3
$\begingroup$

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.

Improve this question
$\endgroup$
12
  • 6
    $\begingroup$ 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 $\zeta(3)$. $\endgroup$
    – Neil Strickland
    Commented May 10, 2017 at 6:44
  • 3
    $\begingroup$ I have got the same results as Neil Strickland with Mathematica 11. $\endgroup$
    – Johannes Trost
    Commented May 10, 2017 at 6:53
  • 1
    $\begingroup$ @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. $\endgroup$
    – Neil Strickland
    Commented May 10, 2017 at 8:20
  • 5
    $\begingroup$ 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. $\endgroup$
    – Douglas Zare
    Commented May 10, 2017 at 13:26
  • 5
    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Commented May 10, 2017 at 17:33

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .