I was going to calculate the Fourier series of
$f(t)=t|t|\sin (n\pi t)$, on the interval $[-1,1]$, with period $T=2$ but the modulus of t got me thinking.
I did
\begin{equation} a_n=\int_{-1}^1t|t|\sin n\pi tdt= \int_{-1}^1t^2\sin n\pi tdt=\frac{4}{n^2\pi^2}\big(-1\big)^n \end{equation}
Then I did
\begin{equation} b_0=\int_{-1}^1t|t|\sin n\pi tdt= \int_{-1}^1t^2\sin n\pi tdt=0 \end{equation}
and finally
\begin{equation} a_0=\int_{-1}^1t|t|\sin n\pi tdt=\int_{-1}^1t|t|\cdot 0dt=0 \end{equation}
and got therefore:
\begin{equation} \mathscr{F}\bigg[f(t)=t|t|\sin (n\pi t)\bigg]_{-1}^1=4\sum_{n=1}^\infty \frac{\big(-1\big)^n}{n^2\pi^2} \end{equation}
But I fear that the simplification $t|t|=t^2$ is not correct. Any ideas ?
Thanks