We know that the Fourier transform of the sinc function is the rectangular function (or top hat). However, I'm at a loss as to how to prove it. Most textbooks and online sources start with the rectangular function, show that
$$\int_{-\infty}^\infty \text{rect}(x)e^{i\omega x}dx=\int_{-1/2}^{1/2}e^{i\omega x}dx=\left.\frac{e^{i\omega x}}{i\omega}\right\vert_{-1/2}^{1/2}=\text{sinc}(\omega/2)$$
and then just invoke duality and claim that the Fourier transform of the sinc function is the rectangular function. Is there any way of deriving this directly? i.e., starting with the sinc function?
I've tried, but I'm not sure as to how to proceed. I know that the sinc is not Lebesgue integrable and only improper Riemann integrable. Some vague recollection of these being important in the Fourier transform hinders my thought process. Can someone clear things up for me?
Best Answer
Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\ &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega - \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;. \end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' + \int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.