sum_{n=1}^\infty 4ne^{-2nx}x^2\,dx\\[1ex] &=\sum_{n=1}^\infty4\int_0^\infty ne^{-2nx}x^2\,dx\\[1ex] &=\sum_{n=1}^\infty\frac{1}{n^2}. \end{align}$$

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    $\begingroup$ This generalises to $\int_0^\infty\frac{x^pdx}{\sinh^2x}=\frac{p!}{2^{p-1}}\zeta(p)$ for $p\ge2$.Generalizing the exponent of $\operatorname{csch}x$ has a more complicated effect, obtaining a linear combination of zeta constants. $\endgroup$
    – J.G.
    Nov 20, 2020 at 23:12
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    $\begingroup$ In fact J.G. suggested the solution for this generalization here: math.stackexchange.com/q/4080397/5051 $\endgroup$
    – heiner
    May 1, 2021 at 19:02

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