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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 $n$?

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  • $\begingroup$ @CarloBeenakker you are right, simplifying the question I accidentally deleted it. $\endgroup$ – Solid State Physicist Feb 15 at 22:44
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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 $\alpha\gg 1$.

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  • $\begingroup$ thank you, that's interesting. Do you have any conjectures about case 2 whether the result will be of "order 1" or "tend to zero"? $\endgroup$ – Solid State Physicist Feb 15 at 15:13
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    $\begingroup$ my "conjecture" would be to substitute $\gamma n=\text{constant}$ in the formula for $\nu$, which would give $\nu=0$ in case 2; incidentally, the plots you show do not take into account the repulsion between the $x_i$'s with $i$ unequal to 1, so I don't think one can draw any conclusion from those. $\endgroup$ – Carlo Beenakker Feb 15 at 15:35
  • $\begingroup$ thank you for your prompt response. This sounds very plausible. I will update the question accordingly. $\endgroup$ – Solid State Physicist Feb 15 at 15:48
  • $\begingroup$ very interesting, so in Case 3: we do not see $\gamma$ at all in the limit and in this regime $2$, although the value is of order $1$, it vanishes as $\alpha$ tends to zero. $\endgroup$ – Solid State Physicist Feb 16 at 15:16
  • $\begingroup$ yes, this somehow makes sense if you think of case 2 approaching case 3 when $\alpha\rightarrow\infty$ (hence $\nu\rightarrow 1/\sqrt 2$), and approaching case 1 if $\alpha\rightarrow 0$ (hence $\nu\rightarrow 0$). $\endgroup$ – Carlo Beenakker Feb 16 at 15:30

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