The shock process $s(\nu)=\nu$ where $\nu=k_1t$ and the decay process is $g(t) = k_2t^{(\gamma - \delta - 1)}$. The total stress in system at any point is defined as $\int_{0}^{t}s(\nu) g(t-s) ds$. Here, $\gamma < 1, \delta<1$, $\gamma > \delta$.
Therefore, the total stress in the system between two periods $T_1$ and $T_2$ is:
$$f(t) = \int_{T_1}^{T_2}\nu k_2(T_2-s)^{(\gamma - \delta - 1)} ds$$
$$= \frac{\nu k_2}{(\gamma - \delta)}(T_2 - T_1)^{(\gamma - \delta)}$$
$$= \frac{k_1 k_2}{(\gamma - \delta)}(T_2 - T_1)^{(1 + \gamma - \delta)} \tag{1}$$
There are no new shocks from $T_2$ to $T_3$ but the previous shocks from $T_1$ to $T_2$ will continue to decay until $T_3$. My question is how do I estimate the total shock in system at $T_3$? Would it be just the below integral $(2)$? $$\int_{T_1}^{T_3}\nu k_2(T_3-s)^{(\gamma - \delta - 1)} ds$$ $$= \frac{k_1 k_2}{(\gamma - \delta)}(T_2 - T_1)(T_3 - T_1)^{(\gamma - \delta)} \tag{2}$$
If so, then let's assume $T_2-T_1 = T_3-T_2 = T$ and $s(v)$ is applied through out the period $0$ to $2T$. Then total stress in the system using $(1)$ would be, $$ \frac{k_1 k_2}{(\gamma - \delta)}(2T)^{(1 + \gamma - \delta)} \tag{3} $$
Alternatively, total stress in the system can be defined as (Stress from $T$ to $2T$) + (Decayed outcome of shock from $0$ to $T$ at $2T$). Using equation $(1)$ and $(2)$ respectively: $$ \frac{k_1 k_2}{(\gamma - \delta)}(2T - T)^{(1 + \gamma - \delta)} + \frac{k_1 k_2}{(\gamma - \delta)}(T-0)(2T-T)^{(\gamma - \delta)}$$ $$= 2\frac{k_1 k_2}{(\gamma - \delta)}(T)^{(1 + \gamma - \delta)} \tag{4} $$
I surely missed something such that $(3) \ne (4)$