Re: ...can't help, but let me muddy the waters a bit. ;-) ...
- To: mathgroup at smc.vnet.net
- Subject: [mg23253] Re: [mg23222] ...can't help, but let me muddy the waters a bit. ;-) ...
- From: "Mark Harder" <harderm at ucs.orst.edu>
- Date: Sat, 29 Apr 2000 22:04:58 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
In mg 23222, Zeno 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. > I tried working on this, and came up with the following interesting results, however I know nothing about the mathematically correct analysis of this problem, so I am showing my work for comment. Apparently Mathematica is following a general rule for finding indefinite integrals of this form: In:= genl = Integrate[Power[x^2 (x - 1), n], x] // InputForm Out//InputForm= (x*((-1 + x)*x^2)^n*Hypergeometric2F1[1 + 2*n, -n, 2 + 2*n, x])/ ((1 + 2*n)*(1 - x)^n) .. and although it involves the Hypergeometric Function, we *can* do things with this, using replacement rules: In:=genl /. x -> 0 Out=0 Now evidently Mathematica is doing some algebra on the general expression when evaluating, because the expression genl above should be indeterminate for x=0 and x=1, (test this by replacing n->1/3 *first*, & *then* x->1 ; or just try to evaluate Zeno's expression at x=1 !!). Here's what I get when I replace x in genl by 1: In:=genl /. x -> 1 Out= (Gamma[1 + n]*Gamma[2 + 2*n])/((1 + 2*n)*Gamma[2 + 3*n]) Now let n=1/3, we get: In:=(% /. n -> 1/3) // InputForm Out//InputForm=(3*Gamma[4/3]*Gamma[8/3])/10 Now evaluate numerically: In:=N[%] Out=0.403067 I have no idea if this answer is valid or not!! Has Mathematica helped me out by performing a rearrangement, which casts the function in a form without a singularity at x=1, *or* has Mathematica helped me to deceive myself in letting me generate a general solution to the integral which for some values of x & n gives integrals that evaluate when they shouldn't ?????