
Re: FriCAS failure on Goursat(?) pseudoelliptics
Posted:
Sep 29, 2012 9:03 AM


clicliclic@freenet.de wrote: > > clicliclic@freenet.de schrieb: > > > > Waldek Hebisch schrieb: > > > > > > clicliclic@freenet.de wrote: > > > > > > > > One of the pseudoelliptic 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 fs. > > > > > > > Thanks for trying. My GradshteynRyzhik has three such integrands and > > adds more detail (without specifying the source). In particular, the > > substitution z*x = SQRT(x*(1x)*(1k^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*(1x)*(1k^2*x)) > > with f(x) + f((1k^2*x)/(k^2*(1x))) = 0 and with > > f(x) + f((1x)/(1k^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((1x^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

