$2\text{D}$ Fourier Transform of Laplacian in polar coordinates Consider a typical function written in standard $2\text{D}$ polar form:
\begin{equation}
f(\underline{r})=f(r,\theta)=\sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta}
\end{equation}
executing the Laplacian of f, we have:
\begin{equation}
\begin{split}
\nabla^2 f(\underline{r})=& \Bigg(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}\Bigg) \sum_{n=-\infty}^{\infty} f_n(r) e^{in\theta}\\
=& \sum_{n=-\infty}^{\infty}  \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg) e^{in\theta}\\
=&\sum_{n=-\infty}^{\infty} \nabla^2_n f_n\ e^{in\theta}
\end{split}
\end{equation}
where:
\begin{equation}
\nabla_n^2=\frac{d^2}{d r^2}+\frac{1}{r}\frac{d}{d r}-\frac{n^2}{r^2}
\end{equation}
the 2D Fourier Transform of $\nabla^2 f(\underline{r})$ is given by the following series:
\begin{equation}
\mathbb{F}_{2D}\Big\{ \nabla^2 f(\underline{r})\Big\}=\sum_{n=-\infty}^{\infty} 2\pi i^{-n} e^{in\psi} \int_{0}^{\infty} \Bigg(\frac{d^2 f_n}{d r^2}+\frac{1}{r}\frac{df_n}{d r}-\frac{n^2 f_n}{r^2}\Bigg)J_n(\rho r)\ rdr
\end{equation}
A simple application of integration by parts along with the definition of a
Bessel function gives:
\begin{equation}
\int_{0}^{\infty} \nabla^2_n f_n J_n(\rho r)\ rdr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\ rdr\ \ \ \ \ (*)
\end{equation}
how could i proof the equation (*)?
 A: The differential equation for the Bessel functions $J_n$,
$$x^2 J_n''(x) + x J_n'(x) + (x^2-n^2) J_n(x) = 0$$
can be written as
$$\left( x J_n'(x) \right)' + (x - \frac{n^2}{x}) J_n(x) = 0,$$
which we will use later.
The given left hand expression can be split into a couple of integrals:
$$\begin{align}
\int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr 
&= \int_{0}^{\infty} \left( f_n''(r) + r^{-1} f_n'(r) - n^2 r^{-2} f_n(r) \right) \, J_n(\rho r) \, r \, dr \\
&= \int_{0}^{\infty} \left( r f_n''(r) + f_n'(r) - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\
&= \int_{0}^{\infty} \left( (r f_n'(r))' - n^2 r^{-1} f_n(r) \right) \, J_n(\rho r) \, dr \\
&= \int_{0}^{\infty} (r f_n'(r))'  \, J_n(\rho r) \, dr - n^2 \int_{0}^{\infty} r^{-1} f_n(r) \, J_n(\rho r) \, dr
\end{align}$$
For the first integral we use integration by parts. The boundary terms all vanish so
$$\begin{align}
\int_{0}^{\infty} (r f_n'(r))'  \, J_n(\rho r) \, dr
&= - \int_{0}^{\infty} r f_n'(r)  \, \frac{\partial}{\partial r} \left(J_n(\rho r)\right) \, dr
\\
&= - \int_{0}^{\infty} r f_n'(r)  \, \rho J_n'(\rho r) \, dr
\\
&= - \int_{0}^{\infty} f_n'(r)  \, \rho r J_n'(\rho r) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \frac{\partial}{\partial r} \left(\rho r J_n'(\rho r)\right) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \rho \frac{\partial}{\partial (\rho r)} \left(\rho r J_n'(\rho r)\right) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, \rho ( \frac{n^2}{\rho r} - \rho r) J_n(\rho r) \, dr
\\
&=   \int_{0}^{\infty} f_n(r) \, ( \frac{n^2}{r} - \rho^2 r) J_n(\rho r) \, dr
\end{align}$$
where the rewritten differential equation for $J_n$ has been used.
Adding the second integral from the first calculation results in
$$\int_{0}^{\infty} \nabla^2_n f_n \, J_n(\rho r) \, r \, dr = -\rho^2 \int_{0}^{\infty} f_n J_n(\rho r)\, r \, dr.$$
