I'm trying to reconcile the following two statements:

1. Quantum Mechanics gives physical predictions which are different than the predictions that are obtained in the $\hbar \rightarrow 0$ limit, that is, the predictions of quantum mechanics generally depend on $\hbar$.
2. In the phase space formulation of QM, the Moyal product and the quantization map are related by the following formulae:
$$ Q(a) Q(b) = Q(a \* b),$$
$$ \text{tr} Q(a) = \frac{1}{2 \pi \hbar} \int dx \int dp a(x, p). $$

Here $Q(a)$ is a quantization of the phase space function $a(x, p)$, and $a \* b$ is the Moyal product given by
$$ a \* b = a e^{\frac{i \hbar}{2} \omega^{ij} \overleftarrow{\partial\_i} \overrightarrow{\partial\_j}} b, $$
the arrow notation means that $\overleftarrow{\partial}$ acts on $a$ only and $\overrightarrow{\partial}$ acts on $b$ only.

Now consider a fairly generic problem from classical mechanics. Let's say I have an observable $E(x, p)$, and a probability distribution $\rho(x, p)$ (it can be for example the energy of the system and the Gibbs distribution for some value of $\beta$). The question is: what is the expectation value of $\left< E \right>$? According to classical physics,

$$ \left< E \right> = \int dx \int dp E(x, p) \rho(x, p). $$

Now if we try to compute the quantum corrections to this formula, we end up with

$$ \left< E \right>\_{\hbar} = 2 \pi \hbar \text{tr} (Q(E) Q(\rho)) = \int dx \int dp E(x, p) \* \rho(x, p). $$

It seems as if the corrections are encoded in the Moyal product. However, that can't be true: it is easy to show that for any $a, b$,

$$ \int dx \int dp \; a \* b = \int dx \int dp \; a b. $$

(Demonstrate this by integrating by parts and observing that each non-zero-order term contains a product of antisymmetric $\omega^{i j}$ and symmetric $\partial\_i \partial\_j b$ and therefore vanishes).

So there are actually no quantum corrections in $\left< E \right>\_{\hbar}$ defined above.

On the other hand, it is well known that there **are** quantum corrections to classical expectation values. Take the expectation value of energy a quantum oscillator for example. Computing it in the Hamiltonian eigenstate basis yields a value $\left<E \right>(\beta, \hbar)$ that depends on $\hbar$ explicitly.

Therefore we have a contradiction, so I must conclude that some of the assumptions I've made before must be incorrect. In particular, I see two possibilities:

1. The trace formula isn't correct:
$$ \text{tr} ( Q(a) Q(b) ) \neq \frac{1}{2 \pi \hbar} \int dx \int dp \; a \* b. $$
2. The quantization of the probability distribution isn't $Q(\rho)$ but somehow acquires extra corrections.

Both possibilities seem strange to me.
What's going on here?