I would like to compute the expansion of the following integrals near $N = + \infty$ up to $\mathcal{O}(1/N^2)$: $$ \int_{0}^{\omega_0} \frac{\sin \left( \frac{2 N + 1}{2} \omega \right) \cos(\omega n)}{\sin \left( \omega / 2 \right)} \ d \omega $$ and $$ \int_{\omega_0}^{\pi} \frac{\sin \left( \frac{2 N + 1}{2} \omega \right) \sin(\omega n)}{\sin \left( \omega / 2 \right)} \ d \omega $$ with $n \in \{ -N, -N + 1, \dots, N \}$ and $0 \leq \omega_0 \leq \pi$. I cannot compute the first order coefficient in the usual way since I believe that I cannot apply the Leibniz integral rule, and thus I do not know how to evaluate the derivative (w.r.t. 1/N) of each integral for $1/N \rightarrow 0$.
(I am asking two questions at a time since I expect the computations to be similar.)
Thank you.
EDIT: Since I was asked to clarify my attempt, these are my calculations for the first integral.
- Order 0 term: the limit for $N \rightarrow + \infty$ of the integral can be shown (change variable $\omega' = N \omega$, then expand the denominator etc.) to be $\pi$
- Order 1 term: the derivative w.r.t $1/N$ is $$ - \frac{1}{N^2} \frac{d}{d N} \int_{0}^{\omega_0} \frac{\sin \left( \frac{2 N + 1}{2} \omega \right) \cos(\omega n)}{\sin \left( \omega / 2 \right)} \ d \omega , $$ and the coefficient of the expansion in the limit of this for $N \rightarrow + \infty$. The Leibniz integral rule requires the integrand to have continuous derivative (w.r.t $N$) in $[0, \omega_0]$. This does not happen here, therefore I cannot apply the LIR. But then I do not know how I should compute the limit.
