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


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


1 Answer 1


For something to be a density we need $\int_{\Bbb{R}^n} f(r) dr =1$. We have the following identity:

$$\int_{\Bbb{R}^n} e^{-\frac{r^2}{2\sigma^2}}dr=\sqrt{(2\pi)^3}\sigma^3$$

So we must divide to make sure it is normalized! What a density is, is:

$$P(X \in A) = \int_A f(r) dr$$

So if $A$ is the whole space, we should want that probability to be $1$!

  • $\begingroup$ I understand the first sentence, but I don't know what an identity is, nor what do you mean with "normalized". Still, perhaps leonbloy was right and I should start with simpler questions. I have a long way to go in math. $\endgroup$
    – Michael
    Oct 4, 2015 at 13:29
  • 1
    $\begingroup$ minor oversight, but $n=3$ or else you have to change your exponents to $n$. $\endgroup$
    – Neil G
    Mar 20, 2017 at 4:53

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