## Tricks for inverting a Laplace Transform, part IV: Substitutions

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 other parts here: part I (guesses based on series expansions), part II (products and convolutions), part III, and part V (pole decomposition).

Today’s trick will be **variable substitutions**. Let’s say we need to find so that

, (1)

where , and are constants. This example is not as absurd as it may seem; it actually came up recently in my research on a variant of the ABBM model.

There is a general formula permitting one to invert the Laplace transform above, in terms of an integral in the complex plane:

.

This so-called Fourier-Mellin integral runs along a contour from to . can be chosen arbitrarily, as long as the integral converges and all singularities (poles, branch cuts, etc.) of lie on the right of it. Note that my convention for the Laplace variable is the convention used for generating functions in probability theory, which has just the opposite sign of of the “analysis” convention for Laplace transforms.

The fact that our Laplace transform is written entirely in terms of the function suggests applying a variable substitution. Instead of performing the Fourier-Mellin integral over , we will integrate over . We then have:

. (2)

Solving the definition of for , we get

and

.

Inserting this into (2), we have

.

The only singularity of the integrand is at , for . We can thus choose the contour to go parallel to the imaginary axis, from to . `Mathematica`

then knows to how to evaluate the resulting integral, giving a complicated expression in terms of Kummer’s confluent hypergeometric function . However, a much simpler expression is obtained if one introduces an auxiliary integral instead:

.

`Mathematica`

knows a simple expression for it:

.

is Tricomi’s confluent hypergeometric function, which is equivalent to Kummer’s confluent hypergeometric function but gives more compact expressions in our case.

Using this auxiliary integral, (2) can be expressed as

.

Simplifying the resulting expressions, one obtains our **final result for **:

(3)

Equivalently, the confluent hypergeometric functions can be replaced by Hermite polynomials:

For complicated Laplace transforms such as these, I find it advisable to check the final result numerically. In the figure below you see a log-linear plot of the original expression (1) for , and a numerical evaluation of , with given by (3). You can see that they match perfectly!

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

Tricks for inverting a Laplace Transform, part I « inordinatumNovember 7, 2012 at 8:29 pm

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

Tricks for inverting a Laplace Transform, part II « inordinatumNovember 9, 2012 at 9:11 pm

[…] This is a continuation of my articles on methods for inverting Laplace transforms. You can find the previous parts here: part I (guesses based on series expansions), part II (products and convolutions), part III, part IV (substitutions). […]

Tricks for inverting a Laplace Transform, part V: Pole Decomposition | inordinatumApril 15, 2013 at 10:27 pm

[…] have a look at part I (guesses based on series expansions), part II (products and convolutions), part IV (substitutions), and part V (pole […]

Tricks for inverting a Laplace Transform, part III | inordinatumApril 15, 2013 at 10:33 pm

Typo: e^(lambda x) –> e^(-lambda x)

aJuly 6, 2015 at 3:02 pm

I believe that with the definition (1) the sign of is actually correct? Note that in (1) is just in the Wikipedia definition of the Laplace transform https://en.wikipedia.org/wiki/Laplace_transform … So if you want to apply the Wikipedia formula for the Inverse Laplace Transform https://en.wikipedia.org/wiki/Inverse_Laplace_transform you also need the inverse sign, correspondingly…?

inordinatumJuly 7, 2015 at 9:36 pm

Typo?:r(lambda)=1/2 -> r(lambda)=1/(2lambda)

aJuly 6, 2015 at 3:17 pm

You are right… Need to check the +- from Baskara…

aJuly 6, 2015 at 3:22 pm

I’ve re-checked it and I still think that if you start with r(lambda)=1/2(c-\sqrt(c^2-\lambda)) and solve this equation for lambda you will get lambda=4cr-4r^2 as the only solution (no +- ambiguity…)

inordinatumJuly 7, 2015 at 9:25 pm