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 $g(x)$?

howI would do it: the sum looks a lot like a Riemann sum. So write the Riemann sum and the integral to which it is an approximation side by side and carefully estimate the differences. – Anthony Quas Jun 12 '12 at 4:26