On $\int_0^{2\pi }\frac{\prod_{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod_{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx$ OP of this post evaluated a lot of remarkable polylog integrals (without proof), from which I conjecture a generalized one ($a_k\in \mathbb N$):
$$\int_0^{2 \pi } \frac{\prod _{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod _{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx=\pi  \left((-1)^{m-1} \prod _{k=1}^m \eta (a_k)+\prod _{k=1}^m \zeta (a_k)\right)$$
This passed numeric verification but I have no solution up to now. I'd like you to help me on this interesting problem. Thanks in advance!
 A: Let $f : [0, 2\pi] \to \mathbb{C}$ has the Fourier series $\sum_{n=-\infty}^{\infty} c_n e^{inx}$ such that
$$\sum_{n=-\infty}^{\infty} n \left| c_n \right| < \infty \tag{*}.$$
Then $f$ is continuous and the Fourier series of $f$ converges to $f$. Now
\begin{align*}
\int_{0}^{2\pi} \frac{f(x) - f(-x)}{e^{ix} - e^{-ix}} \, \mathrm{d}x
&= \sum_{n=-\infty}^{\infty} c_n \int_{0}^{2\pi} \frac{e^{inx} - e^{-inx}}{e^{ix} - e^{-ix}} \, \mathrm{d}x \\
&= \sum_{n=-\infty}^{\infty} c_n \int_{0}^{2\pi} \left( \sum_{k=0}^{n} e^{(2k-n+1)ix} \right) \, \mathrm{d}x \\
&= \sum_{n=-\infty}^{\infty} c_n (2 \pi \mathbf{1}_{\{n\text{ odd}\}}) \\
&= \pi \sum_{n=-\infty}^{\infty} c_n (1 - (-1)^n) \\
&= \pi (f(0) - f(\pi)).
\end{align*}
Returning to the original problem, let $a_1, \dots, a_m > 1$. Then we note that
$$x \mapsto \prod_{k=1}^{m} \operatorname{Li}_{a_k}(re^{ix})$$
satisfies the assumption $\text{(*)}$ for all $r \in (0, 1)$. (It is not clear  as to whether it satisfies the condition at $r=1$, so we adopt limiting argument.) So we have
$$ \int_{0}^{2\pi} \frac{\prod_{k=1}^{m} \operatorname{Li}_{a_k}(re^{ix}) - \prod_{k=1}^{m} \operatorname{Li}_{a_k}(re^{-ix})}{e^{ix} - e^{-ix}} \, \mathrm{d}x
= \pi \left( \prod_{k=1}^{m} \operatorname{Li}_{a_k}(r) - \prod_{k=1}^{m} \operatorname{Li}_{a_k}(-r) \right). $$
Now letting $r \uparrow 1$ shows that the desired identity holds.
