Equating the integrals of 1/(1-xy) and 2/(1+xy) by elementary calculus?
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?
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.