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Consider: $\int \frac{\sin(n+\frac{1}{2})x}{\sin(\frac{x}{2})} = x + \sum_{k=-n}^{n}\frac{2}{k} \sin(kx)$

from what we do see that $\int_{-\pi}^{\pi}\frac{\sin(n+\frac{1}{2})x}{\sin(\frac{x}{2})} dx = 2\pi$ for all $n$

now $\int_{-\pi}^{\pi}|\frac{\sin(n+\frac{1}{2})x}{\sin(\frac{x}{2})}| dx=I_n$(1) does not converge(I mean it holds $\lim_{n\to \infty} I_n = +\infty$) while it shall be $\int_{-\pi}^{\pi}|\frac{\sin(n+\frac{1}{2})x}{\sin(\frac{x}{2})}| dx = 2\int_{0}^{\pi} \frac{\sin(n+\frac{1}{2})x}{\sin(\frac{x}{2})}dx$(and so, this is $2\pi$), which converges.

I need two things: what I did wrong and why this integral(1) diverges? Who does not believe that (1) does not converge, take wolframalpha and compute with 3,5,7,.... and you will see it grows with n.

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The integrand function is not always positive on $(-\pi,\pi)$.

The $L^1$ norm of the Dirichlet kernel is divergent since $$ \left\|D_n\right\|_1 \geq 4\,\text{Si}(\pi)+\frac{8}{\pi}\log(n)\tag{1}$$ as mentioned by Wikipedia. That inequality is obtained by partitioning $(-\pi,\pi)$ into many sub-intervals whose endpoints are given by roots of the integrand function, then applying a Riemann-sum argument near the origin and Jensen's inequality far from the origin.

$(1)$ is the reason behind Gibbs' phenomenon, for instance.

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    Thanks, Jack, after 10 minutes of thinking I got that the integrand is not always positive, but thanks anyways! – 2017-01-13
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    can you tell me where I can find the proof for $||D_n||_1 \geq 4Si(\pi) + \frac{8}{\pi} \log(n)$? – 2017-01-13
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    @nikola: you should find it here by a careful search. I added that line on Wikipedia after solving a problem posted on MSE a while ago. – 2017-01-13
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    A proof is given here, too: math.arizona.edu/~friedlan/teach/456/gibbs.pdf – 2017-01-13