# inordinatum

Physics and Mathematics of Disordered Systems

## Tricks for inverting a Laplace Transform, part I: Guesses based on Series

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

Working with probability distributions one frequently meets the problem of inverting a Laplace transform, which is notoriously difficult. An example that I met recently is the following:

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

The built-in InverseLaplaceTransform function from Mathematica 8 fails to give a result. However, a closed formula for $P(x)$ can be obtained using a trick: Consider integer $v$, for which the right-hand side becomes a rational function. Take its Laplace inverse, and then extend the result analytically for non-integer $v>0$.

Since this may be useful for other cases, too, here a detailed account of how one proceeds:

For small integer values of $v$, one obtains (e.g. using Mathematica):

$v=0: P(x)=\delta(x)$
$v=1: P(x)=q \delta(x)+(1-q)e^{-x}$
$v=2: P(x)=q^2 \delta(x)+2q(1-q)e^{-x}+q(1-q)^2 x e^{-x}$
$v=3: P(x)=q^3 \delta(x)+3q^2(1-q)e^{-x}+3q(1-q)^2 x e^{-x}+\frac{1}{2}(1-q)^3 x^2 e^{-x}$
$v=4: P(x)=q^4 \delta(x)+4q^3(1-q)e^{-x}+6q^2(1-q)^2 x e^{-x}+2q(1-q)^3 x^2 e^{-x}+\frac{1}{6}(1-q)^4 x^3 e^{-x}$
$v=5: P(x)=q^5 \delta(x)+5q^4(1-q)e^{-x}+10q^3(1-q)^2 x e^{-x}+5q^2(1-q)^3 x^2 e^{-x}+\frac{5}{6}q(1-q)^4 x^3 e^{-x}+\frac{1}{24}(1-q)^5 x^4 e^{-x}$

By trying out, one can now guess a general formula for integer $v$:

$P(x)=q^v \delta(x)+e^{-x}\sum_{k=1}^v \frac{1}{(k-1)!}\binom{v}{k}q^{v-k} (1-q)^k x^{k-1}$

For example using Mathematica’s Sum command, one can re-write this in terms of a hypergeometric function:

$P(x)=q^v \delta(x) + e^{-x}(1-q)q^{-1+v}v \,_1 F_1(1-v,2,-\frac{1-q}{q}x)$

This expression now has a well-defined meaning for any $v>0$, and is the result we looked for. Actually, Mathematica even knows how to take its Laplace transform, which can be used to check its validity.

More tricks for similar problems to come later!

Written by inordinatum

July 28, 2011 at 6:44 pm