# inordinatum

Physics and Mathematics of Disordered Systems

## Tricks for inverting a Laplace Transform, part III

This is a continuation of the series of articles on Laplace transforms… You can also have a look at part I (guesses based on series expansions), part II (products and convolutions), part IV (substitutions), and part V (pole decomposition).

This time we will cover the following inverse Laplace problems:

$\displaystyle \int_0^\infty e^{\lambda x}P(x)\,dx = e^{\frac{1}{a+b \lambda}}$

The basic result that can be used for such problems is the Laplace transform of the Bessel function:

$\int_0^\infty e^{\lambda x}\left[\frac{1}{\sqrt{x}}I_1(2\sqrt{x})+\delta(x)\right]dx = e^{\frac{1}{\lambda}}$

Now, using

$\displaystyle LT\{f(x)e^{cx}\}(s)=\int_0^\infty \mathrm{d}x\,e^{s x}f(x)e^{c x} = LT\{f(x)\}(s+c)$,

and by rescaling $\lambda$, one finds that $\int_0^\infty e^{\lambda x}P(x)\,dx = e^{\frac{1}{a+b \lambda}}$ is solved by:

$\displaystyle P(x)= \left[\frac{1}{\sqrt{bx}}I_1\left(2\sqrt{\frac{x}{b}}\right)+\delta(x)\right]e^{-\frac{a}{b}x}$

Have fun!