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[44]:= genl = Integrate[Power[x^2 (x - 1), n], x] // InputForm

Out[44]//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[31]:=genl /. x -> 0
Out[31]=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[34]:=genl /. x -> 1
Out[34]=
(Gamma[1 + n]*Gamma[2 + 2*n])/((1 + 2*n)*Gamma[2 + 3*n])

Now let n=1/3, we get:

In[49]:=(% /. n -> 1/3) // InputForm
Out[49]//InputForm=(3*Gamma[4/3]*Gamma[8/3])/10

Now evaluate numerically:
In[36]:=N[%]
Out[36]=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 ?????

```

• Prev by Date: Re: Applying position-dependent functions on lists.
• Next by Date: Re: fastest way to do pair-sum / make pair-list
• Previous by thread: Special thanks!
• Next by thread: more on vector unions