I have a confusion between the Laplace transform of a circuit and the frequency domain AC analysis. Suppose we have a series RLC resonant circuit connected to an ideal sinusoidal voltage source.
We can write $$V_m \sin(\omega t)=i R + \frac{1}{C} \int i dt + L \frac{d i}{dt} $$
now if we take the Laplace transform (assume initial inductor current to be zero),
$$\frac{V_m \omega }{s^2+\omega^2} =I(s) \big(R + \frac{1}{s C} + s L \big) \Rightarrow I(s) = \frac{V_m \omega }{(s^2+\omega^2) \big(R + \frac{1}{s C} + s L \big)} \tag{A}$$
On the other hand, using frequency domain analysis we can derive current as, $$ I= \frac{V_m }{R +j \omega L + \frac{1}{j\omega C}} \tag{B}$$
My question is: How can we deduce (B) directly from (A) without inverse Laplace transform? just putting \$s=j \omega\$ will not simply reproduce (A).
To further elaborate my question: If we have a current of a branch as the form of $$I_n=\frac{ V_n (s \sin\theta+\omega_1 \cos\theta)}{s^2+\omega_1^2} \frac{1}{Z_1(s)} $$
can we directly deduce (without taking the inverse) that particular branch is effectively acting as it is connected to a voltage source of \$ V_n \sin(\omega_1 t + \theta) \$ and impedance of the branch is \$ Z_1(j \omega) \$?