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 $p\geq q$ :


*

*for $p$ odd (even) and $q$ even(odd)


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


*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.
 A: 
Lemma: If $\,q\,$ is a real value then : 
  \begin{align}
\tag{1} C(q):=\int_0^\infty \frac{\cos(x)}{x^q}\;dx=\sin\left(\frac{\pi}2q\right)\,\Gamma(1-q),\quad 0<q<1\\
\tag{2} S(q):=\int_0^\infty \frac{\sin(x)}{x^q}\;dx=\cos\left(\frac{\pi}2q\right)\,\Gamma(1-q),\quad 0<q<2\\
\end{align}

(ref DLMF $5.9.6$ and $5.9.7$ or table $150$ of de Haan's $1867$ "Nouvelles tables d'intégrales définies"
or use the substitution $\,x:=it\,$ and analytic continuation : $\;\displaystyle\Gamma(1-q)=\int_0^\infty \frac{e^{-t}}{t^q}\,dt=-i\int_0^{i\infty} \frac{e^{ix}}{x^q}i^q\,dx=-i\;e^{\large{\frac {\pi}2q\,i}}\int_0^{i\infty} \frac{e^{ix}}{x^q}\,dx\; $)
Expanding powers of $\;\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ with the binomial theorem gives us :
\begin{align}
\tag{3}\sin(x)^{2n}&=\frac 2{2^{2n}}\sum_{k=0}^n (-1)^{n+k}\binom{2n}{k}\cos\left((2n-2k)x\right)\\
\tag{4}\sin(x)^{2n+1}&=\frac 2{2^{2n+1}}\sum_{k=0}^n (-1)^{n+k}\binom{2n+1}{k}\sin\left((2n+1-2k)x\right)\\
\end{align}
allowing us to generalize $(2)$ to $\;\displaystyle F_p(q):=\int_0^\infty \frac{\sin(x)^p}{x^q}\;dx\ $ with $p$ a positive integer :
\begin{align}
\text{for}\ p=2n&:&F_{p}(q)&:=\frac 2{2^p}\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}\int_0^\infty\frac{\cos\left((p-2k)x\right)}{x^q}\;dx\\
\tag{5}&& F_{p}(q)&=\frac 2{2^p}\sin\left(\frac{\pi}2q\right)\,\Gamma(1-q)\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}(p-2k)^{q-1}\\
\\
\text{for}\ p=2n+1&:&F_{p}(q)&:=\frac 2{2^p}\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}\int_0^\infty\frac{\sin\left((p-2k)x\right)}{x^q}\;dx\\
\tag{6}&& F_{p}(q)&=\frac 2{2^p}\cos\left(\frac{\pi}2q\right)\,\Gamma(1-q)\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}(p-2k)^{q-1}\\
\end{align}
But these are simply the imaginary (for $p$ even) or real (for $p$ odd) part of :
$$\tag{7}G_{p}(q)=\frac 2{2^p}i^q\,\Gamma(1-q)\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}(p-2k)^{q-1},\quad n:=\left\lfloor\frac p2\right\rfloor$$
To evaluate the wished $\;\displaystyle I(p,q,r):=\int_0^\infty \frac{\ln(x)^r\sin(x)^p}{x^q}\;dx\;$ for nonnegative integer values $r$ and since $\;\displaystyle\frac d{dq} x^{-q}=-\ln(x)\,x^{-q}\;$ we need only to use :
$$
\tag{8}I(p,q,r)=\int_0^\infty \frac{\ln(x)^r\sin(x)^p}{x^q}\;dx=\left(-\frac d{dq}\right)^{\large r} \int_0^\infty \frac{\sin(x)^p}{x^q}\;dx$$
and evaluate the imaginary (for $p$ even) or real (for $p$ odd) part of :
$$\tag{9}\left(-\frac d{dq}\right)^{\large r} G_{p}(q)=\frac 2{2^p}\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}\,\left(-\frac d{dq}\right)^{\large r}\left(\Gamma(1-q)((p-2k)\,i)^{q-1}\right)$$
Setting $\;K:=(p-2k)\,i\;\ $ let's study $\quad\displaystyle 
f_r=f_r(k,p,q)=\left(-\frac d{dq}\right)^{\large r}\; \left(\Gamma(1-q)\,K^{q-1}\right)$.
The digamma function $\psi$ defined by $\;\psi(x):=(\ln\Gamma(x))'\;$ allows us to rewrite $\;\displaystyle \left(-\frac d{dq}\right)\Gamma(1-q)=\Gamma(1-q)\,\psi(1-q)\;$ and provides these first results :
\begin{align}
f_1&=\Gamma(1-q)\,K^{q-1}\left[\psi(1-q)-\ln(K)\right]\\
f_2&=\Gamma(1-q)\,K^{q-1}\left[\left(\psi(1-q)-\ln(K)\right)^2+\psi^{(1)}(1-q)\right]\\
f_3&=\Gamma(1-q)\,K^{q-1}\left[\left(\psi(1-q)-\ln(K)\right)^3+3\left(\psi(1-q)-\ln(K)\right)\psi^{(1)}(1-q)+\psi^{(2)}(1-q)\right]\\
f_4&=\Gamma(1-q)\,K^{q-1}\left[\left(\psi(1-q)-\ln(K)\right)^4+6\left(\psi(1-q)-\ln(K)\right)^2\psi^{(1)}(1-x)+4\left(\psi(1-q)-\\\ln(K)\right)\psi^{(2)}(1-q)+3\left(\psi^{(1)}\right)^2+\psi^{(3)}(1-q)\right]\\
\end{align}
The expressions between brackets are obtained from the previous ones using the recurrence : $\,x_{n+1}(t)=x_1(t)\;x_n(t)+x_n(t)'\;$ with $\,x_1(t)=\psi(t)-\ln(K)$.
($t\,$ is actually $\,1-q$) 
We obtain thus the complete Bell polynomials (with $x_1(t)=\psi(t)-\ln(K),\,x_2(t)=\psi^{(1)}(t),\cdots,$ $x_r(t)=\psi^{(r-1)}(t),\cdots$) and may write 
$$\tag{10}f_r(k,p,q)=\Gamma(1-q)\,((p-2k)\,i)^{q-1}\;B_r(x_1(1-q),x_2(1-q),\cdots,x_r(1-q))$$
allowing us to use the different expressions from the Wikipedia link and obtain the wished result :
$$\tag{11}I(p,q,r)=\begin{cases}p\ \text{odd}&|\quad\Re\\p\ \text{even}&|\quad\Im\end{cases}\ \frac 2{2^p}\sum_{k=0}^n (-1)^{n+k}\binom{p}{k}\,f_r(k,p,q)$$ 
The exponential generating function for the terms between brackets gives us for example :
\begin{align}
\tag{12}g(z)&:=\exp\left((\psi(t)-\ln(K))\frac{z}{1!}+\psi^{(1)}(t)\frac{z^2}{2!}+\psi^{(2)}(t)\frac{z^3}{3!}+\cdots\right)\\
&=1+\left(\psi(t)-\ln(K)\right)\frac z{1!}+\left((\psi(t)-\ln(K))^2+\psi^{(1)}(t)\right)\frac{z^2}{2!}+\cdots\\
\end{align}
$\qquad\qquad$(we have simply $\;\displaystyle g(t)=\frac{\Gamma(t)}{K^t}\,$ when $z=t$)
A: 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
