# Computing the Fourier transform of three distributions - one last part

In my analysis class, we are now covering distribution theory. We take the Schwartz space of functions $$S(\mathbb{R})$$ and its continuous dual space, $$S'(\mathbb{R})$$, the space of tempered distributions. For $$T \in S'(\mathbb{R})$$ and $$\phi \in S(\mathbb{R})$$, we use the notation $$\langle T,\phi \rangle = T(\phi)$$ and when $$T(x)$$ is a function, we define $$\langle T,\phi \rangle = \int_{-\infty}^{\infty} T(x)\phi(x)dx$$ For the Foruier transform on tempered distributions, we define $$\langle \mathcal{F}T,\phi \rangle = \langle T,\mathcal{F}\phi \rangle$$ We are asked to compute the Fourier transform of the following functions as distributions

1. $$H(x)=\chi_{[0,\infty)} (x)$$, the Heaviside step function.
1. $$xH(x)$$.
1. $$x^2 \sin(x)$$

Our textbook says that, in the sense of distributions, one has the derivative $$H'(x)=\delta_0(x)$$ so I thought I could take the Fourier transform on both sides and get, using the derivative identity $$ip \hat{H}=\frac{1}{\sqrt{2\pi}}$$ but I do not know if I can "divide" with distributions, given other answers on this site and elsewhere, when dividing we need to take a constant multiple of $$\delta$$ which I do not know why, and I do not how to compute the constant. For the second part, I thought I could do the derivative and get using the product rule (which I think holds for distributions) $$(xH)'=H+xH'=H+x\delta_0$$ but I do not know how to compute the Fourier transform of $$x\delta_0$$. I have managed part 3. I would therefore appreciate any help on 1 and 2, and I thank all helpers.

************* Progress: I have managed part 3 but parts 1 and 2 are still a mystery to me. I do not know how to "divide" in the sense of distributions. I thank all helpers.

• For the first question: you can divide, but you need to take a principal value, and also add a multiple of $\delta$. This gives the result up to a constant, which you will have to compute; see cs.uaf.edu/~bueler/M611heaviside.pdf. For the second, remember that the Fourier transform maps derivatives into multiplications and vice-versa. For the last, use Euler’s formula. Commented Nov 15, 2020 at 4:07
• @GiuseppeNegro thank you, I think I understand the first part. For part two, I do not think I quite know how to do it, and for part 3 I do not know what you mean by Euler's formula. Commented Nov 15, 2020 at 4:11
• @GiuseppeNegro for the Euler formula, do you mean the complex exponential? Also, could you explain a bit on the division and why we need a multiple of delta and how to compute its coefficient? Commented Nov 15, 2020 at 4:20
• @GiuseppeNegro and how do we compute the constant? Commented Nov 15, 2020 at 4:31

I'm using $$\mathcal{F}\{f(x)\} = \int_{-\infty}^{\infty} f(x)e^{-ix\xi} \, dx.$$

First we take the Fourier transform of the $$\operatorname{sign}$$ function using $$\operatorname{sign}'=2\delta$$: $$2 = \mathcal{F}\{2\delta\} = \mathcal{F}\{\operatorname{sign}'\} = i\xi \mathcal{F}\{\operatorname{sign}\} .$$ From this we can conclude that $$\mathcal{F}\{\operatorname{sign}\} = \operatorname{pv}\frac{2}{i\xi} + C\delta(\xi).$$ Here I have used the facts that $$u(x)=\operatorname{pv}\frac{1}{x}$$ is the odd solution to $$xu(x)=1$$ and that $$x\delta(x)=0.$$

But $$\operatorname{sign}$$ is an odd function, and so must be its Fourier transform, so $$C=0.$$ Thus, $$\mathcal{F}\{\operatorname{sign}\} = \operatorname{pv}\frac{2}{i\xi}.$$

Now, $$H=\frac12(\operatorname{sign}+1),$$ so $$\mathcal{F}\{H\} = \mathcal{F}\{\frac12(\operatorname{sign}+1)\} = \frac12(\mathcal{F}\{\operatorname{sign}\}+\mathcal{F}\{1\}) = \frac12(\operatorname{pv}\frac{2}{i\xi} + 2\pi\delta(\xi)) = \operatorname{pv}\frac{1}{i\xi} + \pi\delta(\xi) .$$

Then, $$\mathcal{F}\{xH\} = i\frac{d}{d\xi}\mathcal{F}\{H\} = i\frac{d}{d\xi}(\operatorname{pv}\frac{1}{i\xi} + \pi\delta(\xi)) = -\operatorname{fp}\frac{1}{\xi^2} + i\pi\delta'(\xi) .$$

Finally, $$\mathcal{F}\{x^2\sin x\} = (i\frac{d}{d\xi})^2 \mathcal{F}\{\sin x\} = -\frac{d^2}{d\xi^2} \mathcal{F}\{ \frac{1}{2i}(e^{ix}-e^{-ix}) \} = -\frac{1}{2i} \frac{d^2}{d\xi^2} 2\pi(\delta(\xi-1)-\delta(\xi+1)) \\ = i\pi (\delta''(\xi-1)-\delta''(\xi+1)) .$$