Indefinite integral involving normal cdf [duplicate]

Question

In this answer, the following integral identity is used, where $\Phi(t)$ is the cdf of the standard normal variable.

$$\int_{-\infty}^{\infty}t\frac d{dt}\Phi(t)^2dt=1/\sqrt{\pi}.$$

I tried to simplify to understand how. \begin{align*} \int_{-\infty}^{\infty}t\frac d{dt}\Phi(t)^2dt=\int_{-\infty}^{\infty}2t\Phi(t)\phi(t)dt=\int_{-\infty}^{\infty}2t\Phi(t)d\Phi(t)=\int_0^12x\Phi^{-1}(x)dx \end{align*}

How do I proceed to get the answer? Is there a simpler way?

Answer 1

Note that: $$ \int_{-\infty}^{t}u\phi(u){du}=-1/\sqrt{2\pi}e^{-t^2/2}=-\phi(t). $$ Using integration by parts, we get: $$ \int_{-\infty}^{\infty}2t\Phi(t)\phi(t)dt=-2\phi(t)\Phi(t)|_{-\infty}^\infty+\int_{-\infty}^\infty 2\phi(t)^2dt=\frac 1{\sqrt \pi}\int_{-\infty}^\infty\frac 1{\sqrt\pi}e^{-t^2}dt=\frac 1{\sqrt \pi}. $$