Pi is transcendental

Pi is transcendental is a famous claim regarding the nature of the number $\pi$. It was first proven by Ferdinand von Lindemann in 1882, buildung on methods developed by Charles Hermite, who was able to prove that the number $e$ is transcendental roughly 10 years earlier in 1873. With the help of Karl Weierstrass their findings could be generalized to the Lindemann-Weierstrass theorem in 1885.

Proof

The proof sketched out here is not the original proof by Lindemann but a later proof by David Hilbert which is more accessible.

Suppose $\pi$ is algebraic. Set ${\alpha }_1=i\pi$ where $i$ is the imaginary unit $(i^2=-1)$. Then ${\alpha }_1$ is also algebraic, so it is the root of an $n$-th degree polynomial with integer coefficients. Let ${\alpha }_2\text{, . . . , }{\alpha }_n$ be the other roots of this polynomial.

Since $1+e^{i\pi }=0$ (Euler's identity) we have $(1+e^{{\alpha }_1})(1+e^{{\alpha }_2})\text{ . . . }(1+e^{{\alpha }_n})=1+e^{{\beta }_1}+e^{{\beta }_2}+\text{ . . . }+e^{{\beta }_N}=0$ for some ${\beta }_1\text{, . . . , }{\beta }_N$. These ${\beta }_i$ are also algebraic, since they are the sums of other algebraic numbers (the ${\alpha }_i$). Disregarding the ${\beta }_i$ that are equal to $0$, we end up with $a+e^{{\beta }_1}+e^{{\beta }_2}+\text{ . . . }+e^{{\beta }_M}=0$ for some positive integer $a$ (since $e^0=1$). Because the ${\beta }_1\text{, . . . , }{\beta }_M$ are algebraic, they are the roots of a polynomial $f(z)=b{z}^M+b_1{z}^{M-1}+\text{ . . . }+b_{M-1}z+b_M$ with integer coefficients $b\text{, }{b}_1\text{, . . . , }{b}_M$, and because no ${\beta }_i$ is equal to zero $b_M\ne 0$ also.

First Part

Now we will multiply our equation with the integral ${\int }_0^{\mathrm{\infty }}{z}^{\rho }{[f(z)b^M]}^{\rho +1}{e}^{-z}dz$, where $\rho$ is some positive integer. We will write ${\int }_0^{\mathrm{\infty }}$ as a shorthand. This allows us to split the sum into two Parts:

$P_1=a{\int }_0^{\mathrm{\infty }}+e^{{\beta }_1}{\int }_{{\beta }_1}^{\mathrm{\infty }}+\text{ . . . }+e^{{\beta }_M}{\int }_{{\beta }_M}^{\mathrm{\infty }}$

$P_2=0+e^{{\beta }_1}{\int }_0^{{\beta }_1}+\text{ . . . }+e^{{\beta }_M}{\int }_0^{{\beta }_M}$

where ${\int }_0^{{\beta }_i}$ is a line integral along the line segment from $0$ to ${\beta }_i$ in the complex plane, and ${\int }_{{\beta }_i}^{\mathrm{\infty }}$ is also a line integral obtained by integrating over the line from ${\beta }_i$ to $\mathrm{\infty }$ parallel to the real axis in the complex plane.

If ${\beta }_i$ is a real number these integrals are just normal Riemann integrals over the real numbers, and it's easy to see that it's valid to split them up in this way. So let ${\beta }_i$ be a complex number with imaginary part $\text{Im}\left({\beta }_i\right)\ne 0$. As above, ${\int }_0^{{\beta }_i}$ is the integral along the line segment from $0$ to ${\beta }_i$, ${\int }_{{\beta }_i}^{R^{\prime }}$ along the line segment parallel to the real axis from ${\beta }_i$ to a number $R^{\prime }$ with real part $\text{Re}\left(R^{\prime }\right)=R$, ${\int }_{R^{\prime }}^R$ along the line segment parallel to the imaginary axis from $R^{\prime }$ to the real number $R$, and ${\int }_R^0$ along the real axis from $R$ to $0$.

So ${\int }_0^{{\beta }_i}+{\int }_{{\beta }_i}^{R^{\prime }}+{\int }_{R^{\prime }}^R+{\int }_R^0$ is an integral along a closed path in the complex plane, which is not self-intersecting if we choose $R$ large enough ($R>\text{Re}\left({\beta }_i\right)$). Since the function which is being integrated is analytical on the whole complex plane this integral evaluates to zero according to the Cauchy integral theorem.

Note that if we let $R⟶\mathrm{\infty }$ the lenght of the line segment from $R^{\prime }$ to $R$ stays the same while the function $z^{\rho }{[f(z)b^M]}^{\rho +1}{e}^{-z}$ tends towards zero on that line segment. Using the estimation lemma for contour integrals it follows that $\left|{\int }_{R^{\prime }}^R\right|\le |R-R^{\prime }|\cdot \underset{[R^{\prime };R]}{max}{z}^{\rho }{[f(z)b^M]}^{\rho +1}{e}^{-z}⟶0$.

$\underset{R\to \mathrm{\infty }}{lim}({\int }_0^{{\beta }_i}+{\int }_{{\beta }_i}^{R^{\prime }}+{\int }_{R^{\prime }}^R+{\int }_R^0)=0\Rightarrow {\int }_0^{{\beta }_i}+{\int }_{{\beta }_i}^{\mathrm{\infty }}+0+{\int }_{\mathrm{\infty }}^0=0\Rightarrow {\int }_0^{{\beta }_i}+{\int }_{{\beta }_i}^{\mathrm{\infty }}={\int }_0^{\mathrm{\infty }}$

Second Part

Now we will show that ${\int }_0^{\mathrm{\infty }}={\int }_0^{\mathrm{\infty }}{z}^{\rho }{[f(z)b^M]}^{\rho +1}{e}^{-z}dz$ is an integer that is divisible by $\rho !$ by using the fact that ${\int }_0^{\mathrm{\infty }}{z}^{\rho }{e}^{-z}dz=\rho !$.

${\int }_0^{\mathrm{\infty }}{z}^{\rho }{[f(z)b^M]}^{\rho +1}{e}^{-z}dz=b^{M(\rho +1)}{\int }_0^{\mathrm{\infty }}{(b{z}^M+\text{ . . . }+b_M)}^{\rho +1}{z}^{\rho }{e}^{-z}dz=b^{M(\rho +1)}{\int }_0^{\mathrm{\infty }}(c{z}^{M\rho +M}+\text{ . . . }+c_{M\rho +M})z^{\rho }{e}^{-z}dz$

$=b^{M(\rho +1)}(c{\int }_0^{\mathrm{\infty }}{z}^{M\rho +M+\rho }{e}^{-z}dz+\text{ . . . }+c_{M\rho +M}{\int }_0^{\mathrm{\infty }}{z}^{\rho }{e}^{-z}dz)=b^{M(\rho +1)}(c(M\rho +M+\rho )!+\text{ . . . }+c_{M\rho +M}\rho !)$

$=b^{M\rho +M}\rho !(c\frac{(M\rho +M+\rho )!}{\rho !}+\text{ . . . }+c_{M\rho +M})$

where the $c_i$ are integers; if $f$ is a polynomial with integer coefficients so is $f^{\rho +1}$. The fractions are also integers because $\frac{(\rho +k)!}{\rho !}=\frac{(\rho +k)(\rho +k-1)\text{ . . . }\cdot 2\cdot 1}{\rho (\rho -1)\text{ . . . }\cdot 2\cdot 1}=(\rho +k)\text{ . . . }(\rho +2)(\rho +1)$.

So the whole expression is an integer that is divisible by $\rho !$. Note that in the parentheses every term is divisible by $(\rho +1)$ except the last one $c_{M\rho +M}$. Hence, modulo $(\rho +1)$ we have the congruence ${\int }_0^{\mathrm{\infty }}\equiv b^{M\rho +M}\rho !b_M^{\rho +1}$.

Third Part

Claim

Statement of the claim	Pi is transcendental
Level of certainty	Proven
Nature	Theoretical
Counterclaim	Pi is algebraic
Dependent on
Dependency of