Re: Complex path integral wrong

*To*: mathgroup at smc.vnet.net*Subject*: [mg131353] Re: Complex path integral wrong*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Mon, 1 Jul 2013 05:46:58 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130630072902.1229C69F4@smc.vnet.net>

I don't think this is a bug -- at least with the currently implemented and documented Mathematica functionality. If in the second argument of Integrate one uses something of the form {var, val1, val2, . . ., valn} where n > 2, evidently all Mathematica does is separately calculate the integral along each segment -- finding an indefinite integral on the segment and evaluating at endpoints. And that has nothing to do, really, with evaluating a contour integral around a closed curve -- at least such a closed curve that includes singularities. If you look up Integrate in the Documentation Center, I don't think you'll even find mention of, or example of, this extended kind of 2nd argument. So while Mathematica lets one use this extended form, it's obviously still dangerous to do so. And so for a contour integral, one must resort to the Residue Theorem, as you have done, when the function is not holomorphic on the complex domain. On Jun 30, 2013, at 3:29 AM, Dr. Wolfgang Hintze <weh at snafu.de> 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 --- 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