To find Laplace transform of $\sin{bt}$, I get
How to find the domain of convergence.
Attempt:
It seems to me that it will be for all $s-ib>0$ (the domain for $\mathscr{L}(e^{bt})$ is $\forall ~s-a>0$) but the actual answer I know $s>0$.
Also $\mathscr{L}(e^{at}\sin{bt})=\frac{b}{(s-a)^2+b^2}$ then what will be domain of convergencce of $\mathscr{L}(e^{at}\sin{bt})$?
Note that the Laplace Transform of $\sin(bt)$ is given by
$$\mathscr{L}\{\sin(bt)\}(s)=\int_0^\infty \sin(bt)e^{-st}\,dt \tag 1$$
We have the estimate
$$\left|\int_0^L \sin(bt)e^{-st}\,dt\right|\le \int_0^L e^{-\text{Re}(s)t}\,dt=\frac{1-e^{-\text{Re}(s)L}}{\text{Re}(s)}$$
Evidently, the integral in $(1)$ converges whenever $\text{Re}(s)>0$.
If $\text{Re}(s)\le 0$, then it is easy to see that the integral diverges.
Therefore, the domain of convergence is $\text{Re}(s)>0$ as was to be shown!