# inordinatum

Physics and Mathematics of Disordered Systems

## Tricks for inverting a Laplace Transform, part V: Pole Decomposition

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).

## 1. Result

In this post I will explain how to find the inverse $P(x)$ of the following Laplace transform:

$\displaystyle \int_0^\infty \mathrm{d}x\, e^{-sx}P(x) = \frac{\tanh \sqrt{2s}}{\sqrt{2s}}$.   (1)

The solution is given in terms of the Jacobi theta function $\theta_2$ (which can be re-expressed in terms of Jacobi elliptic functions, though I won’t discuss that here in detail):

$\displaystyle P(x) = \frac{1}{2}\theta_2 \left(0; e^{-x\pi^2/2}\right)$.   (2)

By taking derivatives or integrals with respect to $s$, which are equivalent to multiplication or division of $P$ by $x$, one can obtain many other Laplace transform identities, including for example

$\displaystyle \int_0^\infty \mathrm{d}x\, e^{-sx}\frac{1}{2x}\theta_2 \left(0; e^{-x\pi^2/2}\right) = \ln \cosh \sqrt{2s}$.

If anyone manages to find a systematic list of those, I’d be very grateful. But for now let’s just see how one obtains (2).

## 2. Derivation

First, we use the classic decomposition of trigonometric functions in infinite products due to Euler:

$\displaystyle \begin{array}{rl} \sinh z &= z \prod_{n=1}^{\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right) \\ \cosh z &= \prod_{n=1}^{\infty}\left(1+\frac{z^2}{\pi^2 \left(n-\frac{1}{2}\right)^2}\right) \end{array}$.

From the second identity, we can obtain a partial fraction decomposition of $\tanh z$ (following this post on StackExchange):

$\displaystyle \begin{array}{rl} \tanh z = & \partial_z \ln \cosh z = \partial_z \sum_{n=1}^{\infty} \ln \left(1+\frac{z^2}{\pi^2 \left(n-\frac{1}{2}\right)^2}\right) \\ = & \sum_{n=1}^{\infty} \frac{\frac{2z}{\pi^2 \left(n-\frac{1}{2}\right)^2}}{1+\frac{z^2}{\pi^2 \left(n-\frac{1}{2}\right)^2}} \\ = & 2z \sum_{n=1}^{\infty} \frac{1}{\frac{\pi^2(2n-1)^2}{4}+z^2}. \end{array}$.

Applying this to the right-hand side of (1), we obtain a sum over simple poles:

$\displaystyle \int_0^\infty \mathrm{d}x\, e^{-sx}P(x) = \frac{\tanh \sqrt{2s}}{\sqrt{2s}} = 2\sum_{n=1}^{\infty} \frac{1}{\frac{\pi^2(2n-1)^2}{4}+2s}$

The Laplace inverse of a simple pole is just an exponential, $\int_0^\infty \mathrm{d}x\, e^{-sx}\,e^{-p x}=\frac{1}{p+s}$. By linearity of the Laplace transform, we can invert each summand individually, and obtain an infinite sum representation for $P(x)$:

$\displaystyle P(x) = \sum_{n=1}^{\infty} \exp\left(-\frac{\pi^2(2n-1)^2}{8}x\right)$

This sum can now be evaluated with Mathematica‘s Sum command, or by hand using the representation of theta functions in terms of the nome, for argument $z=0 \Leftrightarrow w=1$ and $q=e^{-\pi^2 x/2}$. This finally gives the solution as claimed above:

$\displaystyle P(x) = \frac{1}{2}\theta_2 \left(0; e^{-x\pi^2/2}\right)$.   (2)

Have fun!

Written by inordinatum

April 15, 2013 at 10:27 pm

Posted in Maths, Technical Tricks

## 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