Dirac Delta Function & Sinc Representation

The Dirac delta function is a function that comes in useful when deriving the equations for Fourier transforms and evaluating other physics calculations. It is defined as a function, $\delta(x)$, which is equal to $0$ for all $x$ except at $x=0$, and if integrated over all $x$, gives a value of $1$. In other words, the Dirac delta function is an infinitely thin, infinitely tall spike at $x=0$ with an area of $1$. Often, the Dirac delta function is written containing a constant, $b$, as in $\delta(x-b)$. This way, the function can be translated horizontally, with the "spike" located at $x=b$.

Symmetrical and Sifting Properties

Two simple but useful properties that the Dirac delta function carries are its symmetrical and sifting properties. The former, extending from the idea that $y=\delta(x)$ is symmetrical about the $y$-axis, states that $\delta(x)=\delta(-x)$. The sifting or selector property of the Dirac delta function is encountered when the definite integral of $f(x)\delta(x-b)$ is taken with respect to $x$, where $f(x)$ is an arbitrary function. Because $\delta(x-b)=0$ for all $x$ except $x=b$, the product of $f(x)$ and $\delta(x-b)$ also gives $0$ for all $x$ except $x=b$. Furthermore, at $x=b$, the graph of $y=f(x)\delta(x-b)$ has $f(b)$ times the area enclosed between it and the $x$-axis than does the graph of $y=\delta(x-b)$. Therefore, if $a < b < c$, then the sifting property of the Dirac delta function states that $\int_a^cf(x)\delta(x-b)dx=f(b)$.

Nascent Sinc Representation

Because the Dirac delta function cannot be written with everyday mathematical expressions, there is no set equation that defines it. Instead, there are many expressions, called "approximate" or "nascent" representations that can be used to model the Dirac delta function. Among these is the sinc representation which comes in useful when deriving the normalization constants for the Fourier transform equations. This representation is given by the following:

\begin{equation}
\delta(x) = \lim_{\epsilon\to0} \frac{\sin{ x/\epsilon}}{\pi x}
\end{equation}

From simply visualizing the graph that $y= \left( \sin{x/\epsilon} \right)/(\pi x)$ would make as $\epsilon$ approaches $0$ and comparing it to that expected from the Dirac delta function, it is sufficient to conclude that $\delta(x) \propto\lim_{\epsilon\to0} \left(\sin{x/\epsilon}\right) / (\pi x)$. In order to prove that $\lim_{\epsilon\to0} \left( \sin{x/\epsilon} \right) / (\pi x)$ is normalized to be equal to $\delta(x)$, the integral of $\lim_{\epsilon\to0} \left(\sin{x/\epsilon} \right) / (\pi x)$ over all $x$ must be shown to give a value of $1$.

\begin{equation}
\int_{-\infty}^\infty\left. \frac{\sin{x/\epsilon}}{\pi x} \right]_{\epsilon=0}dx = 1
\end{equation}
Let $u=x/\epsilon$.
\begin{equation}
\left.\int_{-\infty}^{\infty} \frac{\sin{u}}{\pi \epsilon u}\epsilon du \right]_{\epsilon=0}=1
\end{equation}
\begin{equation}
\left[\int_{0}^{\infty} \frac{2\sin{u}}{\pi u}du \right]_{\epsilon=0} = 1
\end{equation}

To evaluate the integral, consider the following equations:

\begin{equation}
\int_0^{\infty} \int_0^{\infty} e^{-su}ds\sin{(u)} du= \int_0^{\infty}\int_0^{\infty} e^{-su}\sin{(u)}duds
\end{equation}

That inside integral on the right side of the above equation looks brutal, so let's try to whittle it down with integration by parts:

$$\int_0^{\infty} e^{-su}\sin(u)du=\left[-e^{-su}\cos{u} -s\int{e^{-su}\cos(u)}du\right]_{u=0}^{u= \infty}$$
$$\int_0^{\infty} e^{-su}\sin(u)du=\left[-e^{-su}\cos{u} -s\left(e^{-su}\sin{u} + s\int{e^{-su}\sin(u)}du \right) \right]_{u=0}^{u=\infty}$$
$$\int_0^{\infty} e^{-su}\sin(u)du=\left[-e^{-su}\cos{u} -se^{-su}\sin{u}\right]_{u=0}^{u=\infty}-s^2\int_0^{\infty}e^{-su}\sin(u)du$$
$$\left(1+s^2\right)\int_0^{\infty} e^{-su}\sin(u)du = \left[-e^{-su}\cos{u} - se^{-su}\sin{u} \right]_{u=0}^{u=\infty}$$
$$\int_0^{\infty} e^{-su}\sin(u)du=\left. \frac{e^{-su}(-s\sin{u} - \cos{u})}{s^2 + 1}\right]_{u=0}^{u=\infty}$$

So equation $(5)$ becomes:

\begin{equation}
\int_0^{\infty} \left. \frac{e^{-su}}{-u} \right]_{s=0}^{s=\infty} \sin{(u)} du= \int_0^{\infty} \left. \frac{e^{-su}(-s\sin{u} - \cos{u})}{s^2+1} \right]_{u=0}^{u=\infty} ds
\end{equation}
\begin{equation}
\int_0^{\infty} \frac{\sin{u}}{u} du= \int_0^{\infty} \frac{1}{s^2+1} ds
\end{equation}
\begin{equation}
\int_0^{\infty} \frac{\sin{u}}{u} du= \left.\tan^{-1} s\right]_0^{\infty}
\end{equation}
\begin{equation}
\int_0^{\infty} \frac{\sin{u}}{u} du=\frac{\pi}{2}
\end{equation}

The above equation makes equation $(4)$ true and therefore proves that:

$$\delta(x) = \lim_{\epsilon\to0} \frac{\sin{x/\epsilon}} {\pi x}$$