Number Theory: An Introduction to Mathematics

372 IX The Number of Prime Numbers

It follows at once from Proposition 5 thatζ(s)=0forσ>1, since the infinite
product is convergent and each factor is nonzero.

Proposition 6−ζ′(s)/ζ(s)=

∑∞

n= 1 Λ(n)/n

sforσ> 1 ,whereΛ(n)denotes von

Mangoldt’s function.

Proof The seriesω(s)=

∑∞

n= 1 Λ(n)n

−sconverges absolutely and uniformly in any

half-planeσ≥ 1 +ε,whereε>0, since

0 ≤Λ(n)≤logn<nε/^2 for all largen.

Hence

ζ(s)ω(s)=

∑∞

m= 1

m−s

∑∞

k= 1

Λ(k)k−s

=

∑∞

n= 1

n−s

∑

d|n

Λ(d).

Since

∑

d|nΛ(d)=logn, by (4), it follows that

ζ(s)ω(s)=

∑∞

n= 1

n−slogn=−ζ′(s).

Sinceζ(s)=0forσ>1, the result follows. However, we can also prove directly that
ζ(s)=0forσ>1, and thus make the proof of the prime number theorem independent
of Proposition 5.
Obviously ifζ(s 0 )=0forsomes 0 withRs 0 >1thenζ′(s 0 )=0, and it follows
by induction from Leibniz’ formula for derivatives of a product thatζ(n)(s 0 )=0for
alln≥0. Sinceζ(s)is holomorphic forσ>1 and not identically zero, this is a
contradiction.

Proposition 6 may be restated in terms of Chebyshev’sψ-function:

−ζ′(s)/ζ(s)=

∫∞

1

u−sdψ(u)=

∫∞

0

e−sxdψ(ex) forσ> 1. (6)

We are going to deduce from (6) that the functionζ(s)has no zeros on the lineRs=1.
Actually we will prove a more general result:

Proposition 7Let f(s)be holomorphic in the closed half-planeRs≥ 1 , except for a
simple pole at s= 1 .If,forRs> 1 ,f(s)= 0 and

−f′(s)/f(s)=

∫∞

0

e−sxdφ(x),

whereφ(x)is a nondecreasing function for x≥ 0 ,then

f( 1 +it)= 0 for every real t= 0.