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