integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx} $ Evaluate this integrate
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx} $$
My friend tian_275461 proposed this integrate,but I have no idea about it.
 A: Let $u = \ln \tan x$, so that $\frac{\pi}{4} < x< \frac{\pi}{2}$ is mapped to $u>0$, $x=\arctan(\exp(u))$ and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \cosh(u)}$. Then
$$
   \int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\ln (u)}{\cosh(u)} \mathrm{d}u = \frac{1}{2} \lim_{s \to 0^+}\frac{\mathrm{d}}{\mathrm{d} s} \int_0^\infty \frac{u^s}{\cosh(u)} \mathrm{d}u
$$
The latter parametric integral is evaluated by expanding $\cosh(u)$ into exponential and using Euler's gamma-integral:
$$\begin{eqnarray}
  \int_0^\infty \frac{u^s}{2\cosh(u)} \mathrm{d}u &=& \int_0^\infty u^s \frac{\exp(-u)}{1+\exp(-2u)}\mathrm{d}u = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^{s} \exp(-(2n+1)u) \,\mathrm{d}u \\
  &=& \Gamma(s+1)  \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^{s+1}} = \Gamma(s+1) 2^{-2s-2} \left( \zeta\left(s+1,\frac{1}{4}\right) - \zeta\left(s+1,\frac{3}{4}\right) \right)
\end{eqnarray}
$$
Near $s=0$:
$$
    \zeta(1+s,a) = \frac{1}{s} - \psi(a) - \gamma_1(a) s + \mathcal{o}(s)
$$
where $\psi(a)$ is the digamma function, and $\gamma_1(a)$ is the first generalized Stieltjes constant. Differentiating and taking the limit we have
$$ \begin{eqnarray}
 \int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x &=& \frac{1}{4} \left( \left(\log(4) + \gamma\right)\left(\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right) - \left(\psi_1\left(\frac{1}{4}\right)-\psi_1\left(\frac{3}{4}\right)\right)\right) \\
  &=& \frac{\pi}{4} \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \approx -0.260443
\end{eqnarray}
$$
where the latter equality is given my Mathematica.
A: Or find it in Gradshteyn-Rizhyk 4.371
$$
\int_0^\infty \frac{\log x}{\cosh x}\;dx = \pi\log\left[
\frac{\sqrt{2\pi}\;\Gamma(3/4)}{\Gamma(1/4)}\right]
$$
half of that is your answer, agreeing with Sasha.
