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The following integrals (inspired here) are both equal to $\pi^2/6$: $$\int_0^1\!\int_0^1 \frac{1}{1-xy}\,dx\,dy = \int_0^1\!\int_0^1 \frac{2}{1+xy}\,dx\,dy.$$

According to conjecture 1 of Kontsevich and Zagier's article on periods, it should be possible to go from one to the other via other integrals of algebraic functions, using only the three rules of

  • additivity
  • change of variables
  • the Newton-Leibniz formula (a.k.a. the fundamental theorem of calculus),

What sequence of those rules, if any, yields the above identity?

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1 Answers1

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It is enough to set $x=u^2, y=v^2$ in the first integral, then using $$ \frac{4uv}{1-u^2 v^2} = 2\left(\frac{1}{1-uv}-\frac{1}{1+uv}\right).$$ This proves $$ \iint_{(0,1)^2}\frac{du\,dv}{1-uv} = \iint_{(0,1)^2}\frac{2\,du\,dv}{1+uv} $$ through elementary algebraic manipulations.
That is exactly the same as proving $$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} = 2\,\eta(2),$$ but does not require to expand $\frac{1}{1\pm uv}$ as a geometric series, then apply termwise integration.

Jack D'Aurizio
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  • I remember that $\frac{1}{1-xy}$ considered as a kernel, has interesting properties, but I am unable to find back sources about it. – Jean Marie Jan 18 '17 at 16:05
  • @JeanMarie: Beuker and Viola showed we may use diffeomorphisms of the unit square/cube to derive fast-converging series for $\zeta(2)$ and $\zeta(3)$. See, for instance, https://eudml.org/doc/74995 – Jack D'Aurizio Jan 18 '17 at 16:11
  • Thanks for the references. Besides, very interesting and short proof ! – Jean Marie Jan 18 '17 at 16:13