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Topic: cauchy principal value double integral
Replies: 5   Last Post: Dec 6, 2012 4:56 AM

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Brambilla Roberto Luigi (RSE)

Posts: 25
Registered: 2/21/12
Re: cauchy principal value double integral
Posted: Dec 6, 2012 12:09 AM
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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 Value-False.

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](1-e0^(2(1-a))/(1-a)


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: []
Inviato: marted=EC 4 dicembre 2012 10.10
Oggetto: Re: cauchy principal value double integral

On Monday, December 3, 2012 2:18:53 AM UTC-6, 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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