While idly browsing I came across a rather nice paper of Zhou and Markov (2009), which led to the first time I have ever learnt a proof of the famous and slightly random fact that \pi is irrational.

The idea is to construct a sequence A_n of expressions involving integrals, which are on the one hand (integrating by parts) given by a recurrence relation which guarantees that they are positive integers, but which on the other hand must clearly tend to zero, and of course this gives a contradiction. The rabbit from the analyst’s hat seems to be to consider the functions f_n(x) = (\pi x - x^2)^n and then integrate through by a sine function. This then very quickly just works. Ok, let’s do it.

Suppose \pi = \frac{a}{b}, and consider A_n = \frac{1}{n!} \int_{0}^\pi (a x - b x^2)^n \sin x \rightarrow 0. However, we will also show that the A_n are positive integers. Indeed, A_0 = 2, A_1 = 4b and for n\geq 2, integrating by parts gives that

A_n = b(4n-2) A_{n-1} - a^2 A_{n-2} \in \mathbb{N}.