The Dirac Delta Function, denoted as \( \delta(x) \), is a generalized function that is infinite at \( x = 0 \) and zero everywhere else, with an integral of 1 over the entire real line.
\[ \delta(x) = \begin{cases} +\infty & \text{if } x = 0 \\ 0 & \text{otherwise} \end{cases} \]
\[ \int_{-\infty}^{\infty} \delta(x) \, dx = 1 \]
It is often used in physics and engineering to model an idealized point mass or impulse.
The Dirac Delta Function can be represented as the limit of two common functions:
Consider a Gaussian function centered at \( x = 0 \) with height \( \frac{1}{\sigma\sqrt{2\pi}} \) and standard deviation \( \sigma \):
\[ f_\sigma(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{x^2}{2\sigma^2}} \]
As \( \sigma \to 0 \), the Gaussian becomes infinitely tall and narrow, approaching the Dirac Delta Function:
\[ \delta(x) = \lim_{\sigma \to 0} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{x^2}{2\sigma^2}} \]
Consider a rectangular function of height \( \frac{1}{\epsilon} \) and width \( \epsilon \) centered at \( x = 0 \):
\[ f_\epsilon(x) = \begin{cases} \frac{1}{\epsilon} & \text{if } |x| \leq \frac{\epsilon}{2} \\ 0 & \text{otherwise} \end{cases} \]
As \( \epsilon \to 0 \), the rectangular function becomes infinitely tall and narrow, approaching the Dirac Delta Function:
\[ \delta(x) = \lim_{\epsilon \to 0} f_\epsilon(x) \]
The Dirac Delta Function has several important properties:
\[ \int_{-\infty}^{\infty} f(x) \delta(x - a) \, dx = f(a) \]
This is the most commonly used property, often used to extract the value of a function at a specific point.
\[ \delta(ax) = \frac{1}{|a|} \delta(x) \]
\[ \delta(-x) = \delta(x) \]
\[ \delta(x) = \frac{d}{dx} H(x) \]
where \( H(x) \) is the Heaviside step function.
\[ f(x) * \delta(x - a) = f(x - a) \]
Evaluate the integral:
\[ \int_{-\infty}^{\infty} (3x^2 + 2x + 1) \delta(x - 2) \, dx \]
Using the sifting property of the Dirac Delta Function:
\[ \int_{-\infty}^{\infty} f(x) \delta(x - a) \, dx = f(a) \]
Here, \( f(x) = 3x^2 + 2x + 1 \) and \( a = 2 \).
\[ f(2) = 3(2)^2 + 2(2) + 1 = 12 + 4 + 1 = 17 \]
Therefore, the integral evaluates to 17.
Evaluate the integral:
\[ \int_{-\infty}^{\infty} \delta(3x - 6) \, dx \]
Using the scaling property of the Dirac Delta Function:
\[ \delta(ax) = \frac{1}{|a|} \delta(x) \]
Rewrite the argument of the delta function:
\[ \delta(3x - 6) = \delta(3(x - 2)) = \frac{1}{3} \delta(x - 2) \]
Now, evaluate the integral:
\[ \int_{-\infty}^{\infty} \frac{1}{3} \delta(x - 2) \, dx = \frac{1}{3} \int_{-\infty}^{\infty} \delta(x - 2) \, dx = \frac{1}{3} \cdot 1 = \frac{1}{3} \]
Therefore, the integral evaluates to 1/3.