Convergence of hankel transform for polynomials The hankel transform is related to fourier transforms that have some kind of spherical symmetry. The simplest is related to the 2D radially symmetric fourier transform. This transform $F(k)$ of $f(r)$ is defined below as
$$
F(k) = \int_0^\infty f(r) J_0(kr) r {\rm{d}}r, \qquad {\rm{(limit\:for\:upper\:bound\:implied)}}
$$
Since $J_0\sim 1\big/\sqrt{r}$ for large $r$, we can imagine that a convergence condition exists where we just replace the bessel function with the asymptotic limit.
$$
\int_0^\infty f(r) \sqrt{r} {\rm{d}}r < M
$$
And wikipedia lists this is so for invertibility. However, the statement doesn't appear to be bijective and there's an entry in the wiki article that lists hankel transform pairs where the function doesn't meet the condition. I'm interested in these polynomial conditions,
$$
r^3 \rightarrow 9k^{-5} \quad {\rm{and}} \quad r \rightarrow -k^{-3}
$$
I could attempt to crunch it out and see what happens for $r^3$, but I'm worried about convergence of other functions that are less than $r^3$ but greater than or equal to $r^{-1/2}$ or what if $f(r)$ isn't strictly real valued? Is it that the wiki isn't including that there are restrictions on $k$ where this converges?

I did try to do it to see if along the way it would become apparent why it works for $r$ and $r^3$ but not other functions that don't satisfy the invertibility condition.
I use some properties of bessel functions $\partial_z z^\nu J_\nu = z^\nu J_{\nu-1}$ and $\partial_z J_0 = -J_1$ and integration by parts, to obtain,
$$
\begin{align}
F(k; r^n) &= \int r^n J_0(kr) r {\rm{d}}r \\
&= \frac{r^n}{k} J_1(kr) r + \frac{nr^n}{k^2} J_0(kr)r - \frac{n^2}{k^2}\int r^{n-1} J_0(kr) r {\rm{d}}r \\
&= \frac{r^n}{k} J_1(kr) r + \frac{nr^n}{k^2} J_0(kr)r - \frac{n^2}{k^2}F(k; r^{n-2})
\end{align}
$$
From here I've gotten a way to go from $r^3$ to $r$ but I can't imagine the ``surface'' terms from $r^3$ cancelling the ones from $r$ because the additional factor of $k$. What am I missing?
 A: It is not my personal answer but one that someone I know showed me, it is certainly not a rigorous answer but it shows that the behavior of oscillations are what causes the decay. We will use the integral form of the 0th bessel function,
$$
J_0(x) = \frac{1}{2\pi} \int_0^{2\pi} \exp(ix\cos\varphi) {\rm{d}}\varphi
$$
and the transform, $F_n(q)$ of $f(b)=b^n$ becomes,
$$
F_n(q) = \int_0^{\infty}b{\rm{d}}b \frac{1}{2\pi} \int_0^{2\pi}{\rm{d}}\varphi \exp(iqb\cos\varphi) b^n
$$
For a mathematician, this next part is an issue especially since it's so close to what I am trying to show, but I exchange the order of integration.
$$
F_n(q) = \frac{1}{2\pi} \int_0^{2\pi}{\rm{d}}\varphi \int_0^{\infty} b^{n+1}{\rm{d}}b \exp(iqb\cos\varphi)
$$
Let's look at the second integral, we will use $n+1=m-1$ to cite a different result for a later expression.

$$
G_m(z)=\int_0^\infty b^{m-1} \exp(izb) {\rm{d}}b, \quad z\equiv q\cos\varphi + i\eta
$$
where we introduce eta as a convergence parameter and we will take a limit of it vanishing later. Let $x=(\eta+iz)b$ and the previous expression becomes.
$$
G_m(z)=\int_0^\infty \left(\frac{x}{\eta+iz}\right)^{m-1} \exp(-x) \frac{{\rm{d}}x}{\eta+iz}
$$
extracting the terms independent of $x$ gives the integral form of the gamma function for integer argument, so that
$$
G_m(z) = \frac{(m-1)!}{(\eta+iz)^m} = \frac{(m-1)!}{i^m(z-i\eta)^m}
$$
coming back to where we were

We find that
$$
\begin{align}
F_n(q) &= \frac{1}{2\pi} \int_0^{2\pi} {\rm{d}}\varphi G_{n+2}(z, \eta \rightarrow 0) \\
 &= \frac{(n+1)!}{2\pi i^{n+2} q^{n+2}} \int_0^{2\pi}\frac{{\rm{d}}\varphi}{(\cos\varphi - i \eta/q)^{n+2}}, \quad \eta\rightarrow 0
\end{align}
$$
the integral can be done as a contour integral. Taking $z=\exp(i\varphi)$ and factoring the resulting polynomial in the denominator has roots at $z_0=i\eta\pm i\sqrt{1+\eta^2}$, redefining $\eta$ and assuming $q\neq0$, giving that the integral is now
$$
-i2^{n+2} \oint z^{n+1} \frac{\rm{d}}{(z-z_{0+})^{n+2}(z-z_{0-})^{n+2}}
$$
which can be evaluated with residue theorem as all of singularities are removable telling us that the integral is finite for integer $n\geq-1$.
