Can we evaluate the following integral ?
$$\int_0^\infty x e^{-x^2} \Phi(ax+b)\,\mathrm dx$$
Here $\Phi(\cdot)$ is the cumulative probability distribution function of a standard normal random variable, and the question is supposed to be
finding $E[\frac 12\Phi(aX+b)]$ where $X$ is a Rayleigh random variable
with density function $2xe^{-x^2}\mathbf 1_{x: x \geq 0}$.
Best Answer
$$ \int_0^{\infty} \left(-\frac{1}{2}\dfrac{d}{dx}\mathrm{e}^{-x^2}\right)\phi(ax+b)dx $$ use integration by parts $$ \left[-\frac{1}{2}\mathrm{e}^{-x^2}\phi(ax+b)\right]_0^{\infty} +\int_0^{\infty} \frac{1}{2}\mathrm{e}^{-x^2}\phi'(ax+b) dx $$ where $$ \phi'(ax+b) = \dfrac{d}{dx}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{ax+b}\mathrm{e}^{-s^2}ds = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-(ax+b)^2}a $$ thus you end up getting $$ \left[-\frac{1}{2}\mathrm{e}^{-x^2}\phi(ax+b)\right]_0^{\infty} +\frac{a}{2\sqrt{2\pi}}\int_0^{\infty} \mathrm{e}^{-x^2}\mathrm{e}^{-(ax+b)^2} dx $$ so you should be able to finish it off. (if I haven't made a mistake).