Question

Fourier transform question for fun. I'll give my solution/hints tomorrow if anyone bites. :P

Answer 1

Given the definition of a Fourier transform of some function \(g(x)\) is defined as:
\[\mathcal{F}[g](\xi) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty g(x) e^{-ix\xi}dx\]

What is the Fourier transform of the Gaussian, \(g(x) = e^{-x^2/2}\)?

Answer 2

a gaussian

Answer 3

.

Answer 4

Thanks god there's Wolfram|Alpha!

Answer 5

Haha

Answer 6

Alright I'll go ahead and give the answer.

First I assume you know how to integrate the Gaussian, there's a famous and cute trick for changing to polar, so this integral can be worked out real fast:

\[\sqrt{\frac{\pi}{a}} = \int_{-\infty}^\infty e^{-ax^2} dx\]

But let's think about it, you can shift where the Gaussian is centered to any point on the real line and it doesn't alter the integral one bit!

\[\int_{-\infty}^\infty e^{-ax^2} dx = \int_{-\infty}^\infty e^{-a(x+b)^2} dx\]

We expand the quadratic and notice that there's an extra constant term now:

\[\int_{-\infty}^\infty e^{-a(x+b)^2} dx =e^{-ab^2} \int_{-\infty}^\infty e^{-ax^2} e^{-2abx} dx\]

So let's goahead and summarize what we have (which took no effort at all):

\[e^{ab^2}\sqrt{\frac{\pi}{a}} = \int_{-\infty}^\infty e^{-ax^2} e^{-2abx} dx\]

We are tempted to immediately plug in now, \(a=\frac{1}{2}\) and \(b=i\xi\) and that gets us the result we anticipate:

\[e^{-\xi^2/2} = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-x^2/2} e^{-ix\xi}dx\]

BUT we have left out an important step. The translational symmetry we used to add \(b\) in to the integral was along the real line, and we have used an imaginary number with the choice of \(b= i \xi\).

Answer 7

So how do we know this is right? Well we use Cauchy's residue theorem! We can look at this improper integral before taking the limit as the bounds go to infinity and look at only some finite piece,

\[\lim_{c \to \infty} \int_{-c}^c e^{-ax^2}dx\]

The integral we want is similar,

\[\lim_{c \to \infty} \int_{-c}^c e^{-a(x+bi)^2}dx\]

If we look at this integral in the complex plane, both are horizontal lines, except this one we want is shifted up by \(bi\) in the complex plane. Here is the domain:

Answer 8

Do a closed loop integral, and now we can relate everything together with Cauchy's residue theorem. Take the limit now as \(c \to \infty\) and turns out we were justified in our trick after all.

Anyone who's interested in working this out, give it a try and I'll help you work it out; it's not too complicated and it's a sorta fun and useful result.