# yet another integral

• August 24th 2009, 05:52 PM
Random Variable
yet another integral
Show that $\int^{\frac{\pi}{2}}_{\frac{\pi}{4}} \ln\big(\ln \ (\tan x)\big)\ dx = \frac{\pi}{2} \ln \Big(\sqrt{2 \pi} \ \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\Bi g)$

This integral is called Vardi's Integral (presumably named after someone named Vardi ).
• August 24th 2009, 07:58 PM
luobo
Quote:

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

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,

$
\int_0^{\tfrac{\pi}{2}} x^2 \ln^2(2\cos x) \ dx \ = \ \frac{11 \pi^5}{1440}
$
• August 25th 2009, 08:34 AM
Random Variable
EDIT:

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

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

$u = \ln (\tan x )$

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

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

$- \frac{1}{2}\int^{1}_{0} \int^{1}_{u} \frac{1}{t \cosh u} \ dt \ du + \frac{1}{2}\int^{\infty}_{1} \int^{u}_{1} \frac{1}{t \cosh u} \ dt \ du$ $= \frac{1}{2} \int^{\infty}_{0} \frac{\ln u}{\cosh u} \ du$ ?
• August 25th 2009, 04:48 PM
luobo
Quote:

Originally Posted by Random Variable
EDIT:

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

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

$u = \ln (\tan x )$

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

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

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

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.