## 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 of single complex variable functions. Thus it is recommended to make sure that you can evaluate and understand all the identities in this post by yourself. Also, make sure that you can recall all words in italics.

## Problem 1

For real $t$, find the limit by

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 may use the identity

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

If we put $\varphi_A(t)=\int_{\Gamma_A}\frac{1}{2iz}e^{i(t+1)z}dz$, 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

and we are cool with the divisor $2\pi i$.

Now, finish 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 $A$ actually. For simplicity we denote the two paths by $\Gamma_U$ and $\Gamma_L$ Again by the Cauchy theorem, the first case gives us a integral with value $0$, thus by Cauchy’s theorem,

Notice that

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

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

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

which implies that the residue at $0$ is $1$. By the residue theorem,

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

Thus if $t>0$, since $\sin\theta>0$ when $0<\theta<\pi$, we get

But as already shown, $I_A(t)=\varphi_A(t+1)-\varphi_A(t-1)$, thus to conclude,

### What can we learn from this integral

Since $\psi(x)=\left(\frac{\sin{x}}{x}\right)$ is even, by 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

Note that $\hat\psi(t)$ is not continuous, let alone uniformly continuous. ‘Therefore’, $\psi(x) \notin L^1$ since 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

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

## Problem 2

For real $t$, compute

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 $h(x)$. But still we will be using the limit form by

where

And we are still using the contour integration as above (we are still using $\Gamma_A$, $\Gamma_U$ and $\Gamma_L$). For this we get

Therefore it suffices to discuss the function

since we have

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

Integrate $\frac{e^{itz}}{z^2}$ over $\Gamma_L$, we see

Since we still have

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

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

Therefore for $J_A$ we have

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

### What can we learn from this integral

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

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

Thereafter you are able to evaluate the improper integral of $\left(\frac{\sin{x}}{x}\right)^n$. Using Fubini’s or Tonelli’s theorem will be almost infeasible. But using the contour integral as such will force you deal with $n$ binomial coefficients, which might be tedious still. It’s even possible to discuss the convergence of the sequence $(I_n)$ where