I want to make the following estimate: $$\int_{-\pi} ^ {\pi} \left| \frac{\cos\left(\frac{x}{2}\right) -\cos\left(\frac{(2N+1)x}{2}\right) }{\sin\left(\frac{x}{2}\right)} \right| dx \le c \log (N) $$ The integrand is the absolute value of the so-called conjugate Dirichlet kernel defined by: $$D_N (x) = \sum _{|n|<N} sign(x) e^{inx} $$
Any hints are appreciated. Thanks in advance.
First observe that $$\begin{aligned}|D_N(x)|&=\left|\sum_{n=1-N}^{N-1}e^{inx}\right|=\left|\frac{e^{i(1-N)x}(1-e^{i(2N-1)x})}{1-e^{ix}}\right|\\ &=\left|\frac{e^{iNx}-e^{i(1-N)x}}{e^{ix}-1}\right|=\left|\frac{e^{i(N-\frac{1}{2})x}-e^{i(\frac{1}{2}-N)x}}{e^{\frac{1}{2}ix}-e^{-\frac{1}{2}ix}}\right|\\ &=\left|\frac{\sin\left((N-\frac{1}{2})x\right)}{\sin\frac{x}{2}}\right|\leq\sum_{n=1-N}^{N-1}|e^{inx}|=2N-1 \end{aligned}$$ It follows that $$\begin{aligned} \int_{-\pi}^\pi|D_N(x)|dx&=\int_{-\pi}^\pi\left|\frac{\sin\left((N-\frac{1}{2})x\right)}{\sin\frac{x}{2}}\right|dx\\ &=2\int_{0}^\pi\left|\frac{\sin\left((N-\frac{1}{2})x\right)}{\sin\frac{x}{2}}\right|dx\\ &=4\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin\left((2N-1)t\right)}{\sin t}\right|dt\\ \end{aligned}$$ Now recall that $$|\sin t|\geq\frac{2t}{\pi},\quad\forall t\in[0,\frac{\pi}{2}].$$ It follows that $$\begin{aligned} \int_{0}^{\frac{\pi}{2}}\left|\frac{\sin\left((2N-1)t\right)}{\sin t}\right|dt&\leq \int_{0}^{\frac{\pi}{4N-2}}\left|\frac{\sin\left((2N-1)t\right)}{\sin t}\right|dt+\int_{\frac{\pi}{4N-2}}^{\frac{\pi}{2}}\left|\frac{\pi\sin\left((2N-1)t\right)}{2t}\right|dt\\ &\leq \int_{0}^{\frac{\pi}{4N-2}}(2N-1)dt+\int_{\frac{\pi}{4N-2}}^{\frac{\pi}{2}}\frac{\pi}{2t}dt\\ &\leq C\log N \end{aligned}$$ for appropriate choice of $C>0$.