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I am computing the following integrals by numerical integration and this takes a lot of time, although I'm sure there is a general closed-form formula but I can't find it.

Let $t$ be a vector of $\mathbb R_{+}^{d}$. For any integer $d \ge 1$, define $K_d$ as a convex subset of $\mathbb R^{d}$ by :

$$x \in K_d \iff \forall\, j \in {1,..,d}, \;x_j \ge 0 \text{ and }\sum\limits_{j=1}^d t_j x_j \le 1$$

I think $K_d$ is usually called a simplex, but I am not sure.

Let now $i$ be a vector of integers in $\mathbb{N}^{d}$, and consider the integral :

$$I_{i}^{d} = \int\limits_{K_{d}} \;\;\prod_{j=1}^{d} x_j^{i_j} \;\;\partial x_1,...,\partial x_d$$

Can we compute an expression for $I_{i}^{d}$? Maybe some recursion on $d$ or on $i$ can be found?

Edit:

I founded a paper that solves the problem, Lasserre - Simple formula for integration of polynomials on a simplex. it gives a formula a little more general, that reduces to the following :

$$\text{If and only if t_j = 1 for all j, }I_{i}^{d} = \frac{\prod\limits_{j=1}^{d} i_j}{(d+\lvert i \lvert)!}$$

Then, as a comment showed, generalisation can be done via

$$I_i^d(t) = \bigl(\prod_{j = 1}^d t_j^{-(i_j + 1)}\bigr)I_i^d(1)$$

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    $\begingroup$ I think that we have $I_i^d = \int_0^{1/t_d} x_d^{i_d}(1 - t_d x_d)^{\sum_{j = 1}^{d - 1} i_j}I_i^{d - 1}\mathrm dx_d$, but, if not that, then something very like it. $\endgroup$ – LSpice Sep 17 at 15:42
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    $\begingroup$ Nevermind, there is a formula there for exactly that : arxiv.org/pdf/1908.06736.pdf Thanks anyway ! $\endgroup$ – lrnv Sep 17 at 15:58
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    $\begingroup$ Glad you found it. You might consider posting it as an answer, if it totally addresses your question. \\ Your arXiv link, de-PDFd (and with the pun in its title unfurled for us to appreciate): Lasserre - Simple formula for integration of polynomials on a simplex. $\endgroup$ – LSpice Sep 17 at 17:19
  • $\begingroup$ Actualy i'm still stuck on the change of variables... I'll Edit, and maybe you could help ? $\endgroup$ – lrnv Sep 17 at 19:25
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    $\begingroup$ I think your last line should be $I_i^d(t) = \bigl(\prod_{j = 1}^d t_j^{-(i_j + 1)}\bigr)I_i^d(1)$. $\endgroup$ – LSpice Sep 18 at 17:10
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If $\Sigma_t=\{x_i \geq 0: x_1+\cdots +x_d=t\}$and $d\sigma_t$ is its surface measure, then $I_t=\int_{\Sigma_t}\prod_{i=1}^d x_i^{\alpha_i}\, d\sigma_t=t^{|\alpha|+d-1}I_1$ (change variable $x_i=ty_i$). Then for $\alpha_i>-1$ $$ \prod_{i=1}^d \Gamma (\alpha_i+1)=\int_{[0,\infty[^d} \prod_{i=1}^d x_i^{\alpha_i} e^{-(x_1+\cdots +x_d)}dx=\int_0^\infty \frac{e^{-t}}{\sqrt n} I_tdt=I_1\int_0^\infty t^{|\alpha|+d-1}\frac{e^{-t}}{\sqrt n} dt. $$ This gives $I_1=\frac{\sqrt{n} \prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d)}$ and $$\int_{\{x_i \geq 0, x_1+\cdots +x_d \leq 1\}} \prod_{i=1}^d x_i^{\alpha_i}dx=\int_0^1 \frac{dt}{\sqrt n}I_t=\frac{\prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d+1)}.$$

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