Not sure if this is too much physics to be here...
Consider $$H:\mathbb{R}^{2N+1}\rightarrow\mathbb{R}$$ of class $C^2$, let $H(x,y,z)$ such that $x\in\mathbb{R}^N$, $y\in\mathbb{R}^N$ and $z\in\mathbb{R}$. Let $\varphi$ be the flow associate with the Hamiltonian system $$\dot{x}_i=-\frac{\partial H}{\partial y_i}$$ $$\dot{y}_i=\frac{\partial H}{\partial x_i}$$ $$\dot{z}=1$$ I have to prove that if $\eta$ is a 1-form given by $\eta=\sum_{i=1}^Nx_i \ dy_i-H \ dz$ and $c$ is a closed curve in $\mathbb{R}^{2N+1}$, then for all $s$ we have $$\int_{\varphi(s,c)}\eta=\int_c\eta.$$
Thank you in advance!
By Stokes $\int_c\eta = \int_D d \eta$ for any disc $D$ with $\partial D = c$ (which exists because all curves are contractible in $\mathbb{R}^{2n+1}$), and then also $\int_{\phi(s,c)} \eta = \int_{\phi(s,D)} d \eta$
Thus it suffices to show that Lie drivative of $d \eta$ is $0$.
By Cartan's magic formula
$$L_X d \eta = d (i_X (d\eta)) = d (i_X \omega -i_X (dH \wedge dz))$$
We have $i_X dz =1$.
We then compute $i_X \omega = dH - \frac{\partial H}{\partial z} dz$
and thus $d H = i_X \omega + \frac{\partial H}{\partial z} dz$ which contracted with $X$ again gives
$i_X dH = i_X (\frac{\partial H}{\partial z} dz)= \frac{\partial H}{\partial z}.$
Plugging this in we get
$$L_X d \eta=d (i_X \omega -i_X (dH \wedge dz)) = d( dH - \frac{\partial H}{\partial z} dz + (i_X dH) \wedge dz - d H \wedge (i_X dz))= d (0)=0.$$