A CDF for a normal standard is the following:
$$N(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\phi^2/2} d\phi$$
I have the following relation in my notes which I am not very sure how they arrived at:
$$N(x) = \frac{1}{2}+\frac{1}{2}\text{erf}(x/\sqrt{2})$$
Not sure how they can have $x$ value in $N(\cdot)$ and $x/\sqrt{2}$ in $\text{erf}(\cdot)$. Here is what I got:
$$N(x) = \frac{1}{\sqrt{2\pi}}\left( \int_{-\infty}^0 e^{-x^2/2} dx + \int_0^x e^{-\phi^2/2} d\phi \right)$$
I reasoned that since $$\int_{-\infty}^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$$
then
$$\int_{-\infty}^0 e^{-x^2/2} dx = \frac{\sqrt{2\pi}}{2}$$
Therefore:
$$N(x) = \frac{1}{\sqrt{2\pi}}\left( \frac{\sqrt{2\pi}}{2} + \frac{\sqrt{\pi}}{2}\text{erf}(x) \right)$$
Which is:
$$N(x) = \frac{1}{2} + \frac{1}{2\sqrt{2}}\text{erf}(x)$$
And not quite what I have in notes (due to $\text{erf}(x)$)
The definition of an error function is:
$$\text{erf}(x) = \frac{\sqrt{2}}{\pi}\int_{0}^x e^{-s^2}ds$$