Dirac $\delta$ Function

We can firstly focus on a defined $\delta_l$ function as follow:

$\delta_l(x)= \cases{ 0, \quad x<-\frac{l}{2} \\ \frac{1}{l}, \quad -\frac{l}{2}<x<\frac{l}{2} \\ 0, \quad x>\frac{l}{2} }$

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:

$\delta(x) = \lim_{l \rightarrow 0} \delta_l (x)$

Its integral must has a property of

$\int_{-\infty}^{\infty} \delta(x) \text{d}x = 1$

We could easily find that, when $x$ is approaching $0$, $\delta(x)$ will be approaching infinity. That is,

$\delta(x)= \cases{ 0, \quad x \neq 0 \\ \infty, \quad x=0 }$

This $\delta$ function is so-called Dirac $\delta$ function.

Kronecker $\delta$ 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:

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

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.

The other important properties of $\delta$ function

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.