cjh@petsd.UUCP (Chris Henrich) (02/14/85)
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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
UUCP: ..!(cornell | ariel | ukc | houxz)!vax135!petsd!cjh
US Mail: MS 313; Perkin-Elmer; 106 Apple St; Tinton Falls, NJ 07724
Phone: (201) 870-5853leimkuhl@uiucdcsp.UUCP (02/18/85)
Actually, there is another simple way to get this:
Make the substitution y1 = 1-x1
y2 = 1-x1-x2
...
yn = 1-x1-...-xn
Rewrite the integral under this transformation and integrate
by parts repeatedly. The answer pops right out after a few steps.
This problem and others very similar have been quite common on the
Putnam exam, and they also occur frequently in probability. I am
curious, where did the poser find the problem?