Fix $x>0$ and $c\in\mathbb{N}$. Let $f(x):=\frac{c}{4c-2+2x^2}$ and
$$m\_N(x):=\frac{1}{N} \sum\_{i=0}^{f(x)N} \log(\frac{c N}{2}-i(2c-1))$$


I'm pretty sure $m\_N(x)\to\infty$ as $N\to\infty$.


I would like to know the asymptotic behaviour of $m\_N(x)$, and I expect to find something like


$$m\_N(x)=f(x) \log{N} + g(x) + o(1)\ \ \text{ as }N\to\infty$$ 


Can you confirm this result? If this is the case, can you help me to compute the constant $g(x)$?



Asymptotic behaviour of a mean
Let us consider the sum
$$
 m\_N(x)=\frac{1}{N}\sum\_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right).
$$
The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain
$$
 m\_N(x)=[f(x)]\log N+\frac{1}{N}\sum\_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right).
$$
When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that
$$
 m\_N(x)=[f(x)]\log N+[f(x)]\log[z(x)].
$$
We can take $z(x)=\frac{c}{2}-t\[f(x)\](2c-1)$ being $t\in (0,1)$. 


Indeed, we can define a partition with $x\_i=x\_{i-1}+\frac{1}{\[f(x)\]N}$ and so
$$
 \frac{1}{N}\sum\_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum\_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i\[f(x)\](2c-1)\Delta x\right)
$$
being $\Delta x=\frac{1}{\[f(x)\]N}$. But this, in the given limit, is nothing else than
$$
 \int\_{\frac{c}{2}}^{\frac{c}{2}-\[f(x)\](2c-1)}\log(z)dz<\infty
$$
as it should.


