# 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.

• Your function is like $\left| \frac{1}{x} \right|$ so I'm not sure the integral converges in the usual sense. I'm not sure what exactly you did in Mathematica, but it might be worth it to try the integral again with some kind of "convergence factor" like $e^{-\eta x^2}$ and then take the limit that the new function approaches the one you're interested in. – Kevin Driscoll Oct 12 '13 at 21:20
• Yeah, if you go to wolframalpha and type "fourier transform abs(sin x / x)" you'll get computation time exceeded – TheSeamau5 Oct 12 '13 at 22:02

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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• Awesome! I shall bookmark this answer – TheSeamau5 Oct 13 '13 at 3:45
• I would upvote twice if I could: one for the great job on computing the Fourier transform and presenting it; and another one for pointing to Iverson's bracket and so, ultimately, to this interesting article by Donald Knuth. – Giuseppe Negro Oct 14 '13 at 17:41
• +1 great, which software did u use for the, uh, plots? – user153330 Nov 22 '15 at 15:52
• @user153330: I used Mathematica to do the computation and plotting. I did some scaling of the image using Graphic Converter. – robjohn Nov 22 '15 at 17:18


\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: $$k = 0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots$$
• Did you miss the absolute modulus? Whatever problems Mathematica has, the Fourier transform of $\dfrac{\sin x}{x}$ is extremely unlikely to be one of them. The Fourier transform of $\left\lvert \dfrac{\sin x}{x}\right\rvert$ is an entirely different beast. – Daniel Fischer Oct 13 '13 at 0:42
• @DanielFischer Between "without absolute modulus" and "with absolute modulus" I left to have dinner. Sorry about that. – Felix Marin Oct 13 '13 at 1:32
• Erm, $\lvert f(x)\rvert$ and $[2\Theta(x)-1]f(x)$ are not the same. If we were dealing with complex valued functions, it would be more complicated, but for real-valued $f$ you'd still have $\lvert f(x)\rvert = [2\Theta (f(x)) - 1] f(x)$. – Daniel Fischer Oct 13 '13 at 1:39
• @DanielFischer You are right. It was my horrible mistake. Dinner conspiracy !!!. I'll check everything. Thanks. – Felix Marin Oct 13 '13 at 1:44

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:

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?

• Between you and me, in the spirit of your reasoning, why would the Fourier transform exist for $f_1(x) = \frac{sin x}{x} dx$ but not for $f_2(x) = |\frac{sin x}{x}| dx$? – user96815 Oct 13 '13 at 3:05
• sure, $\frac{sin x}{x}$ has a closed form Fourier transform but the absolute value is weird. The problem is that the function $\frac{sin x}{x}$ has weird intervals where it changes from positive to negative. It sounds like it shouldn't be a problem, but because of it, you cant cleanly multiply the sin part with the complex exponential part. (even after you write the sine as a complex exponential) Look at this: math.stackexchange.com/questions/281870/… Scroll a bit until you get to a long response. At one point, sin is split in order to split the integrals – TheSeamau5 Oct 13 '13 at 3:06
• I understand that you haven not claimed a complete proof. Even so, this condition of "closed form integral" might be too restrictive here. Let us simplify the whole thing, by putting $f(x)=|x|$ and $g(x)=0$. Then according to the method you cite, $f(x)$ does not have an integral in a closed form expression. But we can write down an expression without too much trouble, regardless. – user96815 Oct 13 '13 at 3:17
• @TheSeamau5 It looks to me like the question you referred us to did offer the Fourier coefficients -- I haven't gone over it in detail to see if I agree. What you consider "closed form" depends on who you are. – Betty Mock Oct 13 '13 at 3:17
• @TheSeamau5: No need to feel dumb. It was an interesting discussion. Rather than a piece-wise result, you might obtain something involving the generalized functions such as Dirac delta. Have a look here for the computation of Fourier transform of $|x|$ : thefouriertransform.com/pairs/absT.php ... – user96815 Oct 13 '13 at 3:44