Mehta integral and orthogonality 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, 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 $n$?
 A: Case 3:
Let me define $t=x\sqrt{2\gamma}$, then it is known from random-matrix theory (see, for example, Forrester's book) 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 (2014), see also arXiv: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 $\alpha\gg 1$.
