We can firstly focus on a defined $\delta_l$ function as follow:
As we can see, if this function is integrated from infinity to infinity, we yield $1$. Because the area which under $\delta_l$ function is $1$. Hold this value, decrease $l$, which means pressing $l$ to $0$，then we have the definition of $\delta$ function:
Its integral must has a property of
We could easily find that, when $x$ is approaching $0$, $\delta(x)$ will be approaching infinity. That is,
This $\delta$ function is so-called Dirac $\delta$ function.
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:
This $\delta$ function would be $1$ only when $n=0$. Thus this $\delta$ function must have a “definition” like
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
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
This property holds true in discrete space:
We have to mention that, this integral has to be interpreted as follow:
So then by utilizing the properties listed above, we could infer the same result.