Let $f(x) = |\sin(x)|$ be a Schwartz function on $\mathbb{R}$ and $F$ is a corresponding functional in $\mathcal{S}'(\mathbb{R})$. Find its Fourier transform. My attempt: $$\hat{F}(\phi) = F(\hat{\phi}) = \int_\limits{\mathbb{R}}|\sin(x)|\frac{1}{\sqrt{2\pi}}\left(\int\limits_{\mathbb{R}} e^{-ixy}\phi(y)dy\right)dx = $$ $$=\frac{1}{\sqrt{2\pi}}\sum_{k=-\infty}^{+\infty}\left(\int\limits_{2\pi k}^{2\pi k+\pi}\sin(x)\int\limits_{\mathbb{R}} e^{-ixy}\phi(y)dydx - \int\limits_{2\pi k + \pi}^{2\pi k+2\pi}\sin(x)\int\limits_{\mathbb{R}} e^{-ixy}\phi(y)dydx\right) = $$ $$= \frac{1}{\sqrt{2\pi}}\sum_{k=-\infty}^{+\infty}\left(\int\limits_{\mathbb{R}}\phi(y)\int\limits_{2\pi k}^{2\pi k+\pi} e^{-ixy}\sin(x)dxdy - \int\limits_{\mathbb{R}}\phi(y)\int\limits_{2\pi k + \pi}^{2\pi k+2\pi} e^{-ixy}\sin(x)dxdy\right) = $$ $$= \frac{1}{\sqrt{2\pi}}\sum_{k=-\infty}^{+\infty}\int\limits_{\mathbb{R}}\phi(y) \frac{-2e^{-iy(2\pi k + \pi)} - e^{-2iy\pi k}- e^{-iy(2\pi k+2\pi)}}{y^2-1}dy = $$ $$= -\frac{2}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}\phi(y)\left(\sum_{k=-\infty}^{+\infty}\frac{e^{-iy(2\pi k + \pi)} + e^{-2iy\pi k}}{y^2-1}\right)dy$$ Hence, $\hat{F}$ is a corresponding functional to $-\sqrt{\frac{2}{\pi}}\sum\limits_{k=-\infty}^{+\infty}\frac{e^{-ix(2\pi k + \pi)} + e^{-2ix\pi k}}{x^2-1}$. Clearly, this is not the right answer, this function is not continuous at $x=1$.

I use Fubini theorem to swap the integrals but not sure if it's correct here. Could you please tell me what should I do to solve this problem?

  • $\begingroup$ what have you tried so far? $\endgroup$ Jun 1 at 4:24
  • 1
    $\begingroup$ Why do you think the answer should be continuous at $1$? $\endgroup$ Jun 1 at 4:26
  • $\begingroup$ @RyRytheFlyGuy I also tried to use the representation $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$, also considering the integrals on the segments $\endgroup$ Jun 1 at 4:31
  • $\begingroup$ @NinadMunshi it should belong $\mathcal{S}(\mathbb{R})$, shouldn't it? $\endgroup$ Jun 1 at 4:33
  • $\begingroup$ That's not right, the Fourier transform should be in $\mathcal{S}'(\Bbb{R})$, too $\endgroup$ Jun 1 at 4:51

1 Answer 1


First, notice that $|\sin x|$ is a periodic function. So you should expect its Fourier transform to be a sum of Dirac deltas (exercise left to you!).

But instead of computing that Fourier transform, let's first look at that of $\text{sgn}(\sin x)$. Again $\text{sgn}(\sin x)$ is periodic and its Fourier series is easy to compute. Indeed, $$\int_{-\pi}^{\pi} \text{sgn}(\sin x) e^{inx}dx = \left\{ \begin{array}{ll} \frac{4i}{n} & \text{if $n$ is odd}\\ 0 & \text{if $n$ is even} \end{array} \right. $$ Thus the Fourier series in question is $$\text{sgn}(\sin x) = -\frac {2i} \pi \sum_{n\geq 1 \text{ odd}} \frac{e^{inx}-e^{-inx}}n=-\frac {2i} \pi \sum_{m\geq 0}\frac{e^{(2m+1)ix}-e^{-(2m+1)ix}}{2m+1}$$ And the Fourier transform of that function is $$\widehat {\text{sgn}(\sin)}(\omega) = -4i \sum_{m\geq 0}\frac{\delta(\omega - (2m+1))-\delta(\omega +(2m+1))}{2m+1}\tag{1}$$ We also know that $$\widehat {\sin}(\omega) = \frac{\delta(\omega-1)-\delta(\omega+1)}{2i}\tag{2}$$

And because $|\sin x|=\sin(x) \cdot\text{sgn}(\sin x)$, you can deduce the Fourier transform of $|\sin x|$ by taking the convolution of the fourier transforms:

$$ \widehat{|\sin|}(\omega) = \frac 1 {2\pi} \widehat {\sin}\ast \widehat {\text{sgn}(\sin)}(\omega)$$

Using $(1)$ and $(2)$, can you finish?

  • $\begingroup$ +1 This is beautiful, especially since Wolfram Alpha gave up on this $\endgroup$ Jun 1 at 5:13
  • $\begingroup$ Thank you. Doing my best :) $\endgroup$ Jun 1 at 5:14
  • $\begingroup$ @StefanLafon grateful for the help $\endgroup$ Jun 1 at 6:15

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