I am working on a project, where the following integral shows up, and I don't see how to solve this, or to show if it has a closed form solution.
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(z)z \text{exp}(-z^2/2)dz$$ It is in the context of probability theory, so an equivalent problem is for $Z$ standard normal distributed with CDF $\Phi$, then to calculate $$\mathbb{E}_z[\Phi(z)z]$$
I will appreciate your help!
$z\exp(-z^2/2)=-(\exp(-z^2/2))'$, so you can integrate by parts and obtain
$$\begin{align} &\int_{-\infty}^\infty \Phi(z)z\exp(-z^2/2)\,\mathrm dz \\[1ex]=~&~{\left[-\Phi(z)\exp(-z^2/2)\right]}_{-\infty}^\infty+\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}(\exp(-z^2/2))^2\,\mathrm dz \\[1ex]=~&~0+\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \exp(-z^2)\,\mathrm dz \\[1ex]=~&~\frac1{2\sqrt{\pi}}\int_{-\infty}^\infty \exp(-y^2/2)\,\mathrm dy \\[1ex]=~&~\frac1{\sqrt2}\end{align}$$
The rest is trite.