Re: Integrate[[1/(1+x+x^5),{x,-Infinity,Infinity}]

*To*: mathgroup at smc.vnet.net*Subject*: [mg25396] Re: Integrate[[1/(1+x+x^5),{x,-Infinity,Infinity}]*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Fri, 29 Sep 2000 01:06:30 -0400 (EDT)*Organization*: Universitaet Leipzig*References*: <8qlol1$q86@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, it is easy to find that (1+x+x^5) has a simple pole on the real axes at {x -> 1/3 - (2/(25 - 3*Sqrt[69]))^(1/3)/3 - ((25 - 3*Sqrt[69])/2)^(1/3)/3} and you will probably Infinity, with any system. The indefined integral is calculated by Integrate[1/((x - a)*(x - b)*(x - c)*(x - d)*(x - e)), x] /. Thread[{a, b, c, d, e} -> (x /. Solve[1 + x + x^5 == 0, x])] Regards Jens Zak Levi wrote: > > Dear Mathematica experts, > > There was some discussion of this integral in sci.math > without clear answer to the question: > > How to calculate > > Integrate[[1/(1+x+x^5),{x,-Infinity,Infinity}] > > in Mathematica? -preferably NOT in M4 version, > simply because not all users have the latter one. > > Thanks a lot, > ZL