# Discrete Laplacian of Gaussian (LoG)

In Image processing, one often uses the discrete Laplacian of Gaussian (LoG) to do edge detection with

$LoG(x,y) = -\frac{1}{\pi \sigma^4}[1-\frac{x^2+y^2}{2\sigma^2}]\cdot e^{-\frac{x^2+y^2}{2\sigma^2}}$

Various sources here, here or here give discrete Kernels of the LoG to be convoluted with the input image to yield the filtered version. However, I do not understand the derivation of this Kernel from the function.

One possible Kernel for $\sigma = 1.4$ is

$K = \begin{bmatrix}0.0& 1.0& 1.0& 2.0& 2.0& 2.0& 1.0& 1.0& 0.0\\ 1.0& 2.0& 4.0& 5.0& 5.0& 5.0& 4.0& 2.0& 1.0\\ 1.0& 4.0& 5.0& 3.0& 0.0& 3.0& 5.0& 4.0& 1.0\\ 2.0& 5.0& 3.0& -12.0& -24.0& -12.0& 3.0& 5.0& 2.0\\ 2.0& 5.0& 0.0& -24.0& -40.0& -24.0& 0.0& 5.0& 2.0\\ 2.0& 5.0& 3.0& -12.0& -24.0& -12.0& 3.0& 5.0& 2.0\\ 1.0& 4.0& 5.0& 3.0& 0.0& 3.0& 5.0& 4.0& 1.0\\ 1.0& 2.0& 4.0& 5.0& 5.0& 5.0& 4.0& 2.0& 1.0\\ 0.0& 1.0& 1.0& 2.0& 2.0& 2.0& 1.0& 1.0& 0.0\end{bmatrix}$

My question is, how these discrete Kernels are derived from the $LoG(x,y)$ function. Is there any scaling factor involved? Is it an integration over the discrete pixel area?

My calculated values for the central point (-40.0) are:

$LoG(0,0) \approx -0.1624$

and

$\iint\limits_{-0.5}^{0.5}{LoG(x,y)} dx dy \approx -0.0761$

The answer is scaling. The matrix $K$ is simply a $n\times n$ matrix with $n$ being odd and
$K(i,j)=LoG(i-\frac{n-1}{2},j-\frac{n-1}{2})$.