How to use Laplace method to get the asymptotic expansion of multiple integral

Question

I meet difficulty when I try to get the asymptotic behaviour of multiple integral as x tends to plus infinity. And $-1<$p$<1$ $$\int_x^{+\infty}\int_x^{+\infty}e^{-{\frac{1}{2\sigma^2(1-p^2)}\ \ (u^2+2puv+v^2)}}dudv$$First, I can transform the variable to move x into integrant.$$\int_0^{+\infty}\int_0^{+\infty}e^{-{\frac{1}{2\sigma^2(1-p^2)}\ \ (u^2+2puv+v^2+(2+2p)x^2+(2+2p)(u+v)x)}}dudv$$ Then,apply transform: $$ \ \left\{ \begin{aligned} &a=u+v\\ &b=v \end{aligned} \right. $$ I get $$\int_0^{+\infty}\int_0^{a}e^{-{\frac{1}{2\sigma^2(1-p^2)}\ \ (a^2+(2p-2)ab+(2-2p)b^2)+(2+2p)ax}+(2+2p)x^2}dbda$$ To apply Laplace method, I have to view the first part of the integrant as a function g(a), and show that g is not too bad. But it seems that I cannot prove that. Or maybe I have made things worse.
We can try another transform:
$$ \ \left\{ \begin{aligned} &a=u+pv\\ &b=v \end{aligned} \right. $$ Then get
$$\int_x^{+\infty}\int_{pb+x}^{+\infty}e^{-{\frac{1}{2\sigma^2(1-p^2)}\ \ (a^2+(1-p^2)b^2)}}dadb$$ The asymototic behavior of the error function may be inappropriate here, since pb+x may fail to tend to infinity. But when $p\ge0$, we have $pb+x\ge x$, thus the integral can be controled by $$\int_x^{+\infty}\int_{x}^{+\infty}e^{-{\frac{1}{2\sigma^2(1-p^2)}\ \ (a^2+(1-p^2)b^2)}}dadb$$, then we can apply the asymptotic expension of error function.
When $p<0$, it is a problem.
However, the spirit of Laplace method is to calculate the integral near the extremum point of the integrant. So I think the first way to deal with the problem is clear.
Any suggestion would be appreciated.

Answer 1

Ok, if nobody picks it up, I'll do.

Make a transformation $$(u,v) \to (w,t),\ w=u/x,\ t=v/x$$ Then you have an integral from unity to infinity in each coordinate. The maximum of the integrand is at $(1,1)$.

Now make a first order Taylor expansion of the function in the exponent around $(1,1)$. The rest can be neglected. This can shown estimating the error term (see the references below for proof).

Integrate this first order Taylor expansion and you get the asymptotic form of the integral.

This is a simple case of a result by H. Ruben: An asymptotic expansion for the multivariate normal distribution and Mill's ratio, J. Res. Nat. Bur. Standards B ,68(1), 3-11, 1964.

See also: K. Breitung and M. Hohenbichler. Asymptotic approximations for multivariate integrals with an application to multinormal probabilities, J. of multivariate Analysis, 30, 80-97, 1989.