Calculations for 2D Gaussian

2D Gaussian distribution

$$f(x,y)=C*\frac{1}{\sqrt{2\pi\sigma_x^2}}{\rm exp}(-\frac{x^2}{2\sigma_x^2})\frac{1}{\sqrt{2\pi\sigma_y^2}}{\rm exp}(-\frac{y^2}{2\sigma_y^2})$$

For 2D image, assume $\sigma_x=\sigma_y=\sigma$, and

$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)=1$$

Then,

$$C*\frac{1}{2\pi\sigma^2}\int_{-\infty}^{+\infty}{\rm exp}(-\frac{x^2}{2\sigma^2})dx\int_{-\infty}^{+\infty}{\rm exp}(-\frac{y^2}{2\sigma^2})dy=1$$ $$\frac{C}{2\pi\sigma^2}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}{\rm exp}(-\frac{x^2+y^2}{2\sigma^2})dxdy=1$$

Under polar coordinates $(r, \theta)$, we have $r^2=x^2+y^2$, $rdrd\theta=dxdy$. So,

$$\frac{C}{2\pi\sigma^2}\int_{0}^{+\infty} \int_{0}^{2\pi}{\rm exp}(-\frac{r^2}{2\sigma^2})rdrd\theta=1$$ $$\frac{C}{2\pi\sigma^2}\int_{0}^{2\pi}d\theta \int_{0}^{+\infty}{\rm exp}(-\frac{r^2}{2\sigma^2})rdr=1$$ $$\frac{C}{\sigma^2}\int_{0}^{+\infty}{\rm exp}(-\frac{r^2}{2\sigma^2})rdr=1$$ $$-C\int_{0}^{+\infty}{\rm exp}(-\frac{r^2}{2\sigma^2})d(-\frac{r^2}{2\sigma^2})=1$$ $$-C{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{+\infty}=1$$ $$C=1$$

So, different energy at corresponding R is presented by $-{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{\sigma}$:

$$-{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{\sigma}=0.3935$$ $$-{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{2\sigma}=0.8647$$ $$-{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{3\sigma}=0.9889$$ $$-{\rm exp}(-\frac{r^2}{2\sigma^2})|_{0}^{4\sigma}=0.999996$$