how is the inverse Hankel transform defined

Question

The finite Hankel transform in my notes is defined as $$ f^*(k_i) = \int_{0}^{\infty} {r f(r)J_0(rk_i)dr}, $$ where $k_i$ is one of the positive roots of $J_0$. However, my notes don't say anything about the inverse transform (?!). This is what I found online: $$ f(r) = \int_{0}^{\infty}{k_iJ_0(k_ir)f^*(k_i)dk_i} $$ This doesn't make sense to me as $k_i$ are a discrete set. So how I can integrate over them? Shouldn't it be a sum instead?

Answer 1

The Hankel transform is (when it is invertible) the same as its inverse. It is an involution.

Hence, your formula is correct, just your interpretation of $k_i$ is incorrect. It is just a variable and has nothing to do with zeros of the Bessel function.

EDIT: There is another related mathematical construct, called the Fourier Bessel series http://mathworld.wolfram.com/Fourier-BesselSeries.html, which has coeffs

$$A_i = \frac{2}{J_{n+1}(k_i)^2} \int_0^1 r f(r) J_n(k_i r) \mathrm{d} r$$

and

$$f(r) = \sum_i A_i J_n(k_i r), \quad \text{ for } \quad 0 \leq x < 1$$ There the $k_i$ are the positive roots of $J_n$.

Apparently the paper you are citing, defines this Fourier Bessel Series as finite Hankel transform (with slight coefficient change $f^*(i) = \frac{J_{n+1}(k_i)^2}{2} A_i$, but nontheless the same mathematical idea).

I never have heard of name "finite Hankel transform" before, and knew this as Fourier Bessel series. I agree that the coefficient change makes it very similar to the normal Hankel transform, which just has the upper integral bound as $\infty$ instead of $1$.

But note that in the original post you have $\infty$ as upper bound, which makes that, what you present, the normal Hankel transform and not the finite transform.

Answer 2

Found a paper here that describes the finite Hankel transform and its inverse: http://projecteuclid.org/euclid.pjm/1102867723

My notes were right :-)