# 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.