Normal distribution in an interval

Question

I am interested in the cumulative function of the normal distribution, because I want to get the distribution which has the support $[0,2\pi]$, and it looks like a normal distribution. Therefore, if I get the closed form of $F(2\pi)$ and $F(0)$, I can get the new distribution which has the pdf: $$\frac1{F(2\pi) -F(0)}\frac1{\sqrt{2\pi\sigma^2}}\exp{(-\frac{(x-\mu)^2}{2\sigma^2})},~x\in[0,2\pi]$$

For example, can we calculate the following integral: $$\int_0^{2\pi}\frac1{\sqrt{2\pi\sigma^2}}\exp{(-\frac{(x-\mu)^2}{2\sigma^2})}$$

Is there exists any method which can be used to get the closed form of it. Or can we get any similar distribution?

Answer 1

There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.

Answer 2

In fact the mostly used CDF in this case is $1-Q(\dfrac{x-\mu}{\sigma})$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function