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The Hankel Transform is given by

Integrate[f[x] x BesselJ[0,x t],{x,0,Infinity}]

It is self-inverse, so

Integrate[F[t] t BesselJ[0,x t],{t,0,Infinity}]

gives the back-transformation.

I tried out a simple case:

Integrate[UnitBox[x/2] x BesselJ[0,x t],{x,0,Infinity}]

which promptly results in

BesselJ[1,t]/t

This is correct. However, if I do the back-transformation:

Integrate[BesselJ[1,t]/t t BesselJ[0,x t],{t,0,Infinity}]

the integration takes noticably longer (which is expected since the function oscillates) and the result is

ConditionalExpression[0,x>1]

While that single condition would be correct - UnitBox[x/2] == 0 for x>1 - the rest of the function won't show up.

Is there any way to make the obviously missing parts show up?

(I am not sure what tags to use. Please retag as appropriate)

Related but not quite what I'm asking:
Hankel Transform integrals won't work in Mathematica
Strange result when integrating BesselJ functions

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1 Answer 1

Try :

Integrate[BesselJ[1, t]/t t BesselJ[0, x t], {t, 0, Infinity}, 
   Assumptions -> { x > 0}]
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Why not Integrate[BesselJ[1, t] BesselJ[0, x t], {t, 0, Infinity}, Assumptions -> {x > 0}]? –  m_goldberg Feb 2 '13 at 14:45
    
That apparently works. Weird. I think I previously had a case where it didn't. –  kram1032 Feb 2 '13 at 14:55
    
@m_goldberg of course, that bit of simplifying would make it a bit faster but it's less general. I had this problem with more complex functions as input as well. Ones that wouldn't simplify so readily. –  kram1032 Feb 2 '13 at 14:56
    
Integrate[(Sin[t]-t Cos[t])/t^2 BesselJ[0,t x],{t,0,Infinity},Assumptions->x>0] should return something equivalent to UnitBox[x/2]Sqrt[1-x^2] (at least for x>0) but doesn't get evaluated at all. Without assumptions, it returns ConditionalExpression[0,x>1 || x<-1] - basically the same problem but not fixable with that simple assumption. –  kram1032 Feb 2 '13 at 15:05
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