# Integral of product of Standard Normals

Hopefully someone can help me understand the solution to the following integral which is the final step in a problem I'm looking at:

$$\int^{\infty}_{0}\phi(x)\Phi(x)dx=\left[\frac{1}{2}\{\Phi(x)\}^2\right]^{\infty}_{0}=\frac{1}{2}\left[1^2-\left(\frac{1}{2}\right)^2\right]=\frac{3}{8}$$

$$\phi(x)$$ is the probability density function of a Standard Normal and $$\Phi(x)$$ the distribution function. Obviously $$\Phi(x)$$ is the integral of $$\int\phi(x)dx$$ but I don't follow the first step in the evaluation of the above integral. The solution it's trivial. If anyone could explain the above it would be appreciated.

Thanks

This is just standard integral substitution (sometimes called $$u$$-substitution). The fact that $$\phi$$ is the density and $$\Phi$$ is the cumulative distribution for a standard normal implies $$\Phi'(x) = \phi(x).$$ Therefore the integral can also be written as $$\int \Phi(x) \Phi'(x) \, dx.$$ And with the choice of substitution $$u = \Phi(x), \quad du = \Phi'(x) \, dx,$$ we see it becomes $$\int u \, du = \frac{u^2}{2} + C = \frac{(\Phi(x))^2}{2} + C.$$