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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) *)
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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
 ]

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

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