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I'm looking for a way to symbolically integrate the following expression (the $k_i$ are integer constants):

Using a generalized form of the identity

I was able to express it without the cumbersome area condition of the original expression:

I've tried expanding the expression

using the multinomial theorem, in order to eventually pull the sum outside the integral and make an integration step, but it didn't look very promising, the formulas just kept getting larger and larger, with not obvious pattern emerging which I could use to iterate the process for all the integrals.

I realize that in the end, all that will ever come out of these formulas are multivariate polynomials, so in theory, it should be no problem to integrate this stuff symbolically. I just can't find a way to express the result concisely in standard mathematical notation.

Given that the original expression seems simple enough, I figured that there might be an easier way of approaching this problem, or that maybe it has already been solved.

So, does anyone know how to symbolically integrate the given expression?

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You used $n$ in two different meanings. –  joriki Aug 23 '12 at 23:06
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3 Answers 3

up vote 2 down vote accepted

One parametrization of the simplex which is useful for this problem is: $$ x_1 = v_1, \quad x_2 = (1-v_1) v_2, \quad \ldots, \quad x_{d-1} =(1-v_1)\cdots (1-v_{d-2}) v_{d-1}, \quad x_d = (1-v_1)\cdots (1-v_{d-1}) $$ where $0<v_k<1$ for all $1 \leqslant k \leqslant d-1$. It is easy to check that $\sum_{k=1}^d x_k = 1$ explicitly. The measure on the simplex then reads: $$ \mathrm{d}S = (1-v_1)^{d-1} (1-v_2)^{d-2} \cdots (1-v_{d-1}) \mathrm{d}v_1 \mathrm{d} v_2 \cdots \mathrm{d} v_{d-1} $$ Hence the integral over the simplex, now becomes an integral over cube, which factors into the product of univariate integrals: $$ \int_{\sum_{i=1}^d x_i = 1} \prod_{i=1}^d {x_i}^{k_i} \mathrm{d}S = \prod_{m=1}^{d-1} \int_0^1 (1-v_m)^{d-m+\sum_{i={m+1}}^{d} k_i} v_m^{k_m} \mathrm{d} v_m = \prod_{m=1}^{d-1} B\left(d-m+1 +\sum_{i=m+1}^{d} k_i, k_m+1 \right) = \frac{k_1! \cdots k_d!}{(k_1+\cdots+k_d+d-1)!} $$

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Let $$ J(k_1,\ldots,k_n) = \int_{S_n} \prod_{j=1}^n x_j^{k_j}\ d^{n-1} x$$ where $S_n$ is the simplex $\sum_{i=1}^n x_i = 1, x_i \ge 0$ and $d^{n-1} x$ is $n-1$-dimensional Lebesgue measure on $S_n$ using any $n-1$ of the variables $x_j$ (or if you prefer, $n-1$-dimensional Hausdorff measure). Note that by the change of variables $x = r y$, $$\int_{r S_n} \prod_{j=1}^n x_j^{k_j}\ d^{n-1} x = r^{n-1+\sum_j k_j} \int_{S_n} \prod_{j=1}^n y^{k_j}\ d^{n-1} y = r^{n-1+\sum_j k_j} J(k_1,\ldots,k_n)$$ Thus $$\eqalign{J(k_1,\ldots,k_n) &= \int_0^1 dx_n \ x_n^{k_n} \int_{(1-x_n) S_{n-1}} \prod_{j=2}^n x_j^{k_j}\ d^{n-2} x \cr &= \int_0^1 dx_n\ x_n^{k_n} (1-x_n)^{n-2+\sum_{j=1}^{n-1} k_j} J(k_1,\ldots,k_{n-1})\cr}$$
and using the Beta function $\int_0^1 dx\ x^\alpha (1-x)^\beta = B(\alpha+1,\beta+1)$ we get $$J(k_1,\ldots,k_n) = \prod_{i=2}^n B(k_i+1,i-1+\sum_{j=1}^{i-1} k_j) = \prod_{i=2}^n B(k_i+1, \sum_{j=1}^{i-1} (k_j + 1))$$ Now for positive integers $B(\alpha,\beta) = (\alpha-1)! (\beta-1)!/(\alpha + \beta-1)!$, so $$\eqalign{J(k_1,\ldots,k_n) &= \dfrac{k_2! k_1!}{(k_1+k_2+1)!} \dfrac{k_3! (k_1+k_2+1)!}{(k_1+k_2+k_3+2)!} \ldots \dfrac{k_n! (k_1+\ldots+k_{n-1}+n-2)!}{(k_1+\ldots+k_n+n-1)!}\cr &= \dfrac{k_1! \ldots k_n!}{(k_1+\ldots+k_n+n-1)!}}$$

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You can solve all the integrals using integration by parts. The result for the identity you want to generalize is $\binom nk^{-1}$, and the generalized result is

$$ \binom{k_1+\dotso+k_n}{k_1,\dotsc,k_n}^{-1}\;. $$

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