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Given that the Dirac delta function is defined as: $$ \delta(t) = \begin{cases} +\infty, & t = 0\\[2ex] 0, & t \neq 0\\[2ex] \end{cases} $$

And that the Heaviside unit step function is defined as: $$ \Theta(t) = \begin{cases} 0, & t < 0\\[2ex] \dfrac{1}{2}, & t = 0\\[2ex] 1, & t > 0\\[2ex] \end{cases} $$

Are these integrals equal?

$$ \int^t_0 e^{-(t-\tau)}\delta(t-\tau)\;d\tau = \int_{-\infty}^{+\infty}e^{-(t-\tau)}\Theta(t-\tau)\delta(t-\tau)\;d\tau = \dfrac{1}{2} $$

I found that the last integral should evaluate to one half while the first integral in my mind evaluates to one, but I have seen derivations where this equality has been used, and I cannot wrap my head around it. Maybe there is some misconception somewhere in this post? Also, can the integration limits in the last integral be set to any one of these; $\{[0, t], [0, +\infty], [-\infty, +\infty]\}$, without changing its evaluation?

Example derivation:

$$ \mathcal{C} = \int_0^t e^{\textbf{A}(t-\tau)} \textbf{B}\delta(t-\tau)\textbf{Q}_\textbf{u}(\tau)\textbf{B}^T\;d\tau = \int_0^{+\infty}\Theta(t-\tau)e^{\textbf{A}(t-\tau)}\textbf{B}\delta(t-\tau)\textbf{Q}_\textbf{u}(\tau)\textbf{B}^T\;d\tau = \dfrac{1}{2}\textbf{BQ}_\textbf{u}(t)\textbf{B}^T $$

JMy
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    That's actually not technically how the dirac delta function is defined but it is a shorthand notation we use when it is introduced. It's actually defined as a functional, which is a function that inputs a vector and outputs a number. In this case, the vector is whatever integrable function is being multiplied by the dirac delta inside the integral and the output is that function's value at 0 – Alan Dec 09 '22 at 20:21
  • Ok, this then gives the integral of 1*dirac(t-tau) dtau in the first integral? Which is equal to 1, what makes this equal to the other integral? – JMy Dec 09 '22 at 20:26
  • The LHS will give you the output of $e^{-(t-\tau)}$ for when $t=\tau$, thus it appears to me to give the value of 1, not $\frac 1 2$. I am rusty on this, so not leaving as an answer – Alan Dec 09 '22 at 20:30
  • Ok, and with the same logic applied to the second integral it seems like the dirac 'takes the value' of the $\Theta(0) = 1/2$, and $1$ from $e^{-0}$, which when integrated should output $1/2$. – JMy Dec 09 '22 at 20:38
  • times the exponent to the 0 which is 1, but yes. So it appears the two sides are not equal – Alan Dec 09 '22 at 20:39
  • Dirac delta is not defined that way! $a \delta(x)$ would also satisfy this definition. – Anixx Dec 09 '22 at 20:52
  • @JMy The Dirac Delta is NOT a function. As a distribution, the functional on which it operates on test functions is NOT an integral. – Mark Viola Dec 09 '22 at 21:29

1 Answers1

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The second integral makes no sense. To be more precise, the dirac "function" is really a linear map defined on the space of continuous functions with compact support: $\delta: \phi \mapsto \phi(0)$.

You can think of it as a generalized function, because any integrable function $g$ also gives a linear map defined on continuous functions with compact support, defined by $L_g: \phi \mapsto \int_{\mathbb R} \phi g$.

The problem is that you are trying to evaluate $\delta( \Theta g)$, where $g(x)=\exp(-x)$. But $\Theta$ is not continuous, so this expression is not well-defined.

Albert
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