Bourbaki

Exercise 16, § 1

Tag 02KS, permanent. Cite it as [Bourbaki, Tag 02KS].

Spectral Theories, II, p. 266

For every complex number ss such that R(s)>1\mathscr{R}(s)>1, put

ζ(s)=∑1ns \zeta (s) =\sum\frac{1}{n^s}

n⩾1n\geqslant 1 (“Riemann zeta function”).

a) The function ζ\zeta is holomorphic on the open subset of C\mathbf{C} formed by the complex numbers ss such that R(s)>1\mathscr{R}(s)>1.

b) For y∈R+∗y\in \mathbf{R}_+^*, put

θ(y)=∑n∈Ze−πn2y \theta (y) =\sum_{n\in\mathbf{Z}}e^{-\pi n^2y}

this series converges absolutely and uniformly on the compact subsets of R+∗\mathbf{R}_+^*. For every ss such that R(s)>1\mathscr{R}(s)>1, one has

π−s/2Γ(s/2)ζ(s)=12∫0+∞θ(y)ys/2dyy \pi^{-s/2}\Gamma (s/2)\zeta (s) =\frac{1}{2}\int_0^{+\infty}\theta (y)y^{s/2}\frac{dy}{y}

(see FVR, VII, p. 10 for the gamma function in the complex plane).

∗−1√ _*-_1\surd

c) For every y∈R+y\in \mathbf{R}_+, one has θ(y\theta (y ) = yθ(y)y\theta (y). (Use the Poisson formula and exercise 1.)

d) For every ss such that R(s)>1\mathscr{R}(s)>1, one has

π−s/2Γ(s/2)ζ(s)=1s(s−1)+12∫1+∞(θ(y)−1)(ys/2+y(1−s)/2)dyy \pi^{-s/2}\Gamma (s/2)\zeta (s) =\frac{1}{s(s- 1)}+\frac{1}{2}\int_1^{+\infty}(\theta (y)-1)(y^{s/2}+y^{(1-s)/2})dyy

e) There exists a unique holomorphic function on C−{1}\mathbf{C}-\{1\} whose restriction to the open set of complex numbers with real part >1>1 coincides with ζ\zeta. This function, still denoted by ζ\zeta, satisfies (s−1)ζ(s)→1(s-1)\zeta (s)\rightarrow 1 when s→1s\rightarrow 1 (in other words, it has a simple pole of residue 1 at s=1s= 1).

f) The function Λ defined by

Λ(s)=π−s/2Γ(s/2)ζ(s) \Lambda (s) =\pi^{-s/2}\Gamma (s/2)\zeta (s)

is holomorphic on C−{0,1}\mathbf{C}-\{0,1\}; it has simple poles of residue 1 at s=0s= 0 and s=1s= 1. One has Λ(1−s)=Λ(s)\Lambda (1-s) = \Lambda (s) for every s∈C−{0,1}s\in \mathbf{C}-\{0,1\}.

g) Now prove that for every s∈Cs\in \mathbf{C} such that R(s)>1\mathscr{R}(s)>1, one has

ζ(s)=∏p(1−p−s)−1 \zeta (s) =\prod_p(1-p^{-s})^{-1}

where the product is over the set of prime numbers and is absolutely convergent (“Euler product”).

h) When σ>1\sigma >1 tends to 1, one has

∑p−σ=\sum p^{-\sigma}= log(1σ−1)+O(1)(\frac{1}{\sigma- 1})+ O(1),

pp

where the sum is over the prime numbers.

i) Every s∈C−{1}s\in \mathbf{C}-\{1\} such that ζ(s)=0\zeta (s) = 0 is either of the form s=−2ks=-2k, where k⩾1k\geqslant 1 is an integer, or satisfies 0⩽R(s)⩽10\leqslant \mathscr{R}(s)\leqslant 1.

j) One has ζ(s)≠0\zeta (s)\not= 0 if R(s)=1\mathscr{R}(s) = 1 and s≠1s\not= 1. (Prove using the Euler product that

ζ(σ)3ζ(σ+it)4ζ(σ+2it)⩾1 \zeta (\sigma )^3\zeta (\sigma +it)^4\zeta (\sigma + 2it)\geqslant 1

for every σ>1\sigma >1 and t∈Rt\in \mathbf{R}, then consider tt such that ζ(1+it)=0\zeta (1 +it) = 0.)

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.