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Thread: 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 $\displaystyle (1+t^2)^{-1}$.

    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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    $\displaystyle \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 $\displaystyle xe^{-x}$)

    switch the order of integration

    $\displaystyle = \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 $

    $\displaystyle = \Gamma(z) \int^{\infty}_{0} x^{1-z} e^{-x} \ dx = \Gamma(z) \Gamma(2-z)= \Gamma(z) (1-z)\Gamma(1-z) = \frac{\pi (1-z)}{\sin(\pi z)}$ (reflection formula)
    Last edited by Random Variable; Jun 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 $\displaystyle (1+t^2)^{-1}$.
    Could you explain in more detail?
    Last edited by Random Variable; Jun 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 $\displaystyle u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}$. We know $\displaystyle \int_0^\infty u(t) dt$ converges absolutely, hence $\displaystyle \int_0^\infty |v(t)|dt$ converges if $\displaystyle \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}|$ 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 $\displaystyle u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}$. We know $\displaystyle \int_0^\infty u(t) dt$ converges absolutely, hence $\displaystyle \int_0^\infty |v(t)|dt$ converges if $\displaystyle \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}|$ is finite, which is when...
    Unless I'm totally mistaken, the original integral is convergent for $\displaystyle 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 $\displaystyle 0<\mbox{Re }z<1$, and indeed the integral is convergent when $\displaystyle 0< \mbox{Re }z < 2$.
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