I know that Laplace transform of $$e^{-\alpha t}\sin(\omega _0t)u(t)$$ is $\frac{\omega _0}{(s+\alpha )^2+\omega _0^2}$ where $Re(s)>-Re(\alpha )$ and $u(t)=\boldsymbol 1_{t\geq 0}$. Knowing that, how can I find the Laplace transform of $$-e^{bt}\sin(\omega _0t)u(-t) \ \ ?$$
I know it's $\frac{\omega _0}{(s-b)^2+\omega _0^2}$ for $Re(s)<b$, but I really don't know how to get that. I tried to use the fact that $$-e^{bt}\sin(\omega _0t)u(-t)=e^{-(-b)t}\sin(\omega _0 t)\underbrace{(1-u(-t))}_{=\boldsymbol 1_{t\geq 0}}+e^{-(-b)t}\sin(\omega _0t),$$ but I can't conclude. Any idea ?
Let $$f(t) = e^{-bt}\sin(\omega_0t)u(t)$$ Notice that $$f(-t) = -e^{bt}\sin(\omega_0t)u(-t)$$ Laplace transforms have a nice property that $$\mathcal{L}\left\{f(-t)\right\}(s) = \mathcal{L}\left\{f(t)\right\}(-s)$$ Using your known result of $$\mathcal{L}\left\{f(t)\right\}(s) = \frac{\omega_0}{(s+b)^2+\omega_0}$$ we can use the property above to see that $$\mathcal{L}\left\{f(-t)\right\}(s) = \frac{\omega_0}{(-s+b)^2+\omega_0} = \frac{\omega_0}{(s-b)^2+\omega_0}$$