1. ## yet another integral

Show that $\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln\big(\ln \ (\tan x)\big)\ dx = \frac{\pi}{2} \ln \Big(\sqrt{2 \pi} \ \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\Bi g)$

This integral is called Vardi's Integral (presumably named after someone named Vardi ).

2. Originally Posted by Random Variable
Show that $\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln\big(\ln \ (\tan x)\big)\ dx = \frac{\pi}{2} \ln \Big(\sqrt{2 \pi} \ \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\Bi g)$

This integral is called Vardi's Integral (presumably named after someone named Vardi ).
The answer to this may deserve a technical paper (actually you can find it in literature) and may require considerable knowledge in number theory. Here is another one,

$\displaystyle \int_0^{\tfrac{\pi}{2}} x^2 \ln^2(2\cos x) \ dx \ = \ \frac{11 \pi^5}{1440}$

3. 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?

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

4. Originally Posted by 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?

$\displaystyle - \frac{1}{2}\int^{1}_{0} \int^{1}_{u} \frac{1}{t \cosh u} \ dt \ du + \frac{1}{2}\int^{\infty}_{1} \int^{u}_{1} \frac{1}{t \cosh u} \ dt \ du$ $\displaystyle = \frac{1}{2} \int^{\infty}_{0} \frac{\ln u}{\cosh u} \ du$ ?
The remaining is no easy task. It uses the Dirichlet’s L-functions. There are many beautiful definite integrals with beautiful results out there but the solution or proof procedures are just painstaking.