From: Doug Magnoli <dmagnoli@home.com>
Subject: Re: Integral
Date: Tue, 19 Dec 2000 09:03:59 GMT
Newsgroups: sci.math
Summary: Beta functions

Whenever you see an integral from 0 to 1 of anything that has only
powers of x and 1-x in the integrand, you should think in terms of beta
functions because they're generally pretty easy to evaluate.  The hard
part, of course, is generally getting your integral into the form where
it becomes a beta function.

The beta function is defined:

beta [n,m] = Int [0,1, x^(n-1) (1-x)^(m-1) dx]

and is symmetric, i.e., beta[n,m] = beta[m,n].  The property that makes
it so easy to evaluate is:

beta[n,m] = gamma(n)*gamma(m) / gamma(n+m), where

gamma(x) = Int [0, inf, exp^(-t) t^(x-1) dt] = (x-1) !

So let's take a look at your integral and see if it can be beta
functionized:

Int[0,1, (1-y^4)^(-1/2) dy]

Let x = y^4, dx = 4y^3 dy   --> dy = dx / (4y^3) = dx / (4 x^(3/4))

Int [0,1, (1-y^4)^(-1/2) dy] = (1/4)*Int[0,1, (1-x)^(-1/2) * x^(-3/4)
dx]

And now you've got this into the form of the beta function, so

 = (1/4) beta [1/2, 1/4] = (1/4) gamma(1/2)* gamma(1/4) / gamma(3/4)

Since it's well known (and easy to prove) that gamma(1/2) = sqrt(pi),

 = (1/4) sqrt(pi) * gamma(1/4) / gamma(3/4)

 ~ 1.31103,

where I used Mathematica to evaluate the gamma functions for me.

Also, since n*gamma(n) = gamma(n+1), you might notice that
(1/4)*gamma(1/4) = gamma(5/4), so your solution can also be expressed as
gamma(1/2)*gamma(5/4) / gamma(3/4) (to the extent that you care).

Fairly simple once you know how to go about it.

-Doug Magnoli


Gary Clayton wrote:

> I have been trying to solve this integral:
>
>     / 1
>     |
>     | dy/((1-y^4)^(1/2))
>     |
>     / 0
>
> When solved numerically, it is around 1.311.  But is there a closed
> form solution?  Can the solution be expressed in closed form or some
> combination of terms?
>
> Thanks,
> Gary
>
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