I'm considering a ratio of incomplete [Selberg integral](https://en.wikipedia.org/wiki/Selberg_integral):
$$f\_n(a,b)=\frac{\int\_{\Delta\_a}\prod\_{i=1}^nx\_i^{\alpha-\frac{n+1}{2}}\prod\_{i=1}^n(1-x\_i)^{-1/2}\prod\_{i<j}|x\_i-x\_j|}{\int\_{\Delta\_b}\prod\_{i=1}^nx\_i^{\alpha-\frac{n+1}{2}}\prod\_{i=1}^n(1-x\_i)^{-1/2}\prod\_{i<j}|x\_i-x\_j|}$$
where
$$\Delta\_a=\{(x\_1,...,x\_n):0<x\_1<a,0<x\_2<a,...,0<x\_n<a\}$$
and $1>a>b>0$, $\alpha>\frac{n+1}{2}$.


My question is, how can we upper bound $f\_n(a,b)$? Is there a bound like some powers of $Ca/b$ for some constant $C>0$? 



Ratio of Selberg integral
Denote the numerator by $I(a)$. The change of variables $x\_i=ay\_i$ gives 
$$
I(a)=a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\times \\
\times\int\_{[0,1]^n}\prod\_{i=1}^ny\_i^{\alpha-\frac{n+1}{2}}\prod\_{i=1}^n(1-ay\_i)^{-1/2}\prod\_{i<j}|y\_i-y\_j|.
$$
The latter integral may be estimated using the two-sided estimates $1\leqslant (1-ay\_i)^{-1/2}\leqslant (1-y\_i)^{-1/2}$. So, we get $$c\_1(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\leqslant I(a)\leqslant c\_2(n) a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2},$$
where $c\_1(n)$ and $c\_2(n)$ are corresponding integrals (actually $c\_2(n)=I(1)$).
It implies
$$
\frac{I(a)}{I(b)}\leqslant \frac{c\_2(n)}{c\_1(n)}\cdot (a/b)^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}.
$$


You may get some explicit bounds for $c\_2(n)/c\_1(n)$
, for example as follows. 
We have
$$
c\_2(n)=I(1)=\int\_{[0,1]^n}\prod\_{i=1}^ny\_i^{\alpha-\frac{n+1}{2}}\prod\_{i=1}^n(1-y\_i)^{-1/2}\prod\_{i<j}|y\_i-y\_j| dy\_1\ldots dy\_n,
$$
and $c\_1(n)$ the same integral without $\prod\_{i=1}^n(1-y\_i)^{-1/2}$. In the integral for $c\_2(n)$, change the variables: take $y\_i=1-t\_i^2$. You get
$$
c\_2(n)=2^n\int\_{[0,1]^n}\prod\_{i=1}^n(1-t\_i^2)^{\alpha-\frac{n+1}{2}}\prod\_{i<j}|t\_i^2-t\_j^2| dt\_1\ldots dt\_n.
$$
Use the bounds $1-t\_i^2=(1-t\_i)(1+t\_i)\leqslant 2(1-t\_i)$, $|t\_i^2-t\_j^2|=|t\_i-t\_j|(t\_i+t\_j)\leqslant 2|t\_i-t\_j|$. We get 
$$
c\_2(n)\leqslant 2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2}
\int\_{[0,1]^n}\prod\_{i=1}^n(1-t\_i)^{\alpha-\frac{n+1}{2}}\prod\_{i<j}|t\_i-t\_j| dt\_1\ldots dt\_n=\\
=2^{n(\alpha-\frac{n+1}{2})+\frac{n(n+1)}2} c\_1(n),
$$
the last equality follows from the change of variables $1-t\_i=z\_i$.


Is it ok for you or you need the estimates which are explicit w.r.t. $n$?


