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Math Help - Laplace Transform

  1. #1
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    Laplace Transform

    Hello…

    I’m seeking a detailed derivation of the Laplace transform of the Gamma distribution:

    F(x) = \mathcal{L}\{f(t)\}=<br />
\int_0^\infty g(t)\ e^{-st} dt=<br />
(1 + s/\beta)^{-\alpha}<br />

    Where the PDF of the gamma distribution is:

    <br />
g(t;\alpha, \beta) = \frac{\beta}{\Gamma(\alpha)}\ (\beta t)^{\alpha-1}\ e^{-\beta t}<br />

    Can someone please either give me a link with all the steps or show me how to get to the transform?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by paolopiace View Post
    Hello…

    I’m seeking a detailed derivation of the Laplace transform of the Gamma distribution:

    F(x) = \mathcal{L}\{f(t)\}=<br />
\int_0^\infty g(t)\ e^{-st} dt=<br />
(1 + s/\beta)^{-\alpha}<br />

    Where the PDF of the gamma distribution is:

    <br />
g(t;\alpha, \beta) = \frac{\beta}{\Gamma(\alpha)}\ (\beta t)^{\alpha-1}\ e^{-\beta t}<br />

    Can someone please either give me a link with all the steps or show me how to get to the transform?

    Thanks in advance!
    assuming that \alpha, \beta, and s are positive numbers we have:

    \mathcal{L}\{g(t)\}=\int_0^{\infty} g(t)e^{-st} \ dt=\int_0^{\infty} \frac{\beta}{\Gamma(\alpha)} (\beta t)^{\alpha - 1} e^{-\beta t}e^{-st} \ dt=\frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty}t^{\alpha - 1}e^{-(\beta + s)t} \ dt.

    now let (\beta + s)t = u. then t=\frac{u}{\beta + s} and dt=\frac{du}{\beta + s}. hence:

    \mathcal{L}\{g(t)\}=\frac{\beta^{\alpha}}{\Gamma(\  alpha)} \int_0^{\infty}\left(\frac{u}{\beta + s} \right)^{\alpha - 1}e^{-u} \frac{du}{\beta + s}=\frac{\beta^{\alpha}}{(\beta + s)^{\alpha} \Gamma(\alpha)} \int_0^{\infty}u^{\alpha - 1}e^{-u} \ du. but by definition

    \int_0^{\infty}u^{\alpha - 1} e^{-u}du=\Gamma(\alpha). thus \mathcal{L}\{g(t)\}=\frac{\beta^{\alpha}}{(\beta + s)^{\alpha}}=\left(1 + \frac{s}{\beta} \right)^{-\alpha}. \ \ \ \square
    Last edited by NonCommAlg; May 26th 2008 at 01:53 AM.
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