How to calculate the Gibbs phenomenon for sawtooth function? Let $f(x)$ be he function $f(x)= (\pi-x)/2$ on the interval $(0,2\pi)$ with $f(0)=0$ and extended by periodicity to the real line.
It is quite straightforward to show that the Fourier series of $f$ is
$$S_F(f)(x)=\sum_{n=1}^\infty \dfrac{\sin nx}{n}.$$
I want to show that if $S_{F,N}(f)$ denotes the $N$-th partial sum of the Fourier series then
$$\max_{0<x\leq \pi/N}S_{F,N}(f)(x)-\dfrac{\pi}{2}=\int_0^\pi \dfrac{\sin t}{t}dt-\dfrac{\pi}{2}.$$
I have no idea on how to do this. I mean, I know that
$$S_{F,N}(f)(x)=\dfrac{1}{2}\int_0^x D_n(\xi)-1 \ d\xi,$$
but I don't know how this can help. I mean, I know that we have
$$S_{F,N}(f)(x)-\dfrac{\pi}{2}=\dfrac{1}{2}\int_0^xD_n(\xi)-1\ d\xi-\dfrac{\pi}{2},$$
The only idea I had was to define $\varphi_N(x) = S_{F,N}(f)(x)-\pi/2$, so that we are seeking a maximum of $\varphi_N$.
Now, by the expression above, $\varphi_N$ is differentiable with $\varphi_N'(x)=(D_n(x)-1)/2$. So we can search for $x\in [0,\pi/N]$ with $\varphi_N'(x)=0$, which is equivalent to $D_N(x)=1$.
This gives $x = \pi/N+1$. The solution is indeed inside $[0,\pi/N]$. But how can I know use this to prove the claim?
I mean I need to show that the desired maximum is $\int_0^{\pi} \sin t/t dt -\pi/2$.
How can I use the $x$ I found to do this?

First, the notation in what you want to show needs some clarification. We are interested in the maximum of $S_N(f)(x)$ by allowing both $x$ and $N$ to vary.
For any fixed $N$, to find the maximum when we vary $x$, we want to find $x_0$, such that
$$\sin (N+\frac{1}{2})x_0 = \sin\frac{x_0}{2}$$
where $x_0 \in (0,\pi/N)$.
This equation is equivalent to
$$(N+\frac{1}{2})x_0 + \frac{x_0}{2} = \pi.$$
So the solution is $x_0=\frac{\pi}{N+1}$.
The maximum is thus
$$\int_0^{\frac{\pi}{N+1}} \left( \frac{\sin(N+\frac{1}{2})x}{\sin\frac{x}{2}} - 1 \right)dx.$$
Notice that this value increases as $N$ increases.
Now simply do a change of variable $t = (N+\frac{1}{2})x$ and let $N \rightarrow \infty$.
