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

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)?

• Check the exponent in your first integral, I think you have a missing $-2\pi$ factor. Commented Jul 16, 2017 at 12:22
• @GFauxPas I think that is just convention. For example, one may define $$\hat{f}(k)=\int_{\mathbb{R}^n}f(x)e^{ik\cdot x}\,\mathrm{d}x$$ and we may invert it as $$f(x)=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\hat{f}(k)e^{-ik\cdot x}\,\mathrm{d}k,$$(although I am not sure if even the normalising constant is necessary) i.e. I don't think one needs a $2\pi$ in the exponent, or maybe I'm wrong? Commented Jul 16, 2017 at 12:34
• Jason I agree that the $2\pi$ shouldn't be important, but I don't know if you can switch the sign of the exponent. I'll leave it to someone more knowledgeable. Commented Jul 16, 2017 at 14:00
• The $-2 \pi i \omega x$ vs $-i \omega x$ thing is purely a matter of convention. Fourier analysis people tend to use the former and PDEs people tend to use the latter, although even that isn't absolute. the plus or minus in the exponential is also just convention, but pretty much everyone uses the minus one. Commented Jul 16, 2017 at 17:07

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.
• But if $$f(x,y)=\frac{1}{\pi^2}\int_0^\infty\int_0^{2\pi}\hat{f}(2\omega(\cos\theta,\sin\theta))e^{-2\omega i(\cos\theta\cdot x+\sin\theta\cdot y)}\omega\,\mathrm{d}\theta\,\mathrm{d}\omega$$ then surely we would have to consider multiplication by $\frac{1}{\omega}$ in our inversion formula? Commented Jul 16, 2017 at 22:10
• I don't think so, no. The extra $\omega$ comes from the change in variables for the integration variable in the frequency domain. A change of variables for one integral doesn't necessarily mean the other inversion/transform formula needs to change. In particular you can also convert your integral back using $r = 2\omega \cos\theta, r = 2\omega\sin\theta$ and the $\omega$ will not be there and you'll get the usual inversion formula (up to a constant maybe). Commented Jul 17, 2017 at 14:50