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)
 A: 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.


*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.

*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 $n$.

*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 $c$, $N$.
