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May 21, 2017 at 12:17 comment added Agno I have changed my comment into a follow-up question here: mathoverflow.net/questions/270286/…
May 10, 2017 at 9:13 comment added Agno I have used Maple to evaluate $f(n)$ and it came back with the closed forms. Not exactly sure how it does the evaluation, however I have the strong impression that it first expresses the integrals as a finite series of weighted polylogarithms (+other some components) that only for this specific integrals can be simplified and reduced to the form with only weighted $\zeta(2k+1)$s.
May 10, 2017 at 7:20 comment added Zurab Silagadze It seems the last identity is only correct up to the precision $10^{-10}$ but not exactly: mathoverflow.net/questions/269405/interesting-identity
May 10, 2017 at 5:25 comment added Zurab Silagadze Interesting observation. How do you got them? By the way while checking f(2), due to an accidental error, I got (numerically) $$\int\limits_0^1\frac{x^2(\pi-x)}{\pi \sin{x}}dx=\sin\left(\frac{13\pi}{46}\right)-\sin\left(\frac{6\pi}{53}\right).$$ How this identity can be proved?
May 9, 2017 at 13:23 comment added Agno Just like to share a (maybe trivial) observation on the beautiful integral for $\zeta(3)$ in your arxiv paper. When we slightly alter it into: $$f(n):=\frac{1}{7}\,\int_0^{\pi} x^n\,\frac{\pi-x}{\sin(x)}dx$$ we get closed forms expressed as finite series of $\zeta(2k+1)$ that are all weighted by $\pi^k$ and a rational. E.g. $f(2) = \frac12\pi\zeta(3)$, $f(3) = \frac32\pi^2\zeta(3)-\frac{93}{7}\zeta(5)$, $f(4) = 2\,\pi^3\zeta(3)-\frac{279}{14}\zeta(5)$, etc.
May 8, 2017 at 7:08 history edited Zurab Silagadze CC BY-SA 3.0
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May 6, 2017 at 12:48 history edited Zurab Silagadze CC BY-SA 3.0
typos corrected and some extra text added
May 5, 2017 at 10:37 history answered Zurab Silagadze CC BY-SA 3.0