Re: help evaluating definite integral
From: Faheem Mitha (faheem_at_email.unc.edu)
Date: 06/13/04
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Date: Sun, 13 Jun 2004 20:16:11 GMT
On Sat, 12 Jun 2004 17:14:37 -0400, G. A. Edgar
<edgar@math.ohio-state.edu.invalid> wrote:
>> A modified expression appears below. I hope that resolves any
>> remaining ambiguity.
>>
>> \begin{align*}
>> \int \binom{N}{f_1\dots f_r}\;
>> \prod_{i=1}^{r} p_i^{f_i} \; I_{[\sum_{i}p_i=1]} d\mathbf{p}
>> = \frac{1}{\binom{N+r-1}{r-1}}
>> \end{align*}
>> where $\binom{N}{f_1\dots f_r}$ is the multinomial coefficient and
>> $\sum_i f_i=N$.
>>
>
> Let's try to understand better. Take the case r=3, N=5. Then we
> want f1+f2+f3=5, p1+p2+p3=1. So I should take
>
>
> f1 f2 (5 - f2 - f1)
> p1 p2 (1 - p2 - p1)
> t := 120 ------------------------------------
> f1! f2! (5 - f2 - f1)!
>
> and integrate it p1 from 0 to 1, p2 from 0 to 1-p1, right?
> Maple says:
>
>
> 1 1 - p1
> / / f1 f2 (5 - f2 - f1)
> | | p1 p2 (1 - p2 - p1)
> | | 120 ------------------------------------ dp2 dp1
> | | f1! f2! (5 - f2 - f1)!
> / /
> 0 0
>
>
> = 1/42
>
> which is, indeed, independent of the choice of f1 and f2. Also:
>
>
> 1 1
> -------------------------- = -------------- = 1/21
> binomial(N + r - 1, r - 1) binomial(7, 2)
>
>
> Close. What should I be doing differently?
I think I am missing a term. Consider the following argument.
\begin{align*}
(\sum_{i=1}^n p_i)^N =
\sum_{\sum f_i = N}
\binom{N}{f_1\dots f_r}\;
\prod_{i=1}^{r}
p_i^{f_i}
\end{align*}
Integrating both sides we obtain
\begin{align*}
\int (\sum_{i=1}^n p_i)^N \; I_{[\sum_{i}p_i=1]} \; d\mathbf{p} =
\sum_{\sum f_i = N}
\int\binom{N}{f_1\dots f_r}\;
\prod_{i=1}^{r}
p_i^{f_i} \; I_{[\sum_{i}p_i=1]} \; d\mathbf{p}
\end{align*}
Assuming the value for all
\begin{align*}
\int\binom{N}{f_1\dots f_r}\;
\prod_{i=1}^{r}
p_i^{f_i} \; I_{[\sum_{i}p_i=1]} \; d\mathbf{p}
\end{align*}
is the same, and calling this value I, then (assuming that the number
of distinct r-tuples $[f_i, \dots, f_r]$ such that $[f_i, \dots,
f_r]=N$ is $\frac{1}{\binom{N+r-1}{r-1}}$, we obtain
\begin{align*}
I = \frac{1}{\binom{N+r-1}{r-1}}\int I_{[\sum_{i}p_i=1]} \;
d\mathbf{p}
\end{align*}
Since $\int I_{[\sum_{i}p_i=1]} \; d\mathbf{p}$ is not in general 1,
what I had written earlier is not correct. I think there must be an
easy way of calculating $\int I_{[\sum_{i}p_i=1]} \; d\mathbf{p}$ but
I don't know what it is off the top of my head.
I have no idea how to make any further progress with this line of
argument, though showing that all those integrals have the same value
should be possible.
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.
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