Having trouble understanding the output of the integral 
Exposure is given by $$E=\max(V,0)=\max(\mu+\sigma Z,0)$$ The EE defines the expected value over the positive future values and is therefore:$$\mathbb{E}[E]=\int_{-\mu/\sigma}^{\infty}(\mu+\sigma x)\varphi(x)\,dx=\mu\Phi\left(\frac\mu\sigma\right)+\sigma\varphi\left(\frac\mu\sigma\right)$$ where $\varphi(.)$ represents a normal distribution function and $\Phi(.)$ represents the cumulative normal distribution function. We see that EE depends on both the mean and the standard deviation; as the standard deviation increases so will the EE. In the special case of $\mu = 0$ we have $EE_0 = \sigma\varphi(0) = \frac{\sigma}{\sqrt{2\pi}}\approx0.40\sigma$

I cannot understand how the above integral comes out like this.
 A: $$\max \{\mu +\sigma x\,,\,\,0\}=\left\{ \begin{align}
  & \mu +\sigma x\,\,\,\,,\,\,\,\,x>-\frac{\mu }{\sigma } \\ 
 & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,x\le -\frac{\mu }{\sigma } \\ 
\end{align} \right.$$
$$\mathbb{E}[E]=\int_{-\infty }^{+\infty }{\max \{\mu +\sigma x\,,\,\,0\}}\,\varphi (x)\,dx=\int_{-\frac{\mu }{\sigma }}^{+\infty}{\,\,(\mu +\sigma x})\varphi (x)dx$$
$$\mathbb{E}[E]=\mu\int_{-\frac{\mu }{\sigma }}^{+\infty}\varphi(x)dx+\sigma\int_{-\frac{\mu }{\sigma }}^{+\infty}x\varphi (x)dx\tag 1$$
Now We consider
$$I=\int_{-\frac{\mu }{\sigma }}^{+\infty}\varphi(x)dx$$
set $x=-u$ then $dx=-du$ and we have
$$I=-\int_{\frac{\mu }{\sigma }}^{-\infty}\varphi(-u)du=\int_{-\infty}^{\frac{\mu }{\sigma }}\varphi(-u)du=\int_{-\infty}^{\frac{\mu }{\sigma }}\varphi(u)du=\Phi\left(\frac{\mu }{\sigma }\right)\tag 2$$
Note $\varphi$ is a even function, i.e. , $\varphi(-u)=\varphi(u).$
$$J=\int_{-\frac{\mu }{\sigma }}^{+\infty}x\varphi(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\frac{\mu }{\sigma }}^{+\infty}xe^{-\frac{x^2}{2}}dx$$
$$J=-\frac{1}{\sqrt{2\pi}}\left. {{e}^{-\frac{{{x}^{2}}}{2}}} \right|_{-\frac{\mu }{\sigma }}^{+\infty }=-\frac{1}{\sqrt{2\pi }}\left( 0-{{e}^{-\frac{1}{2}{{\left( -\frac{\mu }{\sigma } \right)}^{2}}}} \right)=\frac{1}{\sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{\mu }{\sigma } \right)}^{2}}}}=\varphi \left( \frac{\mu }{\sigma } \right)\tag 3$$
$(1)$,$(2)$ and $(3)$
$$\mathbb{E}[E]=\mu\Phi\left(\frac\mu\sigma\right)+\sigma\varphi\left(\frac\mu\sigma\right)$$
