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I have the next integral, and I would like to know what is the best way to define it in mathematica:

$\int_{0}^{\infty}{du}\int_{0}^{\infty}{dv} \frac{u}{u^2+v^2}(\int_{0}^{0.8}\int_{0}^{\infty} xe^{-2\sqrt{x^2+y^2}} J_{0}(ux)[cos(\pi y/0.8)]^2 cos(yv) {dy}{dx})^2$

Because I am a beginnner, my first idea was touse the next code:

   NIntegrate[ (u/(u^2 + v^2))*r*x*  Exp[-2*0.2*(Sqrt[r^2 + z^2] + Sqrt[x^2 + y^2])]*BesselJ[0, u*r]*  BesselJ[0, u*x]*Cos[v*z]*Cos[y*v]*(Cos[Pi*z/1.6]*Cos[Pi*y/1.6])^2, {u, 0, Infinity}, {v, 0,Infinity}, {r, 0, Infinity}, {x, 0, Infinity}, {z, 0, 0.8}, {y, 0, 0.8}]

But I get warnings about the oscillatory behavior of the integrand.

What would be another way to define the integral in order to gain time and not warnings

I find another way to calculate it, using the Adative MonteCarlo strategy, but each time that I evaluate he cell, I got a different result. What is the origin of this problem?

Thank you for all your help

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    $\begingroup$ where did the r and y come from? $\endgroup$
    – george2079
    Mar 12, 2018 at 18:28
  • $\begingroup$ instead of the square of the integral, in the code I put the the product of two integrals, one using r and z and the other with x and y. $\endgroup$
    – rafa
    Mar 13, 2018 at 6:47
  • $\begingroup$ that is not a valid transformation $\endgroup$
    – george2079
    Mar 13, 2018 at 11:15
  • $\begingroup$ But I am applying the Fubini theorem $\endgroup$
    – rafa
    Mar 13, 2018 at 11:57
  • $\begingroup$ so edit the question to show exactly what you are doing and what you are actually trying to accomplish. $\endgroup$
    – george2079
    Mar 13, 2018 at 13:43

1 Answer 1

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      Quiet[NIntegrate[(u/(u^2 + v^2))*r*x*
      Exp[-2*0.2*(Sqrt[r^2 + z^2] + Sqrt[x^2 + y^2])]*BesselJ[0, u*r]*
     BesselJ[0, u*x]*Cos[v*z]*
     Cos[y*v]*(Cos[Pi*z/1.6]*Cos[Pi*y/1.6])^2, {u, 0, Infinity}, {v, 0, 
     Infinity}, {r, 0, Infinity}, {x, 0, Infinity}, {z, 0, 0.8}, {y, 0, 
     0.8}, PrecisionGoal -> 4]]
  (*   output*)
    (*2.12864*)
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