# Approximate a double integral

I am struggling to approximate the following integral

$$\sqrt n\int_0^\infty \int_0^\infty (1 + n x^2)^{-1}(1 + y^2)^{-1} \Phi\left(\frac{a}{\sqrt{1 + b + x^2y^2}}\right) \text{d}x \text{d}y,$$ where $$\Phi(u) = \frac{1}{2} \left\{1 + \text{erf}\left(\frac{u}{\sqrt2}\right)\right\}$$ is the standard normal cumulative distribution function (and $$\text{erf}$$ is the error function). I also know that:

• $$n$$ is a large natural number (so studying the integral in the asymptotic regime $$n\rightarrow \infty$$ can be sensible/useful);

• $$-4;

• $$b> 0$$ (and it is close to $$0$$).

So far, I have been considering two directions to approximate this integral (but have been unsuccessful). I started by approximating the integral with respect to $$x$$:

1. using the series representation of the $$\text{erf}$$ function, but a) I feel that I would have to use many terms for the approximation to be accurate, and b) the integral is not much (?) simpler to compute when replacing $$\Phi(\cdot)$$ by the truncated series.

2. integrating by parts: an $$\tan^{-1}$$ term appears as well as the normal density function $$\varphi(\cdot)$$, and there I am stuck again...

Any idea to help me? In particular, I feel that using the fact that $$n$$ is large may help.

Thanks

• In terms of getting an upper bound, since $0 < \Phi < 1$, you can see that $$\int_0^{\infty} \int_0^{\infty} \dfrac{\Phi(...)}{(1+n x^2)(1+y^2)} dxdy \leq \left(\int_0^{\infty} \frac{dx}{1+nx^2}\right)\left(\int_0^{\infty} \frac{dy}{1+y^2} \right) = \frac{\pi^2}{2 \sqrt{n}} \to 0 (n \to \infty)$$ – PierreCarre Mar 18 at 14:52
• Thanks for your help. Indeed, when $n$ is large, this will just go to zero... I would need something more precise for fixed $n$ then... – user79097 Mar 18 at 15:27
• Do you need analytical estimates for general $a,b,n$ or just a reliable numerical integration scheme? – PierreCarre Mar 18 at 15:30
• actually I forgot a $\sqrt n$ multiplicative factor, sorry... please see the edited version, the upper bound doesn't go to zero anymore... And yes, I need a general analytical approximation in terms of $a$, $b$ and $n$. – user79097 Mar 18 at 15:33
• I think I found how to proceed. With a simple change of variable $\tilde{x} = \sqrt n x$, and exchanging the order of the limit $n \rightarrow \infty$ and the integration, I get $\Phi(a/\sqrt{1+b})$ (also using the continuity of $\Phi(\cdot)$). – user79097 Mar 18 at 17:50