Two-dimensional fingularity function: $$ f ( x , y ) = \delta ( y - x \tan \phi ) $$
Has anyone ever computed the two-dimensional Fourier transform of a Dirac delta function like this? I'm trying research the steps on how to perform these but information seems very limited.
My approach was to do the $$F(u,v) = \iint f(x,y) e^{-j2\pi(ux+vy)} \,\mathrm dx\mathrm dy \text.$$
$\tan(\phi)$ would be $\frac{fy}{fx}$ or $\frac v u$. I've never seen the process of evaluating this kind of problem before. Any guidance or resources would be very much appreciated. Nobody seems to have any idea on how to perform this calculation.
Doing the $y$-integration first and using $\int e^{-j2\pi wx} \,\mathrm dx=\delta(w)$ we get $$ F(u,v) = \iint f(x,y) e^{-j2\pi(ux+vy)} \,\mathrm dx \,\mathrm dy = \iint \delta ( y - x \tan \phi ) e^{-j2\pi(ux+vy)} \,\mathrm dx \,\mathrm dy \\ = \int e^{-j2\pi(ux+v(x \tan \phi))} \,\mathrm dx = \int e^{-j2\pi(u+v\tan\phi)x} \,\mathrm dx = \delta(u+v\tan\phi). $$