TITLE: A closed form for an integral expressed as a finite series of $\zeta(2k+1)$, $\pi^m$ and a rational?
QUESTION [22 upvotes]: In this paper the following beautiful integral expression for $\zeta(3)$ is derived: 
$$\zeta(3)=\frac{1}{7}\,\int_0^{\pi} x\,(\pi-x)\csc(x)\, dx$$
In a comment at the end of this question, I mentioned the following tweak: 
$$f(n):= \int_0^{\pi} x^n\,(\pi-x)\csc(x)\,dx$$
Maple and Mathematica both return closed forms for $f(n)$ as a finite series of $\zeta(2k+1)$, weighted by $π^m$ and a rational ($n,k,m \in \mathbb{N}$). This is what I get for $n=1..12$ : 

I then tried to find a closed form for the above and got as far as:
For $n=$ odd:
$$f(n)=\scriptsize \left(\Gamma(n+1)\,\sum_{k=1}^{n}(-1)^{k+1}\,\frac{(4-2^{1-2k})\,k\,\pi^{n+1-2k}}{\Gamma(2+n-2k)}\,\zeta(2k+1)\right)-\left(\frac{i}{2}\right)^{n+1}\,(2^{n+2}-1)\,\Gamma(n+2)\,\zeta(n+2)$$
For $n=$ even:
$$f(n)=\scriptsize \left(\Gamma(n+1)\,\sum_{k=1}^{n}(-1)^{k+1}\,\frac{(4-2^{1-2k})\,k\,\pi^{n+1-2k}}{\Gamma(2+n-2k)}\,\zeta(2k+1)\right)+\left(\frac{i}{2}\right)^{n}\,(2^{n+1}-1)\,\pi\,\Gamma(n+1)\,\zeta(n+1)$$
Note that only the last part differs, which encourages me to believe that a simpler formula might exist. Could this be simplified? Since the CAS-tools evaluate the integral very quickly into the series, I am also keen to better understand which mathematical steps they take.
Thanks.

REPLY [11 votes]: I think the following articles can give a clue:
http://www.tandfonline.com/doi/abs/10.1080/10652460701688125?journalCode=gitr20 (Closed-form evaluation of some families of cotangent and cosecant integrals, by D. Cvijović) and http://www.sciencedirect.com/science/article/pii/S0377042702003588 (Integral representations of the Riemann zeta function for odd-integer arguments, by D. Cvijović and J. Klinowski).
Update. In particular, using $2x^n=E_n(x)+E_n(1+x)$ and $E_n(1+x)=\sum\limits_{k=0}^n\binom{n}{k}E_k(x)$, we get
$$f(n)=\frac{\pi^{n+2}}{2}\left[I_n+\sum\limits_{k=0}^n\binom{n}{k}I_k-I_{n+1}-\sum\limits_{k=0}^{n+1}\binom{n+1}{k}I_k\right]=\frac{\pi^{n+2}}{2}\left[I_n-2I_{n+1}-\sum\limits_{k=1}^n\binom{n}{k-1}I_k\right], \tag{1}$$
where $$I_n=\int\limits_0^1E_n(x)\csc{(\pi x)}\,dx.$$
Odd indices in (1) doesn't contribute because $E_n(1-x)=(-1)^nE_n(x)$ leads to the integrand which is antisymmetric with regard to $x\to 1-x$. For even indices, $I_n$ was calculated in the second paper indicated above as $$I_{2n}=(-1)^n\frac{(4-2^{1-2n})(2n)!}{\pi^{2n+1}}\zeta(2n+1),$$ and we obtain, for even $n=2m$:
$$f(2m)=(-1)^m\frac{\pi}{2}(4-2^{1-2m})(2m)!\,\zeta(2m+1)- \sum\limits_{p=1}^m(-1)^p\;\frac{\pi^{2(m-p)+1}}{2}\;\binom{2m}{2p-1}(4-2^{1-2p})(2p)!\,\zeta(2p+1),$$
and for odd $n=2m-1, m>1$:
$$f(2m-1)=(-1)^{m+1}(4-2^{1-2m})(2m)!\,\zeta(2m+1)-\sum\limits_{p=1}^{m-1}(-1)^p\;\frac{\pi^{2(m-p)}}{2}\;\binom{2m-1}{2p-1}(4-2^{1-2p})(2p)!\,\zeta(2p+1).$$