Integral of cumulative normal

Question

Let $$\Phi(x):=\int_{-\infty}^x \frac{1}{\sqrt{2\pi}} \exp\left({-\dfrac{\omega^2}{2}}\right) d\omega.$$

Question: for what values of $a$, $b$ and for what choices of $f(x)$ would the following integral have a closed form?

$$\int_a^b \Phi(f(x)) dx$$

Answer 1

$f=\Phi^{-1}(g(x))$ for any function $g$ with a closed form taking values between 0 and 1 will work.

Or use integration by parts to obtain

$$b\Phi\bigl(f(b)\bigr)-a\Phi\bigl(f(a)\bigr)-\int_a^b x f'(x) \phi\bigl(f(x)\bigr) dx,$$

which suggests that your problem simplifies to finding values such that the sum of the first two terms has a closed form solution, e.g. if $f(b)=f(a)=0$.