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There is a numerical evidence that the following is approximatelly true $$\int\limits_0^1\frac{x^2(\pi-x)}{\pi\sin{x}}dx\approx\sin{\left(\frac{13\pi}{46}\right)}-\sin{\left(\frac{6\pi}{53}\right)},$$ up to $3\times 10^{-10}$ relative precision. Is this just a coincidence or there is some reason why these two numbers are so close?

I came to the alleged identity accidentally thanks to a comment (to https://mathoverflow.net/a/269054/32389) from https://mathoverflow.net/users/12489/agno who observed that $$\int\limits_0^\pi\frac{x^2(\pi-x)}{\pi\sin{x}}dx=\frac{7}{2}\zeta(3).$$

As comments to the original version of the question have shown, the identity is only approximatelly true. So I reformulated the question.

P.S. J.M. Borwein and D.H. Bailey in "Future Prospects for Computer-Assisted Mathematics" (https://cms.math.ca/notes/v37/n8/Notesv37n8.pdf) provide even more surprising example of a near identity: $$\int\limits_0^\infty \cos{2x}\prod\limits_{n=1}^\infty \cos{\left(\frac{x}{n}\right)}\,dx\approx \frac{\pi}{8},$$ where the two numbers begin to differ after 43(!) correct digits. In this case there exists an explanation: see pp. 219-224 in the book "Experimental mathematics in action" by D. H. Bailey, J.M. Borwein, N. J. Calkin and V. Moll, and references cited there.

Zurab Silagadze
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    I do not believe that this identity is exactly correct. Working in Maple to 100 digit accuracy, I find that the difference between the two sides is about $-1.5\times 10^{-10}$. Maple also gives an exact expression for the integral in terms of $\text{polylog}(k,\pm e^i)$ (for $k=2,3,4$) and $\ln(1\pm e^i)$ and $\zeta(3)$. – Neil Strickland May 10 '17 at 06:44
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    I have got the same results as Neil Strickland with Mathematica 11. – Johannes Trost May 10 '17 at 06:53
  • I used a double precision Fortran program (I do not have Maple or Mathematica) and was fooled by seeing the coincidence up to $10^{-10}$. There should be some reason why these two numbers are so close. – Zurab Silagadze May 10 '17 at 07:26
  • Mathematica 11.1 performs $\frac{7 \zeta (3)}{2} $ .This coincides with the numeric value of the integral $4.2071991610585799989 $ up to 20 digits. The code on demand. – user64494 May 10 '17 at 08:07
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    @ZurabSilagadze if you do not have Maple or Mathematica, you can enter N[Int[x^2 (Pi - x)/(Pi Sin[x]),{x,0,1}],100] at www.wolframalpha.com. – Neil Strickland May 10 '17 at 08:20
  • The numeric value for $\sin{\left(\frac{13\pi}{46}\right)}-\sin{\left(\frac{6\pi}{53}\right) }$ up to 20 digits is $0.42750965527002101981$. The Mathematica code on demand. – user64494 May 10 '17 at 08:25
  • @ZurabSilagadze: What you've is an approximation only, not truly an evaluation. – T. Amdeberhan May 10 '17 at 11:52
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    How did you arrive to the right hand side combination of sines? – Fedor Petrov May 10 '17 at 12:37
  • Via inverse symbolic calculator (standard lookup, not the advanced one). – Zurab Silagadze May 10 '17 at 12:40
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    If you generate a random number uniformly distributed on $[0,1]$, then look it up in the inverse symbolic calculator, then this level of accuracy is not a surprise. – Douglas Zare May 10 '17 at 13:26
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    The number of valid mathematical formulae of length $N$ is roughly exponential in $N$; certainly it is at least $10^N$, since terminating decimals are themselves mathematical formulae. So we would expect a random number (say, between 0 and 1) to be approximable by a length $N$ formula to accuracy at least as good as $C^{-N}$, where $C$ is larger than $10$. As such, I don't find the accuracy of this approximation to be so surprising given the length of your formula. – Terry Tao May 10 '17 at 17:33
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    Given the fact that $1$ is a rather meaningless argument to be passed over to the sine function, as well as the sheer randomness (and precision) of the two rational factors of $\pi$ involved in the expression, I deem it only reasonable to consider the obtained result as nothing more than mere coincidence. – Lucian May 12 '17 at 06:52

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