Multiple Integral = Single Integral Identity (Interesting?)

From: Leroy Quet (qqquet_at_mindspring.com)
Date: 06/07/04 

Date: 7 Jun 2004 10:28:49 -0700

I will post the result below, even though it is probably trivial on some
level.
It seems interesting to me anyway.

Let each f and g be ANY real -> real continuous integrable monotonically
increasing function, where f_k(x) is the inverse of g_k(x)
(ie. f_k(g_k(x)) =x for all x and k)),
and each f(0) = each g(0) = 0.

Let a(x) be any integrable real -> real function (for 0<=x<=infinity),
where a'(x) exists for all x >= 0,
and a(x) -> 0 as x -> infinity.

Then:

For n = positive integer,

  /oo n
  ! ---
- ! a'(x) ( ! ! f_k(x) ) dx
  /0 k=1

=

/oo /oo /oo
! ! !
! ! ... ! a(max(g_1(x_1),g_2(x_2),..,g_n(x_n))) dx_1 dx_2..dx_n
/0 /0 /0

if the integrals converge.

(a'(x) is the derivative of a(x) at x.)

In linear mode:

-integral{0 to oo} a'(x) (product{k=1 to n} f_k(x) ) dx

=

integral{0 to oo} integral{0 to oo}...integral{0 to oo}
  a(max(g_1(x_1),g_2(x_2),..,g_n(x_n))) dx_1 dx_2..dx_n

So, for example,

integral{0 to oo} integral{0 to oo}...integral{0 to oo}
  exp(-max((x_1)^(1/m_1),(x_2)^(1/m_2),..,(x_n)^(1/m_n)))
    dx_1 dx_2..dx_n

=

(m_1 +m_2 +...+m_n)!

for each m = positive integer.

thanks,
Leroy Quet

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