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}]