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 $f(x)=\left|\frac{\sin x}{x}\right|$?

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I tried to find one using Mathematica, but it ran for several hours without producing any result.

Solution 1:

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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Solution 2:

$\newcommand{\abs}[1]{\left\vert #1\right\vert}% \newcommand{\ic}{{\rm i}}$ 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: $$ k = 0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots $$