$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$

Question

Is there any closed-form representation for the following integral? $$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx,$$ where $\mathrm{sech}\,x$ is the hyperbolic secant, $\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}$.

Answer 1

Hint: Use the integral: $$\int_0^\infty x^n\,\text{sech}^2x\,\mathrm dx=(2^{1-n}-4^{1-n})\,\Gamma(n+1)\,\zeta(n)$$ and take the 2nd derivative with respect to $n$.

Answer 2

Related technique: (I), (II), (III). Here is a closed form

$$ \int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx = \int_0^\infty(\log x)^2\left(\frac{2}{e^{x}+e^{-x}}\right)^2\mathrm dx $$

$$= \left( \ln \left( \pi \right) \right) ^{2}- \left( 4\,\ln \left( 2 \right) +2\,\gamma \right) \ln \left( \pi \right) +2\, \left( \ln \left( 2 \right) \right) ^{2}+4\,\ln \left( 2 \right) \gamma-2\,\gamma \left( 1 \right) +\frac{1}{4}\,{\pi }^{2} $$

$$ \sim 1.989349759. $$

Note: $\gamma(n)$ are known as Stieltjes $\gamma$-constants

$$ \gamma(n)= \lim_{m\to \infty}\left(\sum_{k=1}^m \frac{(\ln k)^n}{k}-\frac{(\ln m)^{n+1}}{n+1}\right).$$