# 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!

Written by inordinatum

October 15, 2011 at 11:17 am

### 6 Responses

1. […] After a few less technical posts recently, I now continue the series of articles on tricks for the inversion of Laplace transforms. You can find the previous parts here: part I, part II, part III. […]

2. […] series of posts on Laplace Transform inversion. You can find the subsequent articles here: part II, part III, part IV. […]

3. […] series of posts on Laplace Transform inversion. You can find the subsequent articles here: part I, part III, part IV. […]

4. […] parts here: part I (guesses based on series expansions), part II (products and convolutions), part III, part IV […]

5. Typo: e^c x –> e^(-cx)

a

July 6, 2015 at 10:01 am

• Thanks! I agree that the bracketing of the exponent was wrong, I’ve now fixed that in the post and also added an extra step. However I think the sign was correct…

inordinatum

July 7, 2015 at 9:30 pm