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I am trying to find the all the following Fourier Transforms on a finite time interval $[t_0,\ t_f]$ with $t_0<t_f$:

  1. $$A(iw) = \int\limits_{t_0}^{t_f} f(t)\ \delta(t-t_0)\ e^{-iwt} dt$$
  2. $$B(iw) = \int\limits_{t_0}^{t_f} f(t)\ \delta(t-t_f)\ e^{-iwt} dt$$

where $f(t)$ is a well behaved function which is not zero necessarily at $\{t_0;\ t_f\}$, and $\delta(t)$ is the Dirac's delta function. Please consider that $\theta(t)$ is Heaviside step function. The idea is express the functions as results dependent on other functions and in terms of $\int\limits_{t_0}^{t_f} f(t)\ e^{-iwt} dt$

I tried to solve it by myself but I think I am doing something wrong, since I find different values of $\theta(0)$ if I try to solve:

  1. $$X(iw) = \int\limits_{t_0}^{t_f} f(t) \left(\theta(t-t_0)-\theta(t-t_f)\right) e^{-iwt} dt$$
  2. $$Y(iw) = \int\limits_{t_0}^{t_f} f(t) \theta(t-t_0)\theta(t_f-t) e^{-iwt} dt$$

when conceptually they should be the same. I don't know if I am messing up with Cauchy's principal values, with issues when $0<t_0<t_f$ compared $t_0<0<t_f$, or if something with distribution theory which I don't know. Neither I am sure if $\theta(0)$ could be treated as a constant so it could go inside/outside the Fourier Transforms without making problems.

But otherwise, I do was able to fulfill Parseval's theorem on many examples when using: $$\int\limits_{t_0}^{t_f} f'(t) \left(\theta(t-t_0)-\theta(t-t_f)\right) e^{-iwt} dt:= iw \int\limits_{t_0}^{t_f} f(t)\ e^{-iwt} dt + f(t_f)e^{-iwt_f}-f(t_0)e^{-iwt_0}$$ which I found using my flawed formulas (and what it is shown here for the scenario of the Dirac's Delta function), but if I made the direct replacement $f'(t)\to f(t)$ and $f(t) \to F(t)=\int_{t_0}^t f(u)\ du$ it stops working, and I think it is related.

I hope you could help me to find the formulas for 1 and 2 (if you could go 1-4 even better): If they change if I consider instead the interval $(t_0,\ t_f)$ please show them also.

I hope you could explain if I have something conceptually wrong about the things I mentioned too.

Joako
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  • For your $#3$ and $#4$, the Heaviside function can be removed (treated as constants) as the effect is the same and $f$ is well behaved at the boundaries. For your $#1$ and $#2$, it's not as obvious since you are sifting using a delta function at the boundaries and the sifting property is not defined for boundaries. In engineering fields, one would take a shortcut and take half of the value at the boundaries, but that's not mathematically rigorous. – Srini Oct 01 '24 at 14:23
  • @Srini thanks for commenting. I think is not that trivial as just moving step functions out of the transforms, since the example I shown works (eq. not numbered), and there it behaves different from what would happen by just taking the setp funcrions outside (maybe I am missing something here, but is what I am understanding from it - it is just applying integration by parts by the way) – Joako Oct 01 '24 at 21:57
  • The Wiki link you provided for Heaviside does define $\theta(0) = 1$. Based on that, it seemed like we could take them out. But you may be right, I didn't analyze too carefully. For delta, if the definition is as a limit of a nascent delta that is an even function (like sinc or gaussion), the "taking half" shortcut might be loosely based on that. Anyway, I will wait for the more rigorous analysis from experts. – Srini Oct 01 '24 at 22:08
  • @Srini I am using what others answer to another question for the specific scenario of the Dirac's Delta function, but I have not being able to succed with it in the case of the Heaviside step function due I found inconsistencies on the results I get, as I mentioned on the question. Don't know neither if it have to do with $$\theta(0)\equiv \int_{-\infty}^{\infty}\theta(t)\delta(t)\ dt$$ and if this introduces some subleties. – Joako Oct 02 '24 at 06:51
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    Many times $\theta(0)$ is considered undefined (e.g. Mathematica), but analytic representations of $\theta(t)$ typically lead to $\theta(0)=\frac{1}{2}$. Also, I beleive there are several questions on evaluating for example $\int\limits_0^{\infty} \delta(t), f(t), dt$ on Math StackExchange, and it depends on how you define $\delta(t)$. Some definitions lead to $\int\limits_0^{\infty} \delta(t), f(t), dt=f(0)$ while analytic representations of $\delta(t)$ typically lead to the result $\int\limits_0^{\infty} \delta(t), f(t), dt=\frac{1}{2} f(0)$. – Steven Clark Oct 04 '24 at 15:35

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