For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Deduce:
$$\forall x \in ]-2\pi, 2\pi[ \backslash \{0\}, \quad \frac{ix}{e^{ix}-1} = 1 - \frac{ix}{2} - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} \cdot x^{2k}.$$