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Different results of definite integral over complex exponential.

Posted 10 years ago
4 Replies
5 Total Likes
We are currently using Mathematica 9.0 and came across the problem, that in certain situations, Mathematica will give different results for the same definite integral. Here is a minimal working example
Integrate[Abs[Exp[? ?]]^2,{?,0,2??}]
The first integral gives zero, while the second one holds the correct result 2*pi. Using FullSimplify gives the correct result but is too time consuming for larger expressions.
I would appreciate any help concerning this issue, as these functions are rather often used in our codes here and this really questions the reliability of Integrate and/or Abs.

POSTED BY: Andreas B.
4 Replies
Posted 10 years ago
Great, this seems to solve the issue. With your permission, I would send this to the support, so it gets fixed in the next release.
POSTED BY: Andreas B.
Posted 10 years ago
With that disclaimer out of the way:

I got curious where this happened and Spelunked a bit.
The problem arises in
Integrate`ImproperDump`Dispatcher[f_, {x_, 0, 2 PI}]
where the function is decomposed into a polynomial in Cos and Sin whose coefficients get improperly transformed for complex values. (1^2 + I^2 = 0)

I decided to drop that downvalue from Dispatcher to see what happens:
 (* Make sure definitions are loaded *)
 Integrate[Abs[Cos[x] + I Sin[x]]^2, {x, 0, 2 Pi}]
 dw = DownValues[Integrate`ImproperDump`Dispatcher];
 pos = Position[dw,
         HoldPattern[Integrate`ImproperDump`Dispatcher[f_, {x_, 0, 2 Pi}]]
       ][[1, 1]];
 dw = Drop[dw, {pos}];
 DownValues[Integrate`ImproperDump`Dispatcher] = dw;

Integrate[Abs[Cos[x] + I Sin[x]]^2, {x, 0, 2 Pi}]
(* 2 Pi :) *)

To get back to normal just restart the kernel
POSTED BY: Simon Schmidt
Posted 10 years ago
Scary...  Same thing in v8. It seems to only happen when the region of integration is [0,2Pi]  or [-2Pi, 0]:
  {i Pi, (i + 2) Pi, Integrate[ Abs[ Cos[x] + I Sin[x]]^2, {x, i Pi, (i + 2) Pi}]},
  {i, -100, 100}],
{_, _, Except[2 Pi]}]
(* {{-2 Pi, 0, 0}, {0, 2 Pi, 0}} *)

Perhaps you can change your region of integration as a workaround.
POSTED BY: Simon Schmidt
Posted 10 years ago
Thank you very much for your reply. If the integration region is not exactly [0,2 pi], Mathematica gives the correct result indeed.
But this takes an unreasonably high amount of computation time in comparison to the procedures above. Additionally as this is a really common integration region, it would be interesting to know if there are other cases in which this bug arises.
POSTED BY: Andreas B.
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