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)=m=0(Amcos(Kmx)+Bmsin(Kmx)) 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)=A(K)eiKxdK. f(x) = \int_{-\infty}^\infty A(\mathcal K) e^{i \mathcal K x}\, \mathrm d\mathcal K.

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

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

f(x)eiKxdx=A(K)ei(KK)xdKdx=A(K)ei(KK)xdxdK. \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 ei(KK)xdx\int_{-\infty}^\infty e^{i (\mathcal K-\mathcal K') x} \, \mathrm dx; but we first consider the related integral

nneisxdx=2ssin(sn)=2πδn(s),where δn(s)=sin(ns)πs. \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 δn(s)\delta_n(s) function. For now we integrate it over the real line.

δn(s)ds=1πsin(ns)sds=1πsinttdt,Twice the Dirichlet integralwhere t=ns,dt=nds=1. \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 0sinttdt=π2\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 δn(s)\delta_n(s) does not depend on nn.

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

ei(KK)xdx=limn2πδn(KK)=:2πδ(KK). \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 δn\delta_n over the reals equals 11 regardless of nn. Even though δ\delta does not converge, its integral does.

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

f(x,0)eiKxdx=2πA(K)δ(KK)dK. \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 δn\delta_n. By L’Hôppital’s rule, lims0δn(s)=n/π\lim_{s \to 0} \delta_n(s) = n/\pi. When ss is not near 00, the 1/s1/s term dominates and δn(s)0\delta_n(s) \to 0. As nn\to\infty, δ(s)\delta(s) not near s=0s=0 becomes insignificant compared to δ(s0)\delta(s\to 0). Therefore, in the limit as nn\to\infty, the only contribution to the integral is at s=0s=0, or in this case at K=K\mathcal K = \mathcal K'.

Since the only contribution to the integral is at K=K\mathcal K = \mathcal K' and δ\delta integrates to 11, we conclude that A(K)δ(KK)dK=A(K)\int_{-\infty}^\infty A(\mathcal K) \delta(\mathcal K -\mathcal K') \mathrm d\mathcal K = A(\mathcal K').

A(K)=12πf(x,0)eiKxdx. A(\mathcal K') = \frac{1}{2\pi} \int_{-\infty}^\infty f(x,0)\, e^{-i\mathcal K' x} \mathrm dx.
Note

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

δ(x)=0 where x0δ(x)dx=1. \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 δ\delta to be the limit of a sequence of functions δn\delta_n as nn\to\infty. We see that the second property is preserved for all δn\delta_n, and the first becomes true as nn is very large.