Time derivative of Gibbs entropy (the paradox of the constant fine-grained entropy)

In the context of classical systems, the fine-grained (or Gibbs) entropy is defined as the functional:

$S_G(t)=-kint_{Gamma_t}dqdp rho(p,q,t)ln[rho(p,q,t)]$ (1)

I’ve been told (Wehrl and J. van Lith) that the Liouville’s theorem ensures that this quantity is constant under an evolution governed by Hamilton’s equation. I can prove this statement by using the time evolution as a change of variables in (1), then using that the Jacobian of this change of variables is 1 (one of the forms of Liouville’s theorem) :

$S_G(0)=-kint_{Gamma_0}dq_0dp_0 rho(p_0,q_0,0)ln[rho(p_0,q_0,0)]=-kint_{Gamma_t}dqdp {J(frac{partial q_0,p_0}{partial q,p})}rho(p,q,t)ln[rho(p,q,t)]=S_G(t)$

My doubts:

1. Is the proof that I suggest correct?

2. I am trying to prove it by deriving in (1) but I am not successful:

   $frac{d S_G(t)}{dt}=-kint_{Gamma_t}dqdp partial_t rho(p,q,t)(1+ln[rho(p,q,t)])$

   which is trivially equal to zero just in the case of equilibrium ($partial_t rho(p,q,t)=0$). Any idea on why this is not working?

(This is interesting because it implies that either the Hamiltonian evolution or the Gibbs entropy cannot describe a non-equilibrium evolution in which, according to the second law, the thermodynamic entropy should increase.)