Fourier series of $\sqrt[3]{\sin x}$ So I've done some experiments with how to add distortion to audio, and one of the methods I'm proposing is to take the cube root of an audio signal as a way to add overdrive. As the waveform that you're supposed to start with is a sine wave, I decided to tackle the problem of finding the magnitudes of the harmonics for $\sqrt[3]{\sin\left(2\pi fx\right)}$.
Based on what I've read about Fourier transforms, the coefficient of $\sin\left(nx\right)$ for $\sqrt[3]{\sin x}$is:
$\displaystyle a_n=\frac{1}{\pi}\int^{\pi}_{-\pi}\sin\left(nx\right)\sqrt[3]{\sin x}\,dx$
Since $\sqrt[3]{\sin x}$ contains only odd harmonics, I calculated the values of $a_n$ for odd n. Here's what I got out of the first seven terms:
$\displaystyle\left\{a_{1}, a_{3}, a_{5}, a_{7}, a_{9}, a_{11}, a_{13}\right\}=\frac{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^{3}}{4\pi^{2}\sqrt[3]{2}}\left\{1, \frac{1}{5}, \frac{1}{10}, \frac{7}{110}, \frac{1}{22}, \frac{13}{374}, \frac{26}{935}\right\}$
which leads to the following:
$a_{2n+1}=\displaystyle\left(\frac{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^{3}}{4\pi^{2}\sqrt[3]{2}}\right)\frac{\left(\frac{1}{3}\right)_{n}}{\left(\frac{5}{3}\right)_{n}}\qquad n\geqslant 0,\,n\in\mathbb{Z}$
where $\left(a\right)_{q}$ is the Pochhammer symbol.
Putting this all together, I get:
$\displaystyle\sqrt[3]{\sin\left(2\pi fx\right)}=\frac{2\pi^2\sqrt[3]{2}}{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^3}\sum_{n\geqslant 0,\,n\in\mathbb{Z}}\left(\frac{\left(\frac{1}{3}\right)_{n}}{\left(\frac{5}{3}\right)_{n}}\,\sin\left(\left(4n+2\right)\pi fx\right)\right)$
Is this correct?
EDIT: By "cube root", I am referring to the real solution, where $\sqrt[3]{-x}=-\sqrt[3]{x}$ for $x\geqslant 0$.
 A: The solution to the problem is straighforward, and I'm going to present it here with a minor generalisation.
Remember that the Fourier series to order $n$ of a function $f(x)$ is defined as
$$f_s(x,n) = \sum_{k=-n}^{n}c(k) e^{i k x}\tag{1a}$$
where the Fourier coefficients are defined as
$$c(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-i k x}\;dx\tag{1b}$$
It is more convenient to consider the related function
$$f(x) = |\sin(x)|^a\tag{2}$$
which is defined in the range $-\pi \lt x \lt \pi$.
Here we have introduced a paramter $a \gt -1$ which can be set to $\frac{1}{3}$ in the end.
The integral for the Fourier coefficient can be done in closed form:
$$c(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(x)|^a e^{-i k x}\;dx = 2^{-a} \cos \left(\frac{\pi  k}{2}\right) \binom{a}{\frac{k+a}{2}}\tag{3}$$
Notice that $c(k)$ vanishes for odd $k$.
The Fourier series then become
$$f_s(x,n) = c(0) + 2 \sum_{k=1}^{n} c(k) \cos(k x)\tag{4}$$
For the first few orders $n$ these are
$$
\left(
\begin{array}{cc}
 0 & 2^{-a} \binom{a}{\frac{a}{2}} \\
 2 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x) \\
 4 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x)+2^{1-a} \binom{a}{\frac{a+4}{2}} \cos (4 x) \\
 6 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x)+2^{1-a} \binom{a}{\frac{a+4}{2}} \cos (4 x)\\
&-2^{1-a} \binom{a}{\frac{a+6}{2}} \cos (6 x) \\
\end{array}
\right)
\tag{5}$$
The approximations are shown in the graph

Notice that for odd $n \ge 1$ we have $f_s(x,n) = f_s(n-1,x)$
We can also let $n$ go to infinity in $(4)$ which results in a representation of our function in terms of a hypergeometric function:
$$\begin{align} |\sin(x)|^{a}  =2^{-a} \binom{a} {\frac{a}{2}} \left(-1+\\2 \Re\left(\, _2F_1\left(1,-\frac{a}{2};\frac{a}{2}+1;e^{-2 i x}\right)\right)\right)
\end{align}\tag{5}$$
