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MathGroup Archive 2007

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Re: Integrate with PrincipalValue->True

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80003] Re: Integrate with PrincipalValue->True
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Fri, 10 Aug 2007 01:55:24 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <f9em11$k0v$1@smc.vnet.net>

chuck009 wrote:
> Can someone explain to me how Integrate calculates the following improper integral?
> 
> In[42]:= $Version
> Integrate[Sqrt[x]/(1 - x^2), {x, 0, a}, 
>    PrincipalValue -> True, Assumptions -> 
>      a > 1]
> 
> Out[42]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"
> 
> Out[43]= ArcCoth[Sqrt[a]] - ArcTan[Sqrt[a]]

Charles,

I am not sure whether you are asking about Mathematica internal 
algorithms or about Cauchy principal value integral.

In your case, setting the option PrincipalValue to true is conceptually 
equivalent to computing the improper integral as follows.

In[1]:= $Version
f[x_] = Sqrt[x]/(1 - x^2);
pv = Limit[Integrate[f[x], {x, 0, 1 - epsilon},
          Assumptions -> epsilon > 0] +
    Integrate[f[x], {x, 1 + epsilon, a},
          Assumptions -> epsilon > 0 && a > 1], epsilon -> 0,
      Assumptions -> a > 1]

Out[1]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"

Out[3]= -(\[Pi]/2) + ArcTan[1/Sqrt[a]] + ArcTanh[1/Sqrt[a]]

(You can check numerically or symbolically that this is equivalent to 
your result.)

-- 
Jean-Marc


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