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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?

TwilightSparkleTheGeek
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2 Answers2

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In fact there is no such thing as the Dirac delta function. Moreover the definition $$\delta(t) = \left\{\begin{array}{ll}1 & \quad t = 0 \\ 0 & \quad t \neq 0\end{array}\right.$$ Is inconsistent with $$\int^\infty_{-\infty} \delta (t)\,dt = 1$$ because this evaluates to zero and not to $1$, the value being the integral of a function that is zero almost everywhere. In the realm of linear operators on a space $\mathcal{F}$ of suitably integrable functions other objects than functions can be considered. For instance given any function $g$ the operator $R: f\mapsto \int_{-\infty}^{+\infty} f(x).g(x)\mathrm{d}x$ maps a function into a value in a linear manner. This way any function can be considered as an operator in a function space, but the converse is not necessaraly true. Consider the operators $R_\tau: f\mapsto \int_{-\infty}^{+\infty} f(x).\mathrm{\ rect}_\tau(x)/\tau\mathrm{d}x$ that maps a function $f$ into the value $\int_{-\tau/2}^{+\tau/2} f(x)/\tau\mathrm{\ d}x$. While the limit of the functions in the integrand goes to an absurd "function" that is zero almost everywhere except at the origin where it takes the value $+\infty$, the limit of the operators $\lim_{\tau\rightarrow 0}R_\tau$ make sense, we have $$\lim_{\tau\rightarrow 0}\int_{-\infty}^{+\infty} f(x).\mathrm{\ rect}_\tau(x)/\tau\mathrm{\ d}x=f(0)$$ So another definition of the Dirace delta as an operator is $\delta(f)=f(0)$. Another approach that uses Gauss curves instead of rectangles is mentionned here

Marc Bogaerts
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  • This is from an electrical engineering book. My teacher said so himself that it's not really a function, we use it to model pulses in circuits. I posted it here so I could get a better understanding of it. Thanks for your explanation. :) – TwilightSparkleTheGeek Sep 28 '14 at 15:16
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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.

Matt L.
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  • Yes, I can take it from here.:) As for the 0+ limit, I know that is taking the limit from the right side of 0, but why the right side and not the left side 0-? – TwilightSparkleTheGeek Sep 28 '14 at 15:25
  • @TwilightSparkleTheGeek: Because otherwise you won't get $\delta(t)$. Note that the functions are odd functions in $\epsilon$ and that the limits $\epsilon\rightarrow 0^+$ and $\epsilon\rightarrow 0^-$ are different (for $t=0$). – Matt L. Sep 28 '14 at 15:29
  • @ Matt L. Does that go for the second statement too? – TwilightSparkleTheGeek Sep 29 '14 at 02:48
  • @TwilightSparkleTheGeek: For $t=0$ both functions are proportional to $1/\epsilon$. So the limit $\epsilon\rightarrow 0^-$ is $-\infty$, whereas the limit $\epsilon\rightarrow 0^+$ is $\infty$, the latter being what you want. – Matt L. Sep 29 '14 at 06:57
  • I see. Thanks @Matt L. :) – TwilightSparkleTheGeek Sep 30 '14 at 21:56