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How is proved that for $a\in\mathbb{R}$ is true that $\displaystyle{\int} x(t)\delta(t-a)\ dt = x(a)\ \theta(t-a)+\mathbf{C}$?

Here $\mathbf{C}$ is the integration constant for the indefinite integral, $\theta(t)$ is the Heaviside step function, and $\delta(t)$ is the Dirac delta function, in which Wikipedia webpage the asked property it isn't listed (does it have a name? - this for look for any references).

Even more, if I ask Wolfram-Alpha for $\displaystyle{\int} x(t)\delta(t-a)\ dt$ it shows nothing, but if instead I use an constant value for "$a$", let say as example $a=7$, it indeed shows the property holds $\displaystyle{\int} x(t)\delta(t-7)\ dt = x(7)\ \theta(t-7)+\mathbf{C}$.

I would like to know how and why this is true (demonstration). As example, since $\delta(x)=\delta(-x)$ is symmetric, I will have also that $\displaystyle{\int} x(t)\delta(a-t)\ dt = x(a)\ \theta(t-a)+\mathbf{C}$ which looks weird at first glance, since the results takes a different interval than what is within the Dirac's delta argument.

I need this answer to solve this other question, if you are interested.

Hope you could answer step-by-step, or left a reference to this demonstration. Thanks you very much.

Joako
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  • @gist076923 I am aware of it, but I know the property should be true because of the results into the related question I mentioned, and I want to know why it is that true. This is why I tagged Distribution-theory, since I think it is proved in that sense, and I would like to understand how and why it is as it is, or some reference to the demonstration. – Joako Mar 22 '23 at 16:30
  • For everyone else, the comment I deleted was basically saying that we have to take care about the dirac delta not being a function, and that we don't have derivatives in the usual sense. I deleted because I thought I worded things weird – gist076923 Mar 22 '23 at 16:33
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    What have you tried? Are you using distribution theory to try and prove the statement? – gist076923 Mar 22 '23 at 16:34
  • @gist076923 In the mentioned question, after a comment, I realize it must be true in order to have consistency in the derivatives. But I am not instructed in Distribution Theory (only what I have learned through Wikipedia), and since the property is not listed in the Dirac's delta wiki I don't even know how to start to look for this property. So any explanation or guidance would be helpful – Joako Mar 22 '23 at 16:39
  • Perhaps looking into some more distribution theory would help. You essentially have to show that $x(t)\delta(t-a)$ is the distributional derivative of the distribution represented by $x(a)\theta(t-a)$. – gist076923 Mar 22 '23 at 17:11
  • @gist076923 I think I will fall in a circular argument with which I am asking here: since $$\frac{d}{dt}\left(x(a)\ \theta(t-a)\right)=x(a)\delta(t-a)=x(t)\delta(t-a)$$ due the 7th property here, I know is consistent going downwards differentiation, but going upwards the integration I don't understand how appears those integration constants that apply only in a piece-wise section of the integration domain... I don't know If I made myself clear?, it is more a Why than a How question – Joako Mar 22 '23 at 17:22
  • @gist076923 in other words, why if I have $$x(t)\delta(t-a)=x(a)\delta(t-a)=x(a)\delta(a-t)$$ I end always with $$x(a)\ \int \delta(t-a)dt =x(a)\ \int\delta(a-t)dt =x(a)\ \theta(t-a)$$ in always the same interval... maybe is more a conceptual mistake I have. – Joako Mar 22 '23 at 17:30
  • Hint: compute $\int_{-\infty}^tx(u)\delta(u-a)du$ for $t<a$, then for $t>a$. – J.G. Mar 22 '23 at 17:37
  • The Wolfram Alpha result seems to be consistent with integration by parts $\int f(t), \delta (t-a),dt=f(t) \theta (t-a)-\int \theta(t-a), f'(t),dt$ where $C=-\int \theta(t-a), f'(t),dt$. – Steven Clark Mar 22 '23 at 17:46
  • On second thought, how does one justify $C$ is a constant and not a function of $t$? – Steven Clark Mar 22 '23 at 18:16
  • @StevenClark don't know what you aimming to to: the integration constant rises in every antiderivative due differentiating constants leqds to zero – Joako Mar 22 '23 at 18:29
  • @StevenClark and your matching don't look right since is saying that an integral of an arbitrary piecewise section of an arbitrary function is always constant... don't make any sense to me at least. – Joako Mar 22 '23 at 18:31
  • @J.G. Do I use $\theta(-\infty)\equiv 0$? – Joako Mar 22 '23 at 18:37
  • I don't understand your comments. Are you familiar with integration by parts? With respect to your last question, yes $\theta(-\infty)=0$, but it looks like you're now contemplating a definite integral whereas your original question was about an indefinite integral. – Steven Clark Mar 22 '23 at 18:52
  • Perhaps I was ignoring the integration constant (the Wikipedia article on integration by parts also states "where we neglect writing the constant of integration"), but your result seems to be ignoring the integral $-\int \theta(t-a), f'(t),dt$. – Steven Clark Mar 22 '23 at 19:03
  • @StevenClark Later I realized that using the properties $$x(t)\delta(t-a) = x(a)\delta(t-a)$$ and $$\theta'(t-a) = \delta(t-a)$$ the integral follows directly as $$\int x(t)\delta(t-a)dt = x(a)\theta(t-a)+\mathbf{C}$$ with $\mathbf{C}$ the traditional integration constant. – Joako Mar 22 '23 at 19:15
  • $x(t),\delta(t-a)=x(a),\delta(t-a)$ could just as well be $x(t),\delta(t-a)=x(b),\delta(t-a)$ since the left and right sides of both equalities evaluate to $0$ at $t\in\mathbb{R}\land t\ne a$ and are all undefined for all other values of $t$ (i.e. when $t=a$ or when $\Im(t)\ne 0$). – Steven Clark Mar 23 '23 at 20:56
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    $x(t),\delta(t-a)$ is equivalent to $x(a),\delta(t-a)$ in the sense that $\int\limits_{a-\epsilon}^{a+\epsilon} x(t),\delta(t-a),dt=\int\limits_{a-\epsilon}^{a+\epsilon} x(a),\delta(t-a),dt=x(a)$ for any $\epsilon>0$, so it seems to me it only makes sense to talk about equalities involving $\delta(t)$ in the context of definite integrals over $t\in\mathbb{R}$. – Steven Clark Mar 23 '23 at 20:57

1 Answers1

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Later I realized that using:

  • $x(t)\delta(t-a) = x(a)\delta(t-a)$
  • $\theta'(t-a) = \delta(t-a)$

It follows directly that: $$\int x(t)\delta(t-a)dt=\int x(a)\delta(t-a)dt= x(a)\int\delta(t-a)dt = x(a)\theta(t-a)+\mathbf{C}$$

Joako
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