I'd like to know if one integral expression I have can be shown to be zero for all possible cases. Let me introduce some notation.
Consider $\mathfrak{g}=C^{\infty}(M)$ and the dual $\mathfrak{g}^*=\mathcal{D}(M)$ are distributions on $M$. On $\mathfrak{g}$ we have the a bracket that is not Poisson, the bivector is:
$$\Pi=\frac{\partial}{\partial x_i}\wedge\frac{\partial}{\partial p_i}+B_{jk}\frac{\partial}{\partial p_j}\wedge\frac{\partial }{\partial p_k}$$
(the coefficients $B_{jk}$ are functions $B_{jk}=B_{jk}(x)$).
In the dual we define a Poisson bracket on the distribution.
$$\{\{F,G\}\}(f)=\int{f\{\frac{\delta F}{\delta f},\frac{\delta G}{\delta f}\}d^3xd^3p}$$, the following integral (around a closed periodic orbit of a Hamiltonian) is zero.
I'm considering $M$ a phase space $(x_1,x_2,x_3,p_1,p_2,p_3)$, so $f$ is a distribution function $f=f(x,p)$.
I'd like to prove that for two arbitrary functions $a,b\in \mathfrak{g}$, that is, $C^{\infty}(\mathfrak{g}^*)$, the following integral (on a closed periodic orbit of a Hamiltonian $H$ is zero).
$$\int_{\text{closed orbit of H_f}}{\mathcal{L}_{\left(\frac{\partial B_{jk}}{\partial x_i}\frac{\partial a}{\partial p_j}\frac{\partial b}{\partial p_i}\right)\frac{\partial}{\partial p_k}}f}=0$$
where $H_f$ is the Hamiltonian vector field associated with $f$, $B_{jk}=B_{jk}(x)$ and $a=a(x,p), b=b(x,p)$
