Definite integral over a simplex

Solution 1:

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*}

Solution 2:

Another, elegant way to perform your integration is explained by Jaynes in §18.10 of "Probability Theory: The Logic of Science". It uses a Laplace transform:

Slightly modifying your notation, your integral can be rewritten as

$$I(r) := \int_{[0,\infty)^n} x_1^{a_1} \dotsm x_n^{a_n} \,\mathrm{\delta}(r-x_1-\dotsb-x_n)\, \mathrm{d}x_1\dotsm\mathrm{d}x_n\qquad\text{with $r=1$}.$$

Note that the delta allows us to extend the integration to $\infty$, and renders the integration bounds independent from one another.

Now take the Laplace transform of $I(r)$ and eliminate the delta from it:

$$\int_0^{\infty} \exp(-zr) I(r)\,\mathrm{d}r = \int_{[0,\infty)^n} x_1^{a_1} \dotsm x_n^{a_n} \,\exp[-z(x_1+\dotsb+x_n)]\, \mathrm{d}x_1\dotsm\mathrm{d}x_n \\= \prod_{i=1}^n\int_0^\infty x_i^{a_i}\,\exp(-z x_i)\,\mathrm{d}x_i =\prod_{i=1}^n\frac{a_i!}{z^{a_i+1}}.$$

Now take the inverse Laplace transform of the last expression to obtain back $I(r)$:

$$I(r) = \frac{1}{2\pi\mathrm{i}}\int_{-\mathrm{i}\infty}^{+\mathrm{i}\infty} \exp(rz)\,\prod_{i=1}^n\frac{a_i!}{z^{a_i+1}}\,\mathrm{d}z \\=\frac{\prod_{i=1}^n a_i!}{(n+\sum_ia_i-1)!} r^{n+\sum_ia_i-1}$$

and evaluate it at $r=1$. The path of integration of the last integral "passes to the right of the origin, and is closed by an infinite semicircle over the left half-plane" (Jaynes).

Hope this helps.