1. ## Laplace Transform

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?

2. 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?

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