# Scaling in Mehta's integral

The following expression is known as Mehta's integral and deeply connected to random matrix theory:

$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$

An interesting question is what happens if one assumes $$\gamma$$ to be a function of $$n.$$ For example by choosing $$\gamma=1/n$$ one finds that as $$n$$ tends to infinity, the value of the integral tends to zero whereas for $$\gamma=1/n^2$$ the value of the integral approaches a positive constant value as $$n$$ tends to infinity.

These properties one can deduce from the asymptotics of the product of gamma functions. I would like to ask:

It is not too surprising that for some suitable scaling $$\gamma=1/n^{\alpha}$$ one approaches a constant value, as $$\vert t_i-t_j \vert^{1/n} \xrightarrow 1$$ for fixed $$t_i,t_j$$ and

$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} dt_1 \cdots dt_n =1.$$

Can one also conclude these two properties from the integral directly without evaluating it?

Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, your integral is $$E\prod_{1\le i where the $$X_i$$'s are independent standard normal random variables. Introducing $$N:=n(n-1)/2$$, $$X:=(X_1,\dots,X_n)$$, and $$\|X\|:=\sqrt{\sum_1^n X_i^2}$$, and then using the arithmetic-geometric-mean inequality, we have $$\prod_{1\le i for any real $$C>0$$ and all large enough $$n$$, given that $$\gamma=o(1/n^2)$$.
Note also that $$\|X\|^2$$ has the gamma distribution with parameters $$n/2$$ and $$2$$ and hence $$E\|X\|^{2C}=O(n^C)$$. So, $$E\prod_{1\le i and, similarly, $$E\Big[\Big(\prod_{1\le i Also, obviously, $$t^2/t\to\infty$$ as $$t\to\infty$$. So, we have the uniform integrability.
Also, $$\prod_{1\le i and $$E\Big|\sum_{1\le i so that, by Markov's inequality, $$\sum_{1\le i in probability. So, by the condition $$\gamma=o(1/n^2)$$ and (1), $$\prod_{1\le i in probability. Thus, by the uniform integrability, $$E\prod_{1\le i
• @SolidStatePhysicist : You were right. I had indeed overlooked that the dimension of the integral, $n$, goes to infinity. This is now fixed. Instead of the integration over $\mathbb R^n$, with the variable $n$, we now use the expectation, which is the integration over a fixed background probability space. Instead of dominated convergence, we now use uniform integrability, which complicates the reasoning just a bit. – Iosif Pinelis Jan 23 at 4:00