Fourier transform of $\left|\frac{\sin x}{x}\right|$ Is there a closed form (possibly, using known special functions) for the Fourier transform of the function $f(x)=\left|\frac{\sin x}{x}\right|$?
$\hspace{.7in}$
I tried to find one using Mathematica, but it ran for several hours without producing any result.
 A: $\newcommand{\abs}[1]{\left\vert #1\right\vert}%
\newcommand{\ic}{{\rm i}}$
We didn't calculate the Fourier transform the OP asked for. We just calculate its
derivative. It turns out that it has singularities at $k =0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots$. We believe this is the root of problems which makes hard to evaluates the Fourier transform mentioned above. We hope somebody else can take from our final result.
\begin{align}
\phi\left(k\right)
&\equiv
\int_{-\infty}^{\infty}\abs{\sin\left(x\right) \over x}\,{\rm e}^{-\ic kx}
\,{\rm d}x
=
2\int_{0}^{\infty}{\abs{\sin\left(x\right)} \over x}\,\cos\left(kx\right)
\,{\rm d}x
\\[3mm]
\phi'\left(k\right)
&=
-2\int_{0}^{\infty}\abs{\sin\left(x\right)}\sin\left(kx\right)
\,{\rm d}x\,,
\end{align}
$\abs{\sin\left(x\right)}\quad$ is periodic
$\left(~\mbox{of period}\ \pi~\right)$: $\abs{\sin\left(x\right)}
=
\sum_{n = -\infty}^{\infty}A_{n}\cos\left(2nx\right)$ with
$$
\int_{-\pi/2}^{\pi/2}\cos\left(2mx\right)\cos\left(2nx\right)\,{\rm d}x
=
{\pi \over 2}\,\delta_{mn}
$$
$$
A_{n}
=
{2 \over \pi}\int_{-\pi/2}^{\pi/2}\abs{\sin\left(x\right)}\cos\left(2nx\right)
\,{\rm d}x
=
{4 \over \pi}\,{1 \over 1 - 4n^{2}}
=
-\,{1 \over \pi}\,{1 \over n^{2} - 1/4}
$$
Then,

\begin{align}
\phi'\left(k\right)
&=
-2\sum_{n = -\infty}^{\infty}A_{n}\int_{0}^{\infty}\cos\left(2nx\right)\sin\left(kx\right)
\,{\rm d}x
\\[3mm]&=
-\,\Im\sum_{n = -\infty}^{\infty}
A_{n}\int_{0}^{\infty}\left[%
{\rm e}^{\ic\left(k - 2n\right)x}
+
{\rm e}^{\ic\left(k + 2n\right)x}
\right]\,{\rm d}x
\\[3mm]&=
-\,\Im\sum_{n = -\infty}^{\infty}A_{n}\left[%
{-1 \over \ic\left(k - 2n\right) - 0^{+}}
+
{-1 \over \ic\left(k + 2n\right) - 0^{+}}
\right]
\\[3mm]&=
-\,\Im\sum_{n = -\infty}^{\infty}A_{n}\left(%
{-\ic \over 2n - k - \ic 0^{+}}
+
{\ic \over 2n + k + \ic 0^{+}}
\right)
\\[3mm]&=
{1 \over 2}\Re\sum_{n = -\infty}^{\infty}A_{n}\left(%
{1 \over n - k/2 - \ic 0^{+}}
-
{1 \over n + k/2 + \ic 0^{+}}
\right)
\\[3mm]&=
-\,{1 \over 2\pi}{\cal P}\,k\sum_{n = -\infty}^{\infty}
{1 \over n^{2} - 1/4}\,{1 \over n^{2} - \left(k/2\right)^{2}}
\\[3mm]&=
-\,{1 \over 2\pi}{\cal P}\,\left[%
{16 \over k}
+
2k\sum_{n = 0}^{\infty}
{1 \over n^{2} - 1/4}\,{1 \over n^{2} - \left(k/2\right)^{2}}
\right]
\\[3mm]&=
-\,{1 \over 2\pi}{\cal P}\,\left\{%
{16 \over k}
-
{8k \over k^{2} - 1}\sum_{n = 0}^{\infty}\left[%
{1 \over n^{2} - 1/4} - {1 \over n^{2} - \left(k/2\right)^{2}}
\right]\right\}
\end{align}


\begin{align}
\sum_{n = 0}^{\infty}{1 \over n^{2} - a^{2}}
&=
\sum_{n = 0}^{\infty}{1 \over \left(n + \abs{a}\right)\left(n - \abs{a}\right)}
=
{\Psi\left(\abs{a}\right) - \Psi\left(-\abs{a}\right) \over 2\abs{a}}
\\[3mm]&=
{1 \over 2\abs{a}}\left\{%
\Psi\left(\abs{a}\right) - \Psi\left(1 + \abs{a}\right)
+
\pi\cot\left(\pi\left[-\abs{a}\right]\right)
\right\}
=
{1 \over 2\abs{a}}\left[%
-\,{1 \over \abs{a}}
-
\pi\cot\left(\pi\abs{a}\right)
\right]
\end{align}


$$
\sum_{n = 0}^{\infty}{1 \over n^{2} - a^{2}}
=
-\,{1 \over 2a^{2}}\left[%
1
+
{\pi a \over \tan\left(\pi a\right)}\right]\,,
\qquad
a \not\in {\mathbb Z}
$$


$$
\sum_{n = 0}^{\infty}{1 \over n^{2} - 1/4}
=
-2\,,
\qquad
\sum_{n = 0}^{\infty}{1 \over n^{2} - \left(k/2\right)^{2}}
=
-\,{2 \over k^{2}}\left[1 + {\pi k/2 \over \tan\left(\pi k/2\right)}\right]
$$


\begin{align}
\phi'\left(k\right)
&=
-\,{1 \over 2\pi}{\cal P}\,\left[%
{16 \over k}
+
{16k \over k^{2} - 1}
-
{16 \over k}\,{1 \over k^{2} - 1}
-
{16 \over k}\,{1 \over k^{2} - 1}\,{\pi k/2 \over \tan\left(\pi k/2\right)}
\right]
\end{align}


\begin{align}
\phi'\left(k\right)
&=
-\,{1 \over 2\pi}{\cal P}\,\left[%
{32k \over k^{2} - 1}
-
{16 \over k}\,{1 \over k^{2} - 1}\,{\pi k/2 \over \tan\left(\pi k/2\right)}
\right]
\end{align}

We can observe that $\phi'\left(k\right)$ diverges at $k$ values:
$$
k = 0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots
$$

A: 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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A: The more I think about, the more I think that there's no chance of there being a closed form version of the fourier transform of 
$|\frac{sin x}{x}|$
Look here:
How can you prove that a function has no closed form integral? 
If we follow the steps indicated by the answer to this question, we have to say:
$F(\omega) = \int^{+\infty}_{-\infty} f(x) e^{-2i\pi\omega x} dx$ where $f(x) = |\frac{sin x}{x}| $
In order for this integral to have a closed form, we need to find some rational function $h(x)$ which satisfies:
$h'(x) + h(x)g(x) = f(x)$ 
where $f(x) = |\frac{sin x}{x}| $ and $g(x) = -2i\pi\omega x$
So, if you take $h' + gh = f$ and apply old school integrating factor to it you get the following:
$h(x) = \frac{\int e^{-\pi\omega x^2} |\frac{sin x}{x}| dx}{e^{-\pi\omega x^2}}$ 
and, between you and me, $h(x)$ does not look like a rational function to me: http://en.wikipedia.org/wiki/Rational_function
I hope I'm wrong because $f(x)$ looks so nice and wavy. It just looks so fourierable to me. It's like, if we can't describe this simple wave as a fourier series without discretizing, then what can we describe as a fourier series?
