Uploaded August 2017 | Updated September 2026, 2 hours ago

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This is a very hard demonstration that is solid.
There exists also a weak demonstration with holomorphic functions and Fourier's decomposition.
The goal is to prove that for all non-integer real x, pi*cotan(x) = pi/tan(x) = sum between - infinity to + infinity of 1/(x+n) = sum between - infinity to + infinity of 1/(x-n)

Demonstration of this :
∀x∈R\Z,πcotan(πx)=∑_(n=-∞)^(+∞)▒1/(x-n)=∑_(n=-∞)^(+∞)▒1/(x+n)
f(x)=cotan(πx)
g_n (x)=2x/(x^2-n^2 ),n≥1 and g_0 (x)=1/x
g_n^' (x)=-(2x^2+2n^2)/(x^2-n^2 )^2 ≤0
h(x)=f(x)-g(x)=πcotan(πx)-1/x-∑_(n=1)^(+∞)▒2x/(x^2-n^2 )
∑_(n=1)^(+∞)▒2x/(x^2-n^2 )=∑_(n=1)^(+∞)▒〖1/(x-n)+1/(x+n)〗
∑_(n=1)^(+∞)▒2x/(x^2-n^2 )=∑_(n=1)^(+∞)▒1/(x-n)+∑_(n=1)^(+∞)▒1/(x+n)
∑_(n=1)^(+∞)▒2x/(x^2-n^2 )=∑_(n=1)^(+∞)▒1/(x-n)+∑_(n=-∞)^(-1)▒1/(x-n)
∑_(n=1)^(+∞)▒2x/(x^2-n^2 )=∑_(n=-∞)^(+∞)▒1/(x-n)-1/x=∑_(n=-∞)^(+∞)▒1/(x+n)-1/x
∀x∈R\Z,πcotan(πx)=∑_(n∈Z)▒1/(x-n)=∑_(n∈Z)▒1/(x+n)
∀x∈R\πZ,πcotan(x)=∑_(n∈Z)▒1/(x/π-n)=∑_(n∈Z)▒1/(x/π+n)
∀x∈R\πZ,πcotan(x)=∑_(n∈Z)▒π/(x-nπ)=∑_(n∈Z)▒π/(x+nπ)
∀x∈R\πZ,cotan(x)=∑_(n∈Z)▒1/(x-nπ)=∑_(n∈Z)▒1/(x+nπ)
∀x∈R\πZ,cotan(x)=∑_(n=-∞)^(+∞)▒1/(x-n)=∑_(n=-∞)^(+∞)▒1/(x+n)
∀x∈R\πZ,tan⁡(x)*∑_(n=-∞)^(+∞)▒1/(x-n)=tan⁡(x)*∑_(n=-∞)^(+∞)▒1/(x+n)=1