请问下面这个函数的傅里叶展开式怎么展开啊？

$\displaystyle \sqrt{1-\cos x}=\sqrt{2\sin^2\frac{x}{2}}\\{}\\ =\left\{ \begin{aligned} -\sqrt{2}\sin\frac{x}{2},\ -\pi\leqslant x<0\\ \sqrt{2}\sin\frac{x}{2},\ \ \ \ 0\leqslant x\leqslant\pi \end{aligned} \right.\\{}\\ a_0=\frac{1}{\pi}\int_{-\pi}^0\left(-\sqrt{2}\sin\frac{x}{2}\right)dx+\frac{1}{\pi}\int_0^\pi\sqrt{2}\sin\frac{x}{2}dx\\{}\\ =\frac{2\sqrt{2}}{\pi}\int_0^\pi\sin\frac{x}{2}dx=\frac{4\sqrt{2}}{\pi},\\{}\\ a_n=\frac{1}{\pi}\int_{-\pi}^0\left(-\sqrt{2}\sin\frac{x}{2}\right)\cos nxdx+\frac{1}{n}\int_0^\pi\sqrt{2}\sin\frac{x}{2}\cos nxdx\\{}\\ =\frac{2\sqrt{2}}{\pi}\int_0^\pi\sin\frac{x}{2}\cos nxdx\\{}\\ =\frac{\sqrt{2}}{\pi}\int_0^\pi\left[\sin\left(\frac{1}{2}+n\right)x+\sin\left(\frac{1}{2}-n\right)x\right]dx\\{}\\ =-\frac{4\sqrt{2}}{\pi(4n^2-1)}\ \ (n=1,2,\cdots),\\{}\\ b_n=\frac{1}{\pi}\int_{-\pi}^0\left(-\sqrt{2}\sin\frac{x}{2}\right)\sin nxdx+\frac{1}{\pi}\int_{-\pi}^0\left(\sqrt{2}\sin\frac{x}{2}\right)\sin nxdx\\{}\\ =\frac{\sqrt{2}}{\pi}\int_{-\pi}^0\sin\frac{-x}{2}\sin(-nx)d(-x)+\frac{\sqrt{2}}{\pi}\int_0^\pi\sin\frac{x}{2}\sin nxdx\\{}\\ =-\frac{\sqrt{2}}{\pi}\int_0^\pi\sin\frac{x}{2}\sin nxdx+\frac{\sqrt{2}}{\pi}\int_0^\pi\sin\frac{x}{2}\sin nxdx\\{}\\ =0\\{}\\ -------\\{}\\ \mathrm{in\ the\ space\ of\ the}\ (-\pi,\pi)\\{}\\ f(x)=\sqrt{1-\cos x}=\frac{2\sqrt{2}}{\pi}-\frac{4\sqrt{2}}{\pi}\sum_{n=1}^\infty\frac{1}{4n^2-1}\cos nx\\{}\\ -------\\{}\\ \mathrm{time\ of\ the}\ x=\pi\ \mathrm{or}\ x=-\pi\\{}\\ \frac{f(\pi-0)+f(-\pi+0)}{2}=\frac{\sqrt{2}+\sqrt{2}}{2}=\sqrt{2},\\{}\\ \mathrm{so\ in\ the}\ [-\pi,\pi]\\{}\\ f(x)=\sqrt{1-\cos x}=\frac{2\sqrt{2}}{\pi}-\frac{4\sqrt{2}}{\pi}\sum_{n=1}^\infty\frac{1}{4n^2-1}\cos nx$