# Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$

$$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$

I was told it could be calculated in a closed form.

$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{\heartsuit}$$

Derivation:

After the change of variables $x=\frac{1}{\cosh u}$ the integral becomes
$$\int_0^{\infty}\ln u \frac{\sinh u}{\cosh^2 u}du,$$
as was noticed above by Eric. We would like to integrate by parts to kill the logarithm but we get two divergent pieces. To go around this, let us consider another integral,
$$I(s)=\int_0^{\infty}u^s \frac{\sinh u}{\cosh^2 u}du,$$
with $s>0$. The integral we actually want to compute is equal to $I'(0)$, which will be later obtained in the limit.

Indeed, integrating once by parts one finds that
\begin{align}
I(s)&=s\int_0^{\infty}\frac{u^{s-1}du}{\cosh u}=s\cdot 2^{1-2 s}\Gamma(s)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]=\\ &=2^{1-2 s}\Gamma(s+1)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right],
\end{align}
where $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$ denotes Hurwitz zeta function (in the way we have used its integral representaion (5) from here).

Now to get ($\heartsuit$), it suffices to use
\begin{align}
&\frac{\partial}{\partial s}\left[2^{1-2 s}\Gamma(s+1)\right]_{s=0}=-2\gamma-4\ln 2,\\
&\zeta\left(0,\frac14\right)-\zeta\left(0,\frac34\right)=\frac12,
\\
&\frac{\partial}{\partial s}\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]_{s=0}=-\ln\frac{\Gamma(\frac34)}{\Gamma(\frac14)}.
\end{align}
[See formulas (10) and (16) on the same page].