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From: pmontgom@sdcrdcf.UUCP (Peter Montgomery)
Newsgroups: net.math
Subject: Re: Integration Problem
Message-ID: <1757@sdcrdcf.UUCP>
Date: Wed, 13-Feb-85 11:30:08 EST
Article-I.D.: sdcrdcf.1757
Posted: Wed Feb 13 11:30:08 1985
Date-Received: Fri, 15-Feb-85 02:10:00 EST
References: <308@ho95b.UUCP>
Reply-To: pmontgom@sdcrdcf.UUCP (Peter Montgomery)
Organization: System Development Corp. R+D, Santa Monica
Lines: 104
Summary: 

>>Given n variables  X	such that :
>>		      i
>>
>>
>>	     n
>>	 \--------
>>	  \	       X     =	1
>>	  /		i
>>	 /--------
>>	   i = 1
>>
>>
>>   and n constants   S    such that :
>>			i
>>
>>	     n
>>	 \--------
>>	  \	       S     =	m	for some m > 0
>>	  /		i
>>	 /--------
>>	   i = 1
>>
>>WHAT IS :
>>
>>
>>	   1	    1
>>	  S	   S
>>	  S	   S	 S	S      S	  S
>>	  S	   S	  1	 2	3	   n
>>	  S ....   S	X      X      X	 ....	 X     dX  dX  dX  ... dX
>>	  S	   S	 1	2      3	  n	 1   2	 3	 n
>>	  S	   S
>>	 0	  0
>
>
>This is a generalization of the Beta function.	 Using G(x)
>for the Gamma function [BTW, G(x)=(x-1)!], and letting Si=Ti-1:
>
>
>		     G(T1)*G(T2)* ... *G(Tn)
>	  Integral=  -----------------------
>			G(T1+T2+ ... +Tn)      .
>
>
>The above is actually fairly easy to show (but takes up lots of space).
>After the constraint on the X's is inserted, one must be careful
>of what happens to the limits of integration, but a simple recursion
>relation can be derived.  Note that the answer must be symmetric
>with respect to the Ti's.

>If anybody is *really* interested, I can post or mail a more complete
>derivation.
>
        Here is a short derivation when the exponents are nonnegative
integers.  Let F(n) designate n!, the factorial function.  We first show

		x=1   a	     b
	INTEGRAL     x	(1-x)  dx = F(a)F(b)/F(a+b+1)
		x=0

for all nonnegative integral a and b.  If b=0, both sides reduce to 1/(a+1).
If b > 0, rewrite the integral as

		x=1    a      b-1    a+1     b-1
	INTEGRAL     [x	 (1-x)	  - x	(1-x)	 ] dx  =
		x=0

	F(a)F(b-1)/F(a+b) - F(a+1)F(b-1)/F(a+b+1) =

	[ F(a)F(b-1)/F(a+b+1) ] [ (a+b+1) - (a+1) ] = F(a)F(b)/F(a+b+1) .

The result follows by induction on b.

	For the three variable case, we must evaluate

		x=1	    y=1-x  a  b	       c
	INTEGRAL    INTEGRAL      x  y  (1-x-y)  dx dy .
		x=0	    y=0

The change of variable y = (1-x)w in the inner integral reduces this to

		x=1         w=1   a      b  b      c      c
	INTEGRAL    INTEGRAL     x  (1-x)  w  (1-x)  (1-w)  (1-x) dx dw =
		x=0	    w=0

		 x=1  a	     b+c+1	       w=1  b	   c
	[INTEGRAL    x	(1-x)	  dx] [INTEGRAL	   w  (1-w)  dw ] =
		 x=0			       w=0

		F(a)F(b+c+1)			F(b) F(c)
	[	------------	    ] [		---------	] =
		F(a+b+c+2)			F(b+c+1)


	F(a) F(b) F(c) / F(a+b+c+2).

	The extension to triple and higher integrals follows similarly.
-- 
			Peter Montgomery

	{aero,allegra,bmcg,burdvax,hplabs,
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