from Gregory book i read that
$\int_{-\mu/\sigma}^{\infty}(\mu + \sigma x)f(x)dx = \mu F(\mu/\sigma) + \sigma f(\mu/\sigma)$
where $f(x)$ is the density of a normal distribution and $F(x)$ the cumulative distribution.
Of course I agree with the first term $\mu F(\mu/\sigma)$ coming from the integration of $\int_{-\mu/\sigma}^{\infty}\mu f(x)dx$. But I'm not sure how the second term (i.e. $\sigma f(\mu/\sigma)$ is obtained. I thought at integrating by part, but this doesn't seem correct.
Could someone help? Many thanks
The substitution $y=x^2/2$ shows $\int_{-\mu/\sigma}^\infty xf(x)dx=f(-\mu/\sigma)$ (proof is an exercise).