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 

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