Log of partition function I know that in statistical mechanics, the partition function of a system of non-interacting, distinguishable particles is of the form
$$
Z(T,V,\mu) = \prod_i Z_i(T,V,\mu)
$$
where $Z_i$ is the partition function for $i$-th particle that is generally of the form
$$
Z_i(T,V,\mu) = \iint d^3 \vec{x} \,d^3 \vec{p} \,f_i(\vec{x},\vec{p},T,V,\mu)
$$
$f_i$ being some integrable function. If we take the log of the partition function, it should be of the form
$$
\ln Z = \sum_i \ln Z_i = \sum_i \ln \left(\iint d^3 \vec{x} \,d^3 \vec{p} \,f_i \right)\tag{1}\label{eq:1}
$$
but in some cases, it is given as
$$
\ln Z = \sum_i \iint d^3 \vec{x} \,d^3 \vec{p}\, \ln f_i
\tag{2}\label{eq:2}
$$
How is it possible that both of them are correct and if so, how, when and why should I switch from one to another? An example of \eqref{eq:1} is the partition function for ideal monatomic gases and for \eqref{eq:2} is 
the partition function for relativistic nuclear particles.
 A: I don't know the exact topic of (2), but I think I can give an answer to this.
The two definitions are for two different cases. In particular both $f$ are different.
(1) This is the classical definition for independent particles. Here $f = e^{-\beta H}$ (canonical ensemble). Note that an arbitrary state is given by $(x,p)$, so we have to average over them (therefore we have one integral).
(2) Here we talk about the grand canonical ensemble $e^{-\beta (H+\mu N)}$. However $f$ is not given by $e^{-\beta (H+\mu N)}$, but something else that comes out after some more simplifications.
It is something not easy to understand and I will try to explain as best as I can:
Here the arbitrary state is not given by (x,p) but rather by a collection {$n_{x,p}$} (the number of particles at $(x,p)$), wich is a much bigger space. Hence we have for each combination of $(x,p)$ a sum over $n_{x,p}$. Or in formulas: $$Z = \prod_{(x,p)} \sum\limits_{n_{x,p} = 
0}^\infty e^{-\beta(\epsilon_{x,p} n_{x,p}+ \mu n_{x,p})}\\
 = \prod_{(x,p)}\frac{1}{1-e^{-\beta(\epsilon_{x,p}+\mu)}}$$
Now if you take the $\log$ the product becomes a sum. Since $(x,p)$ is a dense set however we have to use an integral:
$$ \log Z = -\int dxdp \log(1-e^{-\beta(\epsilon_{x,p}+\mu)})$$
Now you can see that in this case $f = 1-e^{-\beta(\epsilon_{x,p}+\mu)}$ which is something completely different than in case (1).
Note that the last formula looks similar to formula (1) in your arxiv link for $\mu = 0$ and $\epsilon_p = \sqrt{m^2+p^2}$.
