# Find a Fourier transform of $|\sin(x)|$ in $\mathcal{S}'(\mathbb{R})$

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?

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

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?

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