Pi is transcendental

Pi is transcendental is a famous claim regarding the nature of the number [math]\displaystyle{ \pi }[/math]. 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 [math]\displaystyle{ e }[/math] 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 [math]\displaystyle{ \pi }[/math] is algebraic. Set [math]\displaystyle{ \alpha_{1}=i\pi }[/math] where [math]\displaystyle{ i }[/math] is the imaginary unit [math]\displaystyle{ \left(i^2=-1\right) }[/math]. Then [math]\displaystyle{ \alpha_{1} }[/math] is also algebraic, so it is the root of an [math]\displaystyle{ n }[/math]-th degree polynomial with integer coefficients. Let [math]\displaystyle{ \alpha_{2}\text{, . . . , }\alpha_{n} }[/math] be the other roots of this polynomial.

Since [math]\displaystyle{ 1+e^{i\pi}=0 }[/math] (Euler's identity) we have [math]\displaystyle{ (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 }[/math] for some [math]\displaystyle{ \beta_{1}\text{, . . . , }\beta_{N} }[/math]. These [math]\displaystyle{ \beta_i }[/math] are also algebraic, since they are the sums of other algebraic numbers (the [math]\displaystyle{ \alpha_i }[/math]). Disregarding the [math]\displaystyle{ \beta_{i} }[/math] that are equal to [math]\displaystyle{ 0 }[/math], we end up with [math]\displaystyle{ a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0 }[/math] for some positive integer [math]\displaystyle{ a }[/math] (since [math]\displaystyle{ e^0=1 }[/math]). Because the [math]\displaystyle{ \beta_1\text{, . . . , }\beta_M }[/math] are algebraic, they are the roots of a polynomial [math]\displaystyle{ f(z)=b z^M+ b_1 z^{M-1}+\text{ . . . }+b_{M-1} z + b_M }[/math] with integer coefficients [math]\displaystyle{ b\text{, }b_1\text{, . . . , }b_M }[/math], and because no [math]\displaystyle{ \beta_i }[/math] is equal to zero [math]\displaystyle{ b_M\neq 0 }[/math] also.

First Part

Now we will multiply our equation with the integral [math]\displaystyle{ \int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz }[/math], where [math]\displaystyle{ \rho }[/math] is some positive integer. We will write [math]\displaystyle{ \int_0^{\infty} }[/math] as a shorthand. This allows us to split the sum into two Parts:

[math]\displaystyle{ P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty} }[/math]

[math]\displaystyle{ P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M} }[/math]

where [math]\displaystyle{ \int_0^{\beta_i} }[/math] is a line integral along the line segment from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \beta_i }[/math] in the complex plane, and [math]\displaystyle{ \int_{\beta_i}^{\infty} }[/math] is also a line integral obtained by integrating over the line from [math]\displaystyle{ \beta_i }[/math] to [math]\displaystyle{ \infty }[/math] parallel to the real axis in the complex plane.

If [math]\displaystyle{ \beta_i }[/math] 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 [math]\displaystyle{ \beta_i }[/math] be a complex number with imaginary part [math]\displaystyle{ \text{Im}\left(\beta_i\right)\neq 0 }[/math]. As above, [math]\displaystyle{ \int_0^{\beta_i} }[/math] is the integral along the line segment from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \beta_i }[/math], [math]\displaystyle{ \int_{\beta_i}^{R'} }[/math] along the line segment parallel to the real axis from [math]\displaystyle{ \beta_i }[/math] to a number [math]\displaystyle{ R' }[/math] with real part [math]\displaystyle{ \text{Re}\left(R'\right)= R }[/math], [math]\displaystyle{ \int_{R'}^{R} }[/math] along the line segment parallel to the imaginary axis from [math]\displaystyle{ R' }[/math] to the real number [math]\displaystyle{ R }[/math], and [math]\displaystyle{ \int_{R}^0 }[/math] along the real axis from [math]\displaystyle{ R }[/math] to [math]\displaystyle{ 0 }[/math].

So [math]\displaystyle{ \int_0^{\beta_i} + \int_{\beta_i}^{R'} + \int_{R'}^{R} + \int_{R}^0 }[/math] is an integral along a closed path in the complex plane, which is not self-intersecting if we choose [math]\displaystyle{ R }[/math] large enough ([math]\displaystyle{ R\gt \text{Re}\left(\beta_i\right) }[/math]). 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 [math]\displaystyle{ R\longrightarrow\infty }[/math] the lenght of the line segment from [math]\displaystyle{ R' }[/math] to [math]\displaystyle{ R }[/math] stays the same while the function [math]\displaystyle{ z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z} }[/math] tends towards zero on that line segment. Using the estimation lemma for contour integrals it follows that [math]\displaystyle{ \left|\int_{R'}^{R}\right|\leq\left|R-R'\right|\cdot\max_{\left[R';R\right]}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}\longrightarrow 0 }[/math].

[math]\displaystyle{ \lim_{R\rightarrow\infty}\left(\int_0^{\beta_i} + \int_{\beta_i}^{R'} + \int_{R'}^{R} + \int_{R}^0\right) = 0 \Rightarrow\int_0^{\beta_i} + \int_{\beta_i}^{\infty} +\text{ } 0 + \int_{\infty}^0 = 0\Rightarrow \int_0^{\beta_i} + \int_{\beta_i}^{\infty} = \int_{0}^{\infty} }[/math]

Second Part

Now we will show that [math]\displaystyle{ \int_0^{\infty}=\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz }[/math] is an integer that is divisible by [math]\displaystyle{ \rho! }[/math] by using the fact that [math]\displaystyle{ \int_0^{\infty}z^{\rho}e^{-z}dz = \rho! }[/math].

[math]\displaystyle{ \int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz = b^{M\left(\rho+1\right)}\int_0^{\infty}\left(bz^M+\text{ . . . }+b_M\right)^{\rho+1}z^{\rho}e^{-z}dz = b^{M\left(\rho+1\right)}\int_0^{\infty}\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)z^{\rho}e^{-z}dz }[/math]

[math]\displaystyle{ = b^{M\left(\rho+1\right)}\left(c\int_0^{\infty}z^{M\rho+M+\rho}e^{-z}dz+\text{ . . . }+c_{M\rho+M}\int_0^{\infty}z^{\rho}e^{-z}dz\right) = b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right) }[/math]

[math]\displaystyle{ =b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right) }[/math]

where the [math]\displaystyle{ c_i }[/math] are integers; if [math]\displaystyle{ f }[/math] is a polynomial with integer coefficients so is [math]\displaystyle{ f^{\rho+1} }[/math]. The fractions are also integers because [math]\displaystyle{ \frac{\left(\rho+k\right)!}{\rho!}=\frac{(\rho+k)(\rho+k-1)\text{ . . . }\cdot2\cdot 1}{\rho(\rho-1)\text{ . . . }\cdot2\cdot 1}=(\rho+k)\text{ . . . }(\rho+2)(\rho+1) }[/math].

So the whole expression is an integer that is divisible by [math]\displaystyle{ \rho! }[/math]. Note that in the parentheses every term is divisible by [math]\displaystyle{ (\rho+1) }[/math] except the last one [math]\displaystyle{ c_{M\rho+M} }[/math]. Hence, modulo [math]\displaystyle{ (\rho+1) }[/math] we have the congruence [math]\displaystyle{ \int_0^{\infty}\equiv b^{M\rho+M}\rho! b_M^{\rho+1} }[/math].

Third Part

Claim

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