Problem in the derivation of replica partition function of Hopfield network. While I was following the derivation of the replica partition function of Hopfield network from here, I have found something strange explanation. Due to the independence of $s_{i}$'s, the following term:
$$\sum_{s=\pm1}\exp\left( \beta\sum_{a=1}^{n}\sum_{i=1}^{N}m_{a}\xi_{i}s^{a}_{i} + \frac{\alpha\beta^2}{2}\sum_{i=1}^{N}\sum_{a\neq b}r_{ab}s^{a}_{i}s^{b}_{i} \right)$$
can be represented in the following form, where the indices are removed:
$$\exp\left[ N\ln\left(\sum_{s=\pm1}\exp\left( \beta\sum_{a=1}^{n}m_{a}\xi s^{a} + \frac{\alpha\beta^2}{2}\sum_{a\neq b}r_{ab}s^{a}s^{b} \right)\right) \right].$$
But how can it be possible? For instance, the following equation is generally true.
$$e^{Na} + e^{Nb} \neq e^\left(N\ln(e^{a} + e^{b})\right).$$
If the first two terms implies the same thing, then, I guess, some kind of approximation should be introduced. Could someone please help me?
 A: I am posting the answer to my question. First of all, the derivation in the book is not perfectly correct. It didn't consider an average over the quenched random variable. Anyways, the right form of the first equation in the question is the following.
\begin{equation}
\sum_{\{s\}}\left[ \int \exp\left( \beta\sum_{a = 1}^{n}\sum_{i=1}^{N}m^{a}\xi_{i}s^{a}_{i} + \frac{\alpha\beta^2}{2}\sum_{a\neq b}^{n}\sum_{i=1}^{N}R_{ab}s^{a}_{i}s^{b}_{i}\right)\prod_{i=1}^{N}p(\xi_{i}) d\xi_{i} \right]
\end{equation}
Then change the order of integral and summation.
\begin{equation}
\int \sum_{\{s\}}\left[\exp\left( \beta\sum_{a = 1}^{n}\sum_{i=1}^{N}m^{a}\xi_{i}s^{a}_{i} + \frac{\alpha\beta^2}{2}\sum_{a\neq b}^{n}\sum_{i=1}^{N}R_{ab}s^{a}_{i}s^{b}_{i}\right)\right]\prod_{i=1}^{N}p(\xi_{i}) d\xi_{i}
\end{equation}
Since every $\xi_{i}$'s are independent random variables, and their probability distributions are identical, we get the following without indicies.
\begin{equation}
\left[\int \sum_{\{ s \}}\left[\exp\left( \beta\sum_{a = 1}^{n}m^{a}\xi s^{a} + \frac{\alpha\beta^2}{2}\sum_{a\neq b}^{n}R_{ab}s^{a}s^{b}\right)\right]p(\xi) d\xi\right]^{N}
\end{equation}
Then taking a logarithm and exponential, we obtain something similar equation in the question but with average over quenched random variable.
