Re: Complex path integral wrong

*To*: mathgroup at smc.vnet.net*Subject*: [mg131399] Re: Complex path integral wrong*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Wed, 3 Jul 2013 04:59:26 -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*: <kqomj6$p76$1@smc.vnet.net> <20130701095219.0C6426ABC@smc.vnet.net>

Am Dienstag, 2. Juli 2013 06:30:05 UTC+2 schrieb Murray Eisenberg: > Hmm.. Documentation Center page on Hypergeometric2F1 says: > > > > "Hypergeometric2F1[a, b, c, z] has a branch cut discontinuity > > in the complex z plane running from 1 to Infinity." > > > > Page > > http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/04/02/ > > says: > > > > "[Hypergeometric2F1[a, b, c, z]] is an analytical function of a, b, c, and z > > which is defined on C^4" > > > > Hence for fixed a, b, and c, that implies it is analytic as a function of z on the complex plane. > > > > Which is it? > > > > On Jul 1, 2013, at 5:52 AM, Kevin J. McCann <kjm at kevinmccann.com> wrote: > > > > > 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 (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 A plot reveals that Mathematica's Hypergeometric2F1 in general has is a cut on the real axis from +1 to +oo as it should be: f = Hypergeometric2F1[1/2, 1/2, 1, z]; Plot3D[Re[f /. {z -> x + I y}], {x, 0, 2}, {y, -1, 1}] Plot3D[Im[f /. {z -> x + I y}], {x, 0, 2}, {y, -1, 1}] Best regards, Wolfgang

**Follow-Ups**:**Re: Complex path integral wrong***From:*Murray Eisenberg <murray@math.umass.edu>

**References**:**Re: Complex path integral wrong***From:*"Kevin J. McCann" <kjm@KevinMcCann.com>