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I am trying to do this integral numerically:

NIntegrate[(1/(x^2 + y^2 + z^2)^(1/2) - 1/((x - 1)^2 + y^2 + z^2)^(
   1/2))^2, {x, -Infinity, Infinity}, {y, -Infinity, 
  Infinity}, {z, -Infinity, 0}]

But for different methods in NIntegrate, I got different results:

GlobalAdaptive: 6.28318

MonteCarlo: 5.66018

AdaptiveMonteCarlo: 6.01981

QuasiMonteCarlo: 5.15901

AdaptiveQuasiMonteCarlo: 4.59813

I am not sure which one is more accurate. Any help will be appreciated!

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  • 1
    $\begingroup$ The value given by GlobalAdaptive is very close to 2 pi. It should be more accurate than the other methods, though it does give warnings suggesting that the answer might be doubtful. $\endgroup$
    – mikado
    May 30, 2019 at 21:29

1 Answer 1

2
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Increase the precision and check what happens:

Table[
      NIntegrate[(1/(x^2 + y^2 + z^2)^(1/2) - 1/((x - 1)^2 + y^2 + z^2)^(1/2))^2
      , {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, {z, -Infinity, 0}
      , Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> n 1000}, WorkingPrecision -> 20, MaxRecursion -> 5 n]
, {n, 1, 8}
]
ListLogPlot[Abs[% - 2 π]]

enter image description here

And so, indeed, it appears that the integral converges to $2\pi$.

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  • $\begingroup$ I found the integral interesting in itself and asked in the Mathematica forum how it would be solved analytically. Adam there solved it beautifully: [1]: math.stackexchange.com/questions/3246406/show-integral-is-2-pi $\endgroup$
    – Dominic
    Jun 1, 2019 at 9:17
  • $\begingroup$ @Dominic Cool, thank you for the link :-) $\endgroup$ Jun 1, 2019 at 11:39
  • $\begingroup$ I have solved it analytically too. Thanks for your help! $\endgroup$
    – lol
    Jun 1, 2019 at 19:25

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