# Integrate product of barycentric coordinates over a simplex

Calculate the following integral for a fixed positive integers $d,n_0,...,n_d$:

$\int_{0}^{1}\int_{0}^{1-x_1}\int_{0}^{1-x_1-x_2}...\int_{0}^{1-x_1-...-x_d}(1-x_1-x_2-...-x_d)^{n_0}x_1^{n_{1}}x_2^{n_{2}}...x_d^{n_{d}}(x_1+x_2+...+x_d)(1-x_1)(1-x_2)...(1-x_d)\mathrm{d}x_d\mathrm{d}x_{d-1}...dx_1$

Note: Observe that the integrand is the product of powers of barycentric coordinates $\xi_v$ with $1-\xi_v$. The region of integration is the standard tetrahedron in $\mathbb{R}^d$.

I calculated the integral for $\mathbb{R}^2$ (the case d=2) and got a closed form in terms of Betas: $B(n_2+1,n_3+1)B(n_1+2,n_2+n_3+3)+B(n_2+1,n_3+1)B(n_1+1,n_2+n_3+5)-B(n_2+3,n_3+1)B(n_1+1,n_2+n_3+5)$

Then I tried to evaluate the general case but it turns out to be very routine with enumerous amount of expansions and computations, I suspect a closed form of the general case involves muti-index notations.

Worth mentioning that the integral of powers of barycentric coordinates was evaluated before (the integrand in that case does not invove $(1-\xi_j)$ terms) in the study of classical Jacobi Polynomials on a simplex.

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Let $x_0 = 1-x_1-x_2-\cdots-x_d$. Then the integral acquires a symmetric form: $$\int_0^1 \int_0^1 \cdots \int_0^1 \prod_{k=0}^d (1-x_k) x_k^{n_k} \delta\left(x_0+x_1+\cdots+x_d -1\right)\mathrm{d} x_0 \mathrm{d} x_1 \cdots \mathrm{d} x_d$$ Let's make a change of variables: $$x_1 = v_1, \quad x_2 = (1-v_1) v_2, \quad x_3 = (1-v_1)(1-v_2) v_3, \quad \cdots \quad \\x_d= (1-v_1) \cdots (1-v_{d-1}) v_{d}, \quad x_0 = (1-v_1)(1-v_2)\cdots (1-v_d)$$ which trivializes the constraint $x_0+x_1+\cdots+x_d=1$, and all $0<v_i<1$. Then, the Jacobian reads: $$\mathrm{d} x_1 \mathrm{d} x_2 \cdots \mathrm{d} x_d = \prod_{k=1}^d (1-v_k)^{d-k} \mathrm{d} v_1 \mathrm{d} v_2 \cdots \mathrm{d} v_d$$ Let's first evaluate a simpler integral: $$I(n_0,n_1,\ldots, n_d) = \int\limits_{\mathbb{R}^d_+} \prod_{k=0}^d x_k^{n_k} \delta\left(x_0+x_1+\cdots+x_d -1\right)\mathrm{d} x_0 \mathrm{d} x_1 \cdots \mathrm{d} x_d = \\ \prod_{k=1}^d \int_0^1 (1-v_k)^{d-k+t-n_k} v_k^{n_k} \mathrm{d} v_k = \prod_{k=1}^d \operatorname{B}\left(d-k+1+t-n_k,n_k+1\right)$$ where $t = n_0+n_1+\cdots+n_{d-1}+n_d$. Now, to the result: $$\int\limits_{\mathbb{R}^d_+} \prod_{k=0}^d (1-x_k) x_k^{n_k} \delta\left(x_0+x_1+\cdots+x_d -1\right)\mathrm{d} x_0 \mathrm{d} x_1 \cdots \mathrm{d} x_d = \\ \sum_{m_0=0}^1 \sum_{m_1=0}^1 \cdots \sum_{m_d=0}^1 (-1)^{m_0+m_1+\cdots+m_d} I\left(n_0+m_0,n_1+m_1,\ldots,n_d+m_d\right)$$
Thank you! I wonder what is the definition of $\delta$ in $\delta\left(x_0+x_1+\cdots+x_d -1\right)\mathrm{d} x_0$? I can see you introduced $x_0$ and thus the region of integration is tranformed to $\mathbb{R}^d$, but I do not know how the new integrand is restricted by that $\delta$? – user31899 Aug 21 '12 at 6:36
I see now, through the Dirac delta and the fact $\int_{-\infty}^{\infty}g(x) \delta (x-a)\mathrm{d}x=f(a)$ – user31899 Aug 21 '12 at 8:14
Could you explain this: There are $d$ variables and thus the index of $(1-v_k)$ in the Jacobian is $d-k$ instead of $d+1-k$?How did you incorporate the effect of $x_0$ into the Jacobian? – user31899 Aug 21 '12 at 11:48
@user31899 Thank you for point out the typo. You are correct. It should be $d-k$. I have edited the most. – Sasha Aug 21 '12 at 11:54