Finding an inverse for Clausen Functions

Question

General

Question: What does an inverse function for the Clausen Function look like?

Background: I'm doing some math with infinite series of trigonometric functions. I keep coming back to simplification in the form of clusen functions. The question then arose how to rearrange equations with them according to their argument or what their inverse function (as an equation) would look like, e.g. in a closed form or a series.

Clarification

When I talk about Clausen functions, I mean the functions defined by the series expansions:

$$ \begin{align*} S_{k}\left( x \right) &= \sum\limits_{n = 1}^{\infty}\left[ \frac{\sin\left( n \cdot x \right)}{n^{k}} \right]\\ C_{k}\left( x \right) &= \sum\limits_{n = 1}^{\infty}\left[ \frac{\cos\left( n \cdot x \right)}{n^{k}} \right]\\ \end{align*} $$

The following applies: $\left\{ x,\, k \right\} \in \mathbb{C}$

The whole thing then has the relation to the polylogarithms:

$$ \begin{align*} S_{k}\left( x \right) &= \frac{1}{2} \cdot \left[ \operatorname{Li}_{k}\left( e^{-x \cdot i} \right) - \operatorname{Li}_{k}\left( e^{x \cdot i} \right) \right] \cdot i\\ C_{k}\left( x \right) &= \frac{1}{2} \cdot \left[ \operatorname{Li}_{k}\left( e^{-x \cdot i} \right) + \operatorname{Li}_{k}\left( e^{x \cdot i} \right) \right]\\ \end{align*} $$

My Attempts

$1$ Research

Clausen functions aren't very common compared to other special functions like hypergeometric functions, so I've had trouble finding them. So I just wanted to work with a relation that contains better known functions, which here are the polylogarithms. The best thing I found for this was a question on this forum Inverse of the polylogarithm. But that's where it stops again.

$2$ Polynomial Clusters

There are some relations given by: $$ \begin{align} \operatorname{S_{1}}\left( \theta \right) &= \frac{1}{2} \cdot \pi - \frac{1}{2} \cdot \theta\\ \operatorname{S_{3}}\left( \theta \right) &= \frac{1}{6} \cdot \pi^{2} \cdot \theta - \frac{1}{4} \cdot \pi \cdot \theta^{2} + \frac{1}{12} \cdot \theta^{3}\\ \operatorname{S_{5}}\left( \theta \right) &= \frac{1}{90} \cdot \pi^{4} \cdot \theta - \frac{1}{36} \cdot \pi^{2} \cdot \theta^{3} + \frac{1}{48} \cdot \pi \cdot \theta^{4} - \frac{1}{240} \cdot \theta^{5}\\ &\dots\\ \operatorname{C_{2}}\left( \theta \right) &= \frac{1}{6} \cdot \pi^{2} - \frac{1}{2} \cdot \pi \cdot \theta + \frac{1}{4} \cdot \theta^{2}\\ \operatorname{C_{4}}\left( \theta \right) &= \frac{1}{90} \cdot \pi^{4} \cdot \theta - \frac{1}{12} \cdot \pi^{2} \cdot \theta^{2} + \frac{1}{12} \cdot \pi \cdot \theta^{3} - \frac{1}{48} \cdot \theta^{4}\\ &\dots\\ \end{align} $$

These polynomials are theoretically solvable, but these relations are only special cases that only exist for natural indices. So that's a problem here too.

$3$ Fourier Series

Clausen functions can e.g. be used to regulate intuitively convergent Fourier Series. There are various formulas for this relationship that could perhaps be used here, but I don't know how.

$4$ Examples

I can invert some special cases by hand, e.g. these Polynomials form "Polynomial Clusters" (at least some of them). But that doesn't really get you any further to a general form now... Not even for $k \in \mathbb{N}$.

$5$ Series Reversion

Series Reversion failed 'cause of constant term.

EDIT:

It work's for a few series... But i don't know enogh.

Answer 1

Not A Complete General Answer But A Patial Answer

With a while more research I have a publication on Clausen functions and formal power series expansions for Clausen functions. Namely On the closed form of Clausen functions by Slobodan B. Tričković and Miomir S. Miomir S. Stanković (page $3$).

For $\alpha > 0$, the authors prove the formulas $[9]$ $$ \begin{align*} S_{\alpha}\left( x \right) &= \sum\limits_{n = 1}^{\infty}\left[ \frac{\sin\left( n \cdot x \right)}{n^{\alpha}} \right] = \frac{\pi \cdot x^{\alpha - 1}}{2 \cdot \Gamma\left( \alpha \right) \cdot \sin\left( \frac{\alpha}{2} \right)} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\left( -1 \right)^{k} \cdot \zeta\left( \alpha - 2 \cdot k - 1 \right) \cdot x^{2 \cdot k + 1}}{\left( 2 \cdot k + 1 \right)!} \right]\\ C_{\alpha}\left( x \right) &= \sum\limits_{n = 1}^{\infty}\left[ \frac{\cos\left( n \cdot x \right)}{n^{\alpha}} \right] = \frac{\pi \cdot x^{\alpha - 1}}{2 \cdot \Gamma\left( \alpha \right) \cdot \cos\left( \frac{\alpha}{2} \right)} + \sum\limits_{k = 0}^{\infty}\left[ \frac{\left( -1 \right)^{k} \cdot \zeta\left( \alpha - 2 \cdot k \right) \cdot x^{2 \cdot k}}{\left( 2 \cdot k \right)!} \right]\\ \end{align*} \tag{2.3}$$ where $\zeta\left( s \right)$ is Riemann’s zeta function [and $\Gamma\left( s \right)$ is the complete gamma function].

Note: I rewrote the formula minimally in the sense of my interpretation, since I use different parentheses.

We can apply the Series Reversion Theory to the power series expansion of $S_{\alpha}$ and thus form an inverse series expansion, but for this we first have to further restrict $\alpha$ to natural numbers larger than $1$ ($\alpha \in \mathbb{N} \setminus \left\{ 1 \right\}$), but then we get: $$ \begin{align*} S_{\alpha}^{-1}\left( x \right) &= \left( \frac{\pi}{2 \cdot \Gamma\left( \alpha \right) \cdot \sin\left( \frac{\alpha}{2} \right)} + A_{\left( k \in \mathbb{N}_{0} \setminus \left\{ \frac{\alpha - 2}{2} \right\} \right) + 1} \right) \cdot x^{\alpha - 1} + \sum\limits_{k \in \mathbb{N}_{0} \setminus \left\{ \frac{\alpha - 2}{2} \right\}}^{\infty}\left[ a_{2 \cdot k + 1} \cdot x^{2 \cdot k + 1} \right]\\ \end{align*} $$ where $a_{k} = A_{k} \cdot \frac{1}{k \cdot A_{1}^{k}} \cdot \sum\limits_{s,\, t,\, u,\, \dots}\left[ \left( -1 \right)^{s + t + u + \dots} \cdot \frac{n \cdot \left( n + 1 \right) \cdots \left( n - 1 + s + t + u + \dots \right)}{s! \cdot t! \cdot u! \cdots } \cdot \left( \frac{A_{2}}{A_{1}} \right) \cdot \left( \frac{A_{3}}{A_{1}} \right) \cdots \right] \wedge s + 2 \cdot t + 3 \cdot u + \dots = k - 1$ and $A_{k}$ is the $k$th coefficient of the power series of $S_{\alpha}$. (this formula was found by Morse and Feshbach)

But that's still not really satisfying. As it is only true under very definite and severe restrictions. But it's something...