How to bound $\int^{\pi}_{-\pi}|D_{n}(t)|\,dt$ from above? Rudin asked me to bound $$\int_{-\pi}^{\pi} |D_{n}(t)|\,dt$$ from above. I need a bound at the level of $o(\log(n))$. 
The background is:
If $s_{n}$ is the $n$-th partial sum of the Pourier series of a function $f\in C(T)$, prove that $$\frac{s_{n}}{\log[n]}\rightarrow 0$$ uniformly. That is, prove that $$\lim_{n\rightarrow \infty}\frac{|s_{n}|_{\infty}}{\log(n)}=0$$On the other hand, if $\lambda_{n}/\log[n]\rightarrow 0$, prove that there exist an $f\in C^(T)$ such that sequence $s_{n}(f,0)/\lambda_{n}$ is unbounded. 
Update: a numerical evaluation for $n=10^{800}$ is inconclusive. 
 A: Well, you can get the value of the integral very explicitely. As you already noted
$$ \int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt
= \int_{-\pi}^\pi \frac{\sin(nt)}{\sin\frac t2}\cos\frac t2dt
= \int_{-\pi/2}^{\pi/2} \frac{\sin(2nt)}{\sin t}2\cos t\,dt\,. $$
Now the fraction can be transformed into
$$ \frac{\sin(2nt)}{\sin t}
= \frac{e^{2nti}-e^{-2nti}}{e^{ti}-e^{-ti}}
= \frac{e^{4nti}-1}{e^{2ti}-1}
= 1+e^{2ti}+e^{4ti}+\cdots+e^{(4n-2)ti} $$
We multiply this with $2\cos t=e^{-ti}+e^{ti}$ and get for the integrand
$$ \frac{\sin(2nt)}{\sin t}2\cos t 
= e^{-ti}+2e^{ti}+2e^{3ti}+\cdots+2e^{(4n-5)ti}+2e^{(4n-3)ti}+e^{(4n-1)ti}. $$
Furthermore, 
$$ \int_{-\pi/2}^{\pi/2} e^{kti}dt
= \frac2k\sin\frac{k\pi}2, $$
and hence we finally get
$$ \int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt
=2+4-\frac43+\frac45-\frac47\pm\cdots+\frac{4}{4n-3}-\frac{2}{4n-1} $$
which is obviously convergent by the Leibniz criterion and hence bounded by a constant. 
A: I'll replace $|\sin t/2|$ by $|t|$ since they are comparable: 
$$\frac{|t|}{\pi}\le \left|\sin\frac{t}{2}\right| \le \frac{|t|}{2} \  \text{ for }t\in [-\pi,\pi]$$
 Claim: for all $\lambda\ge 1$ 
$$\frac{1}{3}\log \lambda \le \int_0^\pi \frac{|\sin \lambda t|}{t}\,dt \le \log \lambda +\log \pi +1.$$
Proof. For the upper bound, split the integral into "small $t$" part and the rest: 
$$\int_0^{\lambda^{-1}} \frac{|\sin \lambda t|}{t}\,dt
\le \int_0^{\lambda^{-1}} \frac{\lambda t}{t}\,dt =1$$
and 
$$\int_{\lambda^{-1}}^\pi \frac{|\sin \lambda t|}{t}\,dt \le \int_{\lambda^{-1}}^\pi \frac{1}{t}\,dt = \log\lambda + \log \pi$$
The lower bound needs a bit more work. Since the integrand is nonnegative, we can restrict the region of integration to the 
set $|\sin \lambda t|\ge 1/2$. This set contains the intervals $I_k=[\pi \lambda^{-1} (k+1/6),  \pi \lambda^{-1} (k+5/6)]$ for 
all integers $k$ such that $0\le k \le \lambda-1$. The integral over $I_k$ is at least 
$$
\int_{I_k} \frac{1/2}{t}\,dt \ge |I_k| \frac{1/2}{\pi \lambda^{-1} (k+1)} = \frac{1/3}{k+1}
$$
Therefore, the integral is bounded from below by $$\frac13 \sum_{k=0}^{\lfloor \lambda-1\rfloor }\frac{1}{k+1} \ge \frac{1}{3} \log \lambda.$$ 

Remark. If $\|D_n\|_{L^1}=o(\log n)$ were true, the estimate in #19 would hold for the Fourier series of any finite measure on $[-\pi,\pi]$. This is not the case. The fact that $s_n$ comes from a continuous function should be used. 
