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 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 what all words in italics means.

To be clear, by Fourier transform we actually mean \[ \hat{f}(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-itx}dx. \] But we omit \(\frac{1}{\sqrt{2\pi}}\) and use \(e^{itx}\) in place of \(e^{-itx}\) because it is easier to compute, and does not change the final result.

Problem 1

For real \(t\), find the limit by \[ \lim_{A \to \infty}\int_{-A}^{A}\frac{\sin{x}}{x}e^{itx}dx. \]

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. 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 may 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 \(\varphi_A(t)=\int_{\Gamma_A}\frac{1}{2iz}e^{i(t+1)z}dz\), 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 \[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{\Gamma_A}\frac{e^{itz}}{z}dz \] and we are cool with the divisor \(2\pi i\).

Now, 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, \[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{-\pi}^{0}\frac{\exp{(itAe^{i\theta})}}{Ae^{i\theta}}dAe^{i\theta}=\frac{1}{2\pi}\int_{-\pi}^{0}\exp{(itAe^{i\theta})}d\theta. \] 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 by \[ J(t)=\lim_{A \to \infty}J_A(t) \] 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, we have \[ \lim_{A \to\infty}\mu_A(t)=\begin{cases} 0 \quad &t>0, \\ -\pi t \quad &t<0. \end{cases} \] 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: \[ \begin{aligned} \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty}\hat{h}(t)e^{itx}dt&=\frac{1}{2\pi}\int_{-\infty}^{\infty}J(t)e^{itx}dt \\ &=\frac{1}{2\pi}\int_{-2}^{2}\pi(1-\frac{1}{2}|t|)e^{itx}dt \\ &=\frac{e^{2ix}+e^{-2ix}-2}{-4x^2} \\ &=\frac{(e^{ix}-e^{-ix})^2}{-4x^2} \\ &=\left(\frac{\sin{x}}{x}\right)^2. \end{aligned} \]

Thereafter you are able to evaluate the improper integral of \(\left(\frac{\sin{x}}{x}\right)^n\). Using Fubini's or Tonelli's theorem is not a good idea. 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 \[ I_n(t)=\lim_{A \to \infty}\int_{-A}^{A}\left(\frac{\sin{x}}{x}\right)^ne^{itx}dx. \]