I'm having trouble understanding what this question is trying to ask me. I understand the "limiting case of the rectangular function" but I don't see how I can show that the following functions satisfy the same requirement.
Here is the problem:
The Dirac Delta function was obtained as a limiting case of the rectangular function, given below: $$\delta(t) = \lim_{\epsilon\to\infty}\frac{1}{\epsilon}rect(\frac{t}{\epsilon})$$
where rect() function is defined as:
$$
rect(\frac{t}{\tau}) = \left\{
\begin{array}{ll}
1 & \quad |t| \leq \frac{\tau}{2} \\
0 & \quad |t| > \frac{\tau}{2}
\end{array}
\right.
$$
where the dirac delta function is defined as:
$$
\delta(t) = \left\{
\begin{array}{ll}
1 & \quad t = 0 \\
0 & \quad t \neq 0
\end{array}
\right.
$$
and defined through integration:
$$\int^\infty_{-\infty} \delta (t)\,dt = 1$$
Show that the Dirac delta function can also be obtained from each of the following functions that satisfy the definition of the Dirac delta function given above. \
(i) $$\lim_{\epsilon\to\infty}\frac{\epsilon}{\pi (t^2+\epsilon^2)}$$
(iI) $$\lim_{\epsilon\to\infty}\frac{2\epsilon}{4\pi^2 t^2+\epsilon^2}$$
How do I show that these two functions satisfy the definition as the rect() function does?
Best Answer
Note that all limits need to be $\epsilon\rightarrow 0^+$ instead of $\epsilon\rightarrow\infty$. The other error ($\delta(0)=1$) has already been pointed out in Nimda's answer.
What you need to show is that the functions defined in (i) and (ii) in your question have an integral of unity, and that for $\epsilon\rightarrow 0^+$ they become zero for $t\neq 0$ and tend to infinity for $t=0$. E.g., for the first function you need to verify
$$\frac{\epsilon}{\pi}\int_{-\infty}^{\infty}\frac{1}{t^2+\epsilon^2}dt=1$$
and
$$\lim_{\epsilon\rightarrow 0^+}\frac{\epsilon}{\pi(t^2+\epsilon^2)}=\begin{cases}\infty,& t=0\\0,&t\neq 0 \end{cases}$$
I'm sure you can take it from here.