I have the following integral:
\begin{align} \int^T_0e^\tau I(\tau+t^0)d\tau. \end{align}
In this integral, $I(t)$ is a function with period $T$. At each time $T, 2T, \ldots$, $I$ is increased by a constant $K$. The starting value of $I$ is such that $I$ decreases by $K$ between the moments of these instantaneous increases.
The definition of $I$ is
\begin{align} I(t) = \tilde{K}e^{-t/\tau_0}, \end{align}
where $0\leq t<T$, $\tilde{K}=\frac{K}{1-e^{-T/\tau_0}}$, and $I(t)=I(t+iT)$ for all integers $i$.
In the integral, $t^0$ can be negative, zero, or positive. When $t^0=0$, evaluating the integral is easy.
But when $t^0$ is negative or positive, the integral crosses discontinuities of the function $I$.
How should I handle this?
Note
Ignoring the periodicity, one could rearrange the integral to be
\begin{align} I(t^0)\int^T_0e^{\tau(1-1/\tau_0)}d\tau \end{align}
and solve that. I don't have much intuition for whether that's legitimate with regard to the periodicity, but my impression is that it's not -- mainly because when I compute the integral this way, my results (using specific values for all the constants) don't match the results in the paper I'm trying to replicate.
So my secondary question is, is this rearrangement of the integral legitimate, and why or why not?
Suppose $t_0 = kT + \alpha$ for some $\alpha \in [0, T), k \in \mathbb{Z}$
\begin{align} &\int_0^T \tilde{K} e^\tau I(\tau + t_0) \mathrm{d}\tau \overset{(a)}= \int_0^{T} \tilde{K}e^\tau I(\tau + \alpha) \mathrm{d}\tau \\ = ~& \int_0^{T-\alpha} \tilde{K} e^\tau I(\tau + \alpha) \mathrm{d}\tau + \int_{T-\alpha}^T \tilde{K} e^\tau I(\tau + \alpha) \mathrm{d}\tau \\ \overset{(b)}= ~& \int_\alpha^{T} \tilde{K} e^{\tau_1 - \alpha} I(\tau_1) \mathrm{d}\tau_1 + \int_{0}^\alpha \tilde{K} e^{\tau' - \alpha + T }I(\tau' ) \mathrm{d}\tau' \\ \end{align}
Where $(a)$ follows from $I(t + KT) = I(t)$, and $(b)$ is the following substitutions:
Now you know $I$ over each of the required intervals, as they all lie within the first period. Integrate away.