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Math Help - when is this integral convergent

  1. #1
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    when is this integral convergent

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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint : use the limit comparison test with (1+t^2)^{-1}.

    To prove the identity is a simple residue calculation using the keyhole contour.
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  3. #3
    Super Member Random Variable's Avatar
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     \int^{\infty}_{0} \frac{t^{z-1}}{(t+1)^{2}} \ dt = \int_{0}^{\infty} \int_{0}^{\infty} t^{z-1} e^{-tx} xe^{-x} \ dx \ dt (Laplace Transform of  xe^{-x})

    switch the order of integration

     = \int_{0}^{\infty} x e^{-x} \int_{0}^{\infty} t^{z-1} e^{-tx} \ dt \ dx = \int^{\infty}_{0} xe^{-x} x^{-z} \Gamma(z)  \ dx

     = \Gamma(z) \int^{\infty}_{0} x^{1-z} e^{-x} \ dx   = \Gamma(z) \Gamma(2-z)= \Gamma(z) (1-z)\Gamma(1-z) = \frac{\pi (1-z)}{\sin(\pi z)} (reflection formula)
    Last edited by Random Variable; June 1st 2010 at 07:28 PM.
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  4. #4
    Super Member Random Variable's Avatar
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    nevermind
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Hint : use the limit comparison test with (1+t^2)^{-1}.
    Could you explain in more detail?
    Last edited by Random Variable; June 1st 2010 at 09:32 PM.
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Random Variable View Post
    Could you explain in more detail?
    Let u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}. We know \int_0^\infty u(t) dt converges absolutely, hence \int_0^\infty |v(t)|dt converges if \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}| is finite, which is when...
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Let u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}. We know \int_0^\infty u(t) dt converges absolutely, hence \int_0^\infty |v(t)|dt converges if \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}| is finite, which is when...
    Unless I'm totally mistaken, the original integral is convergent for  0 < z <2 . I'm not sure how that proof shows that.
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    My mistake! What I wrote only proves convergence when 0<\mbox{Re }z<1, and indeed the integral is convergent when 0< \mbox{Re }z < 2.
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