Integral of Poisson Bracket vanishes This arose when reading a text on linear PDE. It is a step in an argument that I am not sure how to justify. Let $(M, g)$ be a smooth compact Riemannian manifold and $p(x, \xi) = |\xi|_g= \sqrt{g^{ij}(x) \xi ^i \xi ^j}$ be a smooth function on $T^*M \setminus 0$ and $\omega = d\xi \wedge dx$ the canonical symplectic form. Suppose that $ b \in C^\infty (T^*M \setminus 0)$ and fix $\lambda > 0$. Is it necessarily true that $$\int_{p < \lambda}\{p, b \} dx d\xi = 0$$ 
Here $\{p, b\}$ denotes the Poisson bracket and I presume that $dx d\xi = \omega ^n$. I tried to integrate by parts but it didn't seem to work.
 A: Yes. This follows from the general result that if $(S, \omega)$ is any symplectic manifold with boundary $\partial S$, then for any $f,g\in C^\infty(S)$,
$$
\int_S\lbrace f, g\rbrace \omega^n = -\int_{\partial S} g\, i_{X_f}\omega^n.
$$
To see this, note that (defining the Hamiltonian vector field $X_f$ by $i_{X_f}\omega = df$, and Poisson bracket $\lbrace f,g\rbrace := \omega(X_f,X_g)$)
$$
\lbrace f,g\rbrace\omega^n = -X_f(g)\,\omega^n = -\mathcal{L}_{X_f}(g\omega^n) = -d(i_{X_f}g\omega^n) - i_{X_f}{d}(g\omega^n) = -d(g\,i_{X_f}\omega^n),
$$
and applying Stoke's theorem.
Now apply this result with $S = \lbrace (x,\xi) \mid p(x,\xi) < \lambda\rbrace$, $f=p$, and $g=b$, to give
$$
\int_{p<\lambda}\lbrace p,b\rbrace \omega^n = -\int_{p=\lambda}b\,i_{X_p}\omega^n.
$$
However $X_p$ is tangent to the hypersurface $p=\lambda$ (since $X_pp = \lbrace p,p\rbrace = 0)$, and so integral on the left hand side vanishes.
Edit: apologies, I missed the fact that $b\in C^\infty(T^*M\backslash 0)$, rather than $C^\infty(T^*M)$. I believe the result still holds, however: there would be an additional contribution on the right hand side from $\lim_{\epsilon\to 0}\int_{p=\epsilon}b\,i_{X_p}\omega^n$, which also vanishes.
