From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: residue of the integral( [a-b*cos(x)-c*cos(2x)]^-½ )
Date: 5 Mar 1999 23:29:15 GMT
Newsgroups: sci.math
Keywords: Elliptic integrals in disguise

Ferrus Thierry <ferrus@insatlse.insa-tlse.fr> wrote:
>Could someone tell me how to calculate the integral
>
>    _
>   / pi
>   |
>   |          dx
>   |  ---------------------              ???
>   |                        ½
>   |  [a-b*cos(x)-c*cos(2x)]
>   |
>  _/ 
>    -pi 

Intgrate from 0 to pi and double.

Set  u = cos(x), du = -sqrt(1-u^2) dx, and note cos(2x)=2u^2-1. Thus we
have an integral from -1 to 1 of  1/sqrt( (1-u^2)(a+c - b u - 2c u^2) ).
This involves the square root of a quartic, and so it's an elliptic integral,
which means you can't expect to get the antiderivative in closed terms unless
you don't mind an answer which involves, well, elliptic integrals!

I asked Maple to try a specific example (a=14, b=2, c=3) and it managed
to get an explicit but terrible expression involving  EllipticF  and
EllipticK. For some reason it failed to do the general case, which ought
to have a similar formula.

Trust me, you don't want to see this integral...

Similar thread now underway in sci.math.symbolic.

dave