Proof that $\sum_{l=1}^{\infty} \frac{\sin((2l-1)x)}{2l-1} =\frac{\pi}{4}$ when $0<x<\pi$

Question

Proof that $$\sum_{l=1}^{\infty} \frac{\sin((2l-1)x)}{2l-1} =\frac{\pi}{4}$$ when $0<x<\pi$

The chapter we are working on is about Fourier series, so I guess I'd need to use that some how.

My idea was to use $$\sum_{l\in \mathbb Z}c_l e^{i(2l-1)x}= a_0+\sum_{l\geq 1}a_1\cos(lx)+\sum_{l\geq 1} b_l\sin((2l-1)x)$$

Where $a_0$ and $a_l$ would be $0$, $b_l= \frac{1}{2l-1}$.

This would give us $c_l = \frac{1}{4il-2i}$. I don't know how knowing that

$$\sum_{l=1}^{\infty} \frac{\sin((2l-1)x)}{2l-1}= \sum_{l\in \mathbb Z} \frac{1}{4il-2i} e^{i(2l-1)x}$$ would help here though.

Am I even on the right track here? Any hints are much appreciated , thanks in advance :)

Answer 1

Define $f(x)$ to be $\frac{\pi}{4}$ on $(0,\,\pi)$, $0$ at multiples of $\pi$ and $-\frac{\pi}{4}$ on $(-\pi,\,0)$, and extend $f$ to $\Bbb R$ by requiring it to be of period $2\pi$. Since $f$ is odd, its Fourier series is of the form $\sum_{n\ge1}s_n\sin nx$. Now we just need to prove $s_n=\frac{1+(-1)^{n+1}}{2n}$. Indeed $$s_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin nxdx=\frac14\left(\int_0^\pi\sin nxdx+\int_0^{-\pi}\sin nx dx\right)\\=\frac{1}{4n}\left([\cos nx]_\pi^0+[\cos nx]_{-\pi}^0\right)=\frac{1+(-1)^{n+1}}{2n}.$$

Answer 2

For short: $$\sum_{k\geq 0}\frac{\sin((2k+1)x)}{2k+1}=\text{Im}\!\sum_{k\geq 0}\frac{(e^{ix})^{2k+1}}{2k+1}=\text{Im}\,\text{arctanh}(e^{ix})=\frac{1}{2}\text{Im}\log\left(\frac{1+e^{ix}}{1-e^{ix}}\right)=\frac{1}{2}\text{Arg}\left(i\cot\frac{x}{2}\right)=\frac{\pi}{4}. $$

There are a couple of subtleties, related to the evaluation of a power series at a point on the boundary of its disk of convergence, and to the determination of a complex logarithm, but I guess you can figure out the details easily.