$\int_a^\infty \Phi( a - x) f(x) dx$

Question

What is a nicer formula for $$\int_a^\infty \Phi( a - x) f(x) dx?$$

Here, $f$ is the density for a normal distribution with mean $a$, variance $1$. $\Phi$ is the standard normal cumulative.

Answer 1

Here we have $f(x)=\frac{1}{\sqrt{2\pi}}e^{-(x-a)^2/2}$ and $\Phi(a-x)=\int_{-\infty}^{a-x}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}\,dy$, hence

$$ \int_{a}^{+\infty}\Phi(a-x)\,f(x)\,dx = \frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}\Phi(-x) e^{-x^2/2}\,dx $$ does not really depend on $a$ and equals $\frac{1}{\sqrt{2\pi}}\left(\frac{1}{4}\sqrt{\frac{\pi}{2}}\right)=\color{red}{\large\frac{1}{8}}$ by integration by parts.