Let $Z$ denote the random walk in continuous time with the generator $(\kappa+\rho)\Delta$ with respective starting point z.
Let $R_\lambda^{\kappa+\rho}(z)$ denote the related semi-group given as following. Then by resolvent representation of semi-groups we get: $R_\lambda^{\kappa+\rho}(z) =\int_0^\infty \:\mathrm{d}t \: e^{-\lambda t}\:\mathbb{E}_z^Z\exp\bigg\lbrace\gamma\int_0^t \: \delta_0(Z_s) \:\mathrm{d}s \bigg\rbrace=\frac{1}{\lambda}+\gamma\bigg (\int_0^1 \:\mathrm{d}t \: e^{-\lambda t}\:p_{(\kappa+\rho)t}(z)\bigg) R_\lambda^{\kappa+\rho}(0)$, where $p_{(\kappa+\rho)t}(z)=\mathbb{P}^{Z}_0{(Z_t=z)}=\mathbb{P}^{Z}_z{(Z_t=0)}$ is the transition probability.
The quenstion is: how to get the expression in the last equality? I mean, the part $R_\lambda^{\kappa+\rho}(z) =\frac{1}{\lambda}+\gamma\bigg (\int_0^1 \:\mathrm{d}t \: e^{-\lambda t}\:p_{(\kappa+\rho)t}(z)\bigg) R_\lambda^{\kappa+\rho}(0)$.
I think the idea is to make use of the resolvent representation of semi-groups, but somehow I tried a lot but still failed to get the last equality.