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

Question

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.

Answer 1

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