In eq. (1.1) of this paper it is claimed that if $X$ is strongly Markov at a stopping time $\tau$, then it would follow from Dynkin's formula that $$(U_\alpha f)(x)=\operatorname E_x\left[\int_0^\tau e^{-\alpha t}f(X_t)\:{\rm d}t\right]+\operatorname E_x\left[e^{-\alpha\tau}(U_\alpha f)(X_\tau);\tau<\infty\right].\tag1$$ Above, $$U_\alpha:=\int_0^\infty e^{-\alpha t}\kappa_t\:{\rm d}t,$$ where $(\kappa_t)_{t\ge0}$ is the transition semigroup of $X$. Iin operator theoretical sense, it is the resolvent of the generator $A$ of this semigroup.
How do we show this?
Dynkin's formula yields $$\operatorname E_x\left[f(X_\tau)\right]=f(x)+\operatorname E_x\left[\int_0^\tau Af(X_s)\:{\rm d}s\right]\tag2.$$ How does $(1)$ follow from this?
Note that $$(\kappa_tf)(x)=\operatorname E_x\left[f(X_t)\right]\tag3$$ and $$f-e^{-\alpha t}\kappa_tf=(\lambda-A)\int_0^te^{-\lambda s}\kappa_sf\:{\rm d}s\tag4.$$
EDIT: Above we treat $(\kappa_t)_{t\ge0}$ as a contraction semigroup on a suitable Banach space $C\subseteq\{f:E\to\mathbb R\mid f\text{ is }\mathcal E\text{-measurable}\}$, where $(E,\mathcal E)$ is the measurable space $X$ takes its values in. The generator $A$ is then an unbounded linear operator given by $$Af:=\lim_{t\to0+}\frac{\kappa_t f-f}t$$ for every $f\in\mathcal D(A)$, where $\mathcal D(A)$ is preceisely the set of those $f$ for which $Af$ exists (i.e. the orbits $t\mapsto\kappa_tf$ are right-differentiable at $0$).