       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))^(1/3)/3 -
((25 - 3*Sqrt)/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

```

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