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I have modeled the Dirac delta function using limiting value of the rectangular pulse and now proving its area property. In the proof, I need to interchange the order of integration and limit, which does not look possible. I looked out at the similar problems on math stackexchange but could not find it anywhere. The details about the problem are as follows:

To model the delta function, first I took the rectangular pulse function as:

\begin{equation} p_n(t) = \frac{1}{n} \ \textrm{for}\ |t| \le \frac{n}{2}\ \textrm{otherwise}\ 0 \end{equation}

Based on this rectangular pulse, we can define the delta function as follows:

\begin{equation} \delta(t) = \lim_{n \rightarrow 0}{p_n(t)} \end{equation}

So the above equation precisely captures the behavior of the delta function.

Now, I want to prove the area property of the delta function, which is:

\begin{equation} \int_{-\infty}^{+\infty}{\delta(t) dt} = 1 \end{equation}

By writing the above equation using the definition of the delta function, it results into:

\begin{equation} \int_{-\infty}^{+\infty}{\lim_{n \rightarrow 0}{p_n(t)}}dt = 1 \end{equation}

Next, if we look at the integrand of the above integral, it is divergent as the function $\lim_{n \rightarrow 0}{p_n(t)}$ has infinite value, so overall its not integrable. Now, I want to swap the limit and integral operator, which is only possible if the limit is convergent (where it is divergent in this case).

My question is that is there really a way I can swap the order of this limit and integral? If yes then how would I do it?

Note: I also checked out many questions/answers regarding this on math stackexchage, but in all cases they directly take limit outside the integral, i.e., $\lim_{n \rightarrow 0}{\int}_{-\infty}^{+\infty}{p_n(t)} dt$, and it would only be equal to $\int_{-\infty}^{+\infty}{\delta(t) dt}$, if the limits and integral operators can be swapped. Moreover, the integral that I am using is the gauge integral.

I would highly appreciate your earliest response on it.

Thanks in advance.

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  • $\begingroup$ Could you please explain what is "the gauge integral"? $\endgroup$ – John B Dec 29 '17 at 21:44
  • $\begingroup$ @JohnB you can find details about Guage integral at "en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral". $\endgroup$ – Adnan Dec 29 '17 at 23:22
  • $\begingroup$ The main property of $p_\epsilon(t) = \frac{ 1_{|t|< \epsilon/2}}{\epsilon}$ is $\lim_{\epsilon \to 0^+} p_\epsilon \ast \varphi(t) = \varphi(t)$ for any $\varphi$ continuous, where $\ast$ is the convolution, and the convergence is locally uniform. This is stronger than $\lim_{\epsilon \to 0^+} p_\epsilon = \delta$ in the sense of distributions. $\endgroup$ – reuns Dec 30 '17 at 4:54
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Just because the function under the integral contains an infinite value, it does not mean that the function is not integrable. A simple example would be $f(x)=1/\sqrt{|x|}$. When $x\rightarrow 0$ $f$ goes to infinity. You can however integrate in an interval around zero. You can integrate between $-1$ and $1$, and you get $4$.

Since the integration limits do not change with $n$, you can safely move the limit outside of the integration.

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  • $\begingroup$ My immediate question is that to swap the limit and integration, we do need the "dominated convergence theorem (en.wikipedia.org/wiki/Dominated_convergence_theorem)", which says that in order to swap the order, we need $|f_n(x)| \le g(x)$, where $g(x)$ is an integrable function, i.e., $\int{g(x) dx} \lt \infty$. So, how is it possible that the |f_n(x)| containing that infinite value would be less than $\infty$. $\endgroup$ – Adnan Dec 29 '17 at 23:10
  • $\begingroup$ Moreover, can I find this particular swap for delta function in any literature, if you could refer. As I could not find any source depicting this swap for the delta function. Its always taken outside of the integral, which is not right indeed. $\endgroup$ – Adnan Dec 29 '17 at 23:15
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    $\begingroup$ -1: this does not answer the question, since $\delta$ is not a function. Besides, the last sentence seems to imply that limit and the integral can always be exchanged, which is definitely not the case (if it were, we wouldn't need the ever-present Monotone and Dominated Convergence theorems). $\endgroup$ – Martin Argerami Dec 30 '17 at 4:44
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Since the support of $\delta $ is a point, the natural interpretation of the gauge integral of $\delta $ is that it should be zero. To get a gauge integral you need to calculate Riemann sums; for an interval $(u_i,u_{i+1}) $, you need to calculate $\delta (t_i)\, (u_{i+1}-u_i) $. You now want to get $\infty\times (u_{i+1}-u_i)=1$, and there's no arithmetic that will give you that.

So, the question about exchanging limit and integral does not really apply, since $\delta $ is not actually a function.

The usual approach to avoid this problem is the following. The $\delta$ is never used on its own. It used in the sense that $$\tag1\int_{\mathbb R} f(t)\,\delta(t)\,dt=f(0)$$ for any (good enough, say) function $f$.

One way to achieve $(1)$ is to consider the left-hand-side as the limit $$ \lim_n \int_{\mathbb R} f(t) p_n(t)\,dt. $$ This is still not ideal because you are subjecting yourself to the choice of the $p_n$, and in the end what one cares about is the net result, which is the equality $(1)$. The solution to this is to be more abstract, and consider $\delta$ as a linear functional, that to every $f$ assigns the number $f(0)$.

A second approach, that agrees with the previous one, is to consider $\delta$ as a measure, namely $$\delta(E)=\begin{cases} 1,&\ 0\in E\\ \ \\ 0,&\ 0\not\in E\end{cases}$$ Then one can write $$\tag2 \int_{\mathbb R} f(t)\,d\delta(t)=f(0) $$ in the sense of Lebesgue integration.


Edit: why $\delta$ "function"?

On the set $C_0(\mathbb R)$, of continuous functions vanishing at infinity, one considers (bounded) linear functionals, which are linear maps $\varphi:C_0(\mathbb R)\to\mathbb R$. By the Riesz-Markov theorem, for any such $\varphi$ there exists a measure $\mu_\varphi$ such that $$\tag3 \varphi(f)=\int_{\mathbb R} f(t)\,d\mu_\varphi(t),\ \ \ \ f\in C_0(\mathbb R). $$ If $\mu_\varphi$ is absolutely continuous with respect to Lebesgue measure, by Radon-Nikodym there exists a function $g_\varphi$ such that $$\tag4 \varphi(f)=\int_{\mathbb R} f(t)\,g(t)\, dt, \ \ \ \ f\in C_0(\mathbb R). $$

Dirac needed to use the functional $\psi(f)=f(0)$. Doing things "the physicist way", he assumed a $g$ like in $(4)$ would somehow exist, even if now a function, he called it $\delta$ and wrote $(1)$.

Mathematically, what's happening is $(2)$. There is no "$\delta$-function".

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  • $\begingroup$ Thanks for your reply. Talking about the first approach described in your answer, you took the limit outside the integral on the left-hand side. Whereas, in the definition of the delta function, it applies on the pulse function. So it could only be used outside the integral if the swap of integral with limit can be done, which does not look possible in the case of the delta function. This clearly implies that Equation (1) in your answer is not achievable, because mathematically you cannot take limit outside the integral without that swap operation. $\endgroup$ – Adnan Dec 30 '17 at 16:36
  • $\begingroup$ You are still thinking of $\delta$ as an actual function. It isn't. You can define $\delta$ as you did, but then it will not have the area property. I'll add some info to the answer to explain where the "function" thing comes from. $\endgroup$ – Martin Argerami Dec 30 '17 at 16:59
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You should be careful writing \begin{equation*} \lim_{n \to \infty} p_n (t) = \delta (t), \end{equation*} since the usual way to interpret that notation without further qualification is as a pointwise limit. Here, the pointwise limit is zero for every non-zero $t$, and does not exist (as a finite value) for $t = 0$, so it does not make sense to assign it to a function $\delta(t)$ defined on the whole real line. Also, we could modify your construction of $p_n (t)$ to get a sequence of approximations to the Dirac delta which actually converges pointwise to zero everywhere, but still really converges to the Dirac delta in the most meaningful sense.

The convergence to the Dirac delta is perhaps best interpreted in the sense of distributions, which are continuous (with respect to a certain topology) linear functionals on the space of smooth compactly supported functions on $\mathbb{R}$. That is, the Dirac delta distribution is a function from the space of functions $C^\infty_c (\mathbb{R})$ to the real (or complex) numbers, defined by $\delta (f) = f(0)$.

The Dirac delta distribution happens to have compact support (https://en.wikipedia.org/wiki/Distribution_(mathematics)#Support_of_a_distribution), and so can be extended to a continuous functional on $C^\infty (\mathbb{R})$ (i.e. we can remove the requirement that the functions it acts on have compact support). Then, the notation \begin{equation*} \int_{-\infty}^\infty \delta (t) \, dt = \int_{-\infty}^\infty 1 \cdot \delta (t) \, dt \end{equation*} should really be interpreted as the value of the Dirac delta distribution at the constant function $1$ (which is smooth, as required). This is equal to $1$ by the definition of the Dirac delta distribution.

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I suggest that you would be better served by approximating the Delta function by a continuous differentiable/integrable function of time. To that end, I propose the superparabola described here. This function can be described by

$$y(t)=(1-t^2)^p,\quad t\in[-1,1]$$

The area under the curve is given by

$$A=\int_{-1}^1 y~dt=\frac{\sqrt{\pi}~\Gamma(p+1)}{\Gamma(p+3/2)}$$

The model delta function can be described as

$$\tilde \delta(t)=\lim_{p\to\infty} \frac{\Gamma(p+3/2)}{\sqrt{\pi}~\Gamma(p+1)}(1-t^2)^p$$

Clearly,

$$\int_{-\infty}^{\infty} \tilde \delta~dt=1$$

and you have full control of all the mathematical properties of this function, such as integrating to get the Heaviside function, or differentiating to get the doublet, and so on. What I like about this model is that it is confined to the region $t\in[-1,1]$, as opposed the Gaussian model that extends to $t\in(-\infty,\infty)$.

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