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I am looking for an upper bound on the following integral:

$$\int_{X_{\delta}}\prod_{j\neq i=1}^{n}\left ( \frac{\delta}{\min (\max(\epsilon, |a_i-a_j|),\delta)}\right )^{b} \prod_{i=1}^{n} da_{i},$$

where $0<\epsilon<\delta<1$, $0<b<1$ and the set $X_{\delta}$ is the points in the hypercube $[0,1]^{n}$ where (*) each $a_i$ is $\delta$-close to at least one other $a_j$ i.e. $|a_i-a_j|<\delta$.

Specifically, I want a bound of the form $c^{n}\delta^{kn}, c,k>0$ (and if there are terms like $\epsilon^{-a}$, I hope to have $a>0$ be small as possible). I hope to avoid bounds with factorials like $n! c^{n}\delta^{kn}$. Any suggestions are welcome.

PS: The condition (*) arose because the original integral contains

$$\int E\left[\prod_{i\geq 1} (e^{U(a_{i})}-1)\right]\prod_i da_i,$$

where $U(a_{i})$ are normalized Gaussians i.e. $E[e^{U(a_{i})}]=1$ and with short-range correlation $R(s,t)=E[U(s)U(t)]=b\ln\left(\frac{\delta}{\min (\max(\epsilon, |t-s|),\delta)}\right)$.

One reasonable approach is to use the derivations for the Selberg/Dirichlet/Dixon-Anderson integral. However I come across an issue in trying to handle the (*)-condition. For example, one can start by ordering the coordinates

$$a_{1}<a_{2}<...<a_{n}<1.$$

Then the (*) condition forces that $a_{2}-a_{1}<\delta$ and $a_{n}-a_{n-1}<\delta$. So the reasonable thing is to start from the inner integral

$$\int_{(a_{2}-\delta) \vee 0}^{a_{2}}\prod_{j\geq 2}^{n}\left ( \frac{\delta}{\min (\max(\epsilon, a_j-a_1),\delta)}\right )^{b} da_{1}$$

This is also reminiscent of the first approach here of $n$ beads on a circle but further imposing short-range correlation as opposed to long-range.

Moreover, starting from the Gaussian product one can use the Vieta formula

$$E[\prod_{i\geq 1} e^{U(a_{i})}-1]=\sum_{\ell=0}^{n}(-1)^{\ell}\sum_{S\subset [n], |S|=\ell}\exp\{\sum_{i, r\in S}R(a_i,a_r)\}, $$

which reduces the number of singular terms.

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  • $\begingroup$ What do $\vee$ and $\wedge$ mean? $\endgroup$ Commented Jul 21, 2022 at 2:50
  • $\begingroup$ max and min respectively. $\endgroup$ Commented Jul 21, 2022 at 15:51
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    $\begingroup$ So, it means $\min(\max(\epsilon,a_j-a_1),\delta)$? $\endgroup$ Commented Jul 21, 2022 at 22:02
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    $\begingroup$ i replaced it to avoid confusion. $\endgroup$ Commented Jul 22, 2022 at 1:05

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