# The Fourier Transform of exp(-cx^2) and Its Convolution

For $0<c<\infty$, define

We want to compute the Fourier transform

As one can expect, the computation can be quite interesting, as $f_c(x)$ is related to the Gaussian integral in the following way:

Now we dive into this integral and see what we can get.

# Computing the Fourier Transform

Let’s admit, trying to compute the integral straightforward is somewhat unrealistic. So we need to go through an alternative way. For convenience (of writing MathJax codes) we may write $\varphi(t)=\hat{f}_c(t)$.

First of all, $\hat{f}_c(t)$ is always well-defined, this is because

so we can compute it without worrying about anything.

## Integration by Parts and Differential Equation

It’s hard to think about but we do have it. An integration by parts gives

On the other hand, we have

(The well-definedness of the integral can be verified easily.) Combining both, we obtain an differential equation

This differential equation corresponds to an integral equation

And we solve it to obtain

or alternatively,

Now put the initial value back in. As we have shown above, this subjects to the Gaussian integral

Therefore

is exactly what we want.

Before showing another method, we first have an question: can we have $\hat{f}_c=f_c$? Solving an equation with variable in $c$ answers this question affirmatively:

In other words, $f_\frac{1}{2}$ is a fixed point of the Fourier transform. For this class of functions, the fixed point is this and only this one.

## Direct Application of the Gaussian Integral

We can also make use of the Gaussian integral to get what we want.

# Convolution

As a classic property of the Fourier transform, for $f,g \in L^1$, we have

where

By the way, $f \in L^1$ means $\int_{-\infty}^{\infty}|f(x)|dx<\infty$. One can verify that $f \ast g \in L^1$ here as well.

With this result, we can compute $f_a \ast f_b$ easily. Note

We expect that there exist some $\gamma$ and $c$ such that $f_a \ast f_b = \gamma f_c$. In other words, we are looking for $\gamma,c \in \mathbb{R}$ such that

We should have

We also have

Therefore

where $c$ is given above. We do not even have to compute the integral of convolution explicitly.

The Fourier Transform of exp(-cx^2) and Its Convolution

https://desvl.xyz/2022/05/06/exp-fourier/

Desvl

2022-05-06

2023-01-03