TITLE: What is the Fourier transform of $|x|$?
QUESTION [5 upvotes]: I am trying to find the Fourier transform of $|x|$ in the sense of distributions in its simplest form. Here is what I have done so far:
Let
$$f(x)=|x|=\lim_{a\rightarrow 0}\frac{1-e^{-a|x|}}{a},$$
then the Fourier transform is given by
$$\begin{aligned}
\hat{f}(\xi)&=\int_{-\infty}^\infty f(x)e^{-2\pi i x \xi}dx \\
&=\lim_{a\rightarrow 0}\frac{1}{a}\left(\delta(\xi)-\frac{2a}{a^2+4\pi^2\xi^2}\right).
\end{aligned}$$
Using the identity (see here),
$$\delta(\xi)=\lim_{a\rightarrow 0}\frac{1}{\pi}\frac{a}{a^2+\xi^2},$$
we know that
$$2\pi\delta(2\pi\xi)=\lim_{a\rightarrow0}\frac{2a}{a^2+4\pi^2\xi^2}.$$
Hence, using the identity,
$$\delta(b x)=\frac{1}{|b|}\delta(x),$$
we know that
$$\hat{f}(\xi)\stackrel{?}{=}\lim_{a\rightarrow0}\frac{1}{a}[\delta(\xi) - \delta(\xi)].$$
This doesn't seem right... Can you see where I have gone wrong and do you know how to calculate $\hat{f}(\xi)$ in its simplest form?

REPLY [2 votes]: From $\operatorname{sign}'=2\delta$ we get $i\xi\,\widehat{\operatorname{sign}}(\xi)=2,$ from which we can conclude
$$
\widehat{\operatorname{sign}}(\xi) = -2i\operatorname{pv}\frac{1}{\xi}+C\delta(\xi).
$$
Since $\operatorname{sign}$ is odd, so must be $\widehat{\operatorname{sign}},$ which forces $C=0.$
Now, $f(x) = x \operatorname{sign}(x),$ so
$$
\hat{f}(\xi)
= i\frac{d}{d\xi}\widehat{\operatorname{sign}}(\xi)
= i\frac{d}{d\xi}\left(-2i \operatorname{pv}\frac{1}{\xi}\right)
= -2 \operatorname{fp}\frac{1}{\xi^2}.
$$