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}]