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Recently I have encountering the dirac delta function $\delta(x)$ more often. And have been thinking about its antiderivatives:

The function is defined as: $$\delta(x)=\begin{cases} \infty ; x=0\\ 0 ; \text{elsewhere} \end{cases}$$

Its antiderivative can be defined as: The function is defined as: $$\int_{-\infty}^t \delta(x)dx=\begin{cases} 0 ; t <0\\ 1 ; t\geq0 \end{cases}$$

This is the definition of the unit step function. Hence one can claim: The function is defined as: $$\int_{-\infty}^t \delta(x)dx=u(t)$$ But now suppose one took an indefinite integral of the function, would the antiderivative look like: $$\int \delta(x)dx=u(t) +c$$ Or if one went back to the definition: The function is defined as: $$\int_{t_0}^t \delta(x)dx=\begin{cases} 1 ; t\cdot t_0 \leq 0\\ 0 ; t_0\cdot t>0 \end{cases}=c\cdot u(t)$$ where $c$ is the constant of integration. Where do I put the constant of integration?

Boy
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    The Dirac Delta is not a function and has no antiderivative. It is a linear functional that maps smooth, compactly supported functions to their value at $0$.

    If $tt_0\ne 0$, then

    $$\begin{align} \int_{t_0}^t \delta(x),dx&=\int_{-\infty}^\infty (u(t-x)-u(t_o-x))\delta(x),dx\tag1\\ &=u(t)-u(t_0) \end{align}$$

    It is important to note that the unit step function is not a smooth and compactly supported function. However, the Dirac Delta as a measure has compact support ${0}$, which permits the evaluation of the functional in $(1)$.

     – Mark Viola Sep 28 '22 at 14:47
  • $u(t) + c$ is correct. Why are you multiplying by $c$ in the last equation? – Qiaochu Yuan Sep 28 '22 at 15:27
  • I was studying unit impulse response, in which linear differential equation like: $\frac{dy}{dt}+ay=\delta(x)$. So, in this an argument I read was that the highest derivative is the delta function. – Boy Sep 28 '22 at 18:26
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    Formatting tip: in a "cases" format, if you put "&" between the value and the condition, it will be spaced out nicely and easier to read: $$\delta(x)=\begin{cases} \infty ;& x=0\ 0 ; &\text{elsewhere}\end{cases}$$ – Paul Sinclair Sep 29 '22 at 17:45
  • @QiaochuYuan Just curious as to why you are perpetuaitng the notion that the Dirac Delta is a funciton and that it has an integral (unless the Dirac Delta is viewed as a measure, which is almost assuredly beyond the scope of this conversation)? – Mark Viola Sep 29 '22 at 23:19
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    @PaulSinclair It was nice to give the OP a formatting tip, but the example you provided would validate the absurdity that the Dirac Delta is a function that takes on values. – Mark Viola Sep 29 '22 at 23:21
  • @MarkViola - the "example" I gave was simply a reformatting of a part of his post. I did not make nor intend to make any comment of the content itself. I considered that already covered. – Paul Sinclair Sep 30 '22 at 01:18
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    @PaulSinclair Yes. I understand the intention. But it seems to have an unintended consequence that perpetuates a misconception often taught in books and in the classroom. – Mark Viola Sep 30 '22 at 02:23
  • @MarkViola - Sorry, you can fight this battle all you want, but physicists are simply going to keep ignoring you, keep using the Dirac delta function with this same questionable justification, and keep happily getting the right results when they do. They don't care that the concept doesn't meet mathematical standards. They only care that it works. And even mathematicians have long since figured out that it does work, and why it works even though it isn't fully sensible, so it is a pointless battle. Also, linear functionals is not the only mathematical way to handle the Dirac delta. – Paul Sinclair Oct 01 '22 at 14:00
  • @PaulSinclair My background is electromagnetic field theory, a branch of classical physics. And Paul Dirac was an electrical engineering student and theoretical physicist, which you likely know. So it is frustrating that my brethren continue to teach this in such disregard for rigor. What other way than linear functionals were you thinking (e.g., as a measure)? – Mark Viola Oct 01 '22 at 14:44
  • If it works, it is rigorous. They just leave it to mathematicians to figure out why, while they get on with their own work. I've seen a few treatments. The one I like best is partly a private musing. Treat linear functions on germs of continuous functions at $0$ as a set of "infinite" values. Extend $\Bbb R$ by adding these values, forming a vector space over $\Bbb R$. Consider the class of functions that are continuous except on a finite set, where they are allowed to take on these "infinite" values. I like it because it almost matches the naive concept. – Paul Sinclair Oct 01 '22 at 15:33

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