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);

  • $-4<a < -2$;

  • $b> 0$ (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.


  • $\begingroup$ In terms of getting an upper bound, since $0 < \Phi < 1$, you can see that $$ \int_0^{\infty} \int_0^{\infty} \dfrac{\Phi(...)}{(1+n x^2)(1+y^2)} dxdy \leq \left(\int_0^{\infty} \frac{dx}{1+nx^2}\right)\left(\int_0^{\infty} \frac{dy}{1+y^2} \right) = \frac{\pi^2}{2 \sqrt{n}} \to 0 (n \to \infty) $$ $\endgroup$ – PierreCarre Mar 18 at 14:52
  • $\begingroup$ 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... $\endgroup$ – user79097 Mar 18 at 15:27
  • $\begingroup$ Do you need analytical estimates for general $a,b,n$ or just a reliable numerical integration scheme? $\endgroup$ – PierreCarre Mar 18 at 15:30
  • $\begingroup$ 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 $n$. $\endgroup$ – user79097 Mar 18 at 15:33
  • $\begingroup$ 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 $\Phi(\cdot)$). $\endgroup$ – user79097 Mar 18 at 17:50

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