Integrating over Bessel Function erroneous? (Hankel Transform)

Question

Bug introduced in 8.0.4 or earlier and persists through 11.0.1 or later

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?

Integrate[BesselJ[1, t]/t t BesselJ[0, x t], {t, 0, Infinity}] in 11.3, returns this int doesn't coverage, and returns back origin formula. — AsukaMinato, Commented Feb 7, 2021 at 19:33

b.gates.you.know.what · Accepted Answer · 2013-02-02 13:50:30Z

3

Try :

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

answered Feb 2, 2013 at 13:50

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$\begingroup$ Why not Integrate[BesselJ[1, t] BesselJ[0, x t], {t, 0, Infinity}, Assumptions -> {x > 0}]? $\endgroup$
– m_goldberg
Commented Feb 2, 2013 at 14:45
$\begingroup$ That apparently works. Weird. I think I previously had a case where it didn't. $\endgroup$
– kram1032
Commented Feb 2, 2013 at 14:55
$\begingroup$ @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. $\endgroup$
– kram1032
Commented Feb 2, 2013 at 14:56
$\begingroup$ 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. $\endgroup$
– kram1032
Commented Feb 2, 2013 at 15:05

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