Laplace Transform

**paolopiace** · May 25th 2008, 10:45 PM

Hello…

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

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

Where the PDF of the gamma distribution is:

$ g(t;\alpha, \beta) = \frac{\beta}{\Gamma(\alpha)}\ (\beta t)^{\alpha-1}\ e^{-\beta t} $

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

Thanks in advance!

**NonCommAlg** · May 26th 2008, 12:18 AM

Originally Posted by paolopiace

Hello…

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

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

Where the PDF of the gamma distribution is:

$ g(t;\alpha, \beta) = \frac{\beta}{\Gamma(\alpha)}\ (\beta t)^{\alpha-1}\ e^{-\beta t} $

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 $\mathcal{L}\{g(t)\}=\frac{\beta^{\alpha}}{(\beta + s)^{\alpha}}=\left(1 + \frac{s}{\beta} \right)^{-\alpha}. \ \ \ \square$

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