How can one verify the closed form for $\int_{0}^{\infty}{\ln{x}\over \cosh^{2n}{x}}\mathrm dx?$

Question

Consider the general integral $(1)$

$$\int_{0}^{\infty}{\ln{x}\over \cosh^{2n}{x}}\mathrm dx=I_{2n}\tag1$$ $n\ge1$

Wolfram integrator was only able to evaluate $I_2$,$I_4$ and $I_6$, from there we conjectured that $I_{2n}$ has the following pattern

$$I_2=\int_{0}^{\infty}{\ln{x}\over \cosh^2x}\mathrm dx=-\left(\gamma+\ln{4\over \pi}\right)$$

$$I_4=\int_{0}^{\infty}{\ln{x}\over \cosh^4x}\mathrm dx=-{7\zeta(3)\over 3}-{2\over 3}\left(\gamma+\ln{4\over \pi}\right)$$

$$I_6=-{7\zeta(3)\over 3\pi^2}-{31\zeta(5)\over5\pi^4}-{2\cdot4\over 3\cdot5}\left(\gamma+\ln{4\over \pi}\right)$$

$$I_8=-{7\zeta(3)\over 3\pi^2}-{31\zeta(5)\over5\pi^4}-{127\zeta(7)\over 7\pi^6}-{2\cdot4\cdot6\over 3\cdot5\cdot7}\left(\gamma+\ln{4\over \pi}\right)$$

Nature question to ask is, how can we show that ($n\ge2$)

$$I_{2n}=-{7\zeta(3)\over 3\pi^2}-{31\zeta(5)\over5\pi^4}-\cdots-{(2^{2n-1}-1)\zeta(2n-1)\over(2n-1)\pi^{2n-2}}-{(2n-2)!!\over (2n-1)!!}\left(\gamma+\ln{4\over \pi}\right)?\tag2$$

An attempt:

$\cosh^{2n}x=\left({e^x+e^{-x}\over 2}\right)^{2n}$

$(1)$ becomes

$$2^{2n}\int_{0}^{\infty}{\ln{x}\over(e^x+e^{-x})^{2n}}dx\tag3$$

Applying negative binomial series

$$2^{2n}\sum_{k=0}^{\infty}{-n\choose k}\int_{0}^{\infty}e^{2xk}\ln{x}\mathrm dx\tag4$$

Applying integration by parts

$$\int_{0}^{\infty}e^{2xk}\ln{x}\mathrm dx={e^{2xk}\over 2k}\ln{x}|_{0}^{\infty}-{1\over 2k}\int_{0}^{\infty}{e^{2xk}\over x}\mathrm dx\tag5$$

$(4)$ diverges.

How else can we tackle $(1)$?

Answer 1

Starting from $$ \zeta(s) = \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} \tag{1}$$ then applying the (inverse) Laplace transform and integration by parts, we get: $$ \int_{0}^{+\infty}\frac{x^s}{\cosh^2(x)}\,dx = \frac{2(2^s-2)\,\Gamma(s+1)}{4^s}\,\zeta(s)=\int_{0}^{1}\text{arctanh}(x)^s\,dx\tag{2} $$ Your result then follows from integration by parts and $$ \int_{0}^{+\infty}\frac{\log x}{\cosh^2(x)}\,dx = \left.\frac{d}{ds}\left(\frac{2(2^s-2)\Gamma(s+1)\zeta(s)}{4^s}\right)\right|_{s=0}=-\gamma+\log\frac{\pi}{4}.\tag{3}$$