## 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 complex analysis. It is a good thing 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 Fourier transform we actually mean

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

This is a matter of convenience. Indeed, the coefficient \(\frac{1}{\sqrt{2\pi}}\) is superfluous, but without it, when computing the Fourier inverse, one has to write \(\frac{1}{2\pi}\). Instead of making it 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 classic exercises in elementary calculus, \(\frac{\sin{x}}{x} \not\in L^1\) 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 of it over \(\mathbb{R}\) directly since it's not defined in the sense of Lebesgue integral (the reader can safely ignore this if he or she has no background in it at this moment, but do keep in mind that being in \(L^1\) is a big matter). Instead, for given \(A>0\), the integral of it over \([-A,A]\) is defined, and we evaluate this limit to get what we want.

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 divisor \(2\pi i\).

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 \(-A\), which finishes a circle with radius \(A\). 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

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

(You 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 *pole* at \(0\). For such a function we have

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

which implies that the residue 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

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

Thus if \(t>0\), 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 it by abuse of language. Therefore 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 implies the value of the Dirichlet integral since we have

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

We end this section by evaluating the inverse of \(\hat\psi(t)\). This requires a simple calculation.

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

## 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 will be using 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 above (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

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

since we have

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

Dividing \(\mu_A(z)\) 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 \(\Gamma_L\) gives

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

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 did you find the value at \(0\), \(2\) or \(-2\)? \(\mu_A(0)\) is not 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:

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