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MathGroup Archive 2007

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Re: Diferent solution of integral in versions 4 and 5...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73625] Re: Diferent solution of integral in versions 4 and 5...
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Fri, 23 Feb 2007 04:37:36 -0500 (EST)
  • References: <ergr6r$jjg$1@smc.vnet.net>

"vromero" <vromero at ucaribe.edu.mx> wrote:
> Hello,
> I tried the following integral:
>
> Integrate[Exp[-I k3 x3]/(k12^2 + k3^2)^2,{k3,-Infinity,Infinity},
> Assumptions -> k12!=0 && Im[k12] == 0 && Im[k3] == 0 && Im[x3] == 0]
>
> The result in version 5 is
>
> (Abs[k12]*MeijerG[{{1/2}, {}}, {{0, 1/2, 3/2}, {}},
> (-(1/4))*k12^2*x3^2])/(k12^4*Sqrt[Pi])
>
> But in version 4 is
>
> (1 + Abs[k12] Abs[x3]) / Abs[k12]^2 * Exp[- Abs[k12] Abs[x3]]
>
> How can I get last result from version 5?
>
> Thanks in advance...

As Dimitris has already noted, both of the results above are incorrect, so
I'll just address the question of how to get a simple and correct result
from the current version of Mathematica.

The value of your integral must be real. Unfortunately, Mathematica does
not use the assumptions which you specified to full advantage. The first
thing I did which gave a correct answer was to manually replace
Exp[-I k3 x3] with Cos[k3 x3] in your integral. Later, I realized there was
an alternative which required less knowledge from the user: Since your
parameters and variable are real, just use ComplexExpand on your integrand

In[9]:=
Integrate[ComplexExpand[Exp[-I k3 x3]/(k12^2 + k3^2)^2], {k3, -Infinity,
Infinity}, Assumptions -> k12 != 0 && Im[k12] == 0 && Im[k3] == 0 &&
Im[x3] == 0]

Out[9]=
(Pi*(k12*Abs[x3] + Sign[k12]))/(E^(Abs[k12]*Abs[x3])*(2*k12^3))

David W. Cantrell


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