# Ratio of Selberg integral

I'm considering a ratio of incomplete Selberg integral: $$f_n(a,b)=\frac{\int_{\Delta_a}\prod_{i=1}^nx_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-x_i)^{-1/2}\prod_{i where $$\Delta_a=\{(x_1,...,x_n):0 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 $$C>0$$?

• I have changed $\Delta_a$ to be a hypercube instead of the previous hyper rectangle. May 5, 2019 at 15:54
• if $C$ may depend on $n$, then of course we have $C_n(a/b)^M$ upper bound, where $a=n(\alpha-\frac{n+1}2)+{n\choose 2}$. May 5, 2019 at 21:31
• Sorry I do not quite get it. May 7, 2019 at 0:34
• I can't write a detailed answer at this time, so here is instead a sketch. If you normalize by $\Delta_1$ both integrals, then what you are asking is the ratio of probabilities that all eigenvalues are smaller than $a$ and the same with $b$. Since you have a large deviation principle at scale $n^2$ for the empirical measure of eigenvalues, with rate function $I(\mu)$, with a function $I$ that is explicit (involving the non-commutative entropy of $\mu$ - see section 2.6 in Anderson-Guionnet-Zeitouni's book on RMT), you can read of the answer: May 13, 2019 at 10:56
• $n^{-2} \log f_n(a,b)\to -\inf_{\mu: \mu([0,a])=1} I(\mu)+\inf_{\mu:\mu([0,b])=1} I(\mu)$. Just for completeness, the function $I$ is the following: $$I(\mu)=-\int\int \log|x-y|\mu(dx)\mu(dy)-\frac{1}{2}\int \log(x) \mu(dx)$$ May 13, 2019 at 11:00

Denote the numerator by $$I(a)$$. The change of variables $$x_i=ay_i$$ gives $$I(a)=a^{n(\alpha-\frac{n+1}2)+\frac{n(n+1)}2}\times \\ \times\int_{[0,1]^n}\prod_{i=1}^ny_i^{\alpha-\frac{n+1}{2}}\prod_{i=1}^n(1-ay_i)^{-1/2}\prod_{i The latter integral may be estimated using the two-sided estimates $$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 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 $$c_2(n)=2^n\int_{[0,1]^n}\prod_{i=1}^n(1-t_i^2)^{\alpha-\frac{n+1}{2}}\prod_{i Use the bounds $$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 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. $$n$$?
• Yeah... I think I need to bound the ratio $c_1(n)/c_2(n)$... How can we do that? May 8, 2019 at 3:31