The Mehta integral is the following expression:


$$\frac{1}{(2\pi)^{n/2}}\int\_{-\infty}^{\infty} \cdots \int\_{-\infty}^{\infty} \prod\_{i=1}^n e^{-t\_i^2/2}
\prod\_{1 \le i < j \le n} |t\_i - t\_j |^{2 \gamma} dt\_1 \cdots dt\_n =\prod\_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$


To simplify the notation, we introduce a measure $$d\mu\_{n,\gamma}(t):=\prod\_{i=1}^n e^{-t\_i^2/2} 
\prod\_{1 \le i < j \le n} |t\_i - t\_j |^{2 \gamma} \ dt. $$
It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure
\begin{equation}
\label{eq:ortho}
\frac{1}{(2\pi)^{1/2}}\int\_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0.
\end{equation}


Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of


$$\nu:=\lim\_{n \rightarrow \infty}\frac{\int (t\_1^2-1) \ d\mu\_{n,\gamma}(t)}{\sqrt{\int (t\_1^2-1)^2 \ d\mu\_{n,\gamma}(t)}\sqrt{\int 1 \ d\mu\_{n,\gamma}(t)}}$$


where I admit that I use the limit without actually knowing whether it exists.


Case 1:


As one can guess from [this thread](https://mathoverflow.net/questions/350980/scaling-in-mehtas-integral?rq=1), if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$


Case 2:


If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.


Case 3:


If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.


**My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?**



Mehta integral and orthogonality
**Case 3:**


Let me define $t=x\sqrt{2\gamma}$, then it is known from random-matrix theory (see, for example, [Forrester's book](https://press.princeton.edu/books/hardcover/9780691128290/log-gases-and-random-matrices-lms-34)) that for a fixed $\gamma$ the probability distribution $P(x\_1)$ of a single eigenvalue $x\_1$ tends in the limit $n\rightarrow\infty$ to the $\gamma$-independent semicircle
$$P(x)=\frac{1}{\pi n}\sqrt{2n-x^2},\;\;|x|\leq\sqrt{2n}.$$
The desired ratio $\nu$ then evaluates to
$$\nu=\frac{\int (2\gamma x^2-1)P(x)\,dx}{\left[\int (2\gamma x^2-1)^2P(x)\,dx\right]^{1/2}}=\frac{\gamma n-1}{\sqrt{2 \gamma n (\gamma n-1)+1}}\rightarrow \frac{1}{\sqrt 2}\;\;\text{for}\;\;n\rightarrow\infty.$$


**Case 2:**


The case that $n\rightarrow\infty$, $\gamma\rightarrow 0$ at fixed $\gamma n=\alpha>0$ has been studied in [The mean spectral measures of random Jacobi matrices related to Gaussian beta ensembles](https://projecteuclid.org/euclid.ecp/1465320995) (2014), see also [arXiv:1611.09476](https://arxiv.org/abs/1611.09476). The probability distribution $P\_\alpha(t)$ is given in this limit by
$$P\_\alpha(t)=\frac{e^{-t^2/2}}{\alpha\sqrt{2\pi}}\frac{\Gamma(\alpha)}
{|f(t)|^2},\;\;f(t)=\int\_0^\infty x^{\alpha-1}e^{ix t-x^2/2}\,dx.$$
From this the desired $\nu$ can be readily computed, 
$$\nu\_\alpha=\frac{\int (t^2-1)P\_\alpha(t)\,dt}{\left[\int (t^2-1)^2P\_\alpha(t)\,dt\right]^{1/2}}=\frac{\alpha}{\sqrt{\alpha (2 \alpha+3)+2}},$$
so for $\alpha=1$ I find $\nu\_1=1/\sqrt 7$. The value $\nu=1/\sqrt 2$ of case 3 is reached for $\alpha\gg 1$.


