Integral over a simplex Let $C_k$ be the $k$-simplex. I know that
$$\int_{C_k} \prod_{i=1}^k x_i^{\alpha_i-1} dx_i = \frac{\prod_{i=1}^k \Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^k \alpha_i\right)} \equiv B(\alpha_1,\ldots,\alpha_k)$$
where $B$ is called the multinomial beta function.
What is the following integral?
$$\int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}$$
 A: Let's suppose that we are aware of the integral
$$\int_{C_k} \prod_{i=1}^k x_i^{\alpha_i-1} dx_i = \frac{\prod_{i=1}^k \Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^k \alpha_i\right)}  
$$
$(\textbf{1})$
(See here for a proof).
It turns out that only thing we need is to rewrite properly the inner one of the two products:
$$
P(k)=\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1} \text{ =?}
$$ 
Let's try the easy example $k=3$, (assuming the $\alpha_{ij}$ are symmetric)we get
$$
(x_1x_1)^{\alpha_{11}-1}(x_1x_2)^{\alpha_{12}-1}(x_1x_3)^{\alpha_{13}-1}(x_2x_2)^{\alpha_{22}-1}(x_2x_3)^{\alpha_{23}-1}(x_3x_3)^{\alpha_{33}-1}
$$
which equals
$$
x_1^{2\alpha_{11}+\alpha_{12}+\alpha_{13}-4}x_2^{2\alpha_{22}+\alpha_{12}+\alpha_{23}-4}x_3^{2\alpha_{33}+\alpha_{13}+\alpha_{23}-4}
$$
or 
$$
P(3)=\prod_i^3 x_i^{2\alpha_{ii}+\sum_{j\neq i}^3\alpha_{ij}-4}
$$
It is obvious  (or can be shown by induction) that one can generalize this to arbitrary $k$
$$
P(k)=\prod_{i=1}^k x_i^{\alpha_{ii}-1+\sum_{j=1}^k(\alpha_{ij}-1)}
$$
Plugging this expression into our integral (using $\prod_ib_i\prod_i a_i=\prod_i a_i b_i$) we see that it is given by
$$\int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}=\int_{C_k} \prod_{i=1}^k dx_ix_i^{\alpha_{ii}-1+\sum_{j=1}^k(\alpha_{ij}-1)}$$
which allows us to just use $(\textbf{1})$, replacing $\alpha_i$ by $\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)$
$$
\int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}=\frac{\prod_{i=1}^k \Gamma\left(\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)\right)}{\Gamma\left(\sum_{i=1}^k \left[\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)\right]\right)}
$$
Edit: 
It appears that in my original answer, i assumed implicitly that the $a_{ij}$ are symmetric $\alpha_{ij}$=$\alpha_{ji}$. If this is not the case, one has to modify the answer as follows: 
Replace $\sum_{j\neq i}^k\alpha_{ij}$ by $\sum_{j\neq i}^k\left(\Theta(i-j)\alpha_{ij}+\Theta(j-i)\alpha_{ji}\right)$
where $\Theta(x)$ is Heaviside's step function. 
Please note also that this approach is generalizable to integrals like
$$
\int_{C_k} \prod_{i=1}^k dx_i\prod_{i_1, i_2, \dot .......,i_n=i}^k (x_{i_1} x_{i_2}......  x_{i_n})^{\alpha_{i_1 i_2...i_n}-1}
$$
