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? 
 A: Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that 
\begin{equation}
 \gamma n^2\to a
\end{equation}
(as $n\to\infty$) for some real $a\ge0$. 
Your integral is 
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$
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 
$$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}
\le\Big(\frac1N\,\sum_{1\le i<j\le n}|X_i-X_j|^2\Big)^{N\gamma} \\ 
=O\Big(\frac{\|X\|^2}n\Big)^{N\gamma}=O\Big(1+\frac{\|X\|^2}n\Big)^C
$$
for $C:=a/2+1$ and all large enough $n$. 
Note also that $\|X\|^2$ has the gamma distribution with parameters $n/2$ and $2$ and hence $E\|X\|^{2C}=O(n^C)$. So, 
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$
and, similarly, 
$$E\Big[\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2\Big]=O(1).$$
Also, obviously, $t^2/t\to\infty$ as $t\to\infty$. So, we have the uniform integrability. 
So, we only need to establish the convergence of 
\begin{equation*}
 {2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|=2\gamma NU_n
\end{equation*}
in probability, where 
\begin{equation*}
 U_n:=\frac1N\,\sum_{1\le i<j\le n}h(X_i,X_j) 
\end{equation*}
is a so-called U-statistic with kernel $h(X_i,X_j):=\ln|X_i-X_j|$, and still $N=\binom n2=n(n-1)/2$. It is easy to see (cf. e.g. page 20) that $Var\,U_n=O(1/n)=o(1)$, whereas 
$$EU_n=m:=E\ln|X_1-X_2|.$$ 
So, $U_n\to m$ in probability, whence 
\begin{equation*}
 2\gamma NU_n\to am
\end{equation*}
in probability and thus, by the uniform integrability,
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to e^{am}=\exp\{a\,E\ln|X_1-X_2|\}$$ 
as $n\to\infty$. 
