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Im trying to integrate the following:

\begin{equation} \int_{-1}^{1}\int_{-1}^{1}\int_{-1}^{x} J_{3/4}(x^2 z)J_{1/4}(y^2z) dy dz dx \end{equation}

So what I did was this

i1[x_?NumericQ, z_?NumericQ] := 
 i1[x, z] = NIntegrate[BesselJ[1/4, y^2 z], {y, -1, x}]
i2[x_?NumericQ] := 
 i2[x] = NIntegrate[BesselJ[3/4, x^2 z] i1[x, z], {z, -1, 1}]
NIntegrate[i2[x], {x, -1, 1}]

Is this correct? And if so, is there any other way to make this faster?

Thanks in advance.

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  • $\begingroup$ Have you tried it as a multiple integral? How precise do you want the answer? $\endgroup$ – Michael E2 Sep 8 '16 at 11:39
  • $\begingroup$ I haven't tried it. However, my post is just something that looks similar to my original problem and my original problem is something that needs numerical integration. I just want a result that does not show this: Numerical integration converging too slowly; suspect one of the \ following: singularity, value of the integration is 0, highly \ oscillatory integrand, or WorkingPrecision too small. $\endgroup$ – PhilCsar Sep 8 '16 at 11:42
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I believe the integral is 0 exactly, by symmetry (z -> -z):

FullSimplify[
 BesselJ[3/4, x^2 z] BesselJ[1/4, y^2 z] + 
   BesselJ[3/4, x^2 (-z)] BesselJ[1/4, y^2 (-z)],
 {x, y, z} ∈ Reals]
(*  0  *)

Numerical test: Set it up so that the "EvenOddSubdivision" strategy may be applied to the z-integral:

NIntegrate[BesselJ[3/4, x^2 z] BesselJ[1/4, y^2 z],
 {x, -1, 1}, {y, -1, x}, {z, -1, 1}, 
 Method -> {"EvenOddSubdivision", 
   Method -> {"LobattoKronrodRule", "GaussPoints" -> 5}}, 
 MaxRecursion -> 0, AccuracyGoal -> 16]
(*  1.9636*10^-18  *)

(You need to set a finite AccuracyGoal to avoid a convergence warning.)

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  • $\begingroup$ Thanks. But is the code right? $\endgroup$ – PhilCsar Sep 8 '16 at 11:59
  • $\begingroup$ @PhilCsar Do you mean did Mathematica make a mistake? Or are you suggesting I messed up the cut & paste of the integrand? -- The interval -1 <= z <= 1 is symmetric with respect to the reflection z -> -z, so the integral over {z, -1, 0} cancels the integral over {z, 0, 1}....if there's no error in my copying. It agrees with numerical tests, up to the error in numerical integration which is somewhat large (<0.001). $\endgroup$ – Michael E2 Sep 8 '16 at 12:04
  • $\begingroup$ Im just asking if my code is correct because Im not sure if I used NumericQ correctly :D $\endgroup$ – PhilCsar Sep 8 '16 at 12:07
  • $\begingroup$ @PhilCsar It seems right, but it will be slow. $\endgroup$ – Michael E2 Sep 8 '16 at 12:09
  • $\begingroup$ @PhilCsar If an integral is zero, NIntegrate will always give a convergence warning unless you set a finite AccuracyGoal to however many digits you want verified are zero. $\endgroup$ – Michael E2 Sep 8 '16 at 12:20

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