Calculating the fourier series of $f(t)=|t|$

Question

calculate the Fourier series of the $2\pi$-periodic continuation of

$$f(t):=|t|, \quad t\in[-\pi,\pi)\tag{1}$$

We know that

$$f(t)=\sum_{k=-N}^N c_k\cdot e^{ikt}\quad \&\quad c_k=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-ikt}dt\tag{2}$$

So let's calcualte the $c_k$.

$$c_k=\frac{1}{2\pi}\int_{-pi}^\pi|t|\cdot e^{-ikt}dt=\frac{2}{2\pi}\int_0^\pi t\cdot \cos(tk) dt=\frac{1}{\pi}[t\cdot\sin(tk)]_0^\pi+\frac{1}{\pi}[\cos(tk)]_0^\pi=0\frac{1}{\pi}(-1-1)=\frac{-2}{\pi}\tag{3}$$

Whereas we used the identity $\int_{-a}^a|t|e^{-ikt}dt=2\int_0^a t\cos(kt)dt$ Explanation (wheres the sub $t\to -t$ was used in the 2nd step)

So we get

$$f(t)=\sum_{k=-\infty}^\infty \frac{-2}{\pi}e^{ikt}\tag{4}$$

Sadly I don't have any solutions. Is that correct?

Answer 1

Your integration is incorrect. Note that the anti-derivative of $\cos(kt)$ is $\frac{\sin(kt)}{k}$, so we have

\begin{align} \int t\cos(kt) dt &= t \color{red}{\frac{\sin(kt)}{k}} - \int\frac{\sin(kt)}{k}dt \\ &= t\color{red}{\frac{\sin(kt)}{k}} + \color{red}{\frac{\cos(kt)}{k^2}} \end{align}

Inserting the limits gives

$$ c_k = \frac{1}{\pi}\frac{\cos(k\pi)-1}{k^2} = \begin{cases} 0, && k \text{ even} \\ \dfrac{-2}{\pi k^2}, && k \text{ odd} \end{cases} $$

Answer 2

The right side of 2nd step of your substitution might be better expanded, assuming $a \ge 0$, as

$ \int_{-a}^0 |t| (\cos(st) + i \sin(st)) \ dt \\ = \int_{-a}^0 (-t) (\cos(st) + i \sin(st)) \ dt \ [ \ \because |t| = -t, t \le 0 \ ] \\ = \int_0^a t (\cos(-st) + i \sin(-st)) \ dt \ [ \ \text{substituting} \ t \to -t \ ] \\ = \int_0^a t (\cos(st) - i \sin(st)) \ dt $