       Re: Complex path integral wrong

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• Subject: [mg131385] Re: Complex path integral wrong
• From: danl at wolfram.com
• Date: Tue, 2 Jul 2013 00:49:02 -0400 (EDT)
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```On Sunday, June 30, 2013 2:26:30 AM UTC-5, Dr. Wolfgang Hintze wrote:
> I suspect this is a bug
>
> In:= \$Version
>
> Out= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)"
>
>
>
> The follwing path integral comes out wrong:
>
>
>
> R = 3 \[Pi] ;
>
> Integrate[Exp[I s]/(
>
>   Exp[s] - 1 ), {s, 1 + I, 1 + I R, -1 + I R, -1 + I, 1 + I}] // FullSimplify
>
>
>
> Out= 0
>
>
>
> It should have the value
>
>
>
> In:= (2 \[Pi] I) Residue[Exp[I s]/(Exp[s] - 1 ), {s, 2 \[Pi] I}]
>
>
>
> Out= (2 \[Pi] I) E^(-2 \[Pi])
>
>
>
> Without applying FullSimplify the result of the integration is
>
>
>
> In:= R = 3*Pi;
>
> Integrate[
>
>  Exp[I*s]/(Exp[s] - 1), {s, 1 + I, 1 + I*R, -1 + I*R, -1 + I, 1 + I}]
>
>
>
> Out=
>
> I*E^((-1 - I) - 3*Pi)*((-E)*Hypergeometric2F1[I, 1, 1 + I, -(1/E)] +
>
>     E^(3*Pi)*Hypergeometric2F1[I, 1, 1 + I, E^(-1 + I)]) +
>
>  I*E^(-I - 3*Pi)*(Hypergeometric2F1[I, 1, 1 + I, -(1/E)] -
>
>     E^(2*I)*Hypergeometric2F1[I, 1, 1 + I, -E]) +
>
>    I*E^I*(Hypergeometric2F1[I, 1, 1 + I, -E]/E^(3*Pi) -
>
>     Hypergeometric2F1[I, 1, 1 + I, E^(1 + I)]/E) +
>
>  I*E^(-1 - I)*(-Hypergeometric2F1[I, 1, 1 + I, E^(-1 + I)] +
>
>     E^(2*I)*Hypergeometric2F1[I, 1, 1 + I, E^(1 + I)])
>
>
>
> which, numerically, is
>
>
>
> In:= N[%]
>
>
>
> Out= -2.7755575615628914*^-17 + 2.7755575615628914*^-17*I
>
>
>
> i.e. zero.
>
>
>
> 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]]).
>
>
>
> The integration topic seems to be full of pitfalls in Mathematica...
>
>
>
> Best regards,
>
> Wolfgang

The problem can be isolated to the first segment of the graph. Some further analysis indicates that a path singularity is not found at s=1+2*Pi*I. I looked into this and learned the following.

(1) The Integrate code notes that that is a point of interest for the antiderivative.

(2) Taking Limit from both sides of approach on that path gives the same result.

(3) This in turn is because the expansions from Series agree even when we give an assumption on s.

The upshot is that there is a bug probably in Series that causes the error in this path integral. I will file a bug report about that.

Daniel Lichtblau
Wolfram Research

```

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