Definition:
Square function $\Pi$
$$ \Pi(x)=\left \{ \begin{matrix} 1 & -0.5\le x\le0.5 \\ 0 & \text{otherwise} \end{matrix}\right . $$
triangle function $\Lambda$
$$ \Lambda(x)=\left \{ \begin{matrix} 1-|x| & -1\le x\le1 \\ 0 & \text{otherwise} \end{matrix}\right . $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/f18ba8f6-a5aa-498e-9d58-14529fecbaaf/dirac_delta.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/f18ba8f6-a5aa-498e-9d58-14529fecbaaf/dirac_delta.png" width="40px" /> Dirac delta function $\delta(x)$:
🛍️ Definition:
$$ \begin{aligned} \delta(x)&=\lim_{k\to\infin}k\Pi(kx) \\ \text{or} \quad \delta(x)&=\lim_{k\to\infin}k\Lambda(kx) \\ \text{or} \quad \delta(x)&=\lim_{k\to\infin}\sqrt{\frac{k}{\pi}}\,e^{-kx^2} \\ \text{or} \quad \delta(x)&=\lim_{k\to\infin}\frac{k}{\pi}\frac{\sin(kx)}{kx}
\end{aligned} $$
🐉 Properties:
$$ \begin{aligned} \int^{+\infin}{-\infin}\delta(x)\,\text dx&=1 \\ \int^{+\infin}{-\infin}\delta(x-x_0)f(x)\,\text dx&=f(x_0) \\
\end{aligned} $$
</aside>
Let’s link it back to Fourier integrals
We can re-write the Dirac delta function as
$$ \delta(x)=\frac{1}{2\pi}\int^\infin_{-\infin} e^{-ikx} \,\text dk=\frac{1}{2\pi}\int^\infin_{-\infin}e^{ikx}\,\text dk=\delta(-x) $$
From this we can derive the Fourier transform pair as follows:
$$ \begin{aligned} \underbrace{\int^\infin_{-\infin}f(x')\delta(x'-x)\,\text dx'}{f(x)}&=\frac{1}{2\pi}\int^\infin{-\infin} \int^\infin_{-\infin}e^{-ik(x'-x)}f(x')\,\text dk\,\text dx' \\ f(x)&=\frac{1}{2\pi}\int^\infin_{-\infin}\underbrace{\int^\infin_{-\infin} f(x')e^{-ikx'}\,\text dx'}_{F(k)} e^{ikx}\,\text dk
\end{aligned} $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/242729b1-7098-49b8-af86-af4b17a2104f/Fourier_transform_pair.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/242729b1-7098-49b8-af86-af4b17a2104f/Fourier_transform_pair.png" width="40px" /> Fourier transform pair:
Physical:
$$ \begin{aligned} f(x)&=\frac{1}{2\pi}\int^\infin_{-\infin}F(k)e^{-ikx}\,\text dk \\ F(k)&=\int^\infin_{-\infin}f(x)e^{ikx}\,\text dx
\end{aligned} $$
Mathematics:
$$ \begin{aligned} f(x)&=\frac{1}{\sqrt{2\pi}}\int^\infin_{-\infin}F(k)e^{-ikx}\,\text dk \\ F(k)&=\frac{1}{\sqrt{2\pi}}\int^\infin_{-\infin}f(x)e^{ikx}\,\text dx
\end{aligned} $$
</aside>
Lets try and prove the following:
$$ \sum^\infin_{n=-\infin}\delta(x-2n\pi)=\frac{1}{2\pi}\sum^\infin_{n=-\infin}e^{inx} $$
We start with the Fourier series in exponential form
$$ \begin{aligned} f(x)&=\sum^\infin_{n=-\infin}c_n e^{in\pi x/L} \\ &= \sum^\infin_{n=-\infin} c_n e^{inx} \\ c_n&= \frac{1}{2\pi}\int^\pi_{-\pi}f(x)e^{-inx}\,\text dx
\end{aligned} $$
where $x$ has been replaced $x\to \pi x/L$
We evaluate the coefficients of the function
$$ f(x)=\sum^\infin_{n=-\infin}\delta(x-2n\pi) $$
by integrating over one period
$$ c_n=\frac{1}{2\pi}\int^{2\pi}_0 f(x) e^{-inx}\,\text dx=\frac{1}{2\pi} $$
hence
$$ \sum^\infin_{n=-\infin}\delta(x-2n\pi)=\frac{1}{2\pi}\sum^\infin_{n=-\infin}e^{inx} $$
🫠 List of properties of the $\delta(x)$ function
$$ \begin{aligned} \delta(x)&=\delta(-x) \\ x\delta(x)&=0 \\ \frac{\text d}{\text dx}\delta(x)&=-\frac{\text d}{\text dx}\delta(-x)\\ x\frac{\text d}{\text dx}\delta(x)&=-\delta(x) \\ \int^\infin_{-\infin} \delta(x-a)f(x)\,\text dx&=f(a) \\ \int^\infin_{-\infin} \delta'(x-a)f(x)\,\text dx&=f'(a) \\ \delta(ax)&=\frac{1}{|a|}\delta(x) \\ \delta(g(x) &= \sum_i \frac{1}{|g'(x_i)|}\delta(x-x_i) \\ \delta(x^2-a^2) &= \frac{1}{2a}\left [ \delta (x-a)+\delta(x+a) \right ] \\ \delta((x-a)(x-b)) &= \frac{1}{|a-b|}[\delta(x-a)+\delta(x-b)]
\end{aligned} $$
where $a$ and $b$ are constants $g(x_i)=0$ and $g'(x_i)\ne0$.
🗒️ Note: the Dirac delta function is even
🎩 Example 1: Find the Fourier transform of the rectangular function which is bound by $\pm a$
The definition of the rectangular function is
$$ f(x)=\left \{ \begin{matrix} 1 & -a\le x\le a \\ 0 & \text{otherwise} \end{matrix}\right . $$
🗒️ Note: this corresponds to calculating the Fourier transform of $\Pi(x/2a)$
If we use the mathematic transform we get
$$ \begin{aligned} F(k)&=\frac{1}{\sqrt{2\pi}}\int^\infin_{-\infin}f(x) e^{-ikx}\,\text dx=\frac{1}{\sqrt{2\pi}}\int^a_{-a}e^{-ikx}\,\text dx \\ &= \frac{1}{\sqrt{2\pi}} \left ( \frac{1}{-ik} \right )\underbrace{\left [ e^{-ika}-e^{ika} \right ]}_{-2i\sin(ka)} \\ &=\sqrt\frac{2}{\pi}\, \frac{\sin(ka)}{k}=\sqrt{\frac{2}{\pi}}\,a \operatorname{sinc}(ka)
\end{aligned} $$
