I am trying to compute this integral for different a with 0.0001<a<0.5,

 int[a_?NumericQ, r_?NumericQ]:=NIntegrate[Exp[-a/Sqrt[1-x^2]]/(x^2-r^2)^2, {x, 0, r, 1},Method ->"PrincipalValue"]

where 0<r<1. For high a, the calculation don't display errors,

int[100, 0.8]
(*  1.1619041*10^-44  *)

But for a={0.1,0.01,0.0001} I get this message error.

int[0.1, 0.8]

NIntegrate::inumri: The integrand E^(-(0.1/Sqrt[1-Power[<<2>>]]))/(-0.64+(0.8 -x)^2)^2+E^(-(0.1/Sqrt[1-Power[<<2>>]]))/(-0.64+(0.8 +x)^2)^2 has evaluated to Overflow, Indeterminate, or Infinity for all sampling points in the region with boundaries {{0.,2.7247563*10^-30}}.

I still can't control the accuracy of the calculation. I proceeded by increasing WorkingPrecision, PrecisionGoal, MaxRecursion but it gives each time different values, I do not even know if it gives the true value.

Please, how to increase the precision of this computation?

  • $\begingroup$ Maybe this could help: mathematica.stackexchange.com/questions/16254/… $\endgroup$ – Ofek Peretz Oct 3 '18 at 8:18
  • $\begingroup$ I already tried with Method ->"PrincipalValue" included in my code, thank you @Ofek. $\endgroup$ – Gallagher Oct 3 '18 at 8:30
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    $\begingroup$ For a near 0, you are effectively trying to evaluate the integral of $1/(x-r)^2$. No wonder it goes south. $\endgroup$ – J. M.'s ennui Oct 3 '18 at 9:04
  • $\begingroup$ Exectute this code:Integrate[ Exp[-a/Sqrt[1 - x^2]]/(x^2 - r^2)^2 /. r -> 1/2 /. a -> 1/10, {x, 0, 1}] then you see the answer ? $\endgroup$ – Mariusz Iwaniuk Oct 3 '18 at 19:24
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    $\begingroup$ "not the correct value" - do you have a reference for this integral, and why you expect the answer to be finite? $\endgroup$ – J. M.'s ennui Oct 4 '18 at 2:55

Looking at the lowest order terms of the Laurant series of the integrand reveals that there are two issues:

Series[Exp[-a/Sqrt[1 - x^2]]/(x^2 - r^2)^2, {x, r, -1}] // Simplify

$$\frac{e^{-\frac{a}{\sqrt{1-r^2}}}}{4 r^2 (x-r)^2}-\frac{e^{-\frac{a}{\sqrt{1-r^2}}} \left(a r^2+\left(1-r^2\right)^{3/2}\right)}{4 \left(r^3 \left(1-r^2\right)^{3/2}\right) (x-r)}+O\left((x-r)^0\right)$$

The first issue is that the $(x-r)^{-2}$ term is not integrable, even not int he principle values sense: It is even so that terms on the left and on the right of the singularity at $x=r$ cannot cancel. The second issue is that term $(x-r)^{-1}$ term is integrable in the principal values sense but that the integration domain lies entirely to the left of the singularity $x = r$; so cancellation also cannot happen.

The only reason why int[100, 0.8] the terms $e^{-\frac{a}{\sqrt{1-r^2}}}$ are numerically so dominant: At $x = r$, you get numerically $\frac{0}{0}$ and at the first quadrature point to the left of $x =r$ you get numerically $0$ for the integrand. That might be the reason why NIntegrate decided to evaluate to 0. It could also be that NIntegrate uses a quadrature rule in this case that does not use interval boundaries as quadrature points (because you warned it that there might be a singularity).

  • $\begingroup$ Thank you @Schumacher, so this integral never converges? $\endgroup$ – Gallagher Oct 3 '18 at 19:09
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    $\begingroup$ That's what I tried to explain, yes. $\endgroup$ – Henrik Schumacher Oct 3 '18 at 19:10
  • $\begingroup$ Serious problem! $\endgroup$ – Gallagher Oct 3 '18 at 19:15
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    $\begingroup$ lol. Yes. Fortunately, not mine ;) $\endgroup$ – Henrik Schumacher Oct 3 '18 at 19:17
  • $\begingroup$ Thank you @Schumacher. ;) $\endgroup$ – Gallagher Oct 3 '18 at 19:20

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