
Re: Complex path integral wrong
Posted:
Jul 1, 2013 5:45 AM


A followup on my earlier reply.
If I compare the analytical and numerical results from Mathematica, the discrepancy occurs on the first leg of the integration (1+I > 1+I R). It is clear that the numerical integration is correct.
Could this be some kind of branch cut issue with the hypergeometric function evaluation?
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 (64bit) (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 >

