find Fourier transform for a distribution I need some help in how to find Fourier transform for a distribution in $R^2$ for the function $1/(y-x)$ ?
As far as I know it coincides with Fourier transform for the function, but I`m confused how to find it? Any help please. 
Thank you 
 A: It's not clear what the groundrules are for you, but the way that I would genuinely try to understand this is the following.
First, since Fourier transform respects rotations, it suffices to find the Fourier transform of what at first appears to be $u(x,y)=1/x$. That is, this function/distribution is constant in $y$.
But $1/x$ is not locally integrable, so it is not at all clear what we mean by declaring this a distribution, insofar as it is understood (often tacitly) that a function-as-distribution is really "integrate-against-that-function". So a not locally $L^1$ function is ambiguous at best.
For this example, a standard (and appropriate/useful) re-interpretation is via a principal-value integral: the distribution (in one variable) attached to $1/x$ is not integrate against $1/x$, but is
$u(f)=\lim_{\varepsilon\to 0^+} \int_{|x|\ge \varepsilon} {f(x)\over x}\;dx$. Note that this is not a literal integral, but is some sort of extension of integral. It only makes literal sense for $f$ with some sort of good behavior at $0$, etc.
There is another exercise which is to show that the Fourier transform of this integral is (a constant multiple of) the sign function $\sigma$. I can't tell how much of the question this sub-question is meant to be.
Thus, the Fourier transform of $u\otimes 1$ is $\widehat{u}\otimes \widehat{1}=c\cdot \sigma\otimes \delta$ (because the Fourier transform of $1$ is $\delta$), and rotating back to the original, this is 
$c\cdot \sigma(x-y)\otimes \delta(x+y)$. (Modulo details! :)
