Fourier transform of some random variables

In probability theory and Stochastic processes the Characteristic function or Fourier transform of a random variable plays an important role, specially in problems involving summations of independent random variables. It´s much easier to compute analytically these sums in the Fourier space. As with the distribution function and the density function, the characteristic function characterizes completely the r.v. Additionally, some r.v´s don´t have an expression for their density given by elementary functions, and we express them by their characteristic functions (most stable distributions). In this post I want to compute the Fourier transform of two stable distributions, namely the Gaussian and the Cauchy distributions. These two, aside from being the only two symmetric stable distributions with a density given by elementary functions, they also represent THE EXAMPLES of a Thin tailed distribution and a Fat tailed distribution, the former being the thin tailed and the latter the heavy tailed. The Tails of the Cauchy distribution are so heavy that even it´s mean is infinite!

First I will start introducing the Gaussian integral. Then we will compute it´s Fourier transform. Next we will compute the Fourier transform of the Cauchy distribution through two different methods. First via contour integration, and then by real methods for those who don´t know complex methods of integrations. Additionally, throughout the post we will use a variety of interesting techniques for computing integrals, and in the end of post, in the appendix section, we show the Cauchy-Schlömilch transformation which will be useful in this post and future posts.

So lets get started!

The Gaussian Integral

First, lets introduce the Gaussian integral which is defined by the following expression

$I=\int_{-\infty}^{\infty} e^{-x^{2}}dx=\sqrt{\pi}$ (1)

for a probability function to be valid, it should integrate to 1. To get this, we divide both sides of (1) by $\sqrt{\pi}$

$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-x^{2}}dx=1$

Now, note that this integral is even. To show this note the following

$\int_{-\infty}^{\infty} e^{-x^{2}}dx=\int_{-\infty}^{0} e^{-x^{2}}dx+\int_{0}^{\infty} e^{-x^{2}}dx$

let x=-y in the first integral on the right hand side

$\int_{-\infty}^{\infty}e^{-x^{2}}dx =\int_{\infty}^{0}e^{-(-y)^{2}}(-dy)+\int_{0}^{\infty} e^{-x^{2}}dx$

$\int_{-\infty}^{\infty}e^{-x^{2}}dx =2\int_{0}^{\infty} e^{-x^{2}}dx$ (2)

Now, lets proof (1)

$I=2\int_{0}^{\infty} e^{-x^{2}}dx$

let x^2=y

$I=2\int_{0}^{\infty} e^{-x^{2}}dx=\int_{0}^{\infty} e^{-y}y^{-1/2}dy=\Gamma\left(\frac{1}{2} \right)$ (3)

Recall the Beta function

$B(a,b)=2\int_{0}^{\pi/2}\cos^{2a-1}(x)\sin^{2b-1}(x)dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$

if we let $a=\frac{1}{2} \, \text{and} \, b=\frac{1}{2}$

$B\left(\frac{1}{2},\frac{1}{2}\right)=2\int_{0}^{\pi/2}\cos^{0}(x)\sin^{0}(x)dx=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}$

$2\int_{0}^{\pi/2}dx=\Gamma^2\left(\frac{1}{2}\right)$

$2\frac{\pi}{2}=\Gamma^2\left(\frac{1}{2}\right)$

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ (4)

from (1), (2) and (3) we conclude that

$\boxed{\int_{0}^{\infty} e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}}$ (5)

$\boxed{\int_{-\infty}^{\infty}e^{-x^{2}}dx =\sqrt{\pi}}$ (6)

Consider now

$J=\int_{0}^{\infty} e^{-ax^{2}}dx$

let $u=\sqrt{a}x \, \Rightarrow \, dx=\frac{du}{\sqrt{a}}$

$J=\int_{0}^{\infty} e^{-ax^{2}}dx=\frac{1}{\sqrt{a}}\int_{0}^{\infty} e^{-u^{2}}du$

and from (5) we conclude that

$\boxed{J=\int_{0}^{\infty} e^{-ax^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}}$ (7)

Fourier transform of a Gaussian

Aside from a normalization constant, the Fourier transform or characteristic function of the Gaussian distribution is given by the following expression

$\langle e^{ikx} \rangle=\int_{-\infty}^{\infty} e^{ikx}p(x)dx$

$\langle e^{ikx} \rangle=\int_{-\infty}^{\infty} e^{ikx}e^{-ax^{2}}dx$

First note that this integrand is also an even function which allows us to rewrite the integral as follows:

$\int_{-\infty}^{\infty} e^{ikx}e^{-ax^{2}}dx=2\int_{0}^{\infty} \cos(kx)e^{-ax^{2}}dx$ (8)

Now define

$I(k)=\int_{0}^{\infty} \cos(kx)e^{-ax^{2}}dx$

Taking derivative with respect to

$\frac{dI(k)}{dk}=-\int_{0}^{\infty} x\sin(kx)e^{-ax^{2}}dx$

Integrating by parts the last equation gives

$\frac{dI(k)}{dk}=-\frac{k}{2a}\int_{0}^{\infty} \cos(kx)e^{-ax^{2}}dx$

But note that the integral in the right hand side is equal to I(k) , so we get a first order differential equation:

$\frac{dI(k)}{dk}=-\frac{k}{2a}I(k)$

Integrating this ODE leads us to:

$I(k)=ce^{-\frac{k^{2}}{4a}}$

where is a constant to be defined. From (7) we have that

$I(0)=\int_{0}^{\infty} e^{-ax^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$

consequently $c=\frac{1}{2}\sqrt{\frac{\pi}{a}}$ and we finally get that

$I(k)=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{k^{2}}{4a}}$

$\boxed{\int_{-\infty}^{\infty} e^{ikx}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{a}}e^{-\frac{k^{2}}{4a}}}$ (9)

and

$\boxed{\int_{0}^{\infty} \cos(kx)e^{-ax^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{k^{2}}{4a}}}$ (10)

The fourier Transform of the Cauchy distribution

The Cauchy distribution is given by the following function

$f(x)=\frac{a}{\pi} \cdot \frac{1}{a^2+ x^2}$

Now we will compute it´s Fourier transform by two different methods. First through Contour integration and than by Real methods

First way through Contour integration

Consider the following integral

$I=\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx$

$\int_{-\infty}^{0}\frac{e^{ikx}}{a^2+ x^2}dx+\int_{0}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx$

let x=-y in the first integral

$\int_{\infty}^{0}\frac{e^{-ikx}}{a^2+ (-y)^2}(-dy)+\int_{0}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx$

$\int_{0}^{\infty}\frac{e^{-ikx}}{a^2+ x^2}dx+\int_{0}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx=2\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx$

$\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx=2\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx$ (11)

Consider the following integral in the complex plain

$\oint_{C}\frac{e^{ikz}}{a^2+ z^2}dz=\oint_{C}\frac{e^{ikz}}{(z+ia)(z-ia)}dz$

Where are the following contours

The integrand has 2 poles @ $\pm ia$ .

We can write our integral as

$\oint_{C}\frac{e^{ikz}}{a^2+ z^2}dz=\int_{-R}^{R}\frac{e^{ikx}}{a^2+ x^2}dx+\oint_{\Gamma_{R}}\frac{e^{ikz}}{a^2+ z^2}dz$

We consider first the case for k>0, and the contour on the left. By the residues theorem it´s also equal to

$\oint_{C}\frac{e^{ikz}}{a^2+ z^2}dz=2\pi i \mathop{Res}_{z=ia}\left(\frac{e^{ikz}}{a^2+ z^2} \right)$

Let´s calculate first the residues

$2 \pi i\mathop{Res}_{z=ia}\left(\frac{e^{ikz}}{a^2+ z^2} \right)=2 \pi i\lim_{z \longrightarrow ia}(z-ia)\frac{e^{ikz}}{a^2+ z^2}=2 \pi i\frac{e^{ikia}}{2ia}=\frac{\pi e^{-ak}}{a}$

$\boxed{\oint_{C}\frac{e^{ikz}}{a^2+ z^2}dz=\frac{\pi e^{-ak}}{a}}$ (12)

Now lets focus on the integral around the arc $\Gamma_{R}$

$\oint_{\Gamma_{R}}\frac{e^{ikz}}{a^2+ z^2}dz$

let $z=Re^{i \theta}$

$\oint_{\Gamma_{R}}\frac{e^{ikz}}{a^2+ z^2}dz=\int_0^{\pi}\frac{e^{ikRe^{i \theta}}}{a^2+ (Re^{i \theta})^2}iR e^{i \theta} d \theta$

$\bigg|\oint_{\Gamma_{R}}\frac{e^{ikz}}{a^2+ z^2}dz\bigg|<\int_0^{\pi}\frac{|e^{ikRe^{i \theta}}|}{|a^2+ (Re^{i \theta})^2|}|iR e^{i \theta}| | d \theta|$

$\leq\frac{R}{R^2+ a^2}\int_0^{\pi}e^{-kR \sin (\theta)}d \theta$

$=\frac{2R}{R^2+ a^2}\int_0^{\pi/2}e^{-kR \sin (\theta)}d \theta$

$\leq \frac{2R}{R^2+ a^2}\int_0^{\pi/2}e^{-2kR \theta/\pi}d \theta$

$=\frac{2R}{R^2+ a^2}\left( -\frac{e^{-2kR \theta/\pi}}{2 k R} \bigg|_0^{\pi/2}\right)=\frac{ 1-e^{-kR}}{(R^2+ a^2)k}$

And for k>0 , as $R \longrightarrow \infty$ the integral vanishes. And we conclude that

$\oint_{C}\frac{e^{ikz}}{a^2+ z^2}dz=\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx$ (13)

and therefore equating (11) and (12) we get

$\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx=\frac{\pi e^{-ak}}{a}$

Similarly for the case of k<0 we use the contour on the right and get that

$\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx=\frac{\pi e^{ak}}{a}$

summing up both result we get that

$\boxed{\langle e^{ikx} \rangle=\frac{\pi e^{-a|k|}}{a}}$ (14)

from (11) we also get that

$\boxed{\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\frac{\pi e^{-ak}}{2a}}$ (15)

Second way via Real Methods

From (11), consider the integral

$I=\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\int_{0}^{\infty}\cos(kx)\int_{0}^{\infty}e^{-(a^2+ x^2)t}dtdx$

swapping the order of integration

$\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\int_{0}^{\infty}e^{-ta^2}\int_{0}^{\infty}e^{-tx^2}\cos(kx)dxdt$

from (10) above we know that the inner integral is

$\int_{0}^{\infty}e^{-tx^2}\cos(kx)dx=\frac{1}{2}\sqrt{\frac{\pi}{t}}e^{-\frac{k^{2}}{4t}}$

therefore

$\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\frac{1}{2}\sqrt{\pi}\int_{0}^{\infty}e^{-ta^2}e^{-\frac{k^{2}}{4t}}\frac{1}{\sqrt{t}}dt$

Let t=y^2 , then

$\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\sqrt{\pi}\int_{0}^{\infty}e^{-y^2a^2-\frac{k^{2}}{4y^2}}dy$

completing the square in the exponent

$\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\sqrt{\pi}\int_{0}^{\infty}e^{-\left(ya-\frac{k}{2y}\right)^2-ak}dy$

$=\sqrt{\pi}e^{-ak}\int_{0}^{\infty}e^{-\left(ya-\frac{k}{2y}\right)^2}dy$

By the Cauchy Schlömilch transformation, we can rewrite the last integral as

$=\frac{\sqrt{\pi}e^{-ak}}{a}\int_{0}^{\infty}e^{-y^2}dy$

$=\frac{\sqrt{\pi}e^{-ak}}{a}\frac{\sqrt{\pi}}{2}$

$\boxed{\int_{0}^{\infty}\frac{\cos(kx)}{a^2+ x^2}dx=\frac{\pi e^{-ak}}{2a}}$ (16)

Which agrees with (15).And from (11) we get

$\boxed{\int_{-\infty}^{\infty}\frac{e^{ikx}}{a^2+ x^2}dx=\frac{\pi e^{-ak}}{a}}$ (17)

Appendix

The Cauchy-Schlömilch transformation

The Cauchy-Schlömilch transformation is a very usefull trick. Besides the evaluation above, in future posts we will utilize it again to compute integrals.

let a,b>0 , then

$\int_0^{\infty}f\left(\left(ax-\frac{b}{x} \right)^2 \right)dx=\frac{1}{a}\int_0^{\infty}f(x^2)dx$

$I=\int_0^{\infty}f\left(\left(ax-\frac{b}{x} \right)^2 \right)dx$ (A.1)

let $x=\frac{b}{ay} \Rightarrow dx=-\frac{b}{a}\cdot\frac{ dy}{y^2}$

$I=\frac{b}{a}\int_{\infty}^{0}f\left(\left(\frac{ab}{ay}-\frac{bay}{b} \right)^2 \right)\frac{(-dy)}{y^2}$

$I=\frac{b}{a}\int_{0}^{\infty}f\left(\left(\frac{b}{y}-ay \right)^2 \right)\frac{dy}{y^2}$

because of the eveness of the integrand

$I=\frac{b}{a}\int_{0}^{\infty}f\left(\left(ay-\frac{b}{y} \right)^2 \right)\frac{dy}{y^2}$

switch the dummy variable $y \longrightarrow x$

$I=\frac{b}{a}\int_{0}^{\infty}f\left(\left(ax-\frac{b}{x} \right)^2 \right)\frac{dx}{x^2}$ (A.2)

adding (A.1) and (A.2)

$2I=\int_0^{\infty}f\left(\left(ax-\frac{b}{x} \right)^2 \right)dx+\frac{b}{a}\int_{0}^{\infty}f\left(\left(ax-\frac{b}{x} \right)^2 \right)\frac{dx}{x^2}$