Return to Revisions

Riesz transform of the function $f(x,y)$ of two variables ($d=2$), $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv,$$ Take the derivative with respect to $x$, make the desired substitution $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ and you end up with $$[\partial_x\mathcal{R}_xf](x',y')=\frac{4}{\pi}\iint \left(\frac{(2 b v-x'+y')^2}{b^2}-\frac{2 (-2 a u+x'+y')^2}{a^2}\right)$$ $$\qquad{}\times\left(\frac{(-2 a u+x'+y')^2}{a^2}+\frac{(2 b v-x'+y')^2}{b^2}\right)^{-5/2}f(u,v)\,dudv.$$ Any reason why you would want to go through this transformation?