integration - Closed formula for integral

Mathematica can evaluate

$$\int\limits_0^\infty \frac{ \ln^{p} {x} \sin^{q} {x}}{x^r}dx $$

for all $p,q,r \in \mathrm{N}$ when $q \geq r>c, c \equiv q-r \ (\textrm{mod} \ 2) $, but can't give general formula. Any references would be appreciated. Also, any hints on how to do $p=q=r=1$ case? All results are some combination of a fraction, Euler–Mascheroni and logarithms.

So, is there a formula in spirit of :

UPDATE

I changed the notation a bit here to match Raymond Manzoni's first reference: integral converges for $p\geq q$ :

1. for $p$ odd (even) and $q$ even(odd)

$$\int\limits_0^\infty \frac{\ln^{r}\! x \sin^{p} \! x }{x^q} \mathrm{d}x = \frac{(-1)^{r} (-1)^{\lfloor \frac{p-q}{2} \rfloor} }{ \Gamma(q) 2^p} \sum\limits_{i=0}^{r} \binom{r}{i} \frac{(-1)^{r-i} (r-i)!}{2^{r-i}} \sum\limits_{k=0}^{\lfloor \frac{p}{2} \rfloor - \text{mod}(2p-q,2)} \! \! \! \! \! \! \! \! \! \! 2(-1)^k \binom{p}{k}(p-2k)^{q-1} \sum\limits_{t=0}^{\lfloor \frac{r-i+1}{2} \rfloor} \frac{\text{Li}_{2t}(-1)\ln^{r-i+1-2t}\left(\frac{1}{(p-2k)^2}\right) }{(r-i+1-2t)!} \sum\limits_{\lambda \vdash i} \psi_0^{m_1}(q) \psi_1^{m_2}(q)\cdot \ldots\cdot \psi_{i-1}^{m_{l(\lambda)} }(q) \frac{\binom{i}{m_1,m_2,\ldots,m_{l(\lambda)}}}{m_1! m_2! \cdot \ldots\cdot ,m_{l(\lambda)}!} (-1)^{m_1+m_2+\ldots +m_{l(\lambda)}}$$

2. for $p$ odd (even) and $q$ odd(even)

$$\int\limits_0^\infty \frac{\ln^{r}\! x \sin^{p} \! x }{x^q} \mathrm{d}x = \frac{(-1)^{r} (-1)^{ \frac{2p-q + \text{mod}(p,2)}{2}} }{ \Gamma(q) 2^{p-1}} \sum\limits_{i=0}^{r} \binom{r}{i} (-1)^{r-i} (r-i)! \sum\limits_{k=1}^{ \frac{p +\text{mod}(p,2)}{2}} \! \! \! \! \! \! 2(-1)^k \binom{p}{\frac{p +\text{mod}(p,2)}{2}- k}(2k-\text{mod}(p,2))^{q-1} \sum\limits_{t=0}^{\lfloor \frac{r-i+1}{2} \rfloor} \frac{\text{Li}_{2t}(-1) \sum\limits_{m=0}^{\lfloor \frac{r-i-2t}{2} \rfloor} \binom{r-i-2t+1}{2m+1} \left( \frac{\pi}{2} \right)^{2m+1} \ln^{r-i-2t-2m} \left( \frac{1}{2k -\text{mod}(p,2) } \right) }{(r-i-2t+1)!} \sum\limits_{\lambda \vdash i} \psi_0^{m_1 }(q) \psi_1^{m_2 }(q)\cdot \ldots\cdot \psi_{i-1}^{m_{l(\lambda)} }(q) \frac{\binom{i}{m_1,m_2,\ldots,m_{l(\lambda)}}}{m_1! m_2! \cdot \ldots\cdot ,m_{l(\lambda)}!} (-1)^{m_1+m_2+\ldots +m_{l(\lambda)}}$$

where $\text{Li}$ is polylogarithm, $\psi$ is polygamma, and the last sum is over all partitions $\lambda = \left(1^{m_1} 2^{m_2} \ldots \right) $ and $l(\lambda)$ is partition length.

fell free to tidy up and improove if you want, I don't have the will right now.

Answer

Here is the proof and Mathematica code to play with. Took me some time to write all this. Thank you again Raymond Manzoni for that nice Edwards book(s) refference, couldn't have done it without it. It would be nice to see a proof for idenetities (37) and (38) in paper, one proof is in Edwards book but it's not too elegant, one way is to reduce it to hypergeometric identity http://functions.wolfram.com/07.31.03.0032.01 but I don't know how to prove that identity. Also the final expressions are equivalent to ones I wrote above without proof, but more elegant.

You can downolad it here : download link

and the .tex file here: download link

This code is also much nicer than the last one, you can download the Mathematica notebook here download link