If we integrate $delta$ function in a continuous space, it appears to be Dirac $delta$ function. But what if we conduct a discrete space integration, and then add these $delta$ function values together? Then we could have:

$$sum_n^{infty} delta(n) = 1$$

This $delta$ function would be $1$ only when $n=0$. Thus this $delta$ function must have a “definition” like

$$delta(n)= cases{ 0, quad n neq 0 \ 1, quad n=0 }$$

This seems to be different when compared to the $delta$ function we have just discussed above. But they are actually the same. They both have only one variable, and the specific integral holds $1$. However, this discrete-condition definition of $delta$ function is similar to a definition of “another” function, that is

$$delta_{ij}= cases{ 0, quad i neq j \ 1, quad i=j }$$

This $delta$ function is so-called Kronecker $delta$ function, which has two variables: i and j. And it shares a similar form with Dirac $delta$ function. Most of the time, we don’t specifically distinguish these two functions, for the reason that they can be easily transformed to the other.

If we times $f(x)$ with $delta(x)$, and then intergrade it from infinity to infinity, yield

$$int_{-infty}^{infty} f(x) delta(x) text{d}x = f(x) |_{x=0} = f(0)$$

This property holds true in discrete space:

$$sum_n^{infty} f(n) delta(n) = f(0)$$

We have to mention that, this integral has to be interpreted as follow:

$$int_{-infty}^{infty} f(x) delta(x) text{d}x = lim_{l rightarrow 0} int_{-frac{l}{1}}^{frac{l}{1}} f(x) delta_l(x) text{d}x$$

So then by utilizing the properties listed above, we could infer the same result.