On Sat, Oct 21, 2023 at 10:32:55AM -0700, Mild Shock wrote:
> > Also, without ability to simplify hypergeometric answers are of
> > little use, so we need first strong simplifier for hypergeometric
> functions.
>
> I am not sure that the lack of a strong simplifier is to blame that
> I don't see a slope in those systems that can solve the integral.
>
> Now I was trying to compute the slope, by using a definite
> integral over the period instead of a indefinite integral.
<snip>
> The more complex problem with cos(t)^(1/3) gives me:
>
> (2) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = 0..2*pi)
>
> (2) "potentialPole"
>
> How can I tell the system that ^(1/3) should be real root?
(7) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = 0..2*%pi, "noPole")
(7) "failed"
Type: Union(fail: failed,...)
in 99.9999% of cases FriCAS needs indefinite integral to
compute definite one. Since definite integral remains
unevaluated FriCAS can not give you definite one.
There is also another catch: FriCAS works with branches of
analytic functions. But cos(t) changes sign close to 0.
This means there are no analytic real third root close to 0.
If you take real root for small positive t, than for small
negative t analytic root must be complex. That must be
taken into account in cases where FriCAS can compute
indefinite integral.
--
Waldek Hebisch
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