Integral involving Clausen function ${\large\int}_0^{2\pi}\operatorname{Cl}_2(x)^2\,x^p\,dx$ Consider the Clausen function $\operatorname{Cl}_2(x)$ that can be defined for $0<x<2\pi$ in several equivalent ways:
$$\begin{align}\operatorname{Cl}_2(x)&=-\int_0^x\ln\left(2\sin\left(\tfrac t2\right)\right)dt\\&=\sum_{n=1}^\infty\frac{\sin\,(n x)}{n^2}\\&=\Im\operatorname{Li}_2\left(e^{i x}\right)\\&=i\left(\frac{\pi^2}6+\frac{x^2}4-\frac{\pi x}2-\operatorname{Li}_2\left(e^{i x}\right)\right).\end{align}\tag1$$
I'm interested in integrals of the form
$$I(p)=\int_0^{2\pi}\operatorname{Cl}_2(x)^2\,x^p\,dx.\tag2$$
I found that $$I(0)=\frac{\pi^5}{90}\tag3$$ and conjectured next several values based on numeric evidence:
$$I(1)\stackrel?=\frac{\pi^6}{90},\ \ I(2)\stackrel?=\frac{44\,\pi^7}{2835},\ \ I(3)\stackrel?=\frac{23\,\pi^8}{945}.\tag4$$
One might expect that $I(4)$ is a rational multiple of $\pi^9$, but apparently it is not (unless the denominator is huge).
I'm asking for your help in proving conjectured values $(4)$, finding a closed form of $I(4)$, and, if possible, a general formula for $I(p)$.
Update: Values $I(-1)$ and $I(-2)$ are also interesting.
 A: $I(1)$ can be evaluated in the following way also. One has to note that
$$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\begin{cases} 0 & n \neq m \\ \pi^2 & n=m \end{cases}$$
To prove the above, write
$$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\frac{1}{2}\int_0^{2\pi} x\cos((m-n)x) \,dx-\frac{1}{2}\int_0^{2\pi} x\cos((m+n)x)\,dx$$
and then use integration by parts.
Back to the original integral,
$$\begin{aligned}
\int_0^{2\pi} x\text{Cl}_2(x)^2\,dx &=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n^2}\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx \\
& =\sum_{n=1}^{\infty} \frac{\pi^2}{n^4} \\
& =\frac{\pi^6}{90}
\end{aligned}$$
A: We can use the identity 
$$\mathrm{cl}_2(x)^2 = \mathrm{cl}_2(2\pi -x)^2 $$
Hence we have 
$$I(5)=\int_0^{2\pi}\operatorname{Cl}_2(x)^2\,(2\pi -x)^5\,dx$$
Expand 
$$I(5) = 32 \pi^5I(0)-80 \pi^4 I(1)+80 \pi^3 I(2)-40 \pi^2 I(3)+10 \pi I(4)-I(5)$$
Solve for $I(5)$
$$I(5) = 16\pi^5I(0)-40\pi^4 I(1)+40\pi^3 I(2)-20\pi^2 I(3)+5\pi I(4)$$
This simplifies to 
$$I(5) = \frac{544}{2835}π^{10}-960 π^2 \zeta(3)\zeta(5)+480 π^2 \zeta(6,2)$$
This method can be generalized for $I(2q+1)$ given we know all the previous terms
$$I(2q+1) = \int^{2\pi}_0 (2\pi-x)^{2q+1} \mathrm{cl}_2(x)^2\,dx$$
Use the binomial theorem 
$$I(2q+1) = \sum _{ k=1 }^{ 2q+1 }{ { \left( -1 \right)  }^{ k }\binom{2q+1}{k} { 2\pi }^{ 2q+1-k } }\int^{2\pi}_0 x^k \mathrm{cl}_2(x)^2\,dx$$
$$I(2q+1) = \frac{1}{2}\sum _{ k=1 }^{ 2q }{ { \left( -1 \right)  }^{ k }\binom{2q+1}{k} { 2\pi }^{ 2q+1-k } }I(k)$$
A: I can confirm $I(1)$:
Using the series expansion for Clausen's function we have 
$$
I(p)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\int_0^{2 \pi}x^p\frac{\sin(n x)}{n^2}\frac{\sin(m x)}{m^2}\,\mathrm dx
$$
now setting $p=1$.
Furthermore we may use the orthogonality of sine $\int^{2 \pi}_{0}{\sin(n x)\sin(m x)}\,\mathrm dx=\pi\delta_{mn}$ and integrate by parts to get
$$
I(1)=\underbrace{\sum_{m,n=1,n\neq m}^{\infty}\frac{1}{2 (m-n)^2}-\frac{1}{2 (m+n)^2}-\frac{2 m n}{(m^2 - n^2)^2}}_{=0}+\pi^2\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^4}}_{=\zeta(4)}=\frac{\pi^6}{90}
$$
I'm checking if there is a pattern, especially if it holds that the doublesum always cancel. But at least for the first few $p$'st this seems likely, nevertheless the algebra gets a little bit messy but i will try to go further tomorrow.
Edit: Playing the same game for $p=0$ we end up with
$$
I(0)=\pi\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^4}}_{=\zeta(4)}=\frac{\pi^5}{90}
$$
The double sum cancels immediatly in this case, so i confirmed also $I(0)$
Edit2: 
Doing some careful algebra,following the same approach as above, we may find that 
$$
I(2)=\underbrace{4 \pi \sum_{n=1}^{\infty}{\frac{8 \pi ^2 n^3-3  n }{24 n^7}}}_{=\frac{\pi^7}{70}}+8 \pi\underbrace{\sum_{m,n=1,n\neq m}^{\infty}\frac{  1}{m n\left(m^2-n^2\right)^2}}_{J}=8 \pi\underbrace{\sum_{m}^{\infty}\sum_{n=0}^{m-1}\frac{  1}{m n\left(m^2-n^2\right)^2}+\sum_{m}^{\infty}\sum_{m+1}^{\infty}\frac{  1}{m n\left(m^2-n^2\right)^2}}_{J_1+J_2}+\frac{\pi^7}{70}
$$
Maybe someone with more experience in calculating sums like this can take it home from here :)
Edit3:
It turns out that the same off-diagonal sum will give us the result for $p=3$ if we multiply by $24\pi^2$ instead of $8\pi$. So if we have solved $p=2$ we get $I(3)$ for free :)
Nevertheless, for $p>3$ the off-diagonal contribution starts to get messier and yields additionals contribution which may well be not proportional to $\pi^9$
