This is a topic that has been discussed before, but after reading other posts I could not find a solution that applies for my specific case.

I have the following integral, which I need to evalutate at $n=1,2,3,4...$ and $a=0.1$:

$$ I(n)=\int_{-\pi}^{\pi} {\rm{d}}x \frac{\Gamma(n,0,2ix)}{(1+iax)^n}.$$

The problem is that Integrate and NIntegrate seem to yield different results for sufficiently large $n$. Numerical integration with $a=0.1$ leads to

a = 1/10;
Table[NIntegrate[Gamma[n, 0, 2 I x]/(1 + I a x)^n, {x, -\[Pi], \[Pi]}], {n, 1,6}] // Chop
{5.82505, 9.23019, 42.4493, 117.67, -303.323, -4233.64}

However, integrating first and then substituting $a=0.1$ returns the following values:

T = Table[Integrate[Gamma[n, 0, 2 I x]/(1 + I a x])^n, {x, -\[Pi], \[Pi]}, Assumptions->a>0], 
{n, 1, 6}];
{5.82505, 9.22998, 41.7814, 0.951343, -26457.8, -3.06652*10^6}

which only agree with numerical integration for $n=1$ and then they become very different. Interestingly, using Integrate with $a=0.1$ yields another different set of values:

a = 1/10;
TT = Table[Integrate[Gamma[n, 0, 2 I x]/(1 + I a x)^n, {x, -\[Pi], \[Pi]}, Assumptions -> a > 0], {n, 1, 6}];
TT // N // Chop
{5.82505, 9.23058, 42.4145, 164.712 + 1.19844*10^-9 I, 0, 701973.}

All these evaluations occur without mathematica returning any errors. What is happening here? Which values are the correct ones?

Things that I know/ tried:

  1. The integral is real, as the imaginary part of the integrand is an odd function of x.
  2. Assumptions->a>0 is necessary, since for arbitrary $a$ the default solution mathematica returns is valid in the complex plane of $a$ minus the positive real axis.
  3. From external arguments, the values that I am most inclined to trust are those from exact integration for arbitrary a>0 and then numerical evaluation at a=0.1 (the second set of numbers).
  4. The integrand oscillates more and more the higher $n$ is, and I believe NIntegrate may have problems getting the right answer for highly oscillatory integrands. However, since it does not return any error I do not know if this is the issue.
  5. The problem persists if I use GammaRegularized(n,0,2ix) to work with smaller numbers.
  6. The problem persists if I increase numerical precision (I am not very familiar with how numerical precision should be managed, maybe I am doing it the wrong way: I increase $MaxExtraPrecision and WorkingPrecision in NIntegrate)

I've read similar posts in which integration methods for NIntegrate such as ExtrapolatingOscillatory and DoubleExponentialOscillatory where suggested, but as far as I know those only work for integrals with infinite range.

  • 1
    $\begingroup$ Try Chop[N[N[T,100]]]. $\endgroup$
    – user293787
    Oct 3, 2022 at 8:43
  • $\begingroup$ @user293787 That transforms the symbolic integration values into those obtained by numerical integration. How can Chop transform something of order $10^6$ into $4233$? $\endgroup$ Oct 3, 2022 at 8:51
  • $\begingroup$ @user293787 I do not see why Chop has any effect. Screen shot !Mathematica graphics it is clear the results are different. $\endgroup$
    – Nasser
    Oct 3, 2022 at 8:55
  • $\begingroup$ I understand now, this is what happens: the symbolic integration expression for lets say $n=6$ is very, very long. When I evaluated it at $a=0.1$ and used N, only 6 digits of precision were kept in each of the many terms and the total evaluation was wrong. With @user293787 command, 100 digits are kept in the evaluation, which drastically changes the result. Numerical integration was right all along. $\endgroup$ Oct 3, 2022 at 8:59
  • 1
    $\begingroup$ The Chop[N[...]] is not important, that was just to produce a compact output. The main thing is N[T,100]. It seems that there are considerable cancellations among terms that lead to large errors with machine precision N. You could look at N[N[List@@@Expand[T],100]] to see the cancellations. $\endgroup$
    – user293787
    Oct 3, 2022 at 9:01

1 Answer 1


The problem is not Integrate, but the numerical evaluation following Integrate.

Consider for example T[[4]]:

(1/(3 (100 + Pi^2)^3))80000 (48003000 Pi - 
   10520000000 E^20 Pi + 998810 Pi^3 - 315600000 E^20 Pi^3 + 
   5257 Pi^5 - 3156000 E^20 Pi^5 - 10520 E^20 Pi^7 + 
   5260000000 I E^20 ExpIntegralEi[-20 - 2 I Pi] + 
   157800000 I E^20 Pi^2 ExpIntegralEi[-20 - 2 I Pi] + 
   1578000 I E^20 Pi^4 ExpIntegralEi[-20 - 2 I Pi] + 
   5260 I E^20 Pi^6 ExpIntegralEi[-20 - 2 I Pi] - 
   5260 I E^20 (100 + Pi^2)^3 ExpIntegralEi[-20 + 2 I Pi])

When one evaluates this numerically, the result depends strongly on how the expression is organized and on what precision is used:

(* 164.712 +1.19844*10^-9 I *)

(* 256. +0. I *)

(* 117.670138197013970506+0.*10^-19 I *)

The reason is that individual summands are many orders of magnitude larger than the final result:

1.61199*10^17-4.62852*10^6 I
4.7729*10^16-1.37045*10^6 I
4.71066*10^15-135258. I
1.54975*10^14-4449.81 I
1.61199*10^17+4.62852*10^6 I
4.7729*10^16+1.37045*10^6 I
4.71066*10^15+135258. I
1.54975*10^14+4449.81 I

Therefore, higher than machine precision is needed to obtain the correct value, which in this case is the one near 117.67.


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