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Kontsevich and Zagier define a period as an integral of a rational function (over $\mathbb{Q}$) defined on a $\mathbb{Q}$-semialgebraic set. They conjecture that if two periods are equal, then the reason may be only a finite number of change of variables and integrations by parts. The question is whether there is a clever computer system which checks whether two periods are equal and if they are, provide a proof (which is the aforementioned sequence.)

A particular motivation is the following identity \begin{align*} 7\int_0^1\log x\cdot \frac{1+x-x^2+x^3-x^4-x^5}{1-x^7}dx=\\ 12\int_{\frac{\sqrt{21}-5}2}^1 \frac{\log|x|}{2+3x+2x^2}dx. \end{align*}

Both parts are periods as may be easily seen if we expand $$\log |x|=\int_{|x|}^1 y^{-1}dy = p.v.\int_{x}^1 y^{-1} dy$$ and so get integrals of rational functions over triangles (if we are ok with p.v.) or triangle/quadrilateral.

Richard Stanley reminds us in this answer at MathOverflow that this is still a conjecture. It comes from the volumes of hyperbolic tetrahedra and it may be proved that the ratio of two sides is a rational number(!), which is calculated and coincides with 1 with accuracy of about 20000 decimal digits.

It is a bit annoying if neither a general theory nor programs motivated by this theory can provide a way to verify such a seemingly toy identity.

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Fedor Petrov
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  • is there a reason that one log has ||, while the other has not? – Dima Pasechnik Mar 11 '15 at 11:56
  • @DimaPasechnik in LHS x is always positive. Or you mean philosophical reason? – Fedor Petrov Mar 11 '15 at 12:08
  • oops, sorry. It's $\sqrt{21}-5$. I need better glasses. By the way: integrating polynomials over simplices is just differentiation (of the Laplace transform). One might be tempted to say that same holds for rational functions (which will need to be expanded...). I don't know if this is correct though (convergence is not obvious...). – Dima Pasechnik Mar 11 '15 at 12:12
  • How about absolute convergence (a requirement in the definition of a period) of your double integrals on respective triangles? Is it obvious? – Dima Pasechnik Mar 11 '15 at 13:10
  • In the LHS yes, in the RHS principal value becomes absolutely convergent after cancelling the sum $f(x,y)+f(x,-y)$ in small triangle corresponding to negative values of $x$. – Fedor Petrov Mar 11 '15 at 13:28
  • OK, so the RHS is a period over certain 4-gon, lying above the $x$-axis? – Dima Pasechnik Mar 11 '15 at 13:55
  • Yes, it is defined by $|x|\leq y\leq 1$, $(\sqrt{21}-5)/2<x<1$. – Fedor Petrov Mar 11 '15 at 16:29
  • Right. I guess such mysteries are not possible in case one has a "nice" (here: affine) map between domains of integration. Perhaps Kontsevich-Zagier knew this already. – Dima Pasechnik Mar 11 '15 at 20:24
  • You may partition quadrilateral onto two triangles and map one of them to another by affine transform, so integral over quadrilateral becomes an integral of some other function over triangle. – Fedor Petrov Mar 11 '15 at 21:35
  • Ah, right. So this looks like that in the end one will get the integral of a rational function over a triangle, which is, conjecturally, identically 0. – Dima Pasechnik Mar 11 '15 at 21:48

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Too long for a comment.

Mathematica 10.0.2.0 produces the LHS equals $$ \frac{1}{7} \left(-\psi ^{(1)}\left(\frac{1}{7}\right)-\psi ^{(1)}\left(\frac{2}{7}\right)+\psi ^{(1)}\left(\frac{3}{7}\right)-\psi ^{(1)}\left(\frac{4}{7}\right)+\psi ^{(1)}\left(\frac{5}{7}\right)+\psi ^{(1)}\left(\frac{6}{7}\right)\right) $$ whereas the RHS is equal to $$ -\frac{12 i \left(\text{Li}_2\left(-\frac{1}{4} i \left(-3 i+\sqrt{7}\right)\right)-\text{Li}_2\left(\frac{1}{4} i \left(3 i+\sqrt{7}\right)\right)+\text{Li}_2\left(\frac{2 i \left(-5+\sqrt{21}\right)}{-3 i+\sqrt{7}}\right)-\text{Li}_2\left(-\frac{2 i \left(-5+\sqrt{21}\right)}{3 i+\sqrt{7}}\right)+\log \left(\frac{1}{2} \left(5+\sqrt{21}\right)\right) \log \left(-7+7 i \sqrt{3}-5 i \sqrt{7}+3 \sqrt{21}\right)-\log \left(\frac{1}{2} \left(5+\sqrt{21}\right)\right) \log \left(-7-7 i \sqrt{3}+5 i \sqrt{7}+3 \sqrt{21}\right)\right)}{\sqrt{7}} .$$

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