Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2013

[Date Index] [Thread Index] [Author Index]

Search the Archive

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



  • Prev by Date: Re: Rather simple function returns curious result. Explanation
  • Next by Date: Animate Command and Vector Reference
  • Previous by thread: Re: Complex path integral wrong
  • Next by thread: Re: Complex path integral wrong