How to prove the following?
$$\text{EE} = \int_{-\mu/\sigma}^{+\infty} (\mu + \sigma x)\phi(x)\ dx = \mu\Phi(\mu/\sigma) + \sigma\phi(\mu/\sigma)$$
Where $\Phi(x)$ is the standard normal distribution and $\phi(x)$ is the standard normal density function.
It seems like there is some problem in your result. Are you sure of that?
The integral can be split into
$$\mu \int_{-\mu/\sigma}^{+\infty}\phi(x)\ dx + \sigma\int_{-\mu/\sigma}^{+\infty} x\phi(x)\ dx$$
The first one is pretty trivial:
$$\mu\left[\Phi(+\infty) - \Phi(-\mu/\sigma)\right]$$
Assuming $\phi(x)$ is a "good" function such that $Phi(x) = \int \phi(x)\ dx$ and $\Phi(+\infty) = 0$ and also $x\Phi(x) = 0$ as $x\to +\infty$ we conclude the first part:
$$\boxed{-\mu\Phi(-\mu/\sigma)}$$
For the second one, we perform an integration by parts with $f = x$, $g' = \phi(x)$:
$$\sigma \left\{x\Phi(x)\bigg|_{-\mu/\sigma}^{+\infty} - \int_{-\mu/\sigma}^{+\infty} \Phi(x)\ dx\right\}$$
$$\sigma\left[\mu/\sigma\Phi(-\mu/\sigma) - \tilde\Phi(x)\bigg|_{-\mu/\sigma}^{+\infty}\right]$$
Where $\tilde\Phi(x) = \int\Phi(x)\ dx$.
I cannot suppose nothing about $\tilde\Phi(x)$, bue even if it vanished at the infinity we'd remain with
$$\boxed{\mu\Phi(-\mu/\sigma) + \sigma\tilde\Phi(-\mu/\sigma)}$$
Which united to the first part gives the final "incorrect" result:
$$-\mu\Phi(-\mu/\sigma) + \mu\Phi(-\mu/\sigma) + \sigma\tilde\Phi(-\mu/\sigma)$$
That is
$$\sigma\tilde\Phi(-\mu/\sigma)$$
One shall investigate $\phi(x), \Phi(x)$ and $\tilde\Phi(x)$ properties.