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FriCAS failure on Goursat(?) pseudo-elliptics
Posted:
Sep 29, 2012 3:00 AM
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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. The substitution
t = SQRT(x*(1-x)*(1 - k^2*x))/x
or, equivalently,
x = (SQRT((t^2 + (k+1)^2)*(t^2 + (k-1)^2)) + k^2 + t^2 + 1)/(2*k^2)
takes the (pseudo-)elliptic integral
INT((k^2*x^2 - 1)/((a*k^2*x + b)*(b*x + a)
*SQRT(x*(1-x)*(1 - k^2*x))), x)
into the very simple rational one
INT(2/(a^2*k^2 + a*b*(k^2 + t^2 + 1) + b^2), t),
and the original antiderivative consequently is
2/(SQRT(a*b)*SQRT((a+b)*(a*k^2 + b)))
*ATAN(SQRT(a*b)*SQRT(x*(1-x)*(1 - k^2*x))
/(x*SQRT((a+b)*(a*k^2 + b)))).
I hope FriCAS reports digestive trouble rather than implying this to be
non-elementary. If an inability to compute the "splitting field of the
Trager resultant" is the reason, this should be rectified, I think,
unless such a computation would take weeks to perform.
Martin.
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