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](https://arxiv.org/abs/1908.06736). 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


$$I\_i^d(t) = \bigl(\prod\_{j = 1}^d t\_j^{-(i\_j + 1)}\bigr)I\_i^d(1)$$



Is there a closed-form expression for these integrals?
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 $$\int\_{\{x\_i \geq 0, x\_1+\cdots +x\_d \leq 1\}} \prod\_{i=1}^d x\_i^{\alpha\_i}dx=\int\_0^1 \frac{dt}{\sqrt n}I\_t=\frac{\prod\_{i=1}^d \Gamma (\alpha\_i+1)}{\Gamma (|\alpha|+d+1)}.$$


