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Math Help - Catalan's Constant

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Catalan's Constant

    Show  \displaystyle \int_1^\infty \frac{\log x}{x^2+1}dx = G, where  \displaystyle G is Catalan's constant.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting u=\frac{1}{x} the integral becomes...

    \displaystyle \int_{1}^{\infty} \frac{\ln x}{1+ x^{2}}\ dx = - \int_{0}^{1} \frac{\ln u}{1 + u^{2}}\ du (1)

    Now with a little of patience You can demonstrated that in general is...

    \displaystyle \int_{0}^{1} u^{n}\ \ln u\ du = -\frac{1}{(n+1)^{2}} (2)

    ... so that, tacking into account that for |u|<1 is...

    \displaystyle \frac{1}{1+ u^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\ u^{2n} (3)


    is...

    \displaystyle \int_{1}^{\infty} \frac{\ln x}{1+ x^{2}}\ dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} (4)

    This integral is very similar to the integral resolved here...

    http://www.mathhelpforum.com/math-he...ls-151443.html

    Kind regards

    \chi \sigma
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