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Let $\Delta(s_1,s_2,\ldots,s_n) := \prod_{i<j}(s_i-s_j)^2$. Is there a standard way to estimate the decay of the Selberg-type integral $$ I_n:= \frac{1}{n!^2}\int_0^1 \int_0^1\cdots\int_0^1 \frac{\Delta(s_1,s_2,\ldots,s_n) \Delta(t_1,t_2,\ldots,t_n)}{\prod_{i,j}(1-s_i t_j)^2} d s_1\ldots d s_n d t_1 \ldots d t_n$$ as $n\to\infty$ ?

Checking numerically for $n\leq 25$, the decay seems to be like $e^{-2 n^2}$.

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    $\begingroup$ Very interesting question! $\endgroup$ – Marc Palm Mar 17 '12 at 12:48
  • $\begingroup$ Btw, how did you get interested in this problem? $\endgroup$ – Adrien Hardy Jan 2 '18 at 14:16
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You can follow large deviation-type estimates from random matrix theory, starting with Ben Arous & Guionnet's paper: https://link.springer.com/article/10.1007/s004400050119

You will eventually obtain that your Selberg-type integral behaves like $$ \frac{1}{(n!)^2}e^{-n^2 E_*} $$ where $E_*$ is the minimum of the functional \begin{multline} E(\mu,\nu)=\iint \log\frac1{|x-y|}d\mu(x)d\mu(y)\\+\iint \log\frac1{|x-y|}d\nu(x)d\nu(y)\\-2\iint\log\frac1{|1-xy|}d\mu(x)d\nu(y) \end{multline} where $(\mu,\nu)$ ranges over pairs of Borel probability measures supported in $[0,1]$. I guess you can work out the existence and unicity of a minimizer for $E$ and obtain Euler-Lagrange equations to characterize this minimiser and, hopefully, compute explicitly the minimal value $E_*$.

There is quite some technical work to do to fill the gaps, and sorry for the self-advertisement, but if you follow this approach let me tell you that I had to struggle with a similar two type particle problem in the paper http://www.worldscientific.com/doi/abs/10.1142/S2010326312500165 written with Arno Kuijlaars. I hope this can help.

By the way, minimizing $E$ is referred as to a vector equilibrium problem in potential theory.

ADDENDUM: After thinking further, it seems more natural to make the changes of variables $s_i=1/x_i$, which leads to the same asymptotics but with $E_*$ the minimum of \begin{multline} \tilde E(\mu,\nu)=\iint \log\frac1{|x-y|}d(\mu-\nu)(x)d(\mu-\nu)(y)+\int_1^\infty \log(x)d\mu(x) \end{multline} where $\mu,\nu$ are Borel probability measures living on $[1,\infty)$ and $[0,1]$ respectively.

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  • $\begingroup$ Are you saying that the integral without the $n!^{-2}$ factor decays like $e^{-n^2 E_*}$? Is it clear that $E_*$ is positive, and do you have any numerical lower bound for it? $\endgroup$ – GH from MO Jan 2 '18 at 16:46
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    $\begingroup$ Yes indeed. My claim is that, if $I_n$ is the full quantity of interest, including the $n!^{-2}$ term (or not, due to Stirling formula), we have: $\lim_{n\to\infty}\frac1{n^2}\log I_n=-E_*$. It is indeed positive, but I don't see an obvious lower bound for this quantity; this would require further work. Now my turn: Why is this quantity of specific interest? Would you have any reference? $\endgroup$ – Adrien Hardy Jan 2 '18 at 18:21
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    $\begingroup$ Btw, if you want instead a numerical upper bound for $E^*$ you can take any two probability measures $\mu$ and $\nu$ supported on $[1,\infty)$ and $[0,1]$ respectively and compute $\tilde E(\mu,\nu)$. $\endgroup$ – Adrien Hardy Jan 2 '18 at 18:34
  • $\begingroup$ I don't know why this quantity is interesting, I did not ask the question. I also overlooked the fact that $n!^2=e^{o(n^2)}$, sorry about that. $\endgroup$ – GH from MO Jan 2 '18 at 18:36
  • $\begingroup$ It comes from a determinant related to proving the irrationality of ζ(2) and the stated guess has been obtained as an upper bound (by elementary means). However your answer is brilliant cause it will give the exact asymptotics. I have attempted to continue this conversation offline. $\endgroup$ – Krishnan Rajkumar Sep 30 '19 at 7:20

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