I have the following integral series, $$\int \cdots \int \prod_i AR_i ^{\alpha_i-s_i}dR_{k-1} \cdots dR_1$$ Which is part of the beta theorem, and here $R_k$ stands for $1-R_1-\cdots-R_{k-1}$, so we have $$\int_0^1 \cdots \int_0^{1-R_1\cdots R_{k-1}} \prod_i AR_i ^{\alpha_i-s_i}dR_{k-1} \cdots dR_1=A\frac{\prod_{i=1}^k\Gamma(\alpha_i-s_i+1)}{\Gamma(\sum_{i=1}^k(\alpha_i-s_i+1))}$$
I have that $$\alpha_i-s_i=\sum_{k=0}^{\infty}\log(1+ab^kx)-\sum_{k=0}^{\infty}\log(1+b^kx) \\ e^{\alpha_i-s_i}=\prod_{k=0}^{\infty}\frac{1+ab^kx}{1+b^kx}$$
When the above integral is standardised then, $$A=\frac{\Gamma(\sum_{i=1}^k(\alpha_i+1))}{\prod_{i=1}^k\Gamma(\alpha_i+1)}$$
and so I then have when $\sum_is_i=s$ $$A\frac{\prod_{i=1}^k\Gamma(\alpha_i-s_i+1)}{\Gamma(\sum_{i=1}^k(\alpha_i-s_i+1))}=\frac{\prod_{i=1}^k\Gamma(\alpha_i-s_i+1)}{\Gamma(\sum_{i=1}^k(\alpha_i-s_i+1))} \frac{\Gamma(\sum_{i=1}^k(\alpha_i+1))}{\prod_{i=1}^k\Gamma(\alpha_i+1)}=\frac{\left(\frac{\prod_{i=1}^k\Gamma(\alpha_i-s_i+1)}{ \prod_{i=1}^k\Gamma(\alpha_i+1)}\right)}{\left(\frac{\Gamma(\sum_{i=1}^k(\alpha_i-s_i+1))}{\Gamma(\sum_{i=1}^k(\alpha_i+1))}\right)} \\ = \frac{\prod_{i=1}^k \binom{-s_i+\alpha_i}{s_i}}{\binom{-s+\sum_{i=1}^k(\alpha_i+1)-1}{s}}$$