On the closed form of a integral related to $cosh(x)$

Let's start with the integral $\;\displaystyle\int \frac {dx}{\cosh(x)^2}=\frac 2{1+e^{-2x}}+C\;$ (with the choice $C=-2$) and use integration by parts :

\begin{align} I(\alpha) &= \int_0^\infty \frac{x^\alpha}{\cosh^2x}dx\\ &=\left.x^\alpha\left(\frac 2{1+e^{-2x}}-2\right)\right|_0^\infty-\int_0^\infty \alpha\,x^{\alpha-1}\left(\frac 2{1+e^{-2x}}-2\right)\,dx\\ &=2\alpha \int_0^\infty \frac {x^{\alpha-1}\,e^{-2x}}{1+e^{-2x}}\,dx\\ &=2\alpha \int_0^\infty \frac {x^{\alpha-1}}{e^{2x}+1}\,dx\\ &=2\alpha\, 2^{1-\alpha-1}\int_0^\infty \frac {(2x)^{\alpha-1}}{e^{2x}+1}\,d(2x)\\ &= 2^{1-\alpha}\alpha\,\int_0^\infty \frac {t^{\alpha-1}}{e^t+1}\,dt \end{align}

Now a closed variant of $\zeta$ is the Dirichlet eta function which may be defined by : $$\eta(s) := \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{e ^ t + 1}\, dt=\left(1-2^{1-s}\right)\zeta(s) $$

All this gives us : $$I(\alpha) ={2^{1-\alpha}\left(1-2^{1-\alpha}\right)}{\,\Gamma(\alpha+1)}\,\zeta(\alpha)$$

Gradshteyn, Ryzhik: Table of Integrals, Series, and Products 7ed; Item 3.527.3 for $\mu \ne 2$ $$\int_0^{\infty} \frac{x^{\mu-1}}{\cosh^2 ax} dx =\frac{4}{(2a)^{\mu}}(1-2^{2-\mu})\Gamma(\mu)\zeta(\mu-1) $$ and $\frac{\ln 2}{a^2}$ for $\mu=2$.

For a derivation of the formula see K.N. Boyadzhiev and V.H. Moll, The integrals in Gradshteyn and Ryzhik. Part 21: Hyperbolic functions Example 9.2, Mathematical Sciences, Vol. 22 (2011), 109-127 available from Moll's page http://129.81.170.14/~vhm/formula_html/final21.pdf