firstname.lastname@example.org schrieb: > > Waldek Hebisch schrieb: > > > > email@example.com 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
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.