# Integrating over Bessel Function erroreous? (Hankel Transform)

Bug introduced in 8.0.4 or earlier and persists through 10.2.

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 noticeably 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?

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

-

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

Why not Integrate[BesselJ[1, t] BesselJ[0, x t], {t, 0, Infinity}, Assumptions -> {x > 0}]? – m_goldberg Feb 2 '13 at 14:45
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