Partial average of normal distribution suppose we have a normal distribution $ N(\mu,\sigma^2)$, then

$$
CDF(x) = \int_{-\infty}^{x} f_X(t)d_t
$$ 

which can be shown as

$$
CDF(x) = \frac{1}{2}[1+erf(\frac{x-\mu}{\sigma\sqrt{2}})]
$$

I am wondering if there is any function for partial averaging of normal distribution such as:
$$
\int_{-\infty}^{x} tf_X(t)d_t
$$
Sure there is.  When you are taking the expected value in this way, the integral is easy in closed form for $\mu=0$, and contains an erf term for non-zero $\mu$:

$$
f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$

$$E(X | X&lt;x_0) = \int_{-\infty}^{x_0} xf(x)\,dx = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0} x e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx =  \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0-\mu} (t+\mu) e^{-\frac{t^2}{2\sigma^2}}dt = \mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}+
\left[ 
-\frac{\sigma}{\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}
\right]_{-\infty}^{x_0}=
\mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}-\frac{\sigma}{\sqrt{2\pi}} e^{-\frac{x_0^2}{2\sigma^2}}
$$
Sure there is.  When you are taking the expected value in this way, the integral is easy in closed form for $\mu=0$, and contains an erf term for non-zero $\mu$:

$$
f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$

$$E(X | X&lt;x_0) = \int_{-\infty}^{x_0} xf(x)\,dx = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0} x e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx =  \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{x_0-\mu} (t+\mu) e^{-\frac{t^2}{2\sigma^2}}dt = \mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}+
\left[ 
-\frac{\sigma}{\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}
\right]_{-\infty}^{x_0}=
\mu\mbox{ erf }\frac{x_0-\mu}{\sigma\sqrt{2}}-\frac{\sigma}{\sqrt{2\pi}} e^{-\frac{x_0^2}{2\sigma^2}}
$$