I need to find Fourier series for function $$f(x)=\begin{cases} 0 & x \in(-\pi,0] \\ x & x \in(0, \pi) \end{cases}$$
I know the formula look like
$$\frac{a_0}2+\sum_{n=1}^{\infty}a_ncos(nx)+b_nsin(nx),$$ where
$$a_0=\frac1\pi\int_{-\pi}^{\pi} \;f(x)dx\;=\frac1{\pi}\int_{-\pi}^{0} \;0dx\;+\frac1{\pi}\int_{0}^{\pi} \;xdx\;=\frac{\pi}2$$
$$a_n=\frac1\pi\int_{-\pi}^{\pi} \;f(x)cos(nx)dx\;=\frac1{\pi}\int_{-\pi}^{0} \;0*cos(nx)dx\;+\frac1{\pi}\int_{0}^{\pi} \;x*cos(nx)dx\;=\frac1{\pi}\int_{0}^{\pi} \;x*cos(nx)dx\;=\frac{\pi}2sin(\pi n)+\frac1{n^2}cos(\pi n)-\frac1{n^2}$$
$$b_n=\frac1\pi\int_{-\pi}^{\pi} \;f(x)sin(nx)dx\;=\frac1{\pi}\int_{-\pi}^{0} \;0*sin(nx)dx\;+\frac1{\pi}\int_{0}^{\pi} \;x*sin(nx)dx\;=\frac1{\pi}\int_{0}^{\pi} \;x*sin(nx)dx\;=-\frac{\pi}ncos(\pi n)+\frac1{n^2}sin(\pi n)$$
I've made everythinh but the result is not what I expected. Did I make a mistake somewhere? Need help
I will write the integration for $a_n$, $b_n$ can be solved using the same technique: $$ \int_0^{\pi}x\cos(nx) = \frac{x}{n}\sin(nx) - \frac{1}{n}\int_0^{\pi}\sin(nx) = \frac{x}{n}\sin(nx)|_0^{\pi} - (-\frac{1}{n^2}\cos(nx))|_0^{\pi} =\\ \frac{\pi}{n}\sin(n\pi) + \frac{1}{n^2}\cos(n\pi) - \frac{1}{n^2} = \frac{n\pi\sin(n\pi) + \cos(n\pi) - 1}{n^2}$$ Decompose the first integral using integration by parts and from there it's just calculations.