# The Fourier transform of sinx/x and (sinx/x)^2 and more

## In this post

We are going to evaluate the Fourier transform of \(\frac{\sin{x}}{x}\) and \(\left(\frac{\sin{x}}{x}\right)^2\). And it
turns out to be a comprehensive application of many elementary theorems
in real and complex analysis. It is advised to make sure that you can
compute and understand all the identities in this post by yourself in
the end. Also, you are expected to be able to recall what all words in
*italics* mean.

To be clear, by the Fourier transform of \(f\) we actually mean

\[ \hat{f}(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-itx}dx. \]

There is a matter of convenience. Indeed, the coefficient \(\frac{1}{\sqrt{2\pi}}\) is superfluous, but without this coefficient, when computing the Fourier inverse, one has to write \(\frac{1}{2\pi}\) on the other side. Instead of making the transform-inverse unbalanced, we write \(\frac{1}{\sqrt{2\pi}}\) all the time and pretend it is not here.

We say a function \(f\) is in \(L^1\) if \(\int_{-\infty}^{+\infty}|f(x)|dx<+\infty\). As a classic exercise in elementary calculus, one can show that \(\frac{\sin{x}}{x} \not\in L^1\) but \(\left(\frac{\sin{x}}{x}\right)^2 \in L^1\).

## Problem 1

For real \(t\), find the following limit:

\[ \lim_{A \to \infty}\int_{-A}^{A}\frac{\sin{x}}{x}e^{itx}dx. \]

Since \(\frac{\sin{x}}{x}e^{itx}\not\in L^1\), we cannot evaluate the integral over \(\mathbb{R}\) in the ordinary sense (the reader can safely ignore this if he or she has no background in Lebesgue integration at this moment, but do keep in mind that being in \(L^1\) is a big deal). However, for given \(A>0\), the integral over \([-A,A]\) is defined, and we evaluate this limit as \(A \to \infty\) to get what we want (by abuse of language). The reader is highly encouraged to write down calculation and supply images that should've been here.

We will do this using contour integration. Since the complex function
\(f(z)=\frac{\sin{z}}{z}e^{itz}\) is
*entire*, by *Cauchy's theorem*, its integral over \([-A,A]\) is equal to the one over the path
\(\Gamma_A\) by going from \(-A\) to \(-1\) along the real axis, from \(-1\) to \(1\) along the lower half of the unit
circle, and from \(1\) to \(A\) along the real axis (why?). Since the
path \(\Gamma_A\) avoids the origin, we
are safe to use the identity

\[ 2i\sin{z}=e^{iz}-e^{-iz}. \]

Replacing \(\sin{z}\) with \(\frac{1}{2i}(e^{itz}-e^{-itz})\), we get

\[ I_A(t)=\int_{\Gamma_A}f(z)dz=\int_{\Gamma_A}\frac{1}{2iz}(e^{i(t+1)z}-e^{i(t-1)z})dz. \]

If we put \(\varphi_A(t)=\int_{\Gamma_A}\frac{1}{2iz}e^{itz}dz\), we see \(I_A(t)=\varphi_A(t+1)-\varphi_A(t-1)\). It is convenient to divide \(\varphi_A\) by \(\pi\) since we therefore get

\[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{\Gamma_A}\frac{e^{itz}}{z}dz \]

and we are cool with the denominator \(2\pi i\).

Now, close the path \(\Gamma_A\) in two ways. First, by the semicircle from \(A\) to \(-Ai\) to \(-A\); second, by the semicircle from \(A\) to \(Ai\) to \(-A\). For simplicity we denote the two paths by \(\Gamma_U\) and \(\Gamma_L\). Again by the Cauchy theorem, the first case gives us an integral with value \(0\), thus by Cauchy's theorem,

\[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{-\pi}^{0}\frac{\exp{(itAe^{i\theta})}}{Ae^{i\theta}}dAe^{i\theta}=\frac{1}{2\pi}\int_{-\pi}^{0}\exp{(itAe^{i\theta})}d\theta. \]

Notice that

\[ \begin{aligned} |\exp(itAe^{i\theta})|&=|\exp(itA(\cos\theta+i\sin\theta))| \\ &=|\exp(itA\cos\theta)|\cdot|\exp(-At\sin\theta)| \\ &=\exp(-At\sin\theta) \end{aligned} \]

hence if \(t\sin\theta>0\), we have \(|\exp(iAte^{i\theta})| \to 0\) as \(A \to \infty\). When \(-\pi < \theta <0\) however, we have \(\sin\theta<0\). Therefore we get

\[ \frac{1}{\pi}\varphi_{A}(t)=\frac{1}{2\pi}\int_{-\pi}^{0}\exp(itAe^{i\theta})d\theta \to 0\quad (A \to \infty,t<0). \]

(The reader should be able to prove the convergence above.) Also trivially

\[ \varphi_A(0)=\frac{1}{2}\int_{-\pi}^{0}1d\theta=\frac{\pi}{2}. \]

But what if \(t>0\)? Indeed, it would be difficult to obtain the limit using the integral over \([-\pi,0]\). But we have another path, namely the upper one.

Note that \(\frac{e^{itz}}{z}\) is a
*meromorphic function* in \(\mathbb{C}\) with a *simple pole* at
\(0\). Since

\[ \frac{e^{itz}}{z}=\frac{1}{z}\left(1+itz+\frac{(itz)^2}{2!}+\cdots\right)=\frac{1}{z}+it+\frac{(it)^2z}{2!}+\cdots, \]

we see the residue of this function at \(0\) is \(1\). By the *residue theorem*,

\[ \begin{aligned} \frac{1}{2\pi{i}}\int_{\Gamma_L}\frac{e^{itz}}{z}dz&=\frac{1}{2\pi{i}}\int_{\Gamma_A}\frac{e^{itz}}{z}dz+\frac{1}{2\pi}\int_{0}^{\pi}\exp(itAe^{i\theta})d\theta \\ &=1\cdot\operatorname{Ind}_{\Gamma_L}(0)=1. \end{aligned} \]

Note that we have used the *change-of-variable* formula as we
did for the upper one. \(\operatorname{Ind}_{\Gamma_L}(0)\) denotes
the *winding number* of \(\Gamma_L\) around \(0\), which is \(1\) of course. The identity above
implies

\[ \frac{1}{\pi}\varphi_A(t)=1-\frac{1}{2\pi}\int_{0}^{\pi}\exp{(itAe^{i\theta})}d\theta. \]

Therefore, when \(t>0\), since \(\sin\theta>0\) when \(0<\theta<\pi\), we get

\[ \frac{1}{\pi}\varphi_A(t)\to 1 \quad(A \to \infty,t>0). \]

But as is already shown, \(I_A(t)=\varphi_A(t+1)-\varphi_A(t-1)\). To conclude,

\[ \lim_{A\to\infty}I_A(t)= \begin{cases} \pi\quad &|t|<1, \\ 0 \quad &|t|>1 ,\\ \frac{1}{2\pi} \quad &|t|=1. \end{cases} \]

### What we can learn from this integral

Since \(\psi(x)=\left(\frac{\sin{x}}{x}\right)\) is
even, dividing \(I_A\) by \(\sqrt{\frac{1}{2\pi}}\), we actually obtain
the *Fourier transform* of \(\psi\) by abuse of language. We also
get

\[ \hat\psi(t)= \begin{cases} \sqrt{\frac{\pi}{2}}\quad & |t|<1, \\ 0 \quad & |t|>1, \\ \frac{1}{2\pi\sqrt{2\pi}} & |t|=1. \end{cases} \]

Note that \(\hat\psi(t)\) is not
continuous, let alone being uniformly continuous. *Therefore*,
\(\psi(x) \notin L^1\). The reason is,
if \(f \in L^1\), then \(\hat{f}\) is *uniformly continuous*
(proof).
Another interesting fact is, this also gives us the value of the
Dirichlet integral since

\[ \begin{aligned} \int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)dx&=\int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)e^{0\cdot ix}dx \\ &=\sqrt{2\pi}\hat\psi(0) \\ &=\pi. \end{aligned} \]

We end this section by evaluating the inverse of \(\hat\psi(t)\). The calculation is not that difficult. Now you can see why we put \(\sqrt\frac{1}{2\pi}\).

\[ \begin{aligned} \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty}\hat\psi(t)e^{itx}dt &= \sqrt{\frac{1}{2\pi}}\int_{-1}^{1}\sqrt{\frac{\pi}{2}}e^{itx}dt \\ &=\frac{1}{2}\cdot\frac{1}{ix}(e^{ix}-e^{-ix}) \\ &=\frac{\sin{x}}{x}. \end{aligned} \]

## Problem 2

For real \(t\), compute

\[ J=\int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)^2e^{itx}dx. \]

Now since \(h(x)=\frac{\sin^2{x}}{x^2} \in L^1\), we are able to say with ease that the integral above is the Fourier transform of \(h(x)\) (multiplied by \(\sqrt{2\pi}\)). But still we use the limit form

\[ J(t)=\lim_{A \to \infty}J_A(t) \]

where

\[ J_A(t)=\int_{-A}^{A}\left(\frac{\sin{x}}{x}\right)^2e^{itx}dx. \]

And we are still using the contour integration as in problem 1(keep \(\Gamma_A\), \(\Gamma_U\) and \(\Gamma_L\) in mind!). For this we get

\[ \left(\frac{\sin z}{z}\right)^2e^{itz}=\frac{e^{i(t+2)z}+e^{i(t-2)z}-2e^{itz}}{-4z^2}. \]

Therefore it suffices to discuss the function

\[ \mu_A(t)=\int_{\Gamma_A}\frac{e^{itz}}{2z^2}dz \]

since we have

\[ J_A(t)=\mu_A(t)-\frac{1}{2}(\mu_A(t+2)-\mu_A(t-2)). \]

Dividing \(\mu_A(t)\) by \(\frac{1}{\pi i}\), we see

\[ \frac{1}{\pi i}\mu_A(t)=\frac{1}{2\pi i}\int_{\Gamma_A}\frac{e^{itz}}{z^2}dz. \]

An integration of \(\frac{e^{itz}}{z^2}\) over \(\Gamma_L\) yields

\[ \begin{aligned} \frac{1}{\pi i}\mu_A(t)&=\frac{1}{2\pi i}\int_{-\pi}^{0}\frac{\exp(itAe^{i\theta})}{A^2e^{2i\theta}}dAe^{i\theta} \\ &=\frac{1}{2\pi}\int_{-\pi}^{0}\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}d\theta. \end{aligned} \]

Since we still have

\[ \left|\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}\right|=\frac{1}{A}\exp(-At\sin\theta), \]

if \(t<0\) in this case, \(\frac{1}{\pi i}\mu_A(z) \to 0\) as \(A \to \infty\). For \(t>0\), integrating along \(\Gamma_U\), we have

\[ \frac{1}{\pi i}\mu_A(t)=it-\frac{1}{2\pi}\int_{0}^{\pi}\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}d\theta \to it \quad (A \to \infty) \]

We can also evaluate \(\mu_A(0)\) by computing the integral but we are not doing that. To conclude,

\[ \lim_{A \to\infty}\mu_A(t)=\begin{cases} 0 \quad &t>0, \\ -\pi t \quad &t<0. \end{cases} \]

Therefore for \(J_A\) we have

\[ J(t)=\lim_{A \to\infty}J_A(t)=\begin{cases} 0 \quad &|t| \geq 2, \\ \pi(1+\frac{t}{2}) \quad &-2<t \leq 0, \\ \pi(1-\frac{t}{2}) \quad & 0<t <2. \end{cases} \]

Now you may ask, how to find the value of \(J(t)\) at \(0\), \(2\) or \(-2\)? \(\mu_A(0)\) is not even evaluated. But \(h(t) \in L^1\), \(\hat{h}(t)=\sqrt{\frac{1}{2\pi}}J(t)\) is uniformly continuous (!), thus continuous, and the values at these points follows from continuity.

### What we can learn from this integral

Again, we get the value of a classic improper integral by

\[ \int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)^2dx = J(0)=\pi. \]

And this time it's not hard to find the Fourier inverse:

\[ \begin{aligned} \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty}\hat{h}(t)e^{itx}dt&=\frac{1}{2\pi}\int_{-\infty}^{\infty}J(t)e^{itx}dt \\ &=\frac{1}{2\pi}\int_{-2}^{2}\pi(1-\frac{1}{2}|t|)e^{itx}dt \\ &=\frac{e^{2ix}+e^{-2ix}-2}{-4x^2} \\ &=\frac{(e^{ix}-e^{-ix})^2}{-4x^2} \\ &=\left(\frac{\sin{x}}{x}\right)^2. \end{aligned} \]

The Fourier transform of sinx/x and (sinx/x)^2 and more