Calculate the Fourier series
$$ f(x) = \left\{\begin{aligned} & -6, && -\pi < x \le 0\\ & 6, && 0 < t < \pi \end{aligned} \right.$$
My Work
$a_n = \dfrac{1}{L}\int^L_{-L}f(t) \cos\left( \dfrac{n \pi t}{L} \right) dt$
$b_n = \dfrac{1}{L}\int^L_{-L}f(t) \sin\left( \dfrac{n \pi t}{L} \right) dt$
I got $a_n = 0 \forall n \in \mathbb{N}$.
$b_n = \dfrac{1}{\pi} \left[ \int^0_{-\pi} -6\sin(nt) dt + \int^\pi_0 6\sin(nt) dt \right]$
$= \dfrac{6}{\pi n} - \dfrac{6}{\pi n} \cdot \cos(-\pi n) - \dfrac{6}{\pi n} \cdot \cos(\pi n) + \dfrac{6}{\pi n}$
We can see that $\cos(-\pi n) = (-1)^n$ and $\cos(\pi n) = (-1)^n$.
Therefore, we can simplify:
$= \dfrac{12}{\pi n} - \dfrac{6}{\pi n} \cdot (-1)^n - \dfrac{6}{\pi n}(-1)^n$
$= \dfrac{12}{\pi n} \left[ 1 - (-1)^n \right]$
$\left[ 1 - (-1)^n \right] = 0$ when $n$ is even.
Therefore, our Fourier series representation of $f(t)$ is
$f(t) = \dfrac{12}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\left[ 1 - (-1)^n \right] \cdot \sin(nt)}{n}$
$= \dfrac{12}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\left[ \sin(nt) - (-1)^n \cdot \sin(nt) \right]}{n}$
I would greatly appreciate it if people could please take the time to review my calculations and provide feedback.