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Math Help - [Complex Analysis] Integral of a multi-valued function

  1. #1
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    [Complex Analysis] Integral of a multi-valued function

    Hey! I need some help to prove the following identity:

    \int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi b^{a-1}}{\cos(\pi a/2)} , \ 1>a>-1, \ b>0

    I'm supposed to use the formula

    \int_{0}^{\infty} x^{a}F(x)dx= \frac{2\pi i}{1-e^{2\pi ia}}\sum_{z\neq 0} Res(z^{a}F(z))

    So, in this case I get the following:

    \int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{2\pi i}{1-e^{2\pi ia}}(Res_{ib}\frac{x^a}{x^2+b^2}+Res_{-ib}\frac{x^a}{x^2+b^2})

    Res_{ib}\frac{x^a}{x^2+b^2}=\ \frac{(ib)^{a-1}}{2}

    and

    Res_{-ib}\frac{x^a}{x^2+b^2}=\ \frac{(-ib)^{a-1}}{2}

    \int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi i}{1-e^{2\pi ia}}((ib)^{a-1}+(-ib)^{a-1})

    Now

    (ib)^{a-1}+(-ib)^{a-1}=\ b^{a-1}[(i)^{a-1}+(-i)^{a-1}]=\ b^{a-1}\sin(\pi a/2)

    So

    \int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi i}{1-e^{2\pi ia}}b^{a-1}\sin(\pi a/2)

    And I'm stuck... What do I do next?

    Any help will be appreciated.
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  2. #2
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    Re: [Complex Analysis] Integral of a multi-valued function

    i\sin(\pi a/2) = e^{i\pi a/2} - \cos(\pi a/2).

    \dfrac{i\pi b^{a-1} \sin(\pi a/2)}{1-e^{2i \pi a}} = \pi b^{a-1} \dfrac{e^{i\pi a/2} - \cos(\pi a/2)}{1 - e^{2\pi i a}} = -\dfrac{1}{4}e^{-i \pi a}\dfrac{\pi b^{a-1}}{\cos(\pi a/2)}.

    So, you may have done something wrong. You get close to what you want...
    Thanks from Cesc1
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    Re: [Complex Analysis] Integral of a multi-valued function

    Quote Originally Posted by SlipEternal View Post
    i\sin(\pi a/2) = e^{i\pi a/2} - \cos(\pi a/2).

    \dfrac{i\pi b^{a-1} \sin(\pi a/2)}{1-e^{2i \pi a}} = \pi b^{a-1} \dfrac{e^{i\pi a/2} - \cos(\pi a/2)}{1 - e^{2\pi i a}} = -\dfrac{1}{4}e^{-i \pi a}\dfrac{\pi b^{a-1}}{\cos(\pi a/2)}.

    So, you may have done something wrong. You get close to what you want...
    Thanks. I noticed one mistake:i^(a-1)+(-i)^(a-1) =/= sin(pi*a/2)

    i^(a-1)+(-i)^(a-1)= 2sin(pi*a/2)

    But that just replaces that 1/4 by 1/2, doesn't it?. There must be another mistake I can't find.
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  4. #4
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    Re: [Complex Analysis] Integral of a multi-valued function

    Wolframalpha evaluates that integral to \dfrac{1}{2}\dfrac{b^{a-1}\pi}{\cos(\pi a/2)}, so it is getting half of your "expected" identity result.
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    Re: [Complex Analysis] Integral of a multi-valued function

    Quote Originally Posted by SlipEternal View Post
    Wolframalpha evaluates that integral to \dfrac{1}{2}\dfrac{b^{a-1}\pi}{\cos(\pi a/2)}, so it is getting half of your "expected" identity result.
    Oh, you're right. So I'm getting closer to the result now. I'll start from zero to see where's the mistake. Thanks for your help!
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