Fourier Transform of $\log{|x|} \text{sign}(x)$ Is there a way to define the Fourier transform of $\log{|x|} \text{sign}(x)$? Naively I thought that one could define it through a convolution, namely
\begin{equation}
\mathcal{F}(\log{|t|}\text{sign}(t))=\mathcal{F}(\log{|t|})*\mathcal{F}(\text{sign}(t))\,.
\end{equation}
However, we have that the Fourier transform of each distribution is given by 
\begin{equation}
\mathcal{F}\left(\text{sgn}(t) \right)=\mathcal{P}\frac{2}{i \omega}\,,
\end{equation}
\begin{equation}
\mathcal{F}\left(\log{|t|}\right)=-\mathcal{P} \frac{\pi}{|\omega|}-\sqrt{2\pi}\gamma \delta(\omega)\,,
\end{equation}
where $\mathcal{P}$ denotes the Cauchy principal value. That means that we need to compute the convolution of two Cauchy principal value distributions, namely 
\begin{equation}
\mathcal{P}\frac{2}{i \omega}*\mathcal{P} \frac{\pi}{|\omega|}\,.
\end{equation}
However, as far as I could understand, one can only define the convolution of distributions if one of them has compact support. That's the the case for $\delta(\omega)$ but neither of this seems to have compact support so I guess their convolution cannot be defined. Is it because  $\log{|x|} \text{sign}(x)$ is not a tempered distribution. How does one tell if some distribution is tempered or not in general? I do not know much about distribution theory... Just wanted to define this Fourier transform.
 A: The Fourier transform of a tempered distribution $u$ is defined by
$
\langle \mathcal{F}u, \varphi \rangle
= \langle u, \mathcal{F}\varphi \rangle
$
for every $\varphi \in \mathcal{S}$ (Schwartz space).
Many of the same rules apply to Fourier transform of distributions as to Fourier transforms of ordinary functions. For example, $\mathcal{F}\{Qu\} = iD(\mathcal{F}u)$ and $\mathcal{F}\{Du\} = iQ \mathcal{F}u$. Here the operators $Q$ and $D$ are defined by $(Qf)(x) = x\,f(x)$ and $(Df)(x) = f'(x).$
Let $u(x) = \log|x| \, \operatorname{sign}(x)$. It can quite easily be shown that
$Q(Du) = \operatorname{sign}.$
Taking the Fourier transform of both sides gives
$$
-D(Q \mathcal{F}u)(\xi) = -\mathcal{P}\frac{2i}{\xi}.
$$
Therefore,
$(Q \mathcal{F}u)(\xi) = 2i \log|\xi| + A,$
for some constant $A$, and so we get
$$
(\mathcal{F}u)(\xi) 
= 2i \, \mathcal{P}\frac{\log|\xi|}{\xi} + A \, \mathcal{P}\frac{1}{\xi} + B \, \delta(\xi).
$$
Here we can however conclude that $B=0$ since $B \, \delta(\xi)$ is even, while all other terms (in the left hand side and the right hand side) are odd.
I haven't yet found the value of $A$, though. Perhaps someone else can help with that part.
A: 9 months late to the party, but I've been deriving Fourier transforms all day and I stumbled upon this and it's begging for closure.
To obtain the distributional form of the Fourier transform one can to consider a more general integral that converges absolutely ($\epsilon>0$):
$$F_\epsilon(\omega)=\int_{-\infty}^{\infty}\log|x|\text{sgn}|x|e^{-\epsilon|x|-i\omega x}dx=\int_{0}^{\infty}\log x~e^{-(\epsilon+i\omega) x}dx-\int_{0}^{\infty}\log x~e^{-(\epsilon-i\omega) x}dx$$
It is known and easy to derive that
$$\int_0^{\infty}\log x ~e^{-sx}dx=\frac{-\gamma-\log s}{s}~~,~ ~\text{Re}(s)>0$$
With $\Theta=\arg(\epsilon+i\omega)\in(-\pi,\pi)$ we can write
$$F_\epsilon(\omega)=2i\gamma\frac{\omega}{\epsilon^2+\omega^2}+2i\frac{\log(\sqrt{\epsilon^2+\omega^2})}{\sqrt{\epsilon^2+\omega^2}}\sin\Theta+2i\frac{\Theta\cos\Theta}{\sqrt{\epsilon^2+\omega^2}}$$
Now in the limit $\epsilon\to 0^+$, $\Theta\to \frac{\pi}{2}\text{sgn}(\omega)$ and wherever we get $\epsilon$ as a cutoff to a pole we rename the function as the "principal value" equivalent of the function we would get with $\epsilon=0$. With all these understood, we obtain the final result
$$F[\log|x|\text{sgn}(x)]=2i\gamma ~\mathcal{P}\frac{1}{\omega}+2i~\mathcal{P}\frac{\log|\omega|}{\omega}$$
which is odd, as prescribed above. In fact I think that with careful treatment of the convolution integrals (everything is well defined since there are principal values regulating the poles and furthermore the integrals are elementary) you would be able to compute the Fourier transform in that way too. However, I will leave this as an exercise to you. Cheers!
