# Peculiar result of InverseHankelTransform

In using the HankelTransform and its inverse I find that the inverse does not lead to the initial input. I begin with r (the independent variable) in the denominator but end up with r0 (a constant) in the denominator. Can anyone explain this?

ringDelta = 1/(2 π r) DiracDelta[r - r0];

fRingDelta = HankelTransform[ringDelta, r, ρ]

(* (BesselJ[0,r0 ρ] HeavisideTheta[r0])/(2 π) *)

InverseHankelTransform[fRingDelta, ρ, r]

(* (DiracDelta[r-r0] HeavisideTheta[r0])/(2 π r0) *)


Hmmm. I am not 100 % sure because I am not familiar with Hankel transforms. But I read in the docs that HankelTransform implicitly assumes that the input function is supported in $$]0,\infty[$$. So, if ringDelta is nonzero, r0 has to be positive. As I said, this is implicitly assumed; for the return value of InverseHankelTransform, this is stated explicitly by multiplying with HeavisideTheta[r0]. Moreover, notice that 1/(2 π r) DiracDelta[r - r0] equals 1/(2 π r0) DiracDelta[r - r0].

We can check that the result of InverseHankelTransform equals ringDelta in the distributional sense by integrating against a symbolic test function:

Integrate[(ringDelta - InverseHankelTransform[fRingDelta, ρ, r]) ϕ[r],
{r, 0, ∞},
Assumptions -> r0 > 0
]


0

• I think your explanation is entirely correct. – mikado Dec 10 '18 at 21:24
• Thank you, Henrik. I was so surprised by the inverse looking different that I did not notice the equality. – David Keith Dec 11 '18 at 17:52
• You're welcome, David. I guess these integral transforms and DiracDeltas hide enough surprises for all of us. – Henrik Schumacher Dec 11 '18 at 17:54