2
$\begingroup$

I am calculating the following integral $$I=\int_0^1 e^{-c_1x^2+c_2x}\mathrm{erf}(c_3x+c_4)dx,$$ with Integrate and NIntegrate respectively.

There is no analytical solution for this integral and I got two different numerical results, with quite large difference: $-0.117035 + 0.5i$.

My question is, which one is more accurate in this case.

Here is my code:

c1 = Pi;
c2 = SetPrecision[10.1 + 10.1 I, 30];
c3 = Sqrt[Pi];
c4 = SetPrecision[1.1 + 6.1 I, 30];
I1 = NIntegrate[Exp[-c1 x^2 + c2 x] Erf[c3 x + c4], {x, 0, 1}, 
   PrecisionGoal -> 20, WorkingPrecision -> 30];
I2 = Integrate[Exp[-c1 x^2 + c2 x] Erf[c3 x + c4], {x, 0, 1}];
error = N[I1 - I2]
(*error=-0.117035 + 0.5i*)

Thanks in advance!!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

Better define the parameters with infinite precision. Then you can use ultrahigh WorkingPrecision.

c1 = Pi;
c2 = (101 + 101 I)/10;
c3 = Sqrt[Pi];
c4 = (11 + 61 I)/10;

I3 = NIntegrate[Exp[-c1 x^2 + c2 x] Erf[c3 x + c4], {x, 0, 1}, 
         WorkingPrecision -> 100]

(*   5.98422153733426875937314115532253471792710605523929063484400170963328\
        0746439954199937370126209185221*10^10 - 
     4.7006274392441169485465943452912663047232799365447530746425809872967\
        02374014674651258602665889195814*10^13 I   *)
Improve this answer
$\endgroup$
3
  • $\begingroup$ @Akkr14: thanks for the answer^_^! After Rationalize the parameters, I recalculated $I_1$, $I_2$ and $I_3$. It turned out that $I_1$ and $I_3$ via NIntegrate are almost the same, but $I_2$ is still much different than them. Does that mean the result obtained by Integrate is not trustable in this case? $\endgroup$
    – Ellery
    Commented Feb 28, 2020 at 9:52
  • 1
    $\begingroup$ Remove the semicolon after Integrate... to see, that it gives back the input. It does not find an analytical solution, most likely the is no one. But when you apply N[I2] the integral is automaticlly calculated numericaly with NIntegrate, but with standard machine precision, which is 16. This is of course less accurate than with WorkingPrecission->30 . $\endgroup$
    – Akku14
    Commented Feb 28, 2020 at 14:33
  • $\begingroup$ @Akkr14: thanks a lot! it is very clear now. $\endgroup$
    – Ellery
    Commented Feb 29, 2020 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.