# The Fourier transform of exp(-cx^2)

For \(0<c<\infty\), define \[ f_c(x)=\exp(-cx^2). \] We want to compute the Fourier transform \[ \hat{f}_c(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f_c(x)e^{-ixt}dx. \] As one can expect, the computation can be quite interesting, as \(f_c(x)\) is related to the Gaussian integral in the following way: \[ \int_{-\infty}^{+\infty}f_c(x)dx=\frac{1}{\sqrt{c}}\int_{-\infty}^{+\infty}\exp(-(\sqrt{c}x)^2)d\sqrt{c}x=\sqrt\frac{\pi}{c}. \] 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. Here comes how we do that. For convenience (of writing MathJax codes) let's write \(\varphi(t)=\hat{f}_c(t)\).

First of all, \(\hat{f}_c(t)\) is always well-defined, this is because \[ \int_{-\infty}^{+\infty}|f_c(x)e^{-ixt}|dx=\int_{-\infty}^{+\infty}|f_c(x)|dx<\infty \] so we can compute it without worrying anything.

## Integration by Parts and Differential Equation

It's hard to think about but we do have it. An integration by parts gives \[ \begin{aligned} \varphi(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\exp(-cx^2)e^{-itx}dx &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\exp(-cx^2)\frac{1}{-it}de^{-itx} \\ &=\frac{i}{t\sqrt{2\pi}}[\exp(-cx^2)e^{-itx}]|_{-\infty}^{+\infty} \\ &\quad -\frac{i}{t\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-itx}d\exp(-cx^2) \\ &=\frac{-2c}{t\sqrt{2\pi}}\int_{-\infty}^{+\infty}-xi\exp(-cx^2)e^{-itx}dx \end{aligned} \] On the other hand, we have \[ \varphi'(t)=\hat{f'_c}(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}-ixf_c(x)e^{-itx}dx. \] (The well-definedness of the integral can be verified easily.) Combining both, we obtain an differential equation \[ t\varphi(t)+2c\varphi'(t)=0 \] This differential equation corresponds to an integral equation \[ \int2c\frac{d\varphi}{\varphi}=-\int tdt. \] And we solve it to obtain \[ 2c\log\varphi=-\frac{1}{2}t^2+C \] or alternatively, \[ \varphi(t)=C\exp(-\frac{1}{4c}t^2). \] Now put the initial value back in. As we have shown above, this subjects to the Gaussian integral \[ \varphi(0)=\frac{1}{\sqrt{2\pi}}\sqrt{\frac{\pi}{c}}=\frac{1}{\sqrt{2c}}. \] Therefore \[ \varphi(t)=\frac{1}{\sqrt{2c}}\exp\left(-\frac{1}{4c}t^2\right) \] 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: \[ \hat{f}_c=f_c \iff \begin{cases}\frac{1}{\sqrt{2c}}=1 \\ -\frac{1}{4c}=-c \end{cases} \iff c = \frac{1}{2}. \] 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.

# Convolution

As a classic property of the Fourier transform, for \(f,g \in L^1\), we have \[ \widehat{f \ast g}(t)=\hat{f}(t)\hat{g}(t) \] where \[ f \ast g(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x-y)g(y)dy. \] 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 \[ \widehat{f_a \ast f_b}(t)=\hat{f_a}(t)\hat{f_b}(t)=\frac{1}{2\sqrt{ab}}\exp\left[-\left(\frac{1}{4a}+\frac{1}{4b}\right)x^2\right] \] Now let's see if we can have \(f_a \ast f_b = \gamma f_c\) for some \(\gamma\) and \(c\). We should have \[ \frac{1}{4c}=\frac{1}{4a}+\frac{1}{4c} \implies c=\frac{1}{\frac{1}{a}+\frac{1}{b}}=\frac{ab}{a+b}. \] We also have \[ \gamma \frac{1}{\sqrt{2c}}=\frac{1}{2\sqrt{ab}} \implies \gamma = \sqrt\frac{c}{2ab}=\sqrt{\frac{1}{2(a+b)}} \] Therefore we have \[ f_a \ast f_b = \sqrt{\frac{1}{2(a+b)}}f_c \] where \(c\) is given above. We didn't even compute the integral explicitly.

The Fourier transform of exp(-cx^2)