The [replica trick] attempts to calculate the expectation of the logarithm $X=\log(Z)$ of a random variable $Z$. The wikipedia article describes the logarithm as the limit
$$
\log(Z) = \lim_{n\to 0}\frac{Z^n -1}{n}, \quad\text{or}\quad \log(Z) = \lim_{n\to 0} \frac{\partial Z^n}{\partial n}
$$
But I think it is much more natural to calculate the moment generating function of $X$ (and not just the expectation $\mathbb{E}[X]$):
$$
   M(\lambda) = \mathbb{E}[\exp(\lambda X)] = \mathbb{E}[Z^\lambda]
$$
Now for $\lambda = n\in \mathbb{N}$ this is simply calculating the expectation of $n$ "replicas" of $Z$, which is where the name comes from.
The expectation can easily be obtained as $\mathbb{E}[X]=M'(0)$. For this we need to let $\lambda\to 0$
> The crux of the replica trick is that while the disorder averaging is done assuming $n$ to be an integer, to recover the disorder-averaged logarithm one must send $n$ continuously to zero. This apparent contradiction at the heart of the replica trick has never been formally resolved, however in all cases where the replica method can be compared with other exact solutions, the methods lead to the same results. - [Wikipedia][replica trick]

So what is happening is: By calculating $M(n)$ we hope to obtain a function $f$ which coincides with $M$ on the natural numbers and hope that it coincides with $M$ completely, i.e.
$$
f: \mathbb{C} \to \mathbb{C},\quad M(n) = f(n)\quad \forall n\in \mathbb{N}\overset?\implies M\equiv f
$$
Were we view $f$ and $M$ as complex functions because this Wikipedia entry already contains the suggestion to use Carlson's Theorem

> To prove that the replica trick works, one would have to prove that Carlson's theorem holds, that is, that the ratio $\frac{Z^n − 1}{n}$ exponential type less than pi.

Now I am surprised about this phrasing, since that requirement
of Carlson's Theorem is actually straightforward to prove.

So here are the requirements of [Carlson's Theorem][carlson] which we want to apply to $g:= f- M$:

1. $g$ is an entire function of exponential type
2. There exists $C\in \mathbb{R}$ and $c<\pi$ such that
$$
|g(iy)|\le Ce^{c|y|} \quad \forall y\in \mathbb{R}
$$
3. $g(n)=0\quad \forall n\in\mathbb{N}$

which implies $g\equiv 0$ and therefore $M\equiv f$. The third requirement is the heart of the replica trick, the second requirmenent seems to be the one referred to as the problematic one in the [replica trick] wikipedia entry. Now we can either prove this requirement for $g$, or for $f$ and $M$ which is sufficient (by the triangle inequality and replacing $C$ with $2C$). But this property is trivial for $M$, since
$$
|M(iy)| = |\mathbb{E}[\exp(iyX)]| \le \mathbb{E}[|\exp(iyX)|] = 1
$$
So $C=1$ and $c=0<\pi$ does the job. Since $f$ is the function we guess in the replica trick, we would simply need to check this requirement for the guessed function too.

So it seems like the much more difficult problem is proving that $g$ is an entire function of (arbitrary) exponential type (i.e. the first requirement).

In [this paper](https://arxiv.org/pdf/2108.12846.pdf) the first requirement of Carlson's theorem is further relaxed to be: 

> $g$ needs only be holomorphic in the closed right half plane (not even of exponential type)

My questions now are:
- am I overlooking something important?
- What are sufficient requirements on $Z$ such that the moment generating function $M$ of $X=\log(Z)$ is holomorphic (assuming that $M$ holomorphic and $f$ holomorphic implies $g$ is holomorphic - holomorphic functions are a vector space right?)

To prove the replica trick works, practicioners would then only need to check this condition on $Z$ and that their guess of $f$ is holomorphic, and satisfies the second requirement of Carlson's theorem (it can't possibly be a moment generating functions if it doesn't anyway). Additionally they get the entire moment generating function $M$ and not just the expectation. This feels like it is too good to be true.


[replica trick]: https://en.wikipedia.org/wiki/Replica_trick
[carlson]: https://en.wikipedia.org/wiki/Carlson%27s_theorem