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Math Help - Vardi's Integral

  1. #1
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    Vardi's Integral

    Prove...

    \int_0^\infty \frac{1}{1+x^2}\ln \ln \frac{1}{x} dx= \frac{\pi}{2}\ln \left\{ \sqrt{2\pi}\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac  {1}{4})}\right\}

    please help...
    Last edited by Ozymandias; December 30th 2012 at 11:31 PM.
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Vardi's Integral

    The evaluation of such integrals involves a lot of number theory. We will need the Dirichlet L Function

    L(s)=1-\frac{1}{3^s}+\frac{1}{5^s}-\cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}

    The gamma function,

    \frac{\Gamma(s) (-1)^n}{(2n+1)^s}=\int_0^\infty t^{s-1}(-1)^n e^{-(2n+1)t}dt

    \Gamma(s) L(s)=\int_0^\infty t^{s-1} \sum_{n=0}^\infty (-1)^n e^{-(2n+1)t} \ dt

    Now let z=e^{-t}

    \Gamma(s) L(s)=\int_01 \left( \ln\frac{1}{z}\right)^{s-1} \sum_{n=0}^\infty (-1)^{n} z^{-(2n+1)} \ \frac{dz}{z}

    \Gamma(s) L(s)=\int_0^1\left( \ln\frac{1}{z}\right)^{s-1} \left( \frac{1}{1+z^2}\right) \ dz

    Differentiate with respect to the parameter s.

    \frac{d}{ds}\Gamma(s) L(s)=\int_0^1\left( \ln\frac{1}{z}\right)^{s-1} \left( \frac{1}{1+z^2}\right) \ln \ln \frac{1}{z}\ dz

    Hence we conclude

    \int_0^1 \frac{1}{1+x^2}\ln \ln \frac{1}{x} dx= \frac{d}{ds}\Gamma(s)L(s) \Bigg|_{s=1}

    Evaluating d/ds (\Gamma(s) L(s)) is it not a "walk in the park" and requires the help of number theory.
    Last edited by sbhatnagar; December 31st 2012 at 02:00 AM.
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