Express $f(x)=\sin{(x)}$, with $(0 < x< \pi )$ as an even function, $f(x+ 2\pi)=f(x)$
The topic is on Fourier Series.
I have the following so far:
Since $f(x)$ must be an even function, we must only calculate the $a_n$ values, $n=1,2,...$.
- $\displaystyle a_0= \frac{2}{\pi}\int \limits_{0}^{\pi}\sin{(x)}dx = \frac{4}{\pi}$
- $\displaystyle a_n = \frac{2}{\pi}\int \limits_{0}^\pi \sin{(x)}\cos{(\frac{n \pi x}{\pi})} = \frac{2}{\pi} \int \limits_{0}^\pi \sin{(x)}\cos{(nx)}dx$
I am, however, having difficulty integrating the above integral for $a_n$ using the method he wants us to use (The table method). Can someone please show me how to do this integral using the tables?
Once $a_n$ is calculated, I know that I will then have my final answer in the form
$$f(x)=\sin{(x)} \sim \frac{2}{\pi} + \sum_{n=1}^\infty a_n \cos{(nx)}$$
Actually, the table method is the method, when you do integration by parts several times. So here you can just do this and get the result. Lets denote $I = \int \limits _0 ^ \pi \sin x \cos nx \, dx$. Here is described the process of integration by parts (two times):
$I = \left. -\cos x \, \cos nx \, \right| _0 ^ \pi -\left. n \, \sin x \, \sin nx \, \right| _0 ^ \pi+ n^2 \int \limits _0 ^ \pi \sin x \, \cos nx \, dx = (-1)^n + 1 + n^2 \, I$.
So, $I = \frac{(-1)^n + 1}{1 - n^2}$.