# inordinatum

Physics and Mathematics of Disordered Systems

## Tricks for inverting a Laplace Transform, part II: Products and Convolutions

EDIT: In the meanwhile, I have continued the series of posts on Laplace Transform inversion. You can find the subsequent articles here: part I (guesses based on series expansions), part III, part IV (substitutions), part V (pole decomposition). Enjoy!

Following the previous post on inverting Laplace transforms, here is another trick up the same alley. This one actually considers a generalization of the previous case

Find $P(x)$ such that $\int_0^\infty P(x)e^{\lambda x}dx=\left(1-q\lambda\right)^a \left(1-\lambda\right)^b$.

As usual, the built-in InverseLaplaceTransform function from Mathematica 8 fails to give a result. To obtain a closed formula manually, note that each of the factors can be easily inverted:
$\int_0^\infty e^{\lambda x}R_{a,q}(x)\,dx=(1-q\lambda)^a$
has the solution
$R_{a,q}(x)=\frac{e^{-\frac{x}{q}}\left(\frac{x}{q}\right)^{-1-a}}{q \Gamma(-a)}$.

Hence, using the fact that Laplace transforms of convolutions give products, the solution for $P(x)$ can be written as a convolution:
$P(x) = \int_0^x dx' R_{a,q}(x')R_{b,1}(x-x')$.

Computing the integral gives the following expression for $P(x)$ in terms of the hypergeometric function $\,_1 F_1$:

$P(x) = \frac{e^{-x}q^a x^{-1-a-b}}{\Gamma(-a-b)} \,_1 F_1\left(-a,-a-b,\frac{q-1}{q}x\right)$

Enjoy!!

Written by inordinatum

September 3, 2011 at 10:34 am