## $\displaystyle \delta(t)=\lim_{ \Delta \to 0 }\text{rect}_{\Delta}(t)$
* $\displaystyle \text{rect}(t)$ is the [[unit rectangle]]
# Properties
## $\displaystyle \int_{-\infty}^{\infty}f(x)\delta(x) \, \mathrm{d}x=\int_{-\infty}^{\infty}f(x)\delta(-x) \, \mathrm{d}x=f(0)$
* The delta function is symmetric and has the effect of sending an integral of $\displaystyle f(x)$ to $\displaystyle f(0)$
## $\displaystyle \int_{a}^{b} f(x)\delta(x-x_{0}) \, \mathrm{d}x=0 \text{ for } x_{0} \notin [a,b]$
* The delta function in an integral evaluates to 0 if $\displaystyle x_{0}$ is not in the bounds of integration
## $\displaystyle \int_{-\infty}^{\infty} f(x)\delta(x-x_{0}) \, \mathrm{d}x=f(x_{0})$
## $\displaystyle \int_{-\infty}^{\infty} f(x)\delta(k x) \, \mathrm{d}x=\frac{1}{\lvert k\rvert}f(0)$
* This is shown by a u-substitution of $\displaystyle u=kx$
* The absolute value sign occurs because the delta function is symmetric because $\displaystyle \delta(x)=\delta(-x)$
## $\displaystyle \delta(x-x_{0})=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(x-x_{0})} \, \mathrm{d}k$
* Another way of saying this is $\displaystyle \mathcal{F}[1]=2\pi \delta(\omega)$, or the [[Fourier Transform]] of a constant $\displaystyle a$ is $\displaystyle 2\pi a\delta(\omega)$
* Proof:
* ![[Assets/Images/Pasted image 20250520013942.png|500]]