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**Random Variable** EDIT:

If you make the substitution $\displaystyle x = \tan^{-1}(e^{u}) $

then $\displaystyle dx = \frac{e^{u}}{e^{2u}+1} \ du = \frac{1}{e^{u}+e^{-u}} \ du= \frac{1}{2 \cosh u} \ du $

$\displaystyle u = \ln (\tan x ) $

$\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln\big(\ln \ (\tan x)\big)\ dx = \frac{1}{2} \int^{\infty}_{0} \frac{\ln u}{\cosh u} \ du $

From here you could use contour integration. But how about integrating under the integral sign?

$\displaystyle - \frac{1}{2}\int^{1}_{0} \int^{1}_{u} \frac{1}{t \cosh u} \ dt \ du + \frac{1}{2}\int^{\infty}_{1} \int^{u}_{1} \frac{1}{t \cosh u} \ dt \ du $ $\displaystyle = \frac{1}{2} \int^{\infty}_{0} \frac{\ln u}{\cosh u} \ du $ ?