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Math Help - yet another integral

  1. #1
    Super Member Random Variable's Avatar
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    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 ).
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    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,

    <br />
\int_0^{\tfrac{\pi}{2}} x^2 \ln^2(2\cos x) \ dx \ = \ \frac{11 \pi^5}{1440}<br />
    Last edited by mr fantastic; September 18th 2009 at 09:49 AM. Reason: Restored original reply
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  3. #3
    Super Member Random Variable's Avatar
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    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 ?
    Last edited by Random Variable; August 25th 2009 at 12:09 PM.
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    Quote Originally Posted by Random Variable View Post
    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.
    Last edited by mr fantastic; September 18th 2009 at 09:49 AM. Reason: Restored orignal reply
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