I need to transform $ρ^{-αt} \sin (wt)$ but I don't understand the $ρ$ part.
Do I treat it like an exponential function and integrate as normal, i.e multiplying it by $ρ^{-st}$ and integrating from $0$ to $\infty$ ? Or is there some other rule that I need to know?
Well, we have:
$$\rho^{-\alpha\cdot t}=\exp\left(\ln\left(\rho^{-\alpha\cdot t}\right)\right)=\exp\left(-\alpha\cdot t\cdot\ln\left(\rho\right)\right)\tag1$$
So, we get:
$$\mathscr{L}_t\left[\rho^{-\alpha\cdot t}\cdot\sin\left(\omega\cdot t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\rho^{-\alpha\cdot t}\cdot\sin\left(\omega\cdot t\right)\cdot e^{-\text{s}\cdot t}\space\text{d}t=$$ $$\int_0^\infty\exp\left(-\alpha\cdot t\cdot\ln\left(\rho\right)\right)\cdot\sin\left(\omega\cdot t\right)\cdot e^{-\text{s}\cdot t}\space\text{d}t=$$ $$\int_0^\infty\exp\left(-\alpha\cdot t\cdot\ln\left(\rho\right)-\text{s}\cdot t\right)\cdot\sin\left(\omega\cdot t\right)\space\text{d}t=$$ $$\int_0^\infty\exp\left(-t\cdot\left(\alpha\cdot\ln\left(\rho\right)+\text{s}\right)\right)\cdot\sin\left(\omega\cdot t\right)\space\text{d}t\tag2$$