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I see that the values of a distribution at discontinuities are essentialy equal to the mean of the right-hand and left-hand limit values when dealing with distributions in relation to the Fourier Transform, which can have some weird effects (if I am not mistaken):

$x \cdot (PV\frac{1}{x} \cdot \delta(x)) = x \cdot 0 = 0$, while $(x \cdot PV\frac{1}{x}) \cdot \delta(x) = 1 \cdot \delta(x) = \delta(x)$ is a common example.

This makes it seem like the discontinuity 'updates': it always seems to be 'defined' such that the mean between right-hand and left-hand limit values are taken, even in the middle of calculations.

This peaks my curiosity: is this a more general result? For example, take sign(t) as the signum function with sign(0) = 0. In this case, will the value of $(\text{sgn}(t) \cdot \text{sgn}(t)) \cdot \delta(t)$ be equal to $0$ or will it be equal to $\delta(t)$ by using the fact that, using this definition, $\text{sgn}(t) \cdot \text{sign}(\text{t}) = 1$, even for $\text{t} = 0$?

I may be overthinking this a bit: I can also imagine that the PV of the integral can influence this result, in addition I am not really that acquainted with distribution theory to deduce this property myself. Please let me know if I am approaching this the right way.

Another important thing is that in principle we can not use $\delta(t)$ to find the value of a discontinuity: the sifting property only holds with continuous functions at the value where the delta spike is active. But using nascent delta functions makes it seem like it is possible. Perhaps nascent delta functions are not applicable for this?

Example: take $\Pi(t)$ as the unit pulse between $(-\tau,\tau)$ such that $$\int_{-\infty}^{\infty}\frac{1}{\tau}\Pi(t)dt = 1$$ This is a nascent delta function if we take $\tau \rightarrow \infty$. If we then multiply this function with a unit step function, we get the following result: $$\int_{0}^{\infty}\frac{1}{\tau}\Pi(t)dt = \frac{1}{2} \stackrel{?}{\implies} \text{u}(t)\cdot\delta(t) = \frac{1}{2}\delta(t)$$

Any help is appreciated, thanks in advance!

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    $\begingroup$ You're playing fast and loose with your math. $x \cdot PV \frac 1 x$ is not equal to 1 everywhere: at $x = 0$ it is indeterminate ($\frac 0 0$). I'm not sure what, if any, theory there is to formalize doing math with the Dirac delta functional, but if there is it would have to recognize that $\delta(x)$ is a short-hand for some really high-valued, short function, which is, in itself, dealt with using limits. As soon as you start taking limits of two things at once, then the math is going to get complicated and you'll need to work carefully to resolve it. $\endgroup$
    – TimWescott
    Oct 17, 2023 at 0:17
  • $\begingroup$ @TimWescott thanks for the response, I was afraid that was happening. Perhaps it is better to say that if we define doing multiplication in time as doing convolution in frequency and taking the inverse FT (with proper scaling) we could say that discontinuities are the mean of the right and left limits? $\endgroup$ Oct 17, 2023 at 4:20
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    $\begingroup$ I'm sure you know that $f(x)\delta(x-x_0)=f(x_0)$ only holds if $f(x)$ is continuous at $x_0$. So I'm not sure if your question makes sense. But it could also be that I haven't yet fully understood what it is that you're asking here. $\endgroup$
    – Matt L.
    Oct 17, 2023 at 8:08
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    $\begingroup$ Have you found this? $\endgroup$
    – Matt L.
    Oct 17, 2023 at 9:25
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    $\begingroup$ @MattL. that is a very helpful post, thank you very much! The paper by Griffiths linked in the comments indicates that it definitely does not work in general: a nascent delta function should be used instead of directly taking the average between the two points. I will add this as an answer and link the relevant papers. $\endgroup$ Oct 17, 2023 at 10:11

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It seems that the following property is often discussed:

$$\int_{-\infty}^{\infty}\delta(t)f(t)dt = \frac{1}{2}[f(0^+)+f(0^-)]$$ Which would solve most of the problems I stated in my question. This however does not hold in general: more information about f(t) is needed to state this in general (see [1] and [2]). A nascent delta function should be used to properly find this integral, but in this case the result would hold for $f(t) = \text{sgn}(t)$ or $f(t) = \text{sgn}^2(t)$ by using steps similar to the one used in the question.

[1] David Griffiths, Stephen Walborn; Dirac deltas and discontinuous functions. Am. J. Phys. 1 May 1999; 67 (5): 446–447.

[2] M. G. Calkin, D. Kiang, Y. Nogami; Proper treatment of the delta function potential in the one‐dimensional Dirac equation. Am. J. Phys. 1 August 1987; 55 (8): 737–739.

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