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 $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$ 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 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, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$ 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.

Also, $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} =\exp\Big({2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|\Big)\tag{1}$$ and $$E\Big|\sum_{1\le i<j\le n}\ln|X_i-X_j|\Big|\le\sum_{1\le i<j\le n}E|\ln|X_i-X_j|\,|=O(n^2), $$ so that, by Markov's inequality, $\sum_{1\le i<j\le n}\ln|X_i-X_j|=O(n^2)$ in probability. So, by the condition $\gamma=o(1/n^2)$ and (1), $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} \to1$$ in probability. Thus, by the uniform integrability, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to1.$$

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    $\begingroup$ interesting, but how did you estimate this product? Also, the dimension of the integral changes, which is perhaps a bit different from the usual DCT. $\endgroup$ – Solid State Physicist Jan 23 at 1:47
  • $\begingroup$ @SolidStatePhysicist : I have added the requested details. $\endgroup$ – Iosif Pinelis Jan 23 at 1:56
  • $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Jan 23 at 4:00
  • $\begingroup$ @IosifPinelis thank you, that's very nice. $\endgroup$ – Xin Wang Jan 23 at 12:42

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