aff@duke.UUCP (Amr F. Fahmy) (02/11/85)
I am faced with the following integration which I do not know how to solve or even if there is a solution for. Your help will be greatly appreciated. Please send replies to aff@duke, or post it on the net. 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
ran@ho95b.UUCP (RANeinast) (02/11/85)
>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
pmontgom@sdcrdcf.UUCP (Peter Montgomery) (02/13/85)
>>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, ihnp4,psivax,randvax,sdcsvax,trwrb}!sdcrdcf!pmontgom Don't blame me for the crowded freeways - I don't drive.