Pi is trancendental

First the rabbit. The introduction of this function is the part which I don’t know how to motivate. Let f be a polynomial and define

\displaystyle I(z,f)=\int_0^z e^{z-t}f(t)\,dt.

Integration by parts gives the recursion

\displaystyle I(z,f)=e^z f(0)-f(z)+I(z,f')

and therefore we have the formula

\displaystyle I(z,f)=e^z\sum_{j\geq 0} f^{(j)}(0) - \sum_{j\geq 0}f^{(j)}(z).

Now suppose (for want of a contradition) that \pi i \in \overline{\mathbb{Q}}. Let the set of Galois conjugates of \pi i be \{a_1,\ldots,a_k\}. Then we have \prod_{i=1}^k (1+e^{a_i})=0, expand this as \sum_{J\subset[k]}e^{\sum_{j\in J}a_j}=0 and rewrite as

\displaystyle (2^k-d)+\sum_{i=1}^d e^{\theta_i}=0

where the \theta_i are the nonzero exponents.

Now consider

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)\sum_{j\geq 0}f^{(j)}(0)-\sum_{j\geq 0}\sum_{i=1}^df^{(j)}(\theta_i)

Let N be an integer such that N\pi i is an algebraic integer. Let p be a (large) prime and we choose to take

\displaystyle f(x)=N^{dp}x^{p-1}\prod_{i=1}^d(x-\theta_i)^p\in\mathbb{Z}[x].

There are absolute constants A and B (independent of p) such that

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\leq A B^p

(look at the integral definition of I(z,f) and apply the naive estimate).

Now consider \sum_{i=1}^d I(\theta_i,f). It is a Galois-invariant algebraic integer, hence an integer. We have

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)f^{(p-1)}(0)+\text{higher order terms}.

Here higher order means at least p derivatives appearing. Each of these higher order terms is divisible by p!, hence by p. Since p is prime, for large enough p, \sum_{i=1}^d I(\theta_i,f) is not divisible by p, hence nonzero.

Now every term is divisible by (p-1)!, so we get the lower bound

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\geq (p-1)!

As there are infinitely many primes, we can send choose p large enough to get a contradiction, QED.

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