Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site ho95b.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxb!mhuxn!mhuxm!mhuxj!houxm!ho95b!ran From: ran@ho95b.UUCP (RANeinast) Newsgroups: net.math Subject: Re: Integration Problem Message-ID: <308@ho95b.UUCP> Date: Mon, 11-Feb-85 15:38:50 EST Article-I.D.: ho95b.308 Posted: Mon Feb 11 15:38:50 1985 Date-Received: Tue, 12-Feb-85 06:38:50 EST Organization: AT&T-Bell Labs, Holmdel, NJ Lines: 58 >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. -- ". . . and shun the frumious Bandersnatch." Robert Neinast (ihnp4!ho95c!ran) AT&T-Bell Labs