I have been trying to find a closed-form (or partially) solution for the integral of the product of Q function and Normal distribution for an application in wireless communication. It is given by:
$$I = \int _{-\infty }^{\infty }\frac{ \text{Q}\left(x\right) * exp({-\frac{(x-\mu )^2}{2 \sigma ^2})}}{\sqrt{2 \pi } \sigma } dx$$
In order to simplify it, I have tried using Craig's formula for Q function in $I$ which is given by:
$$Q(x) =\int _0^{\frac{\pi }{2}}\frac{ exp({-\frac{x^2}{2 \sin ^2 \left( \theta \right)})}}{\pi }d\theta$$
I have changed the order of integration so that d$\theta$ is on the outside and have been trying to solve the integral using moment generating function approach as the Q function has been reduced to exponential form. This approach is mostly used in wireless fading channels to easily solve for Rayleigh or Nakagami Channels, but I am not sure that it can be used here.
Secondly, I tried to find the solution of integral $I$ by using Craig's Formula in Mathematica, which gives the solution as
$$I = \int_0^{\frac{\pi }{2}} \frac{exp({-\frac{\mu ^2}{2 \left(\sigma ^2+ \sin ^2\theta\right)})}}{\pi \sigma \sqrt{\frac{1}{\sigma ^2}+\frac{1}{ \sin ^2\theta}}} \, d\theta$$
I have tried comparing the numerical solution of this equation obtained through Mathematica and the original equation using direct Q function in Matlab, however, the answer is not the same.
Fig shows the code used in Matlab and the answer in comments
Is there any approximate closed-form solution for integral $I$
Best Answer
Let $X\sim N(\mu,\sigma^2)$ and $Z\sim N(0,1)$ be independent. Then $Q(x)=P(Z>x)$ and your formula for $I$ is evidently $I=P(Z>X)$, because the integrand is $P[Z>X|X=x]=P(Z>x)$ times the marginal density of $X$. But $Z-X\sim N(-\mu,\sigma^2+1)$, so $$I=P(Z>X)=P(Z-X>0)=Q\left(\frac\mu{\sqrt{\sigma^2+1}}\right).$$