Re: help evaluating definite integral

From: Alex Selby (a.p.selby.remove_at_pobox.com)
Date: 06/18/04 

Date: 17 Jun 2004 19:21:00 -0700

Faheem Mitha <faheem@email.unc.edu> wrote in message news:<slrnccpdgb.bnb.faheem@Chrestomanci.home.earth>...
> On Sat, 12 Jun 2004 17:14:37 -0400, G. A. Edgar
> <edgar@math.ohio-state.edu.invalid> wrote:
>
> [...]
>
> By the way, I heard from Alex Selby by email. Apparently the formula
> above is a special case of one of the integrals known as Dirichlet
> integrals. I quote him below.
>
> "
> Your integral is a special case of what is known as a "Dirichlet
> Integral".
>
> See e.g.,
> http://mathworld.wolfram.com/DirichletIntegrals.html
>
> and use formula (4) with alpha_i=f_i+1, f(x)=delta(x-1), their n =
> your r
>
> By the way, there is a missing a factor of (r-1)! from the LHS of your
> equation.
> "
> Faheem.

Sorry I think my previous reply is a bit misleading. I think it is not
very enlightening to quote something like "Dirichlet integral" because
its proof is very short, so it is maybe better to see how to prove it than
to look it up. It is also slightly more general than you would need since
it deals with non-integral powers too.

Rereading your original post, it looks like you are happy proving
the formula [as I think it should be]

Integral over (p1+p2+...+pr=1, p1,p2,...>=0) of (p1^f1.p2^f2...pr^fr)
              = f1!f2!...fr!/(N+r-1)!

by induction. I don't think induction is "cheating" - you have
calculated it.

But if you want a non-inductive proof, then this is a probabilistic one:

Instead of cluttering the notation by writing it out for general r, let's
do examples for r=2 and r=3, then it will be obvious how it works for r>3.

r=2:

Imagine placing f1 red dots, f2 green dots and 1 black dot uniformly at
random on the interval [0,1]. What is the probability that the red dots
are to the left of the black dot, and the green dots are to the right of
it? Let's work it out in two different ways.

(i) Conditioning on the position, p, of the black dot, the chance that all
the red dots lie to the left of it is p^f1, and the chance that the green
dots lie to the right is (1-p)^f2. So the overall probability is integral
from 0 to 1 of p^f1.(1-p)^f2.

(ii) The distribution of dots is invariant under permutation of the
f1+f2+1=N+1 dots, and there is exactly one colouring which satisfies the
condition red-then-black-then-green. The only allowable permutations of
dots which preserve the condition are mixing the reds amongst themselves
and the greens amongst themselves. So the overall probability is
f1!f2!/(N+1)! where N=f1+f2.

Thus Integral of p^f1(1-p)^f2 is f1!f2!/(N+1)!.

r=3:

Now we have f1 red dots, f2 green dots, f3 blue dots, 1 brown dot and 1
black dot all placed at random on [0,1]. We ask for the probability the
dots appear in the order: f1 red then 1 brown then f2 green then 1 black
then f3 blue.

(i) Conditioning on the positions x1 and x2 of the brown and black dots,
the chance is x1^f1.(x2-x1)^f2.(1-x2)^f3. Integrating over 0<x1<x2<1 is
the same as integrating over p1+p2+p3=1 where p1=x1, p2=x2-x1, p3=1-x2. So
the answer is Integral over (p1+p2+p3=1) of (p1^f1.p2^f2.p3^f3).

(ii) There are N+r-1 dots, and again only one colouring which satisfies
the condition. So the answer is f1!f2!f3!/(N+2)!.

Thus Integral over (p1+p2+p3=1) of p1^f1.p2^f2.p3^f3 is f1!f2!f3!/(N+2)!.

[In (i) we needed to make an equivalence between integrating over x_i and
p_i. This amounts to asking what was the original definition of
integrating over p1+...+pr=1. If the original definition was inductive on
r, then we are doomed to having an inductive component in our proof. But a
reasonable definition is that we are using the restriction to
(p1+...+pr=1) of a differential form on R^r which when multiplied by
d(p1+...+pr) gives dp1.dp2....dpr. Well that's OK, 'cos x_i=p1+...+p_i, so
dx1....dx_{r-1}=dp1....dp_{r-1} and
dx1....dx_{r-1}d(p1+...+pr)=dp1....dpr.]

Hope this helps.

Alex