How can we prove this integral inequality ? $\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin{(2n+1)t}}{\sin{t}}\right|dt<\pi\left(1+\frac{\ln{n}}{2}\right)$

Question

Use this $$\dfrac{1}{2}+\sum_{k=1}^{n}\cos{(kx)}=\dfrac{\sin{\left(n+\dfrac{1}{2}\right)x}}{2\sin{\dfrac{x}{2}}},x\neq 2m\pi,m\in\mathbb{Z}$$

to show that $$\int_{0}^{\dfrac{\pi}{2}}\left|\dfrac{\sin{(2n+1)t}}{\sin{t}}\right|dt<\pi\left(1+\dfrac{\ln{n}}{2}\right)\text{ for }n\ge 3$$

Answer 1

For all $x$, $$ \left|\frac{\sin((2n+1)x)}{\sin(x)}\right|=\left|\sum_{k=-n}^{n}e^{i2kx}\right|\le2n+1 $$ Note that $\displaystyle\sum_{k=-n}^{n}e^{i2kx}=1+2\sum_{k=1}^n\cos(2kx)$.

For $0\le x\le\pi/2$, we have $\sin(x)\ge2x/\pi$. Therefore, $$ \left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\le\frac\pi{2x} $$ Thus, $$ \begin{align} \int_0^{\pi/2}\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\,\mathrm{d}x &\le\int_0^{\pi/(4n+2)}(2n+1)\,\mathrm{d}x+\int_{\pi/(4n+2)}^{\pi/2}\frac\pi{2x}\,\mathrm{d}x\\ &=\frac\pi2+\frac\pi2\log\left(2n+1\right) \end{align} $$ For $n\ge3$, $2n+1\le\frac73n$. Therefore, $$ \begin{align} \frac\pi2+\frac\pi2\log\left(2n+1\right) &\le\frac\pi2\left(1+\log\left(\frac73\right)+\log(n)\right)\\[6pt] &\le\pi\left(1+\frac{\log(n)}{2}\right) \end{align} $$