I'm wondering how to upper bound the following ratio of integrals:
$$\frac{\int\_{\Delta\_a}(\prod\_{i=1}^n\lambda\_i)^{p-1}\prod\_{i<j}|\lambda\_i-\lambda\_j|}{\int\_{\Delta\_b}(\prod\_{i=1}^n\lambda\_i)^{p-1}\prod\_{i<j}|\lambda\_i-\lambda\_j|}$$
in terms of $a,b,p,n$, where
$$\Delta\_a=\{(\lambda\_1,....,\lambda\_n):\sum\_{i=1}^n\lambda\_i\leq a,\lambda\_1\geq0,...,\lambda\_n\geq0\}$$
and $1\geq a>b>0, p>0$.
For example, when $n=1$, the ratio can be calculated exactly since $\prod\_{i<j}|\lambda\_i-\lambda\_j|=1$, which yields a bound of $(a/b)^p$. However, for $n\geq 2$, I have no idea how to do. I conjecture something like $(a/b)^{np}$ up to some multiplicative constant but do not know how to prove. 


I encountered this problem when trying to bound the ratio of small ball probability of the trace of matrix valued beta distribution which is needed in a Bayesian analysis.



Upper bound of a ratio of integrals
Let 
\begin{equation}
 J\_a:=\int\_{\Delta\_a}
 \Big(\prod\_{i=1}^nx\_i\Big)^{p-1}\prod\_{i<j}|x\_i-x\_j|\,\prod\_{i=1}^n dx\_i. 
\end{equation}
By the change of variables $x\_i=ay\_i$, we see that 
\begin{equation}
 J\_a=a^{np+n(n-1)/2}J\_1.
\end{equation}
Hence, your ratio of integrals is 
\begin{equation}
 J\_a/J\_b=(a/b)^{np+n(n-1)/2}.
\end{equation}
In particular, for $n=1$ we get your result, $(a/b)^p$. 


