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Re: FriCAS failure on Goursat(?) pseudo-elliptics
Posted:
Sep 29, 2012 9:03 AM
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clicliclic@freenet.de wrote:
>
> clicliclic@freenet.de schrieb:
> >
> > Waldek Hebisch schrieb:
> > >
> > > clicliclic@freenet.de wrote:
> > > >
> > > > One of the pseudo-elliptic examples (here taken from Goursat's
> > > > Cours d'analyse) is given on page 50 of the booklet mentioned.
> > >
> > > AFAICS this example is wrong -- FriCAS result indicates that
> > > there is no elementary integral in this case and I see no
> > > reasons why it should have one. OTOH, if one modifies
> > > the example to have
> > >
> > > f/sqrt((1 - x^2)*(1 - k^4*x^2))
> > >
> > > under integral, then FriCAS can handle it for particular f-s.
> > >
> >
> > Thanks for trying. My Gradshteyn-Ryzhik has three such integrands and
> > adds more detail (without specifying the source). In particular, the
> > substitution z*x = SQRT(x*(1-x)*(1-k^2*x)) is said to turn the
> > integrand into a rational function. I will look into this claim. The
> > other two examples are the same integrand f(x)/sqrt(x*(1-x)*(1-k^2*x))
> > with f(x) + f((1-k^2*x)/(k^2*(1-x))) = 0 and with
> > f(x) + f((1-x)/(1-k^2*x)) = 0.
> >
>
> If you made no mistake, FriCAS is in trouble here: I have confirmed that
> the first example (the one cited by Hardy) is indeed elementary, using
> the rational factor f(x) = (k^2*x^2 - 1)/((a*k^2*x + b)*(b*x + a)) for
> which f(x) + f(1/(k^2*x)) = 0 as required.
I made a mistake: I took the irrational factor to be
sqrt(x*(1 - x)*(1 - k^2*x^2)). Goursat example is clearly
based on symmetry under x -> 1/(k^2*x) map, but that factor
is not symmetric under such map, so has no reason to have
elementary integral. When irrational factor is changed to
sqrt((1-x^2)*(1 - k^2*x^2))) we get required symmetry, and
it works fine. I did not notice that replacing k^2*x^2 by
k^2*x also gives symmetry (and that this is what text
actually contained).
To make things clear: with sqrt(x*(1 - x)*(1 - k^2*x)) FriCAS
can do the integrals.
--
Waldek Hebisch
hebisch@math.uni.wroc.pl
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