Estimating a Selberg-type integral (or a Fredholm determinant)

Question

I am concerned with the asymptotical behavior of integrals like this for large $n$ $$\frac{1}{n!}\intop_{\Omega}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}dx_{j},$$ where $\Omega$ is one of these infinite rectangular regions in $\mathbb{R}^n$: $$\Omega_1^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2<\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\Omega_2^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\ldots$$ $$\Omega_n^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3>\lambda\ldots,\,x_n>\lambda\},$$ and $\lambda$ is a large positive number. Some information about such integrals has already been obtained in https://mathoverflow.net/a/200622/22773

Originally, I am trying to estimate a polynomial sum of integrals over $\Omega_k$. Namely, let $a$ be any number such that $a>1$. For convenience, define

$$\Omega_0^{(n)}=\{(x_1,\ldots,x_n):x_1<\lambda,\,x_2<\lambda\ldots,\,x_n<\lambda\},$$

$$I_k^{(n)}(\lambda)=\frac{1}{n!}\intop_{\Omega_k^{(n)}}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}dx_{j},\;k=0,1,\ldots,n$$

and

$$H^{(n)}=\frac{1}{n!}\intop_{\mathbb{R}^n}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}dx_{j}=(2\pi)^{n/2}2^{-n^2/2}\prod_{k=1}^{n-1}k!$$

(the last is the classical Selberg integral). Now consider the sum

$$S^{(n)}(a,\lambda)=\sum_{k=1}^n a^k\binom{n}{k}I_k^{(n)}(\lambda),$$

so that

$$H^{(n)}=\sum_{k=0}^n \binom{n}{k}I_k=I_0^{(n)}(\lambda)+S^{(n)}(1,\lambda).$$

The conjecture to be proven is that for any $a>1$ (the case $0<a\leq 1$ is trivial) there exist positive numbers $\lambda_0$ and $C$ (which possibly depend on $a$) such that

$$\boxed{\forall n\in\mathbb{N} \;\forall \lambda>\lambda_0 \;\; S^{(n)}(a,\lambda)\leq C(H^{(n)}-I^{(n)}_0(\lambda))= C S^{(n)}(1,\lambda).}$$

It might be useful that the following asymptotics are known for the Selberg integral:

$$H^{(n)}=e^{\zeta'(-1)} \pi^n e^{-3n^2/4} 2^{n-n^2/2} n^{n^2/2-1/12}(1+o(1)),$$

where $\zeta'(-1)=-0.165...$ is the derivative of the Riemann zeta-function at $-1$.

One guess is that if there is an estimate like

$$I_{k+1}^{(n)}(\lambda)\leq\frac{f(\lambda_0)}{n}I_{k}^{(n)}(\lambda),$$

where $f(\lambda)=o(1)$ for large $\lambda$, then the previous statement follows immediately (with, say, $C=2a$).

Edit: Another important fact that I remembered is that $I_k^{(n)}(\lambda)$ is, up to a constant, the probability that a $n\times n$ random matrix from the Gaussian Unitary Ensemble has exactly $k$ eigenvalues on $(\lambda,+\infty)$. When $n\rightarrow\infty$, the analogue of this is also known as $E(k,\lambda)$ and the asymptotics of these functions as $\lambda\rightarrow -\infty$ have been examined in Tracy&Widom "Level-Spacing Distributions and the Airy Kernel". Let's call $J=(\lambda,\infty)$ and define $E_k^{(n)}(\lambda)$ as the probability of a random $n\times n$ matrix in the GUE having $k$ eigenvalues in $J$. Then the following expression is known:

$$E_k^{(n)}(\lambda)=\binom{n}{k}\frac{I_k^{(n)}}{H^{(n)}}=\frac{1}{k!}\left(\frac{\mathrm{d}}{\mathrm{d}z}\right)^k R(z,J)\mid_{z=-1},$$

where

$$R^{(n)}(z,J)\equiv R^{(n)}(z,\lambda)=\frac{1}{n!H^{(n)}}\intop_{\mathbb{R}^n}\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}\left[1+z\chi(x_j)\right]\mathrm{d}x_{j}$$

and $\chi$ is the characteristic function of $J$. It is easy to see that

$$R^{(n)}(-1,\lambda)=\frac{I_0^{(n)}(\lambda)}{H^{(n)}}.$$

Using these relations, it is straightforward to check that

$$S^{(n)}(\lambda)=H^{(n)}(R^{(n)}(a-1,\lambda)-R^{(n)}(-1,\lambda)),$$

therefore the estimate that we seek is equivalent to

$$\boxed{R^{(n)}(a-1,\lambda)-R^{(n)}(-1,\lambda)\leq C(1-R^{(n)}(-1,\lambda)).}$$

So, $R^{(n)}$ is a very common object when studying the Gaussian Unitary Ensemble, but typically one is interested in its values or derivatives at $z=-1$. In my problem, however, values at arbitrary $z$ come up.

It would probably suffice to prove that $R^{(n)}(z,\lambda),\, n=1,2,3,\ldots$ is a convergent sequence of analytic functions in $z$ and it converges uniformly for all large $\lambda$.

It is also worth noting that $R^{(n)}(z,\lambda)$ is the generating function of integrals over $J^k \times \mathbb{R}^{n-k}$ (they are, in fact, integrals over $J^k$ of the $k$-point correlation functions):

$$R^{(n)}(z,\lambda)=\sum_{i=0}^n \frac{\rho_i}{i!}z^i,$$

where $\rho_0=1$ and

$$\rho_k=\frac{1}{(n-k)!H^{(n)}}\intop_{\mathbb{R}^n}\prod_{i=1}^k\chi(x_i)\prod_{1\leq i<j\leq n}(x_{j}-x_{i})^{2}\,\prod_{j=1}^{n}e^{-x_{j}^{2}}\mathrm{d}x_{j}.$$

Moreover, $R^{(n)}(z,\lambda)$ is equal to the determinant of the following $n\times n$ matrix:

$$R^{(n)}(z,\lambda)=\det[G_{ij}]_{i,j=0,1,\ldots,n-1},$$

$$G_{ij}=\frac{1}{h_i}\int_\mathbb{R} p_i(x)p_j(x)e^{-x^2}[1+z\chi(x)]\mathrm{d}x,$$

where $p_i$ is the $i$-th Hermite polynomial orthogonal with the weight $e^{-x^2}$ and $\int p_i^2(x)e^{-x^2}\mathrm{d}x=h_i$.

Answer 1

Calculations for $\boldsymbol n$ up to $\boldsymbol 6$ indicate that, for $1\leq k\leq n$, $0\leq r\leq k-1$, there are polynomials $P_{n,k,r}\in\mathbb Q[\lambda]$ of certain degrees $d_{n,k-r}$ - which are even if $k-r$ is even and odd if $k-r$ is odd - and constants $Q_{n,k}\in\mathbb Q$ such that the following holds true: $$ \boxed{\sum_{l=k}^n \binom{n-k}{l-k}\,I_l^{(n)}(\lambda) = \sum_{r=0}^{k-1}e^{-\left(k-r\right)\lambda^2}\pi^{\frac{n-k+r}2} P_{n,k,r}(\lambda)\,E^r(\lambda) + Q_{n,k}\,\pi^{\frac{n}2}E^k(\lambda)} $$ (For the left-hand side: The integral is extended over the region $\lambda<x_i<\infty$ for $1\leq i\leq k$ and $-\infty<x_i<\infty$ for $k+1\leq i\leq n$.) Here, $E(\lambda)=\frac2{\sqrt{\pi}}\int_\lambda^\infty e^{-y^2}dy$ is the complementary error function. Notice that $E(\lambda)=\frac{e^{-\lambda^2}}{\sqrt{\pi}\,\lambda}\left(1+O\left(\frac1{\lambda^2}\right)\right)$ as $\lambda\to\infty$. In particular, one then has $$ I_k^{(n)}(\lambda) \sim c_{n,k}\,\pi^{\frac{n-k}2}\,e^{-k\lambda^2} \,\lambda^{N_{n,k}} \tag{1} $$ as $\lambda\to\infty$ for some constant $c_{n,k}\in\mathbb Q$, where $N_{n,k}= \max \{d_{n,k-r}-r\,\mid\,0\leq r\leq k-1\}$. For $k=1$, this was already observed in this previous post.

Remark. The $P_{n,k,r}$ appear to depend only on $(n,k-r)$, up to a multiplicative constant.

Answer 2

... I am constantly returning to the thought that the actual solution might be much simpler.

I do not think that you are right. To obtain estimates for fixed $n$ is not hard, the big issue is uniformity with respect to $n$. Of course, it is possible that the specific context allows to play some algebraic tricks which facilitate the computations. But in the end it comes down to getting these integrals estimated, and I am convinced that any argument that does not take into account the fine structure of the problem is doomed to fail.

Here is a heuristic argument to support my point of view: You are seeking to prove an estimate like $$ I_{k+1}^{(n)}(\lambda) \leq \frac{f(\lambda)}{n}\,I_k^{(n)}(\lambda) $$ with $f(\lambda)= o(1)$ as $\lambda\to\infty$. Inserting the asymptotics from (1), so neglecting remainder terms, one gets $$ \frac{c_{n,k+1}}{c_{n,k}}\,e^{-\lambda^2}\lambda^{N_{n,k+1}-N_{n,k}} \leq \frac{C\, f(\lambda)}{n}. $$ The calculations performed before suggest that $d_{n,1}=2n-3$, $d_{n,2}\geq 2n-2$, and $d_{n,k}=d_{n,2}-k+2$ for $k\geq2$. If so, then one has $N_{n,1}=2n-3$ and $N_{n,k}=d_{n,2}-k+2$ for $k\geq2$.

In case $k=1$, this yields $$ \frac{c_{n,2}}{c_{n,1}}\,e^{-\lambda^2}\lambda^{d_{n,2}-2n+3} \leq \frac{C\, f(\lambda)}{n}. $$ Now, if $d_{n,2}\geq 2n + \gamma_0\log n +\gamma_1$ for some constants $\gamma_0>0$ and $\gamma_1\in\mathbb R$ (which is not unlikely to hold), then one would need $$ \frac{c_{n,2}}{c_{n,1}} = O\left(n^{-1-\gamma_0}\right) $$ as $n\to \infty$.

In case $k\geq2$, one obtains $$ \frac{c_{n,k+1}}{c_{n,k}}\,\frac{e^{-\lambda^2}}{\lambda} \leq \frac{C\, f(\lambda)}{n}, $$ so here one would need $$ \frac{c_{n,k+1}}{c_{n,k}} = O\left(n^{-1}\right) $$ as $n\to\infty$, but uniformly so in $2\leq k< n$.

Admittedly, these arguments are not rigorous, but in my opinion they demonstrate that one has to proceed with extreme care when doing the estimates.