# What would the entropy of an infinitely continuous and not discrete system be?

Entropy is defined as the sum of the probabilities of microstates times the logarithm.

$$S = - k_B\sum_i p_i \ln p_i$$

The mean value of something is defined as

$$\bar{x} = \sum_i x_i p_i$$

but for continuous cases is:

$$\bar{x} = \int^{\infty}_{-\infty} x \rho_x \, \mathrm{d} x$$

Would the entropy for an infinitely fine system be the following?

$$S = -k_B \int^{\infty}_{-\infty} \rho_x \ln \rho_x \, \mathrm{d} x$$

I realise that due to quantum mechanics and atomic composition of materials that stuff is discrete and not continuous.

First note that the differential entropy can be negative, because $p(x)$ can become greater than one. An easy example (which I took from the Wikipedia article ;)) is the uniform distribution $p(x) \equiv 2$ on the interval $\left[0,\frac 1 2\right]$: The differential entropy is obviously $-k_B \ln 2$.
More generally we find that the value of such a differential entropy does not have a physical significance, but only differences between values. For example, the Nernst Theorem ($\lim_{T \to 0} S = k_B \ln g_0$ where $g_0$ is the degeneracy of the ground state) will typically not hold for a differential entropy.
Why this is so you can understand by considering differential entropy as a discrete entropy of points, in the limit where the points are so close together that we can describe them as if we had a continuous system. (This is called a Limiting Density of Discrete Points.) As you can see in the Wikipedia article, the entropy $S_N$ of such a system differs from the "naive" differential entropy $s$ by $$S_N = s + \ln\left( N / r \right)$$ where $N$ is the number of points and $r$ the size of the system (assuming the point density is constant). When we consider the differential entropy, we essentially neglect the "infinite but constant" term $\ln\left( N/r \right)$.
Finally, I promised you an example where this is used. If you consider a cavity, you can describe its density matrix using the Husimi Q function $\mathcal Q(q,p)$ which is a probability density. The Wehrl entropy of the system is $$S_W = -k_B \int \mathcal Q(q,p) \ln \mathcal Q(q,p) \frac{\mathrm dq\, \mathrm dp}{2\pi\hbar}.$$ This is an interesting quantity which I wrote my bachelor's thesis about ;) people use it e.g. in entanglement theory and to describe decoherence in quantum optics.
A pretty cool fact is $S_W \geq 1$ which is a refinement of the Heisenberg uncertainty relation for the system. This so-called Lieb conjecture was proven only recently.
Yes, this would be the entropy for a continues system. Note there are also statistical treatments of classical theories. In those, entropy is exacly defined the way you wrote it there. $$S[\rho]= - \int d\Gamma \rho(\Gamma)\log{\rho(\Gamma)}$$