Continuous phase transition with no finite critical exponent I am working with a model in which the energy density as a function of chemical potential $\mu$ and density $n$ is given by
$$E = (e^{-1/n}-\mu)n$$
in appropriate units. This model has a phase transition: $n=0$ for $\mu<0$ and $n>0$ for $\mu>0$. Near the transition we cannot write $n \sim \mu^\nu$ for any $\nu$; rather $n(\mu)$ is given in terms of a Lambert W function.
Is this still a 2nd order phase transition or is it something else? Is there any significance to the non-power-law behavior near the transition, and are there common examples in physics?
 A: In practice, power law exponents are defined/extracted by taking the limit of a logarithmic derivative. For example, if some quantity is expected to scale as $F(r) \sim (r - r_c)^{-\nu}$ as $r \rightarrow r_c$ (with possibly different exponents if the limit is approached from above or below), then we would define
$$\nu \equiv -\lim_{r \rightarrow r_c} (r-r_c)\frac{\partial}{\partial r} \ln F(r).$$
(the minus sign is included since we typically expect a divergence). In the case where $F(r) \sim (r - r_c)^{-\nu}$, this gives the actual exponent $\nu$. However, there are cases where we cannot actually calculate this expansion, but this definition proves to be useful for estimating exponents. One such example is the DLog Pade Approximant method, which has been used to try to estimate critical exponents when only a series expansion of $F(r)$ around some point that is not the critical point is known.
Another common use of this definition, which you may be familiar with, is the heat capacity of the Ising model in $d=2$ or $d > 4$. The heat capacity exponent is conventionally given as $\alpha = 0$, even though the heat capacity actually diverges logarithmically. $\alpha = 0$ is useful, however, for satisfying exponent relations. See, e.g., this answer.
So, in your case, one would attempt to define the exponent $\nu$ by
$$\nu = -\lim_{\mu \rightarrow 0^+} \mu \frac{\partial}{\partial \mu} \ln n(\mu).$$
If I did not make any mistakes in inputing everything into Mathematica, $n(\mu) = -1/(1+W_{-1}(-\mu/e))$ for $\mu \geq 0$, where $W_{-1}(x)$ is the $-1$ branch of the Lambert-W function, which behaves as $\ln(-x)$ for $x \rightarrow 0$.
Barring any mistakes, I find the limit reduces to
$$\nu \equiv -\lim_{\mu \rightarrow 0^+} \mu \frac{-W_{-1}(-\mu/e)}{\mu(1+W_{-1}(-\mu/e))^2} \rightarrow 0.$$
Hence, by this generalized definition for the exponents the critical exponent is formally $0$, similar to the case of the heat capacity exponent in the Ising model.
This said, how you should interpret this will depend on details of your model and your calculation. For instance, if you are working within a mean-field approximation, this result could be modified by fluctuations to give an actual power-law behavior (such as the Ising model heat capacity exponent $\alpha$ becoming non-zero in $d = 3$). A full understanding of the properties of the transition would ideally be done by a renormalization group analysis of the statistical model, but that is much easier said than done.
