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MathGroup Archive 2007

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Re: Integrate with PrincipalValue->True

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80032] Re: Integrate with PrincipalValue->True
  • From: chuck009 <dmilioto at comcast.com>
  • Date: Sat, 11 Aug 2007 02:04:25 -0400 (EDT)

Very nice guys!

I just thought maybe contour integration was being used.  The integral and ones like it can be evaluated via a key-hole contour with the branch-cut over one of the poles.  In this case, have the branch-cut be the negative imaginary axis.  The pole there gives rise to the expression:

g[x]=Sqrt[x]/(1-x^2)  which can then be easily evaluated via the Residue Theorem.

An interesting exercise is to use this method to calculate:

Integrate[Sqrt[x]/(1-x^5),{x,0,Infinity}]


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