
Re: cauchy principal value double integral
Posted:
Dec 6, 2012 12:09 AM


Since the integrand is a positive quantity, obviously the integral has to be positive. Principal value has nothing to do with this integral since there is not the cancellation effect that requires a change in sign of the integer and. Mathematica 5.1 answers (correctly) with the non convergence of integra l. Also Mathematica8 gives the same answer, with the option Principal ValueFalse.
I have divided the integration square in eight triangles by the two diagonals and consider the first one in the first quadrant: (0<x<1}&&(y,0,x) And cut away a small e0 part of x near zero. The integral then becomes
R[a,e0]=8*Integrate[Integrate[1/(x^2 + y^2)^a, {y, 0, x}], {x, e0, 1}]
and performing the integrals (first the internal) is immediate to find that
R[a_,e0_]:=4 F[a](1e0^(2(1a))/(1a)
where
F[a_]:= Hypergeometric2F1[1/2, a, 3/2, 1)
This last term results to be always positive 0<F[a]<1. So convergence/diverence of integral is determined by the other factor, i.e. convergence for e0 >0 requires that a<1; For instance a=1/2 R[1/2,0]=8 ArcSAinh[1]
Cheers, Rob
Messaggio originale Da: daniel.lichtblau0@gmail.com [mailto:daniel.lichtblau0@gmail.com] Inviato: marted=EC 4 dicembre 2012 10.10 A: mathgroup@smc.vnet.net Oggetto: Re: cauchy principal value double integral
On Monday, December 3, 2012 2:18:53 AM UTC6, Alex Krasnov wrote: > I am unclear how to interpret the following result (Mathematica 8.0.4): > > > > In: Integrate[1/(x^2+y^2), {x, 1, 1}, {y, 1, 1}, PrincipalValue > True] > > Out: 4*Catalan > > > > How is Cauchy principal value defined in this case? Since the integrand is > > circularly symmetric around (0,0), excluding a shrinking neighborhood > > around (0,0) is not useful. Perhaps this result is merely an issue with > > option handling for multiple integrals? > > > > Alex
That does indeed look like a bug.
Daniel Lichtblau Wolfram Research
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