I intend to find the Gaussian function, where $p$ is the probability, area under its curve, within the boundary $r_{min}=- r_0$ and $r_{max} = +r_0$.
Consider the following Gaussian function:
$$ f(r) = \frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\frac{r^2}{\sigma^2}}$$
To solve this, I employed the ff:
$$p =\int_{r_{min}}^{r_{max}} \frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\frac{r^2}{\sigma^2}}dr$$
But: $$\int e^{-\alpha r^2} dr = \frac{\sqrt{\pi}erf(\sqrt{\alpha}r)}{2\sqrt{\alpha}}$$
Then: $$p=erf(\sigma \sqrt{2}r_0)$$
So: $$\sigma=erf^{-1}(p)/(\sqrt{2}r_0)$$
Finally
$$ f(r) = \frac{1}{erf^{-1}(p)/(\sqrt{2}r_0) \sqrt{2\pi}}e^{-\frac{1}{2}\frac{r^2}{(erf^{-1}(p)/(\sqrt{2}r_0))^2}}$$
The solution seems sound, but when I graphed it, I dont see that the area under the curve within $r_0$ is actually $p$.
Below are the relevant images of what I am talking about.
The top image is the Gaussian graph I obtained while the bottom image is what I am expecting. The red line is where $r_0$ is.
The plot shown was only half of the curve, so I am expecting to get $p/2$ of the area
The definition of the error function is $$\mathrm{Erf}(x) = \int\limits_{t = -x}^{t=x} \frac1{\sqrt\pi}e^{-t^2} \,dt$$
With that in mind, we have \begin{align} p &= \int\limits_{r = -r_0}^{r=r_0} \frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\frac{r^2}{\sigma^2}}\,dr \qquad \text{, let}~~ t = \frac{r}{\sigma \sqrt 2} \\ &= \int\limits_{r = -r_0}^{r=r_0} \frac1{\sqrt\pi} e^{-( \frac{r}{\sigma \sqrt 2})^2 } \frac{dr}{\sigma \sqrt 2} \\ &= \int\limits_{t = -t_0}^{t=t_0} \frac1{\sqrt\pi}e^{-t^2} \,dt \qquad \text{, note that}~~ t_0 = \frac{r_0}{\sigma \sqrt 2} \\ &= \mathrm{Erf}\Bigl( \frac{r_0}{\sigma \sqrt 2} \Bigr) \end{align} $$ \implies \mathrm{Erf}^{-1}(p) = \frac{r_0}{\sigma \sqrt 2} \qquad \implies \sigma = \frac{r_0}{\sqrt 2 \mathrm{Erf}^{-1}(p)}$$ Then you can proceed to carry out your further plans.