Partition function containing QM? I am wondering about the partition function of the classical microcanonical ensemble. It contains Planck's constant and also an indistinguishability argument about the particles I am looking at and I find this confusing for the following reasons:
If I describe for example an ideal gas using classical mechanics, there is no reason why I should assume that there is a problem to distinguish the gas molecules or why $\hbar$ should occur there. 
I mean, how could people like Gibbs etc. derive this equation if they did not know about QM? 
In some sense, there has to be a reason why we have these two QM properties in there and maybe it is just to get the limit to QM right, but is this really the only reason for this?
I would love to receive a few ideas about this strange definition by this community, in order to get impressions, why we model a (NVE) ensemble this way?
 A: You can make the analogy with quantum field theory where due to a lack of a well defined microscopic theory, you are forced to regularize the theory and hope to be able to eliminate the regularization parameters by absorbing them in a few physical quantities (mass, charge etc.). 
Now, if you start with the correct partition function that takes into account the correct statistics (Fermi Dirac of Bose Einstein), you end up with the Maxwell Boltzmann statistics in the dilute limit, where you take overcounting into account by the N! in the denominator. This is a universal feature in the dilute limit, because the probability of two particles occupying the same state vanishes in this limit.
So, you could say that Gibbs was able to succeed because there exists a universal theory in the dilute scaling limit. You then only have one scaling parameter left (h). In terms of h some quantities like entropy are infinite, but as the ACuriousMind points out in his answer, entropy differences will be independent of h. Now this is not true if you consider changes where in one case you have effectively frozen degrees of freedom (e.g. atoms in a solid) and in the other case you don't (e.g. a gas). 
A: A footnote on Wikipedia actually explains this.
The $h$ appearing in the partitioning of the phase space is not Planck's constant, it is simply the size of a phase space cell in which we do not distinguish single states anymore. It's initially arbitrary. Since Planck's constant provides a natural scale below which we should not apply classical thinking, it is nowadays often taken to be that constant, but nothing in the classical formalism necessitates this.
You are correct that it influences the entropy, since that is
$$ S = k\ln(W)$$
with1
$$ W = \frac{1}{h^n C} \int_{\text{phase space}}f(\frac{H - E}{\omega})$$
but the logarithm property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$ means this drops out of all entropy differences, meaning the choice of $h$ has no physical impact. The overcounting factor $C$ depends on whether swapping individual particles/phase space coordinates will make a difference in the macrostate, and is $C = N!$ for $N$ identical (not really indistiguishable, classically) particles.

1Here, $n$ is the number of generalized coordinates (the number of particles in the usual settings), $f(\frac{H - E}{\omega})$ is an "indicator function" of the Hamiltonian $H$ that peaks at $H = E$ with width $\omega$ (for any suitable notion of width, e.g. it could be taken to be a Gaussian or rectangle function. Often, one takes $\omega \to 0$, leaving $f$ to be the Dirac delta $\delta(H - E)$, which means the integral then collapses to the surface of the object traced out by the equation $H = E$ in phase space, which, for free particles, is a sphere.
