TITLE: How to evaluate $\int_0^\infty \frac{e^{-x}+x-1}{x(e^{2x}-e^{-2x})}dx$?
QUESTION [5 upvotes]: We are unable to verify this this equality
$$
4\int\limits_0^\infty \frac{e^{-x}+x-1}{x\left(e^{2x}-e^{-2x}\right)}\;\mathrm{d}x=\gamma+\ln\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\;.
$$
where $\gamma$ is Euler's constant which is approximately $0.5772156\dots$, and $\Gamma(n)$ is the gamma function and the particular value for $\Gamma\left(\frac{1}{4}\right)=3.625690\dots$.

REPLY [2 votes]: Another approach. We have $$\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{k}}{k!}=e^{-x}
 $$ hence $$e^{-x}-1+x=\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}}{k!}
 $$ and so $$I=4\int_{0}^{\infty}\frac{e^{-x}-1+x}{x\left(e^{2x}-e^{-2x}\right)}dx=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k!}\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx
 $$ now note that $$ \begin{align} 
\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx= & \int_{0}^{\infty}x^{k-1}\sum_{m\geq0}e^{-\left(2m+1\right)2x}dx \\ = & \sum_{m\geq0}\int_{0}^{\infty}x^{k-1}e^{-\left(2m+1\right)2x}dx
  \\ =
  & \frac{\left(k-1\right)!}{2^{k}}\sum_{m\geq0}\frac{1}{\left(2m+1\right)^{k}}
  \\ =
  & \frac{\left(k-1\right)!}{2^{k}}\left(1-\frac{1}{2^{k}}\right)\zeta\left(k\right)
 \end{align}$$ and so $$I=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k2^{k}}\zeta\left(k\right)-4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k4^{k}}\zeta\left(k\right)
 $$ and now we can use the identity, for $x\geq0,
 $ $$\sum_{k\geq2}\frac{\left(-x\right)^{k}}{k}\zeta\left(k\right)=x\gamma+\log\left(\Gamma\left(x+1\right)\right)
 $$ so finally we have $$I=\gamma+\log\left(\frac{16\pi^{2}}{\Gamma\left(\frac{1}{4}\right)^{4}}\right).
 $$