Number Theory: An Introduction to Mathematics

3 Proof of the Prime Number Theorem for Arithmetic Progressions 407

−f′(s)/f(s)

=− 2 ζ′(s)/ζ(s)−L′(s+iα,χ)/L(s+iα,χ)−L′(s−iα,χ)/ ̄ L(s−iα,χ) ̄

= 2

∑∞

n= 1

Λ(n)n−s+

∑∞

n= 1

χ(n)Λ(n)n−s−iα+

∑∞

n= 1

χ( ̄n)Λ(n)n−s+iα

=

∑∞

n= 2

cnn−s,

where

cn={ 2 +χ(n)n−iα+ ̄χ(n)niα}Λ(n)= 2 { 1 +R(χ(n)n−iα)}Λ(n).

Since|χ(n)|≤1and|n−iα|=1, it follows thatcn≥0foralln≥2. If we put

g(s)=

∑∞

n= 2

cnn−s/logn,

theng′(s)=f′(s)/f(s)forσ>1 and so the derivative ofe−g(s)f(s)is

{f′(s)−g′(s)f(s)}e−g(s)= 0.

Thusf(s)=Ceg(s),whereCis a constant. In factC =1, sinceg(σ)→0and
f(σ)→1asσ→+∞.Sinceg(s)is the sum of an absolutely convergent Dirichlet
series with nonnegative coefficients, so also are the powersgk(s)(k = 2 , 3 ,...).
Hence also

f(s)=eg(s)= 1 +g(s)+g^2 (s)/2!+···=

∑∞

n= 1

ann−s forσ> 1 ,

wherean≥0foreveryn. It follows from Proposition 6 that the series

∑∞

n= 1 ann

−σ

must actually converge with sumf(σ)forσ≥ 1 /2. We will show that this leads to a
contradiction.
Ta k en=p^2 ,wherepis a prime. Then, by the manner of its formation,

an≥cn/logn+cp^2 /2log^2 p

={ 2 +χ(p)^2 p−^2 iα+ ̄χ(p)^2 p^2 iα}/ 2 +{ 2 +χ(p)p−iα+ ̄χ(p)piα}^2 / 2

= 2 −χ(p)χ( ̄ p)+{ 1 +χ(p)p−iα+ ̄χ(p)piα}^2 ≥ 1 ,

since|χ(p)|≤1. Hence

f( 1 / 2 )=

∑∞

n= 1

an/n^1 /^2 ≥

∑

n=p^2

an/n^1 /^2 ≥

∑

p

1 /p.

Since

∑

p^1 /pdiverges, this is a contradiction.