Electrical – How do PID controllers effect time domain and frequency domain response

Your equations are correct, assuming initial conditions are zero, but your Vi should be Vi(s) in the first equation. I would guess that the Vin in Wolfram is a step of magnitude Vi, with Laplace transform Vi/s, which gives the Laplace output, Vout(s), in your final equation.

Applying a sinusoidal input to, e.g., a stable, SISO, LTI system will produce an output with a transient component and a steady-state component. The transient does what all transients do - it decays to zero after a certain time. The steady-state component is a sine wave at the same frequency as the input sinusoid but, generally, with a different amplitude and also a phase shift relative to the input

It turns out (see numerous references) that the steady-state sinusoidal response (or 'frequency response') can be obtained from the system transfer function by the simple algebraic substitution: \$\small s\rightarrow j\omega\$. And hence the frequency domain is extremely accessible from the s-domain. The time domain, for example, is not nearly so easy to access as it requires the inverse Laplace transform.

To use your example, if a system has TF \$\small G(s)=\large \frac{1}{s}\$, the frequency domain gain and phase angle would be: \$\frac{1}{\omega}\$ and \$\small -\frac{\pi}{2}\$, respectively.

From a signal, rather than system, perspective, the unit step signal, \$\small H(t)\$, has the same LT as the system, above, viz: \$\small H(s)=\large \frac{1}{s}\$, and in the frequency domain this maps to the positive spectrum \$\small H(j\omega)=\large\frac{1}{j\omega}\$, and has amplitude \$\frac{1}{\omega}\$, and a phase angle of \$\small -\frac{\pi}{2}\$ relative to a notional sine wave, \$\small sin(\omega t)\$