Trying to solve $\frac{1}{\sigma\sqrt{2\pi}}\int_r^\infty xe^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx$ I'm looking to simplify/solve $$\frac{1}{\sigma\sqrt{2\pi}}\int_r^\infty xe^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx$$
, where $\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$ is the pdf of the normal distribution.
I tried to look at solving it with substitution, but I struggle at a certain point.
So if I define $u=-\frac{1}{2}(\frac{x-\mu}{\sigma})^2$, then $du = -\frac{x-\mu}{\sigma^2}dx$
Now, I want to come to:
$$\frac{1}{\sigma\sqrt{2\pi}}\int_r^\infty e^udu$$ but is the following correct?
$$\mu(1-F(r))- \frac{1}{\sigma^3 \sqrt{2\pi}}\int_r^\infty e^udu$$, where $F(.)$ is the CDF of the normal distribution.
I believe I'm making a mistake here, but just don't see what. Any help or hints are appreciated.
EDIT
I’m not looking for a closed form solution but a solution in terms of the pdf and cdf of the normal distribution.
 A: $$
\sqrt{2 \pi} \sigma I = \int_r^\infty x e^{-\frac{1}{2\sigma^2} (x - \mu)^2}\,dx = [u = x-\mu, dx = du] = \int_r^\infty (u +\mu) e^{-\frac{1}{2\sigma^2} u^2}\,du
= \sqrt{2 \pi} \sigma (I_1 + I_2)
$$
$$
 \sigma\sqrt{2 \pi} I_1 = \int_r^\infty u e^{-\frac{1}{2\sigma^2} u^2}\,du = \sigma^2 e^{-\frac{1}{2\sigma^2} (r-\mu)^2}
$$
$$
I_2 = \frac{1}{\sqrt{2 \pi} \sigma}\mu \int_r^\infty e^{-\frac{1}{2\sigma^2} u^2}\,du = \mu\bigg(1-\Phi\Big(\frac{r-\mu}{\sigma}\Big)\bigg)
$$
Putting 1 and 2 together,
$$
I = \sigma\phi\Big(\frac{r-\mu}{\sigma}\Big) +
\mu\bigg(1-\Phi\Big(\frac{r-\mu}{\sigma}\Big)\bigg).
$$
A: Let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ and $\Phi(x)=\int_{-\infty}^x\phi(t)\,\mathrm dt$ denote the standard normal density and distribution, respectively. Now consider the slightly more general problem
$$
I_b=\int_r^bx\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\,\mathrm dx.
$$
If we define $\rho=(r-\mu)/\sigma$, $\beta=(b-\mu)/\sigma$, and $Z=\Phi(\beta)-\Phi(\rho)$ then the problem can be written as
$$
I_b=Z\int_r^bx\frac{\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)}{Z}\,\mathrm dx=Z\mathsf E(X|r\leq X\leq b),
$$
which is just the first moment of the truncated gaussian distribution scaled by $Z$.  Using the first moment of the trucnated gaussian we obtain
$$
I_b=\mu(\Phi(\beta)-\Phi(\rho))+\sigma(\phi(\rho)-\phi(\beta)).
$$
For the specific problem at hand, $b=\infty$ and upon taking taking the limit we obtain
$$
I_\infty=\mu(1-\Phi(\rho))+\sigma\phi(\rho).
$$
