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Re: Contour integral around z_infinity != negative

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106601] Re: [mg106586] Contour integral around z_infinity != negative
  • From: danl at wolfram.com
  • Date: Mon, 18 Jan 2010 02:34:10 -0500 (EST)
  • References: <201001171211.HAA09234@smc.vnet.net>

> If I recall grad school correctly, the contour integral of f around
> z_infinity is not simply the negative of the contour integral of f
> around the origin. I never understood why not. I don't think I could
> pose the question to Mathematica properly, anytime soon. So I thought
> I'd bring it to this list. Would someone formulate these two contour
> integrals as Mathematica expressions, and attempt to draw an equality
> between the two? I'd be curious to see how Mathematica takes it.
>
> If this is nonsense, just leave it in wonderland.
>
> Thanks,
>
> Vince Virgilio

Pretty much by definition they are the same, if you have in mind the same
contour in each case.

What is not the same is a residue computation. For the origin you would
need a contour small enough that it only encloses the pole (if there is
one) at the origin (and of course itself hits no poles). For the residue
at infinity, one requires a contour sufficiently  "close" to infinity,
meaning it must "enclose" all finite poles on the outside of said contour.

In practice the residue at infinity is often computed by transforming
infinity to the origin, via z-->1/w, and then using a small contour. This
gives a factor of -1/w^2 in the integral. After sorting through the
algebra, I think it comes out as contourAtInfinityI below, where, as with
contourI (for the origin case), the distance r is meant to be small.

contourI[f_, r_] :=
 NIntegrate[(f[z] /. z -> r*Exp[I*t])*(r*I*Exp[I*t]), {t, 0, 2*Pi}]

contourAtInfinityI[f_,
  r_] := -NIntegrate[((f[z]*z^2 /. z -> 1/w) /. w -> r*Exp[I*t])*(r*I*
      Exp[I*t]), {t, 0, 2*Pi}]

f[z_] := 1/z + 1/(z - 1) + 1/(z + 1/4) + z^3 + 1/(z - 1/2)^2

In[110]:= contourI[f, 20.]
Out[110]= 1.45519*10^-11 + 18.8496 I

In[111]:= contourI[f, .1]
Out[111]= 0. + 6.28319 I

In[112]:= contourAtInfinityI[f, 1/20.]
Out[112]= -7.27596*10^-12 - 18.8496 I

Daniel Lichtblau
Wolfram Research






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