Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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: $$ \int\limits_{\mathbb{R}^d_+} \prod_{k=0}^d (1-x_k) x_k^{n_k} \delta\left(x_0+x_1+\cdots+x_d -1\right)\mathrm{d} x_0 \mathrm{d} x_1 \cdots \mathrm{d} x_d = \\ \sum_{m_0=0}^1 \sum_{m_1=0}^1 \cdots \sum_{m_d=0}^1 (-1)^{m_0+m_1+\cdots+m_d} I\left(n_0+m_0,n_1+m_1,\ldots,n_d+m_d\right) $$

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.