Re: Please help with a Hypergeometric2F1 problem...

*To*: mathgroup at smc.vnet.net*Subject*: [mg23252] Re: Please help with a Hypergeometric2F1 problem...*From*: Ronald Bruck <bruck at math.usc.edu>*Date*: Sat, 29 Apr 2000 22:04:57 -0400 (EDT)*Organization*: Univ of Southern California*References*: <8e3b5h$kom@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <8e3b5h$kom at smc.vnet.net>, zeno at magicnet.net wrote: > I symbolically integrated the function..(x^2*(x-1))^(1/3) with respect to > x. > > There is in the answer... Hypergeometric2F1[2/3,2/3,5/3,x] > > I can do nothing more with that..it just returns it. A Hypergeometric2F1 > with different parameters like Hypergeometric2F1[2,2,5,x] gives an > answer. I > am using version 3. Is Mathematica unable to compute it? > > I can get the Integral with out the Hypergeometric function on the > TI-92+, > (it gives the answer in a different for using Tan, etc.) but I still > would > like to work with the Mathematica answer. > zeno e-mailed me the TI-92+ solution, which translated to Mathematica is Log[(((x - 1)^(1/3) - x^(1/3))^2/((x*(x - 1))^(1/3) + (x - 1)^(2/3) + x^(2/3)))]/18 - Sqrt[3]*ArcTan[Sqrt[3]*(2*(x - 1)^(1/3) + x^(1/3)) /(3*x^(1/3))]/9 + x^(2/3)*(x - 1)^(4/3)/2 + x^(2/3)*(x - 1)^(1/3)/3 (I've removed a couple of Abs from the expression, since Mathematica doesn't like to differentiate these; as phrased, it's valid for x >= 1). I am **impressed**. While I am unable to coax Mathematica to differentiate this with respect to x and simplify the derivative to the original expression, when I plot the difference on [1,2] all I get is the typical roundoff noise. It seems to be correct. --Ron Bruck -- Due to University fiscal constraints, all .sigs must be only one line.