Lately, I have come accross a difficult inverse Fourier transform computation (or at least for me) and I have no idea how to start or if there is any tricks to use. Let us define the functions of interest as:
\begin{equation*} \tilde{Z}_n(u, v) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} Z_n(x,y) e^{-i (xu+yv)} dx dy \end{equation*}
In this post, I demonstrated that: \begin{equation*} \tilde{Z}_n(u, v) = \tilde{Z}_0\frac{\lambda_1^n + \lambda_2^n}{2} - \tilde{Z}_0\frac{iv}{\sqrt{u^2 - 1}}\frac{\lambda_1^n - \lambda_2^n}{2} \end{equation*} where $\tilde{Z}_0$ is the Fourier transform of $Z_0(x, y) = 2\pi J_0\left(\sqrt{x^2 + y^2}\right)e^{id}$ and $\lambda_{1, 2}$ are defined as: \begin{equation} \begin{cases} \lambda_1 = u - \sqrt{u^2 - 1}, \\ \lambda_2 = u + \sqrt{u^2 - 1}. \end{cases} \end{equation}
Using Binomial Theorem, we can write: \begin{equation*} \tilde{Z}_n(u, v) = \tilde{Z}_0\sum_{k=0}^{n} \begin{pmatrix} n\\ k \end{pmatrix} u^{n-k}\left(u^2 - 1\right)^\frac{k}{2}\left(\frac{1+(-1)^k}{2}+\frac{1-(-1)^k}{2}\frac{iv}{\sqrt{u^2-1}}\right) \end{equation*}
Using the computation done here, we have: \begin{equation} \tilde{Z}_0(u, v) = 2\pi e^{id} \delta(\sqrt{u^2 + v^2}-1), \\ \end{equation}
Is there any way to inverse the transformation and get back $Z_n(x, y)$? Considering the origin of the problem, I guess that Bessel functions should be involved.
I found the solution of my initial problem via another method than the inverse Fourier transform (see this post for the full answer), the solution being: \begin{align*} Z_{\pm n} = \left(\frac{x\pm iy}{x^2+y^2}\right)^{n}\left[r^2 J_0(r) \sum_{k=1}^{u_n}a_k^{n} \left(x^2 + y^2\right)^{k-1} + i^n r J_1(r) \sum_{k=1}^{u_{n+1}}b_k^{n} \left(x^2 + y^2\right)^{k-1}\right] \end{align*} where $n\geq 1$ and $u_n$ is defined as $u_n = \left(2n+(-1)^n -1\right)/4$. The two coefficients can be computed using: \begin{equation*} \forall k\geq 1, n\geq 2k,\quad a_k^{n} = (2i)^{n-2k}\frac{(n-k)!}{(k-1)!k!}\prod_{i=1}^{k-1}(n-(k+i)) \end{equation*}
\begin{equation*} \forall k\geq 1, n\geq 2k-1,\quad b_k^{n} = (-1)^{k+1}2^{n-(2k-1)}\frac{(n-k)!}{(k-1)!(k-1)!}\prod_{i=1}^{k-1}(n-(k-1+i)) \end{equation*} If anyone wants to demonstrate this result using inverse Fourier transform, be my guest...
