# Fourier Transform of the “regular” tempered distribution of $|x|$

As the title states I am trying (without luck) to compute the Fourier Transform (in tempered distributional sense) of $|x|$, meaning ($\mathcal{S}(\mathbb{R})$ the Schwartz space): $$\widehat{|x|}[\varphi]:=\int_\mathbb{R}|x|\widehat{\varphi}(x)dx\qquad \varphi\in\mathcal{S}(\mathbb{R})$$ (The goal of the exercise is to express it in terms of other well known distributions, that it is well defined is clear). So far I've only managed to go rewrite it as: $$-i\int_{\mathbb{R}}(2\theta(x)-1)\widehat{\varphi'}(x)dx$$ where $\theta$ is the Heaviside function or to get expressions involving $\cos$ by splitting up the integral but it does not seem to be a big improvement over the starting situation... Have you got any suggestions?

Many thanks

It's usually more convenient to guess the answer and then show afterwards that it's right. Observe that (in $\mathcal S'$) we have that $$|x|'=\textrm{sign}(x)=\begin{cases} 1 & x>0 & \\ -1 & x<0\end{cases} ,$$ and the Fourier transform of this is $\textrm{PV-}(1/t)$, so $t\widehat{|x|}=c\textrm{PV-}(1/t)$ (I'm not paying attention to the constant here, which depends on your conventions anyway), and this suggests to try $$(\widehat{|x|},\varphi)= \lim_{h\to 0+} \int_{|t|>h} \frac{\varphi(t)-\varphi(0)}{t^2}\, dt$$ (and yes, this is a distribution). We can now check that this works, by comparing it with your second formula. To do this, notice that $\int (2\theta-1)\widehat{\varphi'}=(\textrm{PV-}(1/t),\varphi')$ by the fact I quoted above, and integrate by parts in my formula.
Edit: On second thoughts, it would have been easier to write $|x|=\textrm{sign}(x) x$, so (again, up to a multiplicative constant) $\widehat{|x|}=\textrm{PV-}(1/t)*\delta'$, so this acts on a test function as $$\textrm{PV-}(1/t)(\delta'*\varphi) = -(\textrm{PV-}(1/t),\varphi')=-\lim_{h\to 0+}\int_{|t|>h}\frac{\varphi'(t)}{t}\, dt .$$