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I am interested in evaluating the integral:

$$\small\iint\limits_{[0,2\pi) \times [0,2\pi)} (x^2 + y^2)^{-1/2}((b-x)^2 + (c-y)^2)^{-1/2} J_{1}\left(\rho \sqrt{x^2 + y^2}\right)J_{1}\left(\rho \sqrt{(b-x)^2 + (c - y)^2}\right)\mathrm{d}x\mathrm{d}y,$$

where $b, c, \rho > 0$ are constants, and $J_{\nu}$ denotes the Bessel function of the first kind.

I've tried integrating this with Mathematica, using an input such as:

Integrate[((x^2 + y^2)^(-1/2))*(((b-x)^2 + (c-y)^2)^(-1/2))*BesselJ[1, k*Sqrt[x^2 + y^2]]*
BesselJ[1, k*Sqrt[(b - x)^2 + (y - c)^2]], {x, 0, 2*Pi}, {y, 0, 
2*Pi}]

but this just returns the input. We can bound the integral using the bound $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2},$ but I can't seem to be able to compute this either. Can anyone help me evaluate this integral?

user363087
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  • Usually, when Mathematica returns the Input to you seemingly unaltered, that means that there has been an error in the way you provide the input. The first step should be to carefully check the syntax and details in the documentation. – TransferOrbit Oct 09 '16 at 11:59
  • Sorry, I've fixed that -- it's supposed to be $J_{1}$. – user363087 Oct 09 '16 at 12:09
  • I've tried a little bit and it seems, that Mathematica can't compute this Integral (I've tried Simplifying, Assumptions and even Rubi). But NIntegrate works like a charm for given parameters. – Julien Kluge Oct 09 '16 at 12:18
  • @JulienKluge Could you show me which parameters you used to get a result using NIntegrate? – user363087 Oct 09 '16 at 12:29
  • Well, i think nearly every value for $k,b,c$ should work. – Julien Kluge Oct 09 '16 at 12:36
  • OK, I'll ask it on MathSE. Roughly speaking, this integral should be solvable (it can be shown that taking a sum of these integrals over $b,c \in \mathbb{N}$ yields a result which tells us about the distribution of lattice points in a circle). – user363087 Oct 09 '16 at 12:46
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    @user363087 Generally, for Mathematica it's harder to solve an integral with parameters than one with explicit values inserted. It's easier to solve an integral with parameters if you provide explicit Assumptions about the parameters (look it up in the docs). Definite and indefinite versions are different: it may be able to solve one but not the other. If I see something of this complexity with this many parameters, I am usually not very hopeful. – Szabolcs Oct 09 '16 at 12:55
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    two general suggestions for this sort of thing, 1) see if you get a solution with any ( exact ) values for the parameters 2) see if you get a result for just the inner most 1-d integral. As has been said, I'm not optimistic. rho can be eliminated by a simple change of variable by the way but i doubt that will help. – george Oct 09 '16 at 14:01
  • Are you sure that this is the integral you want to compute? The integration range looks odd to me. – mikado Oct 09 '16 at 14:35

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