In this paper a few steps are missing. I am trying to fill in the gaps to gain understanding. The following integral is given
$$ \int_{0}^\infty e^{-\beta t}\,dD_k(t) $$ where we have a interest rate $\beta \geq 0$ and a departure process $D_k(t)$. This function is piece-wise constant over intervals $[t_{k-1},t_k)$ at some integer value. In the next interval $[t_k,t_{k+1}]$ it then increases by one over the previous values. In other words $$ D_{k+1} = D_k(t) + 1, \quad t \in [t_k,t_{k+1}) $$ I would like to compute the above integral but need to deal with $dD_k(t)$. I do so by reasoning with the Heaveside step function $H_a(t)$ where $a$ is the point at which the step occurs. Let us assume $D_0(0) = 0$ such that $$ D_k(t) = \sum_{k=1}^{\infty} H_{t_k}(t) \cdot \mathbf{1}_{\{t_k \leq t\}} $$ where the inidcator function is simply there as a constraint that will be omitted soon. The derivative of $H_{a}(t)$ is the Dirac delta function $\delta_a(t)$. Then we have $$ \frac{dD_k}{dt} = \sum_{k=1}^{\infty} \delta_{t_k}(t)\cdot \mathbf{1}_{\{t_k \leq t\}} $$ such that the integral becomes the following. $$ \begin{eqnarray} \int_{0}^\infty e^{-\beta t}\,dD_k(t) &=& \int_{0}^\infty e^{-\beta t} \left( \sum_{k=1}^{\infty} \delta_{t_k}(t)\cdot \mathbf{1}_{\{t_k \leq t\}} \right) \, dt\\ &=& \sum_{k=1}^{\infty} e^{-\beta t_k} \end{eqnarray} $$ The above answer seems to be what the original paper obtained. I am just not quite sure if this is how one would go about getting the solution. I am afraid that I have "forced" a solution.
Please do point out whether this is nonsense, where I have gone wrong and how it it should be improved upon.