TITLE: Fourier transform of $\left|\frac{\sin x}{x}\right|$
QUESTION [22 upvotes]: Is there a closed form (possibly, using known special functions) for the Fourier transform of the function $f(x)=\left|\frac{\sin x}{x}\right|$?
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I tried to find one using Mathematica, but it ran for several hours without producing any result.

REPLY [35 votes]: Since $\left|\frac{\sin(x)}{x}\right|$ is not in $L^1$, there is no Fourier transform in the strict sense. However, we can get a Fourier transform in the sense of distributions (via Plancherel's Theorem).
The standard result about the sinc function is that
$$
\int_{-\infty}^\infty\frac{\sin(ax)}{ax}e^{-2\pi ix\xi}\,\mathrm{d}x
=\frac\pi{a}\left[|\xi|\le\frac{a}{2\pi}\right]\tag{1}
$$
where $[\,\cdot\,]$ are Iverson brackets.
$$
\int_{-\infty}^\infty\frac{\sin(2(k+1)\pi^2x)-\sin(2k\pi^2x)}{\pi x}e^{-2\pi ix\xi}\,\mathrm{d}x
=\Big[k\pi\le|\xi|\le (k+1)\pi\Big]\tag{2}
$$
Note that
$$
\mathrm{sgn}\left(\frac{\sin(x)}{x}\right)=\sum_{k=0}^\infty(-1)^k\Big[k\pi\le|\xi|\le (k+1)\pi\Big]\tag{3}
$$
Combining $(2)$ and $(3)$ yields the Fourier transform of $\mathrm{sgn}\left(\frac{\sin(x)}{x}\right)$ in the sense of distributions
$$
\begin{align}
&\sum_{k=0}^\infty(-1)^k\frac{\sin(2(k+1)\pi^2x)-\sin(2k\pi^2x)}{\pi x}\\
&=\frac2{\pi x}\sum_{k=0}^\infty(-1)^k\sin(2(k+1)\pi^2x)\\
&=\frac{\tan(\pi^2x)}{\pi x}\tag{4}
\end{align}
$$
Since the Fourier transform of a product is the convolution of the Fourier transforms, the Fourier transform of $\left|\frac{\sin(x)}{x}\right|$ is the convolution
$$
\left[|\xi|\le\frac1{2\pi}\right]\ast\frac{\tan(\pi^2\xi)}{\xi}
=\mathrm{PV}\int_{\xi-\frac1{2\pi}}^{\xi+\frac1{2\pi}}\frac{\tan(\pi^2t)}{t}\,\mathrm{d}t\tag{5}
$$
using the Cauchy Principal Value in $(5)$.

Plots of the Fourier Transform:
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