Seperable functions have seperable Fourier transforms

Question

How do I show that if $f(x,y)$ is separable into a product of a function of $x$ and a function of $y$, its Fourier transform $F(u,v)$ is also separable into a function of $u$ and a function of $v$?

Answer 1

$\int\int f(x,y)\ e^{i 2\pi (vx+uy)} \ dx dy = \int\int f(x)f(y)\ e^{i 2\pi vx} e^{i 2\pi ux}\ dx dy = \int f(x)e^{i 2\pi vx} dx \int f(y)e^{i 2\pi ux}\ dy = F(u)F(v)$