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 $n=1..12$ :

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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.


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    $\begingroup$ I don't think you can simplify this any further, in view of the linear independence of the list $\{\zeta(2k+1)\}$. $\endgroup$ May 20, 2017 at 16:40
  • 2
    $\begingroup$ @T.A, that linear independence is hypothesized, not proved, right? $\endgroup$ May 21, 2017 at 3:38
  • $\begingroup$ Yes, it is, but overwhelmingly correct. $\endgroup$ May 21, 2017 at 3:43
  • $\begingroup$ @T.Amdeberhan What do you mean with "overwhelmingly correct"? Is there any more evidence to it than the fact we have not found any linear relation between them? $\endgroup$
    – Wojowu
    May 21, 2017 at 10:41
  • $\begingroup$ Clarification: it seems more likely true than not. $\endgroup$ May 21, 2017 at 12:33

2 Answers 2


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 $n=2m-1, m>1$: $$f(2m-1)=(-1)^{m+1}(4-2^{1-2m})(2m)!\,\zeta(2m+1)-\sum\limits_{p=1}^{m-1}(-1)^p\;\frac{\pi^{2(m-p)}}{2}\;\binom{2m-1}{2p-1}(4-2^{1-2p})(2p)!\,\zeta(2p+1).$$

  • $\begingroup$ Many thanks, Zurab. I have tested your now proven formulae and they work perfectly! $\endgroup$
    – Agno
    May 22, 2017 at 9:19
  • $\begingroup$ I found some further simplification is possible using the Heaviside step function: $$\Theta(n)= \begin{cases} 1 &\quad n \ge 0\\ 0&\quad n < 0 \\ \end{cases}$$ For $n=$ even: $$f(m)=\Gamma(m+1)\,\sum_{p=1}^{\frac{m}{2}} \left(\tfrac{m-1}{m}\right)^{\Theta\left(p-\frac{m}{2}\right)} (-1)^{p+1}\,\frac{\pi^{m+1-2p}\,p\,(4-2^{1-2p})}{\Gamma(m+2-2p)}\,\zeta(2p+1)$$ For $n=$ odd: $$f(m)=\Gamma(m+1)\,\sum_{p=1}^{\frac{m}{2}+\frac12}\, \,2^{\Theta\left(p-\frac{m}{2}-\frac12\right)} \,\,(-1)^{p+1}\,\frac{\pi^{m+1-2p}\,p\,(4-2^{1-2p})}{\Gamma(m+2-2p)}\,\zeta(2p+1)$$ $\endgroup$
    – Agno
    May 23, 2017 at 15:02
  • $\begingroup$ Made a typo: For $ n=$ even/odd should of course be: For $m=$ even/odd. $\endgroup$
    – Agno
    May 24, 2017 at 11:11

Not an answer, but... Mathematica evaluates the indefinite integral $$\int x^k \csc x d x$$ in terms of polylogs, from which your integral follows readily, so the obvious question is - how does it do the indefinite integral. The short answer is: The Risch algorithm (or, more precisely, an extension, since the original algorithm only deals with elementary antiderivatives). However, if you want to do it by hand, the two methods which come to mind are:

  1. Make the substitution $u = \exp(i x),$ which transforms your integrand into a power of log times a rational function - this is plausbible, since the arguments of the polylogs have the form $\exp(i x),$ so Mathematica might, indeed, be doing this.
  2. Expand $\csc x$ in its Laurent series around $0,$ and integrate term by term. The Laurent series is explicit (in terms of Bernoulli numbers), so it may be that a simple manipulation gets it to your linear combination of zetas (I haven't tried).
  • 2
    $\begingroup$ Many thanks, Igor. Great suggestions that I will pursue. Mathematica seems to indeed firstly evaluate the integral in terms of polylogs and then to run a FullSimplify to reduce it to the series with the $\zeta(2k+1)$s. I tried to find other integrals that have this "reduction"-property, but so far only found: $$\int_0^{\frac{\pi}{2}} \big(x\,\csc(x)\big)^n\,dx$$ that only for even $n$ reduces to combinations of $\pi^k$, $\ln(2)$ and $\zeta(2k+1)$. E.g. for $n=2$ we get $\pi\,\ln(2)$ and for $n=4$ we get: $\frac{1}{12}\pi\,\left(-\pi^2-18\,\zeta(3)+4\,\pi^2\,\ln(2)+24\,\ln(2)\right)$, etc. $\endgroup$
    – Agno
    May 21, 2017 at 12:37
  • $\begingroup$ @Agno: The formula for that integral is e.g. here : math.stackexchange.com/questions/786281/… $\endgroup$
    – user90369
    Nov 6, 2017 at 11:40
  • $\begingroup$ @user90369 Many thanks for that link. Much appreciated. I had completely missed that discussion on MSE. In the mean time I also derived (a more complicated) formula here: mathoverflow.net/questions/273375/… Does your formula also work for $n < 0$ ? $\endgroup$
    – Agno
    Nov 6, 2017 at 12:05
  • $\begingroup$ Sorry, the formula doesn't work for $n<0$ it was only an answer to the question. :-) Perhaps it's useful to formulate a separate question about $\int\limits_0^z x^m \left(\frac{\sin x}{x}\right)^n dx$ with $m\geq 0$ and $n \in \mathbb{Z}$ . $\endgroup$
    – user90369
    Nov 6, 2017 at 16:46

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