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From: cjh@petsd.UUCP (Chris Henrich)
Newsgroups: net.math
Subject: Integration problem and solution
Message-ID: <445@petsd.UUCP>
Date: Wed, 13-Feb-85 18:07:51 EST
Article-I.D.: petsd.445
Posted: Wed Feb 13 18:07:51 1985
Date-Received: Thu, 14-Feb-85 03:11:44 EST
Organization: Perkin-Elmer DSG, Tinton Falls, N.J.
Lines: 92

[]
	Amr F.Fahmy asks for the integral, over the region
in n-space defined by "X1 + ... + Xn = 1", of

	  S    S       S
	   1    2       n
	X     X   ... X     dX  dX  ... dX  .
	 1     2       n      1   2       n

There is a quibble that can be made here: the manifold of
integration is n-1 dimensional, whereas the differential as
stated is n-dimensional.  Presumably, what is meant is the
hypersurface element on that manifold.  Also, I think it is
intended that the coordinates be all positive, which makes the
region of integration bounded and nice.
	
	Here is a relatively nice way to get the answer.
Let
		dS[n-1]
denote the hypersurface element on that manifold or any hyperplane 
parallel to it.  Define

	P(A, t  , ..., t ) =
              1         n

          INTEGRAL      t -1       t -1
          INTEGRAL       1          n
          INTEGRAL     X    ...   X      dS[n-1].
          INTEGRAL      1          n
          INTEGRAL
              X  + X  + ... + X = A
               1    2          n

Then
                           t  + ... t
                            1        n
     P(A, t , ..., t ) =  A            P(1, t , ..., t )
           1        n                        1        n

Now consider


          INTEGRAL      -(X  + ... + X )   t       t
          INTEGRAL         1          n     1       n
(1)       INTEGRAL     E                  X   ... X   dX  ... dX
          INTEGRAL                         1       n    1       n
          INTEGRAL
           X  , ... X   >= 0
            1        n

and attempt to evaluate it in two ways.  First, the integrand factors
into functions of one coordinate, and the domain of
integration also "factors" the same way; so (1) =

          G(t )G(t )...G(t )
             1    2       n

where G denotes Euler's gamma function (G(n+1) = n! if n is
integer.)  Second, slice the domain of integration into
hyperplanes of constant  X  + ... + X .  The (1) becomes
                          1          n


                          -A
          INTEGRAL       E   P(A, t  ,..., t ) dA
                A >= 0             1        n

                                           t  + ... + t
                                            1          n   -A
        = P(1, t ,..., t )  INTEGRAL      A               E  dA
                1       n          A >= 0


        = P(1, t ,..., t ) G(t +...+ t + 1)
                1       n     1       n


Therefore

                                 G(t ) G(t ) ... G(t )
                                    1     2         n
        P(1, t  ,..., t )  =   -------------------------
              1        n       G(t  + t  + ... + t  + 1)
                                  1    2          n
Regards,
Chris

--
Full-Name:  Christopher J. Henrich
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