How to prove $\int_0^\pi\biggl |\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le c(k+1)$ for some constant $c$? Let $|a_i|\le1$, prove that $$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \le c(k+1)$$ for some constant $c$.
I tried to solve the question as follows
$$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le 
\int_0^\pi \sum_{i=0}^k \biggl| \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \\=
\sum_{i=0}^k|a_{i}|\int_0^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \le
\sum_{i=0}^k\int_0^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx.$$
Then it looks like a classic equality about Lebesgue constant,
$$\frac{1}{\pi}\int_{-\pi}^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx=\frac{4}{\pi^2}\log i+O(1).$$
However, it didn't work. The above process of estimation doesn't seem to be careful enough.
How to make more careful and effective estimates to solve the question?
Thank you so much!
 A: On the other forum, I found the answer to this question.The website link is here, https://www.zhihu.com/question/410940277/answer/1400850373.
I copied the proof roughly here.
$$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx = \int_0^\frac{\pi}{k+1} \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx+\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx$$
It is easy to get that $$\frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x} \le i+\frac12.$$
Then $$\int_0^\frac{\pi}{k+1} \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \frac{\pi}{k+1}\sum_{i=0}^{k}|a_i|\biggl(i+\frac12\biggr) \le \frac\pi2(k+1).$$
According to Cauchy-Schwarz inequality,we have
$$\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \sqrt{\int_\frac{\pi}{k+1}^\pi \biggl(\frac{1}{2\sin \frac{1}{2}x}\biggr)^2 \, dx}\sqrt{\int_\frac{\pi}{k+1}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx}.$$
After calculation,we get that $$\int_\frac{\pi}{k+1}^\pi \biggl(\frac{1}{2\sin \frac{1}{2}x}\biggr)^2 \, dx \lt 2k+2,$$
and $$\int_\frac{\pi}{k+1}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx \le \int_{-\pi}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx=\pi\sum_{i=0}^{k}a_i^2 \le \pi(k+1).$$
Therefore, $$\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \sqrt{2\pi}(k+1).$$
In brief, we come to the conclusion $$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le (\frac\pi2+\sqrt{2\pi})(k+1).$$
