I am trying to calculate the Fourier series of $u(t) = \sin^3(wt)$ in complex form. I am first rewriting $\sin^3(wt)$ using the following equations:
$$e^{ix} = \cos(x) + i\sin(x)$$
$$e^{-ix} = \cos(x) - i\sin(x)$$
$$\sin(x) = \frac{1}{2i}\left(e^{ix}-e^{-ix}\right)$$
$$\sin^3(x) = \frac{i}{8}\left(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}\right)$$
Resulting in the following formula for the Fourier-coefficients:
$$c_k = \frac{iw}{16\pi}\int_0^{\frac{2\pi}{w}}(e^{3iwt}-3e^{iwt}+3e^{-iwt}-e^{-3wt})e^{-iwkt}dt$$
Computed, this gives
$$c_k = \frac{1}{16\pi}\left(\frac{e^{2i\pi(3-k)}}{3-k} - 3\frac{e^{2i\pi(1-k)}}{1-k} - 3\frac{e^{2i\pi(1+k)}}{1+k} -\frac{e^{2i\pi(3+k)}}{3+k} - \frac{1}{3-k} + \frac{3}{1-k} + \frac{3}{1+k} + \frac{1}{3+k}\right)$$
Since $k \in \Bbb{Z}$, the exponential functions all evaluate to $1$ (because $e^{i2\pi k}$ is $1$ for integers $k$) and summed up with the other four terms, calculate to $0$.
I do not think I messed up because my calculator (TI Nspire CX CAS) gives me the same coefficients, except for splitting it up into $\cos(x) + i\sin(x)$ and stopping at the last step because it does not know $k \in \Bbb{Z}$. But it cant be right either because $\sin^3(wt) \neq 0$.
The only mistake I can think of is that I am using the formula wrong. I am using the following definition:
$$c_k = \frac 1 {T_1} \int_0^{T_1} u(t)e^{-ikw_1t} \, dt$$
$$u(t) = \sum_{-\infty}^\infty c_ke^{ikw_1t}$$
Does anyone know why my coefficients are all $0$?
Your closed form for $c_k$ has division by $k-1,k+1,k-3,k+3$. So it doesn't work when $k=\pm1, \pm 3$. Another way of seeing this is that:
$$\frac{1}{T}\int_{0}^{T} e^{ikwx}\,dx = \begin{cases}0&k\neq 0\\1&k=0\end{cases}$$
The key is that your formula for $\sin^3(x)$ is already in Fourier series form, so you get:
$$\sin^3(wt)=\frac{i}{8}e^{3iwt}-\frac{3i}{8}e^{iwt}+\frac{3i}{8}e^{-iwt}-\frac{i}{8}e^{-3iwt}$$