The closed form of $\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$ An integral I discussed last days in a chat, and it looks like a hard nut since 
after some manipulations of the initial form we reach an integral where the integrand is expressed in
 terms of 
digamma function  that can be further reduced to some very resistent series.
 I conjecture that the integral has  a  closed  form expressed in terms of G-Barnes function
(and newer constants?) Well, I might be wrong, but here we have many integration gurus
that could probably  answer that.
$$\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$$ 

The integral to evaluate is
\begin{align}
I(1) = \int_{0}^{\infty} e^{-x} \, \ln(\cosh(x)) \, \frac{dx}{x}.
\end{align}
The evaluation is as follows. Consider 
\begin{align}
\int_{1}^{\infty} e^{-\alpha x} d\alpha = \left[ - \frac{1}{x} e^{-\alpha x} \right]_{1}^{\infty} = \frac{e^{-x}}{x}.
\end{align}
Let
\begin{align}
I(\alpha) = \int_{0}^{\infty} e^{-\alpha x} \, \ln(\cosh(x)) \, \frac{dx}{x},
\end{align}
where $I(1)$ is the integral to evaluate and $I(\infty) = 0$, for which differentiation with respect to $\alpha$ leads to
\begin{align}
\partial_{\alpha} I(\alpha) = - \int_{0}^{\infty} e^{- \alpha x} \ln(\cosh(x)) dx.
\end{align}
Using integration by parts it is seen that
\begin{align}
\partial_{\alpha} I(\alpha) &= \left[ \frac{1}{\alpha} \, e^{- \alpha x} \ln(\cosh(x)) \right]_{0}^{\infty} - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\
&= - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\
&= - \frac{1}{2 \alpha} \left[ \beta\left( \frac{\alpha}{2} \right) - \beta\left( \frac{\alpha + 2}{2} \right) \right] \\
&= - \frac{1}{4\alpha} \left[ 2 \psi\left( \frac{\alpha + 2}{4} \right) - \psi\left( \frac{\alpha}{4} \right) - \psi\left( \frac{\alpha}{4} + 1 \right) \right] \\
&= \frac{1}{\alpha^{2}} - \frac{1}{\alpha} \beta\left(\frac{\alpha}{2}\right),
\end{align}
where $2 \beta(x) = \psi((x+1)/2) - \psi(x/2)$. Since
\begin{align}
\beta\left(\frac{x}{2}\right) = 2 \partial_{x} \ln\left( \frac{\Gamma((x+2)/4)}{\Gamma(x/4)} \right)
\end{align}
then
\begin{align}
\partial_{\alpha} I(\alpha) &= \frac{1}{\alpha^{2}} - \frac{2}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right)
\end{align}
and
\begin{align}
\int_{1}^{\infty} \partial_{\alpha} I(\alpha) \, d\alpha &= 1 - 2 \int_{1}^{\infty} \frac{1}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \\
I(\infty) - I(1) &= 1 - 2 \left[ \frac{1}{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \right]_{1}^{\infty} - 2 \int_{1}^{\infty} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \, \frac{d\alpha}{\alpha^{2}} \\
- I(1) &= 1 + 2 \ln\left( \frac{\Gamma(3/4)}{\Gamma(1/4)} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u+1/2)}{\Gamma(u)} \right) \, \frac{du}{u^{2}}.
\end{align} 
This is
\begin{align}
I(1) &= \ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}}.
\end{align}
It is of note that
\begin{align}
\ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) < \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}} < \ln\left( \frac{\Gamma^{4}(1/4)}{4 \pi^{2} e} \right).
\end{align}
