# Definite integral over a simplex

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.

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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

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*}
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.