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Topic: Complex path integral wrong
Replies: 7   Last Post: Jul 3, 2013 4:50 AM

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Kevin J. McCann

Posts: 146
Registered: 12/7/04
Re: Complex path integral wrong
Posted: Jul 1, 2013 5:44 AM
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If you change Integrate to NIntegrate, you get the answer you expect
from the residue. So, this is a bug.

Kevin


On 6/30/2013 3:26 AM, Dr. Wolfgang Hintze wrote:
> I suspect this is a bug
> In[361]:= $Version
> Out[361]= "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[351]= 0
>
> It should have the value
>
> In[356]:= (2 \[Pi] I) Residue[Exp[I s]/(Exp[s] - 1 ), {s, 2 \[Pi] I}]
>
> Out[356]= (2 \[Pi] I) E^(-2 \[Pi])
>
> Without applying FullSimplify the result of the integration is
>
> In[357]:= R = 3*Pi;
> Integrate[
> Exp[I*s]/(Exp[s] - 1), {s, 1 + I, 1 + I*R, -1 + I*R, -1 + I, 1 + I}]
>
> Out[358]=
> 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[359]:= N[%]
>
> Out[359]= -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
>





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