Partial sum of Fourier series of square wave

Question

Let $f$ be a $2π$ -periodic square wave function so that
$$f\, = -1 \quad -π \le x<0$$ $$f=1 \qquad 0 \le x< π$$
$S_{2n-1}(x)$ is the $(2n-1)st$ Fourier polynomial of $f$. Prove that it can be written as:
$$S_{2n-1}(x)=\frac{1}{nπ}\int_0^{2nx} \frac{\sin t } {\sin {\frac {t} {2n}}} dt$$

It's obvious that the Fourier-Series can be written as: $$F_N(x)=\frac {4}{\pi} \sum_{n=1}^{N}\frac{\sin((2n-1)x)}{2n-1}$$ But thats as much as I can do about it. Can someone explain the ongoing to me further?

Answer 1

You just have to notice that: $$\frac{\sin((2n-1)x)}{(2n-1)}=\int_{0}^{x}\cos((2n-1)u)\,du $$ hence: $$ F_N(x) = \frac{4}{\pi}\int_{0}^{x}\sum_{n=1}^{N}\cos((2n-1)u)\,du =\frac{2}{\pi}\int_{0}^{x}\frac{\sin(2Nu)}{\sin u}\,du=\frac{1}{\pi N}\int_{0}^{2N x}\frac{\sin t}{\sin\frac{t}{2N}}\,dt.$$ For the middle equality, see this classical question.