(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)](https://mathoverflow.net/questions/200690/estimating-a-selberg-type-integral)



Upper bound for a Selberg-type integral over a rectangular region
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](http://dlmf.nist.gov/8.12).


**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.


2. 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.**
3. 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 $n$.
4. 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 $c$, $N$.


