How to interpret $\hat{f}(2\omega(\cos\theta,\sin\theta))$?

Question

As per the title, for $\omega\in\mathbb{R}_{\ge 0}$ and $\theta\in[0,2\pi]$, I was thinking

$$\hat{f}(2\omega(\cos\theta,\sin\theta))=\int_\mathbb{R}\int_\mathbb{R} f(x,y)e^{2\omega i(\cos\theta\cdot x+\sin\theta\cdot y)}\,\mathrm{d}x\,\mathrm{d}y$$

But I am not sure if this is correct and I was thinking of how I would, for instance, invert it:

$$ \begin{aligned} f(x,y)&=\frac{1}{(2\pi)^2}\int_\mathbb{R}\int_\mathbb{R}\hat{f}(k_x,k_y)e^{-i(k_x\cdot x+k_y\cdot y)}\,\mathrm{d}k_x\,\mathrm{d}k_y \\ &=\frac{1}{4\pi^2}\int_0^\infty\int_0^{2\pi}\hat{f}(2\omega(\cos\theta,\sin\theta))e^{-2\omega i(\cos\theta\cdot x+\sin\theta\cdot y)}\bigg|\frac{\partial(k_x,k_y)}{\partial(\omega,\theta)}\bigg|\,\mathrm{d}\theta\,\mathrm{d}\omega \\ &=\frac{4}{4\pi^2}\int_0^\infty\int_0^{2\pi}\hat{f}(2\omega(\cos\theta,\sin\theta))e^{-2\omega i(\cos\theta\cdot x+\sin\theta\cdot y)}\omega\,\mathrm{d}\theta\,\mathrm{d}\omega \end{aligned}$$ where I have used that $$k_x=2\omega\cos\theta,\qquad k_y=2\omega\sin\theta.$$ I would really like to know if the highlighted part is correct though (or if I am, for instance, missing a Jacobian or something)?

Answer 1

The highlighted box should be correct. You're dealing with the 2-dimensional Fourier transform here, which is defined as (I'm using the $2\pi$ convention here): $$\hat{f}(r,s) = \int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y) e^{-2\pi i (x,y)\cdot(r,s)} \, dx \, dy$$ The argument of $\hat{f}$ in your question is simply $r$ and $s$ given in polar form so you can interpret it as $$\hat{f}(2\omega\cos\theta,2\omega\sin\theta) = \int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y) e^{-2\pi i (2\omega\cos\theta,2\omega\sin\theta)\cdot(x,y)}\,dx\,dy$$ $$= \int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y) e^{-4\pi i \omega (x\cos\theta+y\sin\theta)}\,dx\,dy $$ In particular, this is a change of variables in frequency not in your integration variable so no Jacobian is needed.