I am told to find the Fourier series of $f(t)=t^3$ on the interval $-1<t<1$ where $T=2$, thus $L=1$ and $\omega = \pi$.
Since $f(t)$ is odd, I need only find the Fourier Sine Series which is $f(t)=\sum^{\infty}_{n=1} b_n \sin(n\omega t)$ where $b_n =\frac{2}{L}\int^{L}_{0} f(t)\sin(n\omega t)$
Using integration by parts (confirmed using MATLAB and Wolfram) I found $b_n$ to be:
$$2\cos(n\pi)\bigg[\frac{-1}{n\pi}+\frac{6}{n^3\pi ^3}\bigg]$$ $$=\frac{2(-1)^{n+1}}{n\pi}+\frac{12(-1)^n}{n^3\pi^3}$$
Thus, the Fourier series of $f(t)$ is:
$$\sum^{\infty}_{n=1} \sin(\pi nt)\bigg[\frac{2(-1)^{n+1}}{n\pi}+\frac{12(-1)^n}{n^3\pi^3}\bigg]$$
However, when I use Wolfram to find the Fourier Sine Series to check, it shows a similar but different answer (https://www.wolframalpha.com/input/?i=FourierSineSeries%5Bt%5E3,t,5%5D)
For example, for $n=1$, my result would obtain $f(t)=\sin(t)[\frac{2}{\pi}-\frac{12}{\pi^3}]$ whereas Wolfram says it's $2(\pi^2-6)\sin(t)$.
Not sure where I have gone wrong.