TITLE: Integral expressed as a finite sum of $r_n \zeta(2n+1) \pi^{-(2n+1)}$.
QUESTION [8 upvotes]: The integral
\begin{equation}
I_m=\int_0^1 \frac{x^m-x}{\sin \pi x} \:dx
\end{equation}
is expressed as a finite sum of terms of the shape $r_n \zeta(2n+1) \pi^{-(2n+1)}$, where $r_n$ is a rational number that depends on $n$.
In this question Evaluating $\int_0^1 \frac{x^2-x}{\sin \pi x} dx = - \frac{7 \zeta(3)}{\pi^3}.$ The user "user90369" provided a answer for $I_2$. So I followed his steps and generalized the integral. 
Define:
\begin{align}
&f(a) = \int_0^1 x e^{ax} dx= \frac{1+e^a(a-1)}{a^2} \\
&g(a) = \int_0^1 x^m e^{ax} dx = \frac{e^a}{a^{m+1}} \left(  (-1)^mm!  + \sum_{n=1}^{m}  (-1)^{m+n} \frac {a^n}{n!} \right) - \frac{(-1)^mm!}{a^{m+1}} 
\end{align}
Rewrite $I_m$ as:
\begin{align}
I_m &= \int_0^1 \frac{x^m-x}{\sin \pi x} = 2i \int_0^1 \frac{x^m-x}{ e^{i\pi x } - e^{-i \pi x} }\nonumber\\
&= 2i \sum_{k=0}^{\infty} \int_0^1 (x^m-x)e^{-i\pi x(2k+1) } dx \\
&= 2i \sum_{k=0}^{\infty} g(-i\pi(2k+1)) - f(-i\pi (2k+1)) \nonumber
\end{align}
From that I arrived at the following formulas for $I_m$. 
When $m$ is even and $\geq 2$:
\begin{align*}
I_m =  \frac{4m!i^m}{\pi^{m+1}} \left( 1-\frac{1}{2^{m+1}}\right)  \zeta(m+1) + 2 m! \sum_{n\:odd \geq 0}^{m-2} \frac{i\: \zeta (m-n)}{(n+1)!(i\pi)^{m-n}}   \left(  1-\frac{1}{2^{m-n}} \right)
\end{align*}
When $m$ is odd and $\geq 3$:
\begin{align*}
I_m = 2 m! \sum_{n\:even \geq 0}^{m-2} \frac{i\: \zeta (m-n)}{(n+1)!(i\pi)^{m-n}}   \left(  1-\frac{1}{2^{m-n}} \right).
\end{align*}
A few values for $I_m$:
\begin{align*}
& I_2 = - \frac{7\zeta(3)}{\pi^3} \\
& I_3 = -\frac{21 \zeta(3)}{2\pi^3}\\
& I_4 =  \frac{93\zeta(5)}{\pi^5}-\frac{21\zeta(3)}{\pi^3} \\
& I_5 =  \frac{465\zeta(5)}{2\pi^5} - \frac{35\zeta(3)}{\pi^3} \\
& I_6 = -\frac{1}{2} \left(  \frac{5715\zeta(7)}{\pi^7} - \frac{1395\zeta(5)}{\pi^5} + \frac{105\zeta(3)}{\pi^3}  \right) \\
& I_7 = -\frac{1}{2} \left(  \frac{40005\zeta(7)}{2\pi^7} - \frac{3255\zeta(5)}{\pi^5} + \frac{147\zeta(3)}{\pi^3} \right) \\
&I_8 = \frac{160965\zeta(9)}{\pi^9} - \frac{40005\zeta(7)}{\pi^7} + \frac{3255\zeta(5)}{\pi^5}-\frac{98\zeta(3)}{\pi^3}
\end{align*}
It's also possible to integrate by parts $I_m$. Anyway, I found this integral interesting and wanted to share. Perphaps someone can do something cool with it, or not.
Sorry for the weird/sloppy english.

REPLY [6 votes]: Not an answer but too long for a comment.
This is for sure a very interesting integral.
Just by curiosity, I wondered if the approximation
$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad (0\leq t\leq\pi)$$  proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician, could lead to something simple.
Changing variable $x=\frac y \pi$ and considering
$$\int y^m \csc (y)\,dy \approx \frac{y^m \left(5 m y \, _2F_1\left(1,m+1;m+2;\frac{y}{\pi }\right)-4 m y+5 \pi 
   (m+1)\right)}{16 m (m+1)}$$ we can arrive to the incredibly simple result
$$I_m \approx \frac{2 m^2+3 m+5}{16\,m\, (m+1)}-\frac 5 {16} H_m $$
The table below reproduces some values (the so-called exact values being obtained by numerical integration)
$$\left(
\begin{array}{ccc}
 m & \text{approximation} & \text{exact} \\
 2 & -0.270833 & -0.271377 \\
 3 & -0.406250 & -0.407066 \\
 4 & -0.497917 & -0.499007 \\
 5 & -0.567708 & -0.569075 \\
 6 & -0.624256 & -0.625896 \\
 7 & -0.671875 & -0.673782 \\
 8 & -0.713046 & -0.715212 \\
 9 & -0.749330 & -0.751749 \\
 10 & -0.781780 & -0.784444 \\
 20 & -0.995573 & -1.000305 \\
 30 & -1.121081 & -1.127370 \\
 40 & -1.210330 & -1.217854 \\
 50 & -1.279654 & -1.288200 \\
 60 & -1.336350 & -1.345767 \\
 70 & -1.384318 & -1.394494 \\
 80 & -1.425892 & -1.436739 \\
 90 & -1.462578 & -1.474027 \\
 100 & -1.495406 & -1.507399
\end{array}
\right)$$
Edit
More amazing (at least to me) would be
$$I_{m+1}-I_m\approx-\frac{5}{16 m}+\frac{1}{4 (m+1)}-\frac{1}{4 (m+2)}$$