Assume that $f(x,y)$ is a two dimensional rotationally symmetric p.d.f., $f(x,y)=\tilde{f}(\sqrt{x^2+y^2})$. In the polar coordinates, $x=r \cos\varphi$, $y=r\sin\varphi$, we can average out the azimuthal angle to find a one dimensional radial p.d.f as follows, $$ p(r)= r\int_{0}^{2\pi}d\varphi\, \tilde{f}(r)=2\pi r\, \tilde{f}(r).\qquad\qquad\qquad \qquad (1)$$
The generating function of the distribution $f(x,y)$ is written as $\mathbb{E}[e^{i\mathbf{k}\cdot \mathbf{x}}]$ where $\mathbf{x}=(x,y)$ and $\mathbf{k}=(k_x,k_y)$ are two dimensional vectors. Here $\mathbb{E}[\cdots]=\int dx dy \cdots \,f(x,y)$. Then the generating function for a rotationally symmetric distribution is given by $$ \mathbb{E}[e^{i\mathbf{k}\cdot \mathbf{x}}]=\int_{0}^{\infty}\int_{0}^{2\pi}r dr d\varphi\,e^{i k r \cos(\varphi-\varphi_k)}\tilde{f}(r)=2\pi\int_{0}^{\infty}r dr\, \tilde{f}(r)\,J_0(kr)=\int_0^{\infty} dr\, p(r) J_0(k r), $$ where in the above $k=\sqrt{k_x^2+k_y^2}$ and $\varphi_k=\text{atan2}(k_x,k_y)$ and $J_0(x)$ is the modified Bessel-Function of the first kind. If I define the averaging with respect to $p(r)$ as $\tilde{\mathbb{E}}[\cdots]=\int_0^{\infty} dr\cdots p(r)$ then we have the following: $$ \mathbb{E}[e^{i\mathbf{k}\cdot \mathbf{x}}] \equiv \tilde{\mathbb{E}}[J_0(k r)]=1-\frac{1}{4}k^2\tilde{\mathbb{E}}[r^2]+\frac{1}{64}k^4\tilde{\mathbb{E}}[r^4]+\cdots\,.$$
Question: Based on the above, the characteristic function of a 2D rotationally symmetric p.d.f. can be obtained only from even moments ( $\tilde{\mathbb{E}}[r^{2k}]$) of the p.d.f. $p(r)$ which is defined in the range $[0,\infty)$. On the other hand, the characteristic function of $p(r)$ can be written as $\tilde{\mathbb{E}}[e^{i t r}]$ where $t\in(-\infty,\infty)$(What is the characteristic function of a p.d.f with $\mathbb{R}^+$ support?). It simply means that $\tilde{\mathbb{E}}[r^{2k+1}]$ are moments of $p(r)$ too. Now I am a bit confused. If I know all $\tilde{\mathbb{E}}[r^{2k}]$ then I know $f(x,y)$. However, in order to know $p(r)$ I should know about $\tilde{\mathbb{E}}[r^{k}]$ for odd $k$ as well as even $k$. On the other hand, $f(x,y)$ and $p(r)$ are uniquelly related to each other from equation (1).
My Guess: I think I am missing something obvious! My guess is that $p(r)$ and rotatinaly symmetric $f(x,y)$ are not equivalent, and I can use them interchangeably if I replace the characteristic function $\tilde{\mathbb{E}}[e^{i t r}]$ to $\tilde{\mathbb{E}}[J_0(k r)]$ for $p(r)$.