Suppose that $$ V_{n}(x)=(1+e(nx)+e(-nx))K_{n}(x), $$ where $e(x)=\mathrm{e}^{2\pi\mathrm{i}x}$, and $K_{n}(x)=\frac{1}{n}\cdot\frac{\sin^{2}(n\pi x)}{\sin^{2}(\pi x)}$. Now I want to estimates: $$ \int_{-1/2}^{1/2}\left|V_{n}'(x)\right|\mathrm{d}x\le Cn, $$ where $C$ is independent of $n$, Any suggestions?
What I've tried some result till now:
$$ \begin{aligned}\int_{-1/2}^{1/2}\left|V_{n}'(x)\right|\mathrm{d} x & \le\int_{-1/2}^{1/2}\left|(1+e(nx)+e(-nx))'K_{n}(x)\right|\mathrm{d} x+\int_{-1/2}^{1/2}\left|(1+e(nx)+e(-nx))K_{n}'(x)\right|\mathrm{d} x\\ & \lesssim\int_{-1/2}^{1/2}n\left|K_{n}(x)\right|\mathrm{d} x+\int_{0}^{1/2}\left|K_{n}'(x)\right|\mathrm{d} x\\ & \lesssim n\int_{-1/2}^{1/2}C_{K}\mathrm{d} x+\int_{0}^{2\pi/n}\left|K_{n}'(x)\right|\mathrm{d} x+\int_{2\pi/n}^{1/2}\left|K_{n}'(x)\right|\mathrm{d} x\\ & \lesssim Cn+\int_{0}^{2\pi/n}cn^{2}\mathrm{d} x+\int_{2\pi/n}^{1/2}\left|K_{n}'(x)\right|\mathrm{d} x\\ & \lesssim Cn+\int_{2\pi/n}^{1/2}\left|K_{n}'(x)\right|\mathrm{d} x \end{aligned} $$
how to estimate $\int_{2\pi/n}^{1/2}\left|K_{n}'(x)\right|\mathrm{d} x$?
I've checked Fourier Analysis from stein, p. 115-116. the last term can be controlled well. So I'll leave this without re-editting..