Some interesting integrals with dilogarithm Calculating without techniques involving the contour integration
$$a) \ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^2-(\operatorname{Li}_2(e^{i x}))^2}{e^{-i x}-e^{i x}}\textrm{d}x;$$
$$b) \ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^3-(\operatorname{Li}_2(e^{i x}))^3}{e^{-i x}-e^{i x}}\textrm{d}x.$$
I'm working now on such a method. What would your real method inspiration be here?
Supplementary question: Calculate
$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^4-(\operatorname{Li}_2(e^{i x}))^4}{e^{-i x}-e^{i x}}\textrm{d}x.$$
Moreover, may we hope for a generalization of the type below?
$$ I(n)=\int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^n-(\operatorname{Li}_2(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x.$$
Preparing another two generalizations:
$$ i) \ J(n,m)=\int_0^{2\pi} \frac{(\operatorname{Li}_m(e^{-i x}))^n-(\operatorname{Li}_m(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x;$$
$$ ii) \ K(n)=\int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\cdots \operatorname{Li}_n(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\cdots \operatorname{Li}_n(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x.$$
 A: $ \large \text{ Hooray!!!}$ The closed-form of the integral $a)$ is impressive. According to my calculations, 
$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^2-(\operatorname{Li}_2(e^{i x}))^2}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^5}{48}.$$
Including also the trivial case, $n=1$,
$$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})-\operatorname{Li}_2(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^3}{4}.$$
$ \large \text{ Second Hooray!!!}$
$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^3-(\operatorname{Li}_2(e^{i x}))^3}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^7}{192}.$$
$ \large \text{Third Hooray!!!}$
I think I have found a first generalization!
$$ I(n)=\int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^n-(\operatorname{Li}_2(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^{2n+1}}{6^n}\left(1-\left(-\frac{1}{2}\right)^n\right).$$
$ \large \text{Fourth Hooray!!!}$
Guess what?! I'm also done with the generalization $J(n,m)$
$$\ J(n,m)=\int_0^{2\pi} \frac{(\operatorname{Li}_m(e^{-i x}))^n-(\operatorname{Li}_m(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x=\pi(\zeta(m)^n-((2^{1-m}-1)\zeta(m))^n).$$
$ \large \text{Fifth Hooray!!!}$
I computed $2$ cases of the generalization in $K(n)$ and I approach the solution of the generalization. So, 
$$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{5}{48}\pi^3\zeta(3);$$
$$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\operatorname{Li}_4(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\operatorname{Li}_4(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{17}{6912}\pi^7 \zeta(3).$$
$ \large \text{Sixth Hooray!!!}$
Looks like I have been lucky today! Let me put the last generalization I just proved in a nice form
$$K(n)=\int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\cdots \operatorname{Li}_n(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\cdots \operatorname{Li}_n(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x$$
$$=\pi \left(\zeta(2)\zeta(3)\cdots \zeta(n)+(-1)^{n-1} \eta(2)\eta(3)\cdots\eta(n))\right).$$
Extra information:
https://en.wikipedia.org/wiki/Riemann_zeta_function
https://en.wikipedia.org/wiki/Dirichlet_eta_function 
https://en.wikipedia.org/wiki/Polylogarithm
A: Getting an idea by going trough an example
Consider the Integral 
$$
I=\int_0^{2\pi}dx\frac{\text{Li}^2_2(e^{i x})-\text{Li}^2_2(e^{-i x})}{2 i \sin(x)}
$$
using the series representation for the dilogarithm this can be rewritten as
$$
I=\int_0^{2\pi}dx\sum_{n,m>0}\frac{1}{n^2 m^2}\frac{\sin(m+n)x}{\sin(x)}
$$
exchange summation and integration and using the simple fact that $\int_0^{2\pi}dx \frac{\sin(lx)}{\sin(x)}=2 \pi$ for $l \in 2\mathbb{N}+1$ we get
$$
I=2  \pi s^{(2)}_2 =2 \pi\sum_{\substack{n,m>0 \\n+m=odd}}\frac{1}{n^2 m^2}
$$
The closed form solution to $s^{(2)}_2$ is pretty simple to obtain. Observe to fullfil the condition $n+m=odd$ either $n$ has to be odd and $m$ to be even, or vice versa. This means we have $2$ possible combinations of even and odd which yield a contribution to our sum.
$$
 s^{(2)}_2 =2\sum_{\substack{n>0,m\geq 0}}\frac{1}{(2n)^2 (2m+1)^2}=2\frac{\zeta(2)}{4}\frac{3\zeta(2)}{4}=\frac{3}{8}\zeta^2(2)
$$
the strategy for providing a closed form solution will follow the same arguments, except that we additonally need a combinatoric lemma proven in the appendix
The General Case
We now want to investigate
$$
I^{(r)}_n=\int_0^{2\pi}dx\frac{\text{Li}^n_r(e^{i x})-\text{Li}^n_r(e^{-i x})}{2 i \sin(x)}
$$
Going through the same procedure then in the motivating example we might show that

$$
I^{(r)}_n=2 \pi s^{(r)}_n
$$

This means we are interested in a family of Euler like sums, since
$$
s^{(r)}_n=\sum_{\substack{ k_i \geq 1, \\ \sum_{n\geq i \geq 1} k=odd }}\frac{1}{\prod_{ n \geq i\geq1}{k^r_i}}
$$
we now have to take care that we account for all possible partitions of the integers such that the constraint $\sum_{n\geq i \geq 1} k=odd$  is fulfilled. As shown in the Appendix, we have to choose $2l-1$numbers to be odd and $n-2l+1$ to be even. Each of this partitions contains $N_{l,n}=\binom{n}{2l-1}$ equivalent combinations.
This means that
$$
s^{(r)}_n=\sum_{l_{max}(n)\geq l\geq1}N_{l,n}\sum_{k_i\geq 1, K_i \geq 0}
\prod_{2l-1 \geq i\geq1}\frac{1}{{(2K_i+1)^r}}
\prod_{n- 2l+1 \geq i\geq1}\frac{1}{{(2k_i)^r}}
$$
using now the well known identity $\sum_{k\geq 0}(2m+1)^{-r}=(1-1/2^{-r})\zeta(r)$ we can carry out the infinite summations
$$
s^{(r)}_n=\sum_{l_{max}(n)\geq l\geq1}\frac{N_{l,n}}{2^{r(n-2l+1)}}(1-\frac{1}{2^{r}})^{2l-1}\zeta^n(r)=\sum_{l_{max}(n)\geq l\geq1}c_{l,n}\zeta(r)^n
$$
Furthermore the sum over coefficents can be done in closed form by virtue of the Binomial identiy:


$$
s^{(r)}_n=C_{n,r}\zeta(r)^n\,\,,\,\,C_{n,r}=\begin{cases}
\frac{1}{2}\left(1+\frac{2^{n-r}}{4^{r n/2}}(2^{r-1}-1)^n\right)\,\, \text{if} \,\,n \,\, even\\
\frac{1}{2}\left(1+\frac{2^{n-r}}{4^{r(n-1)/2}}(2^{r-1}-1)^n\right)\,\, \text{if} \,\,n \,\, odd \\
\end{cases} 
$$


Note that we get the sums with the constraint $\sum_{n\geq  i\geq1} k_i=even$ for free: 
$$
\bar{s}_n^{r}=\left(1-C_{n,r}\right)\zeta(r)^n
$$
It is also interesting to note, that $\lim_{n\rightarrow\infty}\frac{s^{(r)}_n}{\zeta(r)^n}=\frac{1}{2}$
which can be traced back to fact that for very large $n$ we have to choose roughly $n/2$ odd factors out ouf $\sum_{n\geq 1 i\geq1} k_i$ due to concentration of $N_{l,n}$ around $n/2$.
Last but not least a few examples:
\begin{align*}
 s^{(2)}_2=\frac{3}{8}\zeta^2(2)\,\, ,\,\,\bar{s}^{(2)}_2=\frac{5}{8}\zeta^2(2) \\
s^{(3)}_3=\frac{91}{128}\zeta^3(3)\,\, ,\,\,s^{(3)}_5=\frac{1267}{2048}\zeta^5(3), 
\end{align*}
Appendix: A small detour to combinatorics
Consider the sum of integers
$$
c_m=n_1+n_2+...+n_m
$$
how can we partion $c_m$ into odd and even elements, such that $c_m$ is odd?
Since the odd and even numbers furnish a representation of the group $\mathbb{Z}_2$ it follows trivially 
that we need always an odd number $2l-1$ of the $n_m$'s to be odd. For any fixed $l$ we then have 
$$
N_{l,m}=\binom{m}{2l-1}\,\, ,\,\, l \in
\begin{cases}
\{1,m/2\} \,\, \text{if} \,\,m \,\, \text{even}\\
\{1,\lceil m/2 \rceil\} \,\, \text{if} \,\,m \,\, \text{odd}\\
\end{cases}
$$
equivalent admissible partitions of $c_m$. 
