Need help with $\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx$ I need help with this integral:
$$\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx\approx0.20597312051214...$$
Is it possible to evaluated it in a closed form?
 A: A useful identity here is ${\rm arccosh\,} z = \ln( z + \sqrt{z^2 -1} )$.
Therefore $$\int_0^{\infty} e^{-x} \ln \ln( e^x + \sqrt{e^{2x}-1}) \;dx= \int_0^{\infty} e^{-x}\ln {\rm arccosh\,} e^x \;dx =\int_1^{\infty} \frac{\ln {\rm arccosh\,} y}{y^2} \;dy\, $$
Change integration variable $z={\rm arccosh\,}y$,
$$= \int_0^{\infty} \frac{\ln z \sinh z}{ \cosh^2 z} \;dz\,.$$
Integrate by parts and use the definition of the Euler's $\gamma$ constant and $\Gamma$ function see here,
$$= -\gamma + \ln\left[\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \right]\,. $$
We can further simplify this using $\Gamma(1-z)\Gamma(z)=\pi/\sin \pi z$  and $\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ for $z=3/4$ and $\sin (3\pi/4)=1/\sqrt{2}$. This gives $$=-\gamma - 3\ln 2 - 2 \ln \pi + 4 \ln \Gamma(1/4)\,. $$
A: This is a partial answer only.
Let $x=\log u$; then $dx=du/u$, and
$$
I=\int_{1}^{\infty} \frac{1}{u^2}\log\log(u+\sqrt{u^2-1})du.
$$
Next let $u=\cosh v$; then $du=\sinh v dv$, and $u+\sqrt{u^2-1}=\cosh v + \sinh v=e^v$, so 
$$
I=\int_{0}^{\infty}\frac{\sinh v \log v}{\cosh^2 v} dv=\int_{0}^{a}\log{v}\cdot d\left(1-\frac{1}{\cosh v}\right) - \int_{a}^{\infty}\log v \cdot d\left(\frac{1}{\cosh v}\right),
$$
where we can choose $a\in(0,\infty)$.  Now integrate by parts to get
$$
I=\log{a}\left(1-\frac{1}{\cosh a}\right)-\int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv +\frac{\log a}{\cosh a} +\int_{a}^{\infty}\frac{dv}{v\cosh v}= \\\log a - \int_{0}^{a}\left(\frac{1}{v}-\frac{1}{v\cosh v}\right)dv+\int_{a}^{\infty}\frac{dv}{v \cosh v}.
$$
If we let $a\rightarrow 0$, this eliminates the middle term:
$$
I=\lim_{a\rightarrow 0+}\left[\log{a} + \int_{a}^{\infty}\frac{dv}{v\cosh v}\right].$$
Perhaps someone can take it from here?
