I've read that Poincare's integral invariance can be used as a definition of a Hamiltonian system. That is to say, if $g^t$ is a phase flow satisfying
$$\oint_{\gamma} \omega = \oint_{g^t \gamma} \omega$$
with $\omega = p_{1}dq_{1}+\ldots+p_{N}dq_{N}$ for any closed path $\gamma$, then there exists a function $H = H(q,p)$ such that $\dot{q} =\frac{\partial H}{\partial p}$ and $\dot{p}= - \frac{\partial H}{\partial q}$.
Most 'proofs' I have seen of this statement don't seem to be very rigorous. Do you know of any good proofs?
edit
I have made a bit of progress.
After working on it for a bit, I have a part of the proof. Let $\phi(t)=\oint_{g^t \gamma}\omega$. Then $$\eqalign{ \phi'(0)&=& \lim_{t \to 0}\frac{1}{t} \left(\int_{g^{t} \gamma} \omega -\int_{\gamma} \omega \right) \cr &=& \lim_{t \to 0} \frac{1}{t} \int_{\Sigma_{t}} d\omega }$$
Here, $\Sigma_{t} = \{g^{s} x |x\in\gamma, s\in(0,t)\}$ is the 'tube of trajectories' adjoining $\gamma$ and $g^{t} \gamma$.
Let $v$ be the vector $\left(\matrix{\dot{q} \\ \dot{p}}\right)$ evaluated at every point on $\Sigma_{t}$. Then define the 1-form $\chi$ by the formula: $$\eqalign{ \chi (u) &=& d \omega (v,u) \cr &=& \sum \dot{p}_{k}dq_{k}-\dot{q}_{k}dp_{k}. }$$ To prove Hamilton's equations, we only need to show that $\chi$ is exact. This can be done with the following lemma:
Lemma: $$\int_{\gamma} \chi = \lim_{t\to 0} \frac{1}{t} \int_{\Sigma_t}d\omega$$
This lemma seems reasonable enough. But I'm not sure how to prove it if it is true.
edit 2 (final)
Ok, I've proven the lemma if anyone is interested. It turns out it is really easy to evaluate $\int_{\Sigma_{t}} d \omega$. The idea is that $\Sigma_{t}$ has a parametrization $\Phi:(0,t)\times(0,1) \to \Sigma_{t}$ given by $$\Phi (s, \lambda) = g^{s}(\gamma (\lambda)).$$
Here, $\gamma$ is parametrized by $\lambda$ for $\lambda \in [0,1]$.
Then $$ \int_{\Sigma_{t}} d \omega = \int_{0}^{t} \int_{0}^{1} \frac{\partial (p_{k}\circ \Phi, q_k \circ \Phi)}{\partial (\lambda,s)} d\lambda ds$$
By the fundamental theorem of calculus, $\phi'(0)$ is equal to the value of the inner integral and evaluated with $s=0$. But this is exactly $\oint_{\gamma} \chi$.