Proof that Verlet integration conserves energy

I was reading about different integrators that one might use to solve the system of differential equations which governs the $n$-body problem. I read that the Verlet integrator is time-reversible, and thus conserves energy.

I don't understand why time-reversibility implies energy conservation.

Suppose I know the position and velocity of each of $n$ particles in the system at some time $t_0$. I calculate the energy $E_0$ of the system at time $t_0$. I use the Verlet integrator to calculate the approximate positions and velocity of the particles at time $t_0+dt$ (one time step) and I recalculate the energy of the system and find it to be $E_1$. Is it true that $E_0=E_1$? And is there an easy way to prove this?

Firstly, the Verlet integrator only conserves energy in the limit $\Delta t\to 0$. In practice it produces energy drift, although the long-term energy drift is smaller than for most integrators.
Regarding your question, the gist of the argument is that integrators that are not time-reversible do not exhibit so-called area-preservation. This basically means that the volume of phase space of constant energy $E$ will evolve to a larger volume of the phase space over time, and hence it will necessarily span a region of phase space with different energies.
• @JoshuaBenabou Imagine the volume of phase space that contains every configuration with a fixed energy $E$. Now evolve each point forward through time. This volume should remain unchanged (Liouville's theorem) but if time reversibility does not hold then this volume will change (and increase, typically). The larger this volume is, the more non-$E$ energy states become accessible. – lemon May 15 '17 at 10:13