Dirac Delta Function
Definition
A function that satisfies the following two conditions is called the Dirac delta function.
$$ \delta (x) = \begin{cases} 0, & x\neq 0 \\ \infty , & x=0 \end{cases} $$
$$ \int_{-\infty}^{\infty}{\delta (x) dx}=1 $$
Description
※Be careful not to confuse it with the Kronecker delta.
In engineering, it is called the unit impulse function. Strictly speaking, mathematically, the Dirac delta function is not a function because it diverges to infinity at 0. Griffiths’ textbook explains it as follows: “Since the value of the delta function becomes infinitely large at $x=0$, technically it is not a function. In mathematical literature, it is called a generalized function or distribution.” It might be difficult to understand what the delta function is just by looking at the formula above. The picture below will help understand its geometric meaning.
To explain it more intuitively, it’s like the following. A series of functions whose height is $n$ and width is $\displaystyle \frac{1}{n}$, a rectangle $R_{n}(x)$, or whose height is $n$ and base is $\displaystyle \frac{2}{n}$, an isosceles triangle $T_{n}(x)$, and the limit of such function series
If you have understood what the delta function is, let’s look at its characteristics. If the function $f(x)$ is not a delta function but a general function, the value of $f(x)\delta (x)$ is $0$ everywhere except at $x=0$. (Since $\because$ $\delta (x)$ is $0$ everywhere except at $x=0$) That is, the value exists only at $x=0$. Hence, the following equation holds.
$$ f(x)\delta (x) = f(0) \delta (x) $$
In integral form
$$ \displaystyle{ \int_{-\infty}^{\infty} f(x) \delta (x) dx = f(0) \int_{-\infty}^{\infty} \delta (x) dx = f(0)} $$
To represent a general case, if we move the peak of the delta function from $x=0$ to $x=a$, it is as follows.
$$ \delta (x-a) = \begin{cases} 0, & x\neq a \\ \infty , & x=a \end{cases} $$
$$ \displaystyle{ \int_{-\infty}^{\infty}{\delta (x-a) dx}=1 } $$
$$ f(x)\delta (x-a) = f(a) \delta (x-a) $$
$$ \displaystyle{ \int_{-\infty}^{\infty} f(x) \delta (x-a) dx = f(a)} $$
In 3D, the delta function is as follows.
$$ \int f( \mathbf{r} ) \delta ^3 (\mathbf{r}-\mathbf{a}) d\tau = f(\mathbf{a}) $$
In this case,
$$ \int \delta ^3 (\mathbf{r} ) d\tau =1 $$
$$ \delta ^3 (\mathbf{r})=\delta (\mathbf{x}) \delta (\mathbf{y}) \delta (\mathbf{z}) $$