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.

Let $T^d:=\{(x_1,..,x_d):x_i \geq 0, \sum_{i=1}^{d}x_i \leq 1\}$ be the standard simplex in $\mathbb{R}^d$. Compute the integral $$\int_{T^d} x_1^{\nu_1-1}x_2^{\nu_2-1}...x_d^{\nu_d-1}(1-x_1-...-x_d)^{\nu_0-1}$$ where $\nu_i>0$.

Remark: I know the answer is $$\frac{\prod_{i=0}^{d}\Gamma(\nu_i+1)}{\Gamma(\sum_{i=0}^{d}\nu_i)}$$. I evaluated for the case $d=2$ by using the transformation $(p-1)\iiint\limits_{T^{3}} x^{m-1}y^{n-1}z^{p-2} \mathrm{d}z\mathrm{d}y\mathrm{d}x= \iint\limits_{T^{2}} x^{m-1}y^{n-1}(1-x-y)^{p-1}\mathrm{d}y\mathrm{d}x$

and the substitutions $ \left\{\begin{matrix}x=u^2& &\\y=v^2& &\\z=w^2& &\end{matrix}\right.$and $ \left\{\begin{matrix}u=r\sin\varphi\cos\theta& &\\v=r\sin\varphi\sin\theta& &\\w=r\cos\varphi& &\end{matrix}\right.$. But this method is complex for computing the general case.

share|improve this question
2  
Your indices are off by one. In the present form, the last factor is a (possibly negative) power of $0$, and the answer you suggest doesn't depend on $a_0$. Also note that punctuation at the end of a displayed equation needs to go inside the double dollar signs (preferably set off from the equation by a \; space) since otherwise it ends up on the next line. –  joriki Oct 4 '12 at 12:00
    
i think these are dirichlet integrals, maybe mathworld.wolfram.com/DirichletIntegrals.html –  mike Oct 4 '12 at 14:24
    
What are the elements in the sum when you define $T^d$? –  Davide Giraudo Oct 4 '12 at 20:11
    
Edited. It's the standard simplex. The integral is a generalised Beta function (multinomial type), I have not found any literature about how to compute it –  user31899 Oct 4 '12 at 20:26

2 Answers 2

Put $(n_0,n_1,\ldots, n_d)=:n$, $\sum_{k=0}^d n_k=:|n|$ and for $\lambda\geq0$ define $$Q(n,\lambda):=\int\nolimits_{\lambda T^d}\prod_{1\leq k\leq d} x_k^{n_k-1} \ (\lambda -x_1-\ldots -x_d)^{n_0-1}\ {\rm d}(x)\ .$$ The substitution $x:=\lambda y\ (y\in T^d)$ gives $$Q(n,\lambda)=\lambda^{d+|n|-(d+1)}\int\nolimits_{T^d}\prod_{1\leq k\leq d} y_k^{n_k-1} \ (1 -y_1-\ldots -y_d)^{n_0-1}\ {\rm d}(y)=\lambda^{|n|-1} Q(n,1)\ .$$ Letting the "outer integration" be with respect to the last variable we now obtain $$\eqalign{Q(n,1)&=\int_0^1 x^{n_d-1}\int\nolimits_{(1-x)T^{d-1}}\prod_{1\leq k\leq d-1} x_k^{n_k-1} (1-x_1-\ldots-x_{d-1}-x)^{n_0-1}\ {\rm d}(x')\ dx\cr &= \int_0^1 x^{n_d-1} Q(n',1-x)\ dx=Q(n',1)\int_0^1 x^{n_d-1}(1-x)^{|n'|-1}\ dx\ . \cr}$$ Here the last integral evaluates to $B(|n'|,n_d)$, so that we end up with the recursion $$Q(n,1)=Q(n',1) B(|n'|,n_d)\ .$$ From here it should not be too difficult to arrive at the desired result.

share|improve this answer

For any $a \in \mathbb{R}-\{0\}$ and $m,n \in (0,+\infty)$ one has \begin{align*} \frac{1}{a^{m+n-1}}\int_{0}^{a}y^{m-1}(a-y)^{n-1}\mathrm{d}y&=\frac{1}{a^{m+n-1}}\int_{0}^{a}a^{m+n-2}\left(\frac{y}{a}\right)^{m-1}\left(1-\frac{y}{a}\right)^{n-1}\mathrm{d}y\\&=\frac{1}{a}\int_{0}^{a}\left(\frac{y}{a}\right)^{m-1}\left(1-\frac{y}{a}\right)^{n-1}\mathrm{d}y\\&=\int_{0}^{1}x^{m-1}(1-x)^{n-1}\mathrm{d}x \end{align*} Thus $\int_{0}^{a}y^{m-1}(a-y)^{n-1}\mathrm{d}y=a^{m+n-1}B(m,n)$. With above observation we can integrate $x_1^{\nu_1-1}...x_d^{\nu_d-1}(1-x_1-...-x_d)^{\nu_{0}-1}$ over $T^{d}$ by integrating out variables one at each step \begin{align*} &\mathrel{\phantom{=}} \int_{T^{d}}x_1^{\nu_1-1}...x_d^{\nu_d-1}(1-x_1-...-x_d)^{\nu_{0}-1}\mathrm{d}\boldsymbol x\\ &=\int_{0}^{1}\int_{0}^{1-x_1}...\int_{0}^{1-x_1-...-x_{d-1}}x_1^{\nu_{1}-1}x_2^{\nu_{2}-1}...x_d^{\nu_{d}-1}(1-x_1-...x_d)^{\nu_{0}-1}\mathrm{d}x_d...\mathrm{d}x_2\mathrm{d}x_1\\ &=\int_{0}^{1}x_1^{\nu_{1}-1}\int_{0}^{1-x_1}x_2^{\nu_{2}-1}...\int_{0}^{1-x_1-...-x_{d-1}}x_d^{\nu_{d}-1}(1-x_1-...x_d)^{\nu_{0}-1}\mathrm{d}x_d...\mathrm{d}x_2\mathrm{d}x_1\\&=B(\nu_0,\nu_d)\int_{0}^{1}x_1^{\nu_{1}-1}\int_{0}^{1-x_1}x_2^{\nu_{2}-1}...\int_{0}^{1-x_1-...-x_{d-2}}x_{d-1}^{\nu_{d-1}-1}(1-x_1-...x_{d-1})^{\nu_{0}+\nu_{d}-1}\mathrm{d}x_{d-1}...\mathrm{d}x_2\mathrm{d}x_1\\&=B(\nu_{d-1},\nu_{0}+\nu_{d})B(\nu_0,\nu_d)\int_{0}^{1}x_1^{\nu_{1}-1}...\int_{0}^{1-x_1...-x_{d-2}}x_{d-2}^{\nu_{d-2}-1}(1-x_1-...-x_{d-2})^{\nu_{0}+\nu_{d-1}+\nu_{d}-1}\mathrm{d}x_{d-2}...\mathrm{d}x_1\\&=...\\&=B(\nu_1,\nu_0+\nu_d+\nu_{d-1}+...+\nu_{2})B(\nu_2,\nu_0+\nu_d+\nu_{d-1}+...+\nu_{3})...B(\nu_{d-1},\nu_0+\nu_d)B(\nu_{0},\nu_d)\\&=\frac{\Gamma(\nu_{0})\Gamma(\nu_{1})...\Gamma(\nu_{d})}{\Gamma(\nu_{0}+\nu_{1}+...+\nu_{d})}\\&=\frac{\Gamma(\boldsymbol {\nu})}{\Gamma(|\boldsymbol {\nu}|)} \end{align*}

share|improve this answer

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.