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Reproducing Hankel transform pairs?

John Burkhardt
Posted 2 years ago

There is a classic Hankel transform pair from optics that I am trying to reproduce with Mathematica, without luck, so I can confidently generate an unknown case. The function in the spatial frequency domain, H, is:

H[rho_]:=2/Pi*(ArcCos[rho]-rho*Sqrt[1-rho^2])

with rho < 1. There are several versions of the normalized inverse Hankel transform of the frequency domain function available in the literature with hNorm[r] = h[r]/h[0] where h is the inverse hankel transform of H. I'd be happy to find any of them. For discussion sake one is published to be:

hNorm[r_]:=BesselJ[1,r/2]^2/(r/2)^2

When I try the inversion, however, I only get a result for the rho*Sqrt[1-rho^2] term. The inverse transform for the ArcCos term (hStar) is returned unchanged:

hStar[r_]=Assuming[rho<1,InverseHankelTransform[ArcCos[rho],rho,r]]

yields,

InverseHankelTransform[ArcCos[rho],rho,r,0]

I've tried to enforce various appropriate assumptions and have tried other Bessel function orders in the transform without luck. Any ideas?

Thanks
POSTED BY: John Burkhardt
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InverseHankelTransform[]'s support is still somewhat limited,so don't be surprised if some things don't work yet.

I have only workaround: InverseHankelTransform[ArcCos[rho], rho, r] is:

 InverseFourierCosTransform[
    InverseHankelTransform[FourierCosTransform[ArcCos[a rho], a, s], 
      rho, r, Assumptions -> {s > 0}][[1]], s, a] /. a -> 1 // Expand

(*(\[Pi] BesselJ[0, r/2] BesselJ[1, r/2])/(2 r) - (
I Sqrt[\[Pi]] MeijerG[{{}, {1/2}}, {{0, 1}, {0}}, r^2/4])/r^2*)

but solution is expressed by MeijerG function.

  FullSimplify[HankelTransform[%, r, rho] // FunctionExpand, 
   Assumptions -> 0 < rho < 1]
(*ArcCos[rho]*)
POSTED BY: Mariusz Iwaniuk
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