Suppose that $\Phi$ and $\phi$ are CDF and PDF of a standard Normal random variable.
Is it possible to compute:
$$\int^{b}_{a} \Phi\left( \frac{x-\mu}{\sigma} \right) \phi(x) dx$$
(where a and b are any real numbers) in closed form? What if we set $a=0$?
I know that if $b=\infty$ and $a=-\infty$ then one can show that
$$\int^{\infty}_{-\infty} \Phi\left( \frac{x-\mu}{\sigma} \right) \phi(x) dx = \Phi\left(\frac{-\mu}{\sqrt{1+\sigma^2}}\right)$$
See the link: computations of the above integral
But what if $a>-\infty$ and $b<\infty$?
The way one computes the above integral when $b=\infty$ and $a=-\infty$ is by considering $$ I(\mu)= \int^{\infty}_{-\infty} \Phi\left( \frac{x-\mu}{\sigma} \right) \phi(x) dx $$
Then computing $$I_{\mu}=\int^{\infty}_{-\infty} \frac{\partial}{\partial \mu}\left[ \Phi\left( \frac{x-\mu}{\sigma} \right) \phi(x) \right]dx$$
which can be easily computed, and which gives a pdf of normal distribution. Then integrating with respect to $\mu$ yields the final result.
However, if I follow the above steps I obtain: $$ I_{\mu}= \phi\left(\frac{-\mu}{\sqrt{1+\sigma^2}}\right) \left[ \Phi\left( \frac{b-\frac{\mu}{1+\sigma^2}}{\sqrt{\frac{\sigma^2}{1+\sigma^2}}} \right) - \Phi\left( \frac{a-\frac{\mu}{1+\sigma^2}}{\sqrt{\frac{\sigma^2}{1+\sigma^2}}} \right) \right]$$
which I do not know how to integrate with respect to $\mu$ in order to compute $I(\mu)$.
Is there any hope to somehow compute this integral?