# Why is a $2\pi$ in the gaussian distribution function?

I am using a 3 dimensional gaussian point spread function in the form of

$$\frac{1}{\sqrt{(2\pi)^3}\sigma^3}e^{-\frac{r^2}{2\sigma^2}}$$

being $r^2$ the square of the distances $x^2 + y^2 + z^2$, to distribute a particle with mass on a specific point in space, in a density-probability area

I read somewhere that the $2\pi$ thing is so that the sum of the probabilities is $1$. Could somebody help me, showing how we can go from that $2\pi$ to the $1$?

Thank you very much in advance

• Perhaps you should start with the 1 dimensional gaussian. See eg math.stackexchange.com/questions/28558/… Oct 4, 2015 at 1:00
• Just what I needed, thank you Oct 4, 2015 at 13:26

For something to be a density we need $\int_{\Bbb{R}^n} f(r) dr =1$. We have the following identity:
$$\int_{\Bbb{R}^n} e^{-\frac{r^2}{2\sigma^2}}dr=\sqrt{(2\pi)^3}\sigma^3$$
$$P(X \in A) = \int_A f(r) dr$$
So if $A$ is the whole space, we should want that probability to be $1$!
• minor oversight, but $n=3$ or else you have to change your exponents to $n$. Mar 20, 2017 at 4:53