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


*

*$H(x)=\chi_{[0,\infty)} (x)$, the Heaviside step function.





*$xH(x)$.





*$x^2 \sin(x)$

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