Fourier transform

A Fourier transform is similar to a Fourier series except instead of a sum over discrete frequencies, it is an integral over continuous frequencies. A Fourier transform can represent any function, while a Fourier series can only represent periodic functions. Starting from our Fourier series,

f(x) = \sum_{m=0}^\infty (A_m \cos(\mathcal K_m x) + B_m \sin(\mathcal K_m x))

turn the sum into an integral and replace the sinusoids with a complex exponential

f(x) = \int_{-\infty}^\infty A(\mathcal K) e^{i \mathcal K x}\, \mathrm d\mathcal K.

Instead of discrete frequencies $\mathcal K_m$ , we now have a continuous $\mathcal K$ ranging from $-\infty$ to $\infty$ . Instead of discrete amplitudes $A_n$ and $B_n$ , we have a continuous amplitude function $A(\mathcal K)$ . We will solve for $A(\mathcal K)$ given $f(x)$ .

We multiply both sides by $e^{- i\mathcal K' x}$ and integrate with respect to $x$ . We then change the order of integration on the right side.

\begin{align*} \int_{-\infty}^\infty f(x)\, e^{-i\mathcal K' x} \mathrm dx &= \int_{-\infty}^\infty \int_{-\infty}^\infty A(\mathcal K) e^{i (\mathcal K-\mathcal K') x} \, \mathrm d\mathcal K\, \mathrm dx \\ &= \int_{-\infty}^\infty A(\mathcal K) \int_{-\infty}^\infty e^{i (\mathcal K-\mathcal K') x} \, \mathrm dx \, \mathrm d\mathcal K. \tag{1} \end{align*}

We need to evaluate $\int_{-\infty}^\infty e^{i (\mathcal K-\mathcal K') x} \, \mathrm dx$ ; but we first consider the related integral

\begin{align*} \int_{-n}^n e^{isx} \, \mathrm dx &= \frac 2s \sin(sn) = 2\pi \delta_n(s), \tag{2} \\ \text{where } \delta_n(s) &= \frac{\sin(ns)}{\pi s}. \end{align*}

We will see shortly the significance of this $\delta_n(s)$ function. For now we integrate it over the real line.

\begin{align*} \int_{-\infty}^\infty \delta_n(s) \mathrm ds &= \frac 1\pi \int_{-\infty}^\infty \frac{\sin(ns)}{s} \mathrm ds \\ &= \frac 1\pi \underbrace{\int_{-\infty}^\infty \frac{\sin t}{t} \mathrm dt,}_{\text{Twice the Dirichlet integral}} \text{where } t=ns, \mathrm dt=n\,\mathrm ds \\ &= 1. \end{align*}

Where the Dirichlet integral $\int_0^\infty \frac{\sin t}{t} \mathrm dt = \frac{\pi}{2}$ (techniques for evaluating this integral are given on the linked page). Interestingly, we see that the integral of $\delta_n(s)$ does not depend on $n$ .

We now return to equation (2). We see that

\begin{align*} \int_{-\infty}^\infty e^{i(\mathcal K-\mathcal K')x} \mathrm dx &= \lim_{n \to \infty} 2\pi\delta_n(\mathcal K-\mathcal K') =: 2\pi\delta(\mathcal K-\mathcal K'). \end{align*}

This limit does not exist; but as we have seen, the integral of $\delta_n$ over the reals equals $1$ regardless of $n$ . Even though $\delta$ does not converge, its integral does.

Let’s consider what this means for equation (1):

\begin{align*} \int_{-\infty}^\infty f(x,0)\, e^{-i\mathcal K' x} \mathrm dx &= 2\pi \int_{-\infty}^\infty A(\mathcal K) \delta(\mathcal K -\mathcal K') \mathrm d\mathcal K. \end{align*}

Let’s quickly consider the behavior of $\delta_n$ . By L’Hôppital’s rule, $\lim_{s \to 0} \delta_n(s) = n/\pi$ . When $s$ is not near $0$ , the $1/s$ term dominates and $\delta_n(s) \to 0$ . As $n\to\infty$ , $\delta(s)$ not near $s=0$ becomes insignificant compared to $\delta(s\to 0)$ . Therefore, in the limit as $n\to\infty$ , the only contribution to the integral is at $s=0$ , or in this case at $\mathcal K = \mathcal K'$ .

Since the only contribution to the integral is at $\mathcal K = \mathcal K'$ and $\delta$ integrates to $1$ , we conclude that $\int_{-\infty}^\infty A(\mathcal K) \delta(\mathcal K -\mathcal K') \mathrm d\mathcal K = A(\mathcal K')$ .

A(\mathcal K') = \frac{1}{2\pi} \int_{-\infty}^\infty f(x,0)\, e^{-i\mathcal K' x} \mathrm dx.

Note

$\delta$ is known as the Dirac delta and is defined by its properties

\begin{align*} &\delta(x) = 0 \text{ where } x \ne 0 \\ &\int_{-\infty}^\infty \delta(x) \mathrm dx = 1. \end{align*}

Strictly speaking, no function exists with these properties. We can instead consider $\delta$ to be the limit of a sequence of functions $\delta_n$ as $n\to\infty$ . We see that the second property is preserved for all $\delta_n$ , and the first becomes true as $n$ is very large.