# Upper bound for a Selberg-type integral over a rectangular region

(Cross-posted from math-SE).

I am trying to estimate the values of the following integral for large $n$,

$$\frac{1}{n!}\intop_{\Omega}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}dx_{j},$$ where $\Omega$ is one of these infinite rectangular regions in $\mathbb{R}^n$: $$\Omega_1^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2<\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\Omega_2^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\ldots$$ $$\Omega_n^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3>\lambda\ldots,\,x_n>\lambda\},$$ and $\lambda$ is a large positive fixed number. Based on the context where this question comes from, my guess is that the integral over $\Omega_k$ decays like $b^{-k}$ (with some $b>1$) when $\lambda$ is large enough. Moreover, this $b$ can supposedly be made as large as one wants by increasing $\lambda$.

The original question:

This question was motivated by Estimating a Selberg-type integral (or a Fredholm determinant)

Mathematica can do these integrals explicitly, for small $n$. One gets $$\frac1{n!}\int_{\Omega_1^{(n)}}\prod_{1\leq i<j\leq n}(x_i-x_j)^2 \prod_{i=1}^ne^{-x_i^2}\,dx = c_n\,\pi^{\frac{n-1}2} e^{-\lambda^2}\lambda^{2n-3}\left(1+O\left(\frac1{\lambda^2}\right)\right) \tag{1}$$ with $c_2=\frac14$, $c_3=\frac1{12}$, $c_4=\frac1{30}$, $c_5=\frac3{160}$, and $c_6=\frac{3}{128}$.

Here is a rough upper estimate which reproduces this result, but with too large a constant $c_n'$ (replacing $c_n\,\pi^{\frac{n-1}2}$): Write $x'=(x_2,\dots,x_n)$ with $r=|x'|$ and use $$\prod_{1\leq i<j\leq n}(x_i-x_j)^2\leq 2^{(n-2)(n-1)}(|x_1|+r)^{2n-2}r^{(n-2)(n-1)}. \tag{2}$$ Then, for any $\lambda>0$, \begin{aligned} \int_{\Omega_1^{(n)}}\prod_{1\leq i<j\leq n}(x_i-x_j)^2 \prod_{i=1}^ne^{-x_i^2}\,dx &\leq \int_\lambda^\infty\int_{{\mathbb R}^{n-1}}\prod_{1\leq i<j\leq n}(x_i-x_j)^2 \prod_{i=1}^ne^{-x_i^2}\,dx'dx_1 \\ &\leq \frac{2^{(n-2)(n-1)}n\,\pi^{\frac{n}2}}{\Gamma\left(\frac{n+2}2\right)}\, \int_\lambda^\infty\int_0^\infty(x_1+r)^{2n-2}r^{n(n-2)} e^{-x_1^2-r^2}\,drdx_1. \end{aligned} Now, \begin{aligned} \int_0^\infty (x+r)^{2n-2}r^{n(n-2)} e^{-x^2-r^2}\,dr &\leq 2^{2n-3}e^{-x^2}\left(x^{2n-2} \int_0^\infty r^{n(n-2)} e^{-r^2}\,dr + \int_0^\infty r^{n^2-2} e^{-r^2}\,dr\right) \\ &\leq 2^{2n-4}e^{-x^2} \left(\Gamma\left(\frac{n^2-2n+1}2\right)x^{2n-2} +\Gamma\left(\frac{n^2-1}2\right)\right) \end{aligned} and, therefore, \begin{aligned} & \int_\lambda^\infty \int_0^\infty (x+r)^{2n-2}r^{n(n-2)} e^{-x^2-r^2}\,drdx \\ & \qquad \leq 2^{2n-4}\int_\lambda^\infty e^{-x^2} \left(\Gamma\left(\frac{n^2-2n+1}2\right)x^{2n-2} +\Gamma\left(\frac{n^2-1}2\right)\right) dx \\ & \qquad = 2^{2n-5}\left(\Gamma\left(\frac{2n-1}2,\lambda^2\right) \Gamma\left(\frac{n^2-2n+1}2\right) + \Gamma\left(\frac12,\lambda^2\right) \Gamma\left(\frac{n^2-1}2\right)\right). \end{aligned} Next one needs to bound the incomplete $\Gamma$ function $\Gamma(\alpha,\mu)$ as $\alpha\to\infty$ uniformly in the parameter $\mu$. Such estimates can be found here.

Note. 1. Knowing the asymptotics $$\int_\lambda^\infty \int_0^\infty (x+r)^{2n-2}r^{n(n-2)} e^{-x^2-r^2}\,dr dx = \frac14\,\Gamma\left(\frac{(n-1)^2}2\right) e^{-\lambda^2}\lambda^{2n-3} \left(1+O\left(\frac1\lambda\right)\right)$$ as $\lambda\to\infty$ is less effective, as the dependence of the remainder on $n$ is not exhibited.

1. Without having done these computations, I would say that one has to be more sensitive in estimate (2). For instance, expanding $\prod_{1\leq i< j\leq n}(x_i-x_j)^2$, all the terms $\prod_{i=1}^n x_i^{k_i}$ (notice that $\sum_{i=1}^nk_i=n(n-1)$) with (at least) one $k_i$ being odd contribute nothing to the integral over $(\lambda,\infty)\times{\mathbb R}^{n-1}$. Once the combinatorics is done, the method as outlined should work.

2. With more effort, I belief that one can work out (1) for all $n$ (including a formula for the $c_n$). Still, this does not settle the question on the dependence of the remainder on $n$.

3. Flipping over one $x_i<\lambda$ to $x_i>\lambda$ introduces another factor $c\,e^{-\lambda^2}\lambda^N$ in the asymptotics as $\lambda\to\infty$, for some constant $c>0$ and integer $N$ (with varying $c$, $N$). This probably can be worked out, too, but it will get messy rather quickly if one wants to know these $c$, $N$.

• Thanks! It hadn't occurred to me that Mathematica could shed some light on this. I clarified my question and explained the background for it. The most interesting part is actually how these integrals decrease when we flip more of the inequalities. Moreover, the asymptotics of the classical Selberg integral may provide stronger bounds for your constants $c_n$. Proving the appearance of a $e^{-k \lambda^2}$ factor in the integral with $k$ flipped ineqs would be great. – level1807 Mar 21 '15 at 18:44
• @level1807 As I understand this site, you should leave your question as originally posted, see e.g here. If necessary, ask a new question. – ifw Mar 21 '15 at 23:14
• Thanks for the info, I've separated that into a new question. Your calculation is still a very good starting point! – level1807 Mar 22 '15 at 6:56
• I think you forgot the measure of the n-sphere, which is an additional factor of $2\pi^{\frac{n-1}{2}}/\Gamma(\frac{n-1}{2})$ – level1807 Mar 22 '15 at 13:13