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If $f'(t)=x'(t)\ \theta(T-t)$, What is the accurate formulation of $f(t)$? (antiderivative over distributions/special_functions)

Maybe the question look simple, by I need to understand Why it is as so, not just to know the proper form.

Intro__________________________

If I define a function as: $f(t) = x(t)\ \theta(T-t)+x(T)\ \theta(t-T)$

with $x(t)$ an arbitrary analytic function (for now, just for avoid complications with singular points), $T$ an arbitrary constant time (w.l.g., assume from now on it is such as $0<T<\infty$), and $\theta(t)$ is the Heaviside step function.

Then, by differentiation I should have: $$\begin{array}{r c l} f'(t) & = & x'(t)\ \theta(T-t)+x(t)\ \theta'(T-t)+\underbrace{\require{cancel}\cancel{0\ \theta(t-T)}}_{x(T)\text{ is a constant for }dt} +x(T)\ \theta'(t-T) \\ & = & x'(t)\ \theta(T-t)+\underbrace{x(t)\ \delta(T-t)\ (-1)}_{\text{but }\delta(-x)\ =\ \delta(x) \text{ and also } x(t)\delta(t-a)\ =\ x(a)\delta(t-a)} + x(T)\ \delta(t-T) \\ & = & x'(t)\ \theta(T-t)+\underbrace{-x(T)\ \delta(t-T) + x(T)\ \delta(t-T)}_{=\ 0}\\ & = & x'(t)\ \theta(T-t) \end{array}$$ with $\delta(t)$ the Dirac's delta function.

This is why I believe the mentioned $f(t)$ is the proper form.

Question________________

But in the other way of working it out, if I have that $f'(t) = x'(t)\ \theta(T-t)$ and I apply the indefinite integral in both sides I will have: $$ f(t) = \underbrace{\displaystyle{\int} x'(t)\ \theta(T-t)\ dt}_{\text{function of }x(t)} + \mathit{C}$$ with $\mathit{C}$ the integration constant:

  • How you make appear the term $x(T)\theta(T-t)$ from the integration constant $\mathit{C}$?
  • Which operations you have to make to make appear $f(t)$ from $f'(t)$ applying the indefinite integral operation (step-by-step)? and why "those" steps?
  • Following the same logic, if I have that $g''(t)=x''(t)\ \theta(T-t)$, then what is the proper form of $g(t)$?... my guess is $g(t) = x(t)\ \theta(T-t)+\left(x(T)+x'(T)(t-T)\right)\ \theta(t-T)$, Does it means that it is required that $x'(T)\equiv 0$? (since $g(t)\neq f(t)$)
Joako
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  • Do an integration by parts of the equation. – md2perpe Mar 22 '23 at 07:08
  • @md2perpe it doesn't work $$ \int x'(t)\theta(T-t)dt \ dv=x'(t)dt\quad u=\theta(T-t) \ \Rightarrow = x(t)\theta(T-t)-\int x(t)(-\delta(t-T))dt\ = x(t)\theta(T-t)+\int x(t)\delta(t-T)dt$$ let me with the unknown therm $$\int x(t)\delta(t-T)dt$$: How I evaluate it?... even if I constraint it to use the sifting property $$\int_{-\infty}^{\infty} x(t)\delta(t-T)dt = x(T)$$ I end with a constant for all times and not the term only for $t>T$ – Joako Mar 22 '23 at 13:37
  • @md2perpe if I try $\int x(t)\delta(t-a)dt$ in Wolfram-Alpha shows nothing, but if I use $\int x(t)\delta(t-7)dt$ it indeed solves is equal to $x(7)\theta(t-7)+\mathit{C}$, and I don't know why?!!... hope you could elaborate in an answer, since this property isn't listed in Wikipedia website for Dirac delta function. By the way, thanks for commenting. – Joako Mar 22 '23 at 15:24
  • @md2perpe since big part of what I am asking it is related to prove the indefinite integral over $\delta(t)$ I asked it separately in this other question. – Joako Mar 22 '23 at 16:27
  • @md2perpe from the other question I think it follows $$\int x(t)\delta(t-T)\ dt=\int x(T)\delta(t-T)\ dt=x(T)\int\delta(t-T)dt=x(T)\theta(t-T)$$ which close the integration by parts. thanks you very much. – Joako Mar 22 '23 at 18:42

1 Answers1

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We can find one primitive function by integrating from an arbitrary point. Here it's natural to take the point $t=T$: $$ f_0(t) = \int_T^t f'(s) \, ds = \int_T^t x'(s) \, \theta(T-s) \, ds \\ = \begin{cases} x(t) - x(T), & t<T \\ 0, & t>T \end{cases} \\ = \left( x(t)-x(T) \right) \theta(T-t). $$

We get all solutions by adding an arbitrary constant: $$ f(t) = f_0(t) + C = \left( x(t)-x(T) \right) \theta(T-t) + C. $$

md2perpe
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