Here we present an example of Fourier series:

Example:

Let $f\colon (-\pi,\pi) \to \Reals$ be the ``identity'' function, defined by $$f(x)=x, \text{ for all }x\in (-\pi,\pi).$$ We will compute the Fourier coefficients for this function. Notice that $\cos(nx)$ is an even function, while $f$ and $\sin(nx)$ are odd functions. \begin{eqnarray*} a_0^f & =& \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{1}{2\pi}\int_{-\pi}^{\pi} x dx= 0\\ a_n^f &=& \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx= \frac{1}{\pi}\int_{-\pi}^{\pi}x \cos(nx)dx = 0\\ b_n^f &=& \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx=\frac{1}{\pi}\int_{-\pi}^{\pi} x \sin(nx)dx =\\ &=& \frac{2}{\pi}\int_{0}^{\pi} x\sin(nx) dx= \frac{2}{\pi}\left( \left[-\frac{x\cos(nx)}{n}\right]_0^{\pi}+\left[\frac{\sin(nx)}{n^2}\right]_0^{\pi}= \right)=(-1)^{n+1}\frac{2}{n} \end{eqnarray*} Notice that $a_0^f,a_n^f$ are $0$ because $x$ and $x \cos(nx)$ are odd functions. Hence the Fourier series for $f(x)=x$ is:

\begin{eqnarray*} f(x)=x &=& a_0^f + \sum_{n=1}^{\infty}(a_n^f\cos(nx)+b_n^f\sin(nx)) =\\ &=& \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n} \sin(nx), \quad \forall x\in (-\pi,\pi) \end{eqnarray*} For an application of this Fourier series, see value of the Riemann zeta function at $s=2$