Let $\operatorname{Ai}(x,y)$ be the Airy kernel which is given by \begin{equation}\label{equ2.12} \operatorname{Ai}(x,y)= \begin{cases} \frac{\operatorname{Ai}(x)\operatorname{Ai}'(y)-\operatorname{Ai}(y)\operatorname{Ai}'(x)}{x-y}, & x\ne y, \\ \operatorname{Ai}'(x)^2-x\operatorname{Ai}(x)^2 & x=y. \\ \end{cases} \end{equation} Here $\operatorname{Ai}(x)$ denotes the standard Airy function \begin{align*} \operatorname{Ai}(x)=\frac{1}{2\pi i}\int_{\mathcal{C}}{e^{x\cdot t+t^3/3}\,\mathrm dt}, \end{align*} where $\mathcal{C}$ is a contour running from $\infty e^{- i \pi /3}$ to $\infty e^{ i \pi /3}$, or any other contour which can be deform from this one such that $\operatorname{Re}(t^3) \to -\infty$ along the contour. Using the relation \begin{align}\label{equ2.13} \operatorname{Ai}(x,y)=\int_0^{\infty}{\operatorname{Ai}(t+x)\operatorname{Ai}(t+y)\,\mathrm dt}, \end{align} we have the following double integral representation of the Airy kernel \begin{align*} \operatorname{Ai}(x, y)=\frac{1}{(2\pi i)^2}\int_{-\mathcal{C}}\mathrm dt\int_{\mathcal{C}}\mathrm ds\, \frac{e^{x\cdot s+s^3/3}}{e^{y\cdot t+t^3/3}}\cdot\frac{1}{s-t}, \end{align*}
Now let $\cdots<-\lambda_{k}<\cdots<-\lambda_1<0$ be the zeroes of the Airy function $\operatorname{Ai}(x)$ and let $\cdots<-\mu_{k}<\cdots<-\mu_1<0$ be the zeroes of $\operatorname{Ai}'(x)$ respectively. Choose large enough and close $\lambda_{k}$, $\mu_{j}$ satisfying $$ |\lambda_k-\mu_j|\sim \frac{\lambda_k^{-\frac12}}{|k-j|}. $$ Then one can obtain $$ A(k, j)\triangleq\frac{|\operatorname{Ai}(-\lambda_k, -\mu_j)|}{\sqrt{\operatorname{Ai}(-\lambda_k, -\lambda_k)}\cdot \sqrt{\operatorname{Ai}(-\mu_j, -\mu_j)}}\approx \frac{1}{2|k-j|} $$ Now I need to deal with a more complicated form than the Airy kernel. More precisely, consider \begin{align*} I(x, y)=\frac{1}{(2\pi i)^2}\int_{-\mathcal{C}}\mathrm dt\int_{\mathcal{C}}\mathrm ds\, \frac{e^{x\cdot s+s^3/3}}{e^{y\cdot t+t^3/3}}\cdot\frac{1}{s-t}\cdot\frac{2e^{-2(s-t)}}{1-e^{-(s-t)}}. \end{align*} This can be seen as a "perturbation" of the Airy kernel. I hope to understand the asymptotic behavior of the form $$ T(k, j)\triangleq\frac{|I(-\lambda_k, -\mu_j)|}{\sqrt{\operatorname{Ai}(-\lambda_k, -\lambda_k)}\cdot \sqrt{\operatorname{Ai}(-\mu_j, -\mu_j)}}. $$ Question: Is there a good way to evaluate $T(k, j)$ and is there a similar estimate as $A(k, j)$ above?
I think the key is to evaluate the double integral $I(-\lambda_k, -\mu_j)$. Maybe some kind of saddle point argument is needed but I don't know how to deal with it.
Thanks in advance.