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I'm trying to obtain this 7- dimensional integral:

NIntegrate[(8*Pi)^4/(2*Pi)^9*4*Pi*Sin[thetar]*Sin[thetaw]*w^2*q^2*
  r^2*(ArcTan[
     Sqrt[r^2 - 2*r*q*Cos[thetar] + q^2]/2]/(4*Pi*
      Sqrt[r^2 - 2*r*q*Cos[thetar] + q^2]))*1/((w^2 + 1)^3*(r^2 + 
       1) (1 + q^2)*(r^2 - 2*r*q*Cos[thetar] + q^2 + 1) (w^2 + 
       2*w*q*Cos[thetaw] + q^2 + 1)*(w^2 + 
       2*w*r (Cos[thetar]*Cos[thetaw] + 
          Sin[thetar]*Sin[thetaw]*Cos[phir]*Cos[phiw] + 
          Sin[thetar]*Sin[thetaw]*Sin[phir]*Sin[phiw]) + r^2 + 
       1)), {phir, 0, 2*Pi}, {phiw, 0, 2*Pi}, {thetar, 0, 
  Pi}, {thetaw, 0, Pi}, {w, 0, Infinity}, {q, 0, Infinity}, {r, 0, 
  Infinity}, 
  Method -> {"GlobalAdaptive", Method -> "CartesianRule"}, 
  PrecisionGoal -> 12]

to within 10^-10 or better.

But with such a method, unfortunately, mathematica cannot cope (I waited for 9 days), at least at my weak pc (3.2 GHz, 4Gb), although mathematica calculates 3-dim similar integral up to 10^-17 no longer than 10 minutes.

I've read the article "Advanced Numerical Calculations", but I didn't find solutions of my problem, all MC methods give a small precision.

Who has any ideas? I will be the happiest person in the world!

I beg for help)!

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  • 1
    $\begingroup$ This isn't really a Mathematica problem ... the time required to compute integrals grows exponentially with the dimension. Perhaps you can look for ways to reduce the dimensionality of the problem. $\endgroup$ – Szabolcs Sep 5 '17 at 10:22
  • $\begingroup$ Honestly, dimensionality here was already reduced from 12. But in anyway, thanks for your comment. I wanted to ask about other methods which can give a enough high accuracy. $\endgroup$ – Andrew Kudlis Sep 5 '17 at 10:34

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