Suppose $\phi(\cdot)$ and $\Phi(\cdot)$ are density function and distribution function of the standard normal distribution.

How can one calculate the integral:

$$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$

A more conventional notation is

$$y(\mu, \sigma) = \int\Phi\left(\frac{x-\mu}{\sigma}\right)\phi(x) dx = \Phi\left(\frac{-\mu}{\sqrt{1+\sigma^2}}\right).$$

This can be found by differentiating the integral with respect to $\mu$ and $\sigma$, producing elementary integrals which can be expressed in closed form:

$$\frac{\partial y}{\partial \mu}(\mu, \sigma) = -\frac{1}{\sqrt{2 \pi } \sqrt{\sigma ^2+1}}e^{-\frac{1}{2}\frac{\mu ^2}{\sigma ^2+1}},$$

$$\frac{\partial y}{\partial \sigma}(\mu, \sigma) = \frac{\mu\sigma  }{\sqrt{2 \pi } \left(\sigma ^2+1\right)^{3/2}}e^{-\frac{1}{2}\frac{\mu ^2}{\sigma ^2+1}}.$$

This system can be integrated, beginning with the initial condition $y(0,1)$ = $\int\Phi(x)\phi(x)dx$ = $1/2$, to obtain the given solution (which is easily checked by differentiation).
