Imagine I have a process given by SDE $$ d\lambda_t = \kappa (\lambda_\infty - \lambda_t)dt + \delta_{1} dN_t $$
where $\lambda_\infty$ is a constant and $N_t$ is a poisson counting process.
Solving this SDE using substitution $ \tilde{\lambda}_t = e^{\kappa t} \lambda_t$ I get $$ \lambda_t = \lambda_\infty + (\lambda_0- \lambda_\infty) e^{- \kappa t} + \delta_{1}\sum_{\tau_i<t} e^{- \kappa (t-\tau_i)} $$
Now I am interested in $$\int_0^T\lambda(t) dt = \int_{0}^{T}\lambda_0(t)dt + \int_{0}^{T} \sum_{\tau_i<t}\delta_{1} e^{-\kappa(t-\tau_i)}dt$$ where $\lambda_0(t)$ is a deterministic part of the process and $\tau_i$ is a set of arrival times (i.e. a-priori known point).
The part I don't know how to correctly do is the last integral: $$\int_{0}^{T} \sum_{\tau_i<t}\delta_{1} e^{-\kappa(t-\tau_i)}dt$$
could someone please explain me how this integral affects the sum?
Cheers.
Might be easier if you rewrite the sum in terms of the unit step function $H(t)$: $$\sum_{i:\,\tau_i < t} e^{-\kappa(t-\tau_i)} = \sum_i e^{-\kappa(t-\tau_i)} H(t-\tau_i).$$ The integral of the $i$th term is zero if $\tau_i>T$ and reduces to the integral of $e^{-\kappa(t-\tau_i)}$ from $\tau_i$ to $T$ otherwise: $$\int_0^T \sum_i e^{-\kappa(t-\tau_i)} H(t-\tau_i) dt = \sum_i \frac {1 - e^{-\kappa (T - \tau_i)}} \kappa H(T-\tau_i).$$