But the message is: unlike the use of hypergeometric functions, which are likely to end up being elementary for some values of the parameters, this approach proves that the given integral is indeed non-elementary.
In article <3B5A65D1.ADFEE2DE@home.com>, Doug Magnoli <firstname.lastname@example.org> wrote: :And BTW, Mathematica gives the indefinite integral as: : :Int [(sin x)^(1/3) dx] := - cos(x)*HG2F1[1/2,1/3,3/2,(cos x)^2]*(sin x)^(4/3) / (sin x)^(4/3), : :where HG2F1 is Hypergeometric2F1, defined as: : :HG2F1(a,b;c;z) = sum(k=0,inf) (a_k*b_k / c_k) * z^k / k! : :hth, : :-Doug Magnoli : : :Tim 9-23 wrote: : :> Can this be expressed in elementary functions? I doubt it. :> :> My email address is anti-spammed. If you want to email me, :> remove the 2 B's after hitting email reply. :> :> Tim 9-23 :