The Hankel transform of order $\nu$ of a function $f(r)$ is defined by \begin{equation*} F_\nu(k) = \int_0 ^\infty r dr f(r) J_\nu(kr ), \end{equation*} where $J_v$ is a Bessel function of the first kind of order $\nu$.
I am interested in the scenario where $f(r)=1$, where one must be careful about how to consider convergence. For instance it is commonly cited in the literature that for $f(r)=1$ we have $F_0(k) = \delta(k)/k$, where $\delta$ is the Dirac delta function.
Is there a generalisation of this to arbitrary $\nu$, i.e., what is $F_\nu(k)$ for $f(r)=1$? (My primary interest is in the case of $\nu=2$.) If there is no exact answer, is there anyway to approach this numerically given that the answer will likely involve a Dirac delta?
Motivation:
This question arises in the context of a problem in physics. In particular, I require the Fourier transform of $f(r,\theta) = e^{2 i \theta}$, written in polar coordinates, which can be reduced to solving the Hankel transform of $f(r)=1$ with $\nu=2$.
One way to proceed is to examine the Inverse Second-Order Transform for $F(k)=\frac1{k^2}$. Proceeding, we find that the Second-Order Inverse Hankel Transform of $F$ is
$$\begin{align} \mathscr{H_2}^{-1}\{F\}(r)&=\int_0^\infty \frac{J_2(kr)}{k^2}\,k\,dk\\\\ &=\int_0^\infty \frac{J_2(x)}{x}\,dx\\\\ &=\lim_{L\to \infty\\\varepsilon\to 0^+}\left.\left(-\frac{J_1(x)}{x}\right)\right|_\varepsilon^L\tag1\\\\ &=\frac12 \end{align}$$
where in arriving at $(1)$, we used the recurrence relationship $J_2(x)=\frac{J_1(x)}{x}-J_1'(x)$, along with $J_1(x)=x/2+O(x^2)$.
Therefore, the Second-Order Inverse Hankel Transform of $\frac2{k^2}$ is equal to $1$. And in distribution, we must have
$$\mathscr{H_2}\{1\}(k)=\frac2{k^2}$$