Re: Complex path integral wrong
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- Subject: [mg131381] Re: Complex path integral wrong
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Tue, 2 Jul 2013 00:47:42 -0400 (EDT)
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Am Montag, 1. Juli 2013 11:40:25 UTC+2 schrieb Murray Eisenberg:
> I don't think this is a bug -- at least with the currently implemented
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> and documented Mathematica functionality.
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>
>
> If in the second argument of Integrate one uses something of the form
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>
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> {var, val1, val2, . . ., valn}
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>
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> where n > 2, evidently all Mathematica does is separately calculate the
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> integral along each segment -- finding an indefinite integral on the
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> segment and evaluating at endpoints. And that has nothing to do, really,
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> with evaluating a contour integral around a closed curve -- at least
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> such a closed curve that includes singularities.
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>
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> If you look up Integrate in the Documentation Center, I don't think
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> you'll even find mention of, or example of, this extended kind of 2nd
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> argument. So while Mathematica lets one use this extended form, it's
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> obviously still dangerous to do so.
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>
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> And so for a contour integral, one must resort to the Residue Theorem,
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> as you have done, when the function is not holomorphic on the complex
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> domain.
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> On Jun 30, 2013, at 3:29 AM, Dr. Wolfgang Hintze <weh at snafu.de> wrote:
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>
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> > I suspect this is a bug
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> > In[361]:= $Version
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> > Out[361]= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)"
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> >
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> > The follwing path integral comes out wrong:
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> >
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> > R = 3 \[Pi] ;
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> > Integrate[Exp[I s]/(
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> > Exp[s] - 1 ), {s, 1 + I, 1 + I R, -1 + I R, -1 + I, 1 + I}] // FullSimplify
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> >
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> > Out[351]= 0
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> >
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> > It should have the value
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> >
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> > In[356]:= (2 \[Pi] I) Residue[Exp[I s]/(Exp[s] - 1 ), {s, 2 \[Pi] I}]
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> >
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> > Out[356]= (2 \[Pi] I) E^(-2 \[Pi])
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> >
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> > Without applying FullSimplify the result of the integration is
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> >
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> > In[357]:= R = 3*Pi;
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> > Integrate[
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> > Exp[I*s]/(Exp[s] - 1), {s, 1 + I, 1 + I*R, -1 + I*R, -1 + I, 1 + I}]
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> >
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> > Out[358]=
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> > I*E^((-1 - I) - 3*Pi)*((-E)*Hypergeometric2F1[I, 1, 1 + I, -(1/E)] +
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> > E^(3*Pi)*Hypergeometric2F1[I, 1, 1 + I, E^(-1 + I)]) +
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> > I*E^(-I - 3*Pi)*(Hypergeometric2F1[I, 1, 1 + I, -(1/E)] -
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> > E^(2*I)*Hypergeometric2F1[I, 1, 1 + I, -E]) +
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> > I*E^I*(Hypergeometric2F1[I, 1, 1 + I, -E]/E^(3*Pi) -
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> > Hypergeometric2F1[I, 1, 1 + I, E^(1 + I)]/E) +
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> > I*E^(-1 - I)*(-Hypergeometric2F1[I, 1, 1 + I, E^(-1 + I)] +
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> > E^(2*I)*Hypergeometric2F1[I, 1, 1 + I, E^(1 + I)])
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> >
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> > which, numerically, is
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> >
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> > In[359]:= N[%]
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> >
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> > Out[359]= -2.7755575615628914*^-17 + 2.7755575615628914*^-17*I
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> >
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> > i.e. zero.
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> >
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> > On simpler functions like 1, s and s^2 (instead of Exp[I s]) it works out fine, but not so with e.g. Sin[s] in which case we get 0 again (instead of Sinh[2 \[Pi]]).
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> >
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> > The integration topic seems to be full of pitfalls in Mathematica...
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> >
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> > Best regards,
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> > Wolfgang
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>
>
> ---
>
> Murray Eisenberg murray at math.umass.edu
>
> Mathematics & Statistics Dept.
>
> Lederle Graduate Research Tower phone 413 549-1020 (H)
>
> University of Massachusetts 413 545-2838 (W)
>
> 710 North Pleasant Street fax 413 545-1801
>
> Amherst, MA 01003-9305
Murray,
strictly speaking you are right.
So thank you for pointing out that "contour integral" is not a documented functionalty (I couldn't find it in the documentations center). But I was surprised because I have been using this funcionality for many years (starting in version 5.2) without problems.
Also there are nice examples in Wolfram Demonstrations Projects, e.g. http://demonstrations.wolfram.com/ContourIntegration/.
Other relevant links can be found in Google, including contributions from Paul Abbott and Michael Trott.
So I would say it is at least misleading if a function is available and even has intuitive syntax but is not really meant to work correctly.
As I mentioned it seems to work fine for many integrands (e.g. for s^k/(E^s-1), k=0,1,2,...) but there are exceptions (e.g. Sin[s]/(E^s-1)).
Regards,
Wolfgang