Number Theory: An Introduction to Mathematics

390 IX The Number of Prime Numbers

ζ(s− 1 )/ζ(s)=

∑∞

n= 1

φ(n)/ns.

From the property by which we defined the M ̈obius function we obtain also, for
Rs>1,

1 /ζ(s)=

∑∞

n= 1

μ(n)/ns.

In view of this relation it is not surprising that the distribution of prime numbers is
closely connected with the behaviour of the M ̈obius function. Put

M(x)=

∑

n≤x

μ(n).

Since|μ(n)|≤1, it is obvious that|M(x)|≤xforx>0. The next result is not so
obvious:

Proposition 12M(x)/x→ 0 as x→∞.

Proof The functionf(s):=ζ(s)+ 1 /ζ(s)is holomorphic forσ≥1, except for a
simple pole with residue 1 ats=1. Moreover

f(s)=

∑∞

n= 1

{ 1 +μ(n)}/ns=

∫∞

1 −

x−sdφ(x)forσ> 1 ,

whereφ(x)=x+M(x)is a nondecreasing function. Since

f(s)=

∫∞

0 −

e−sudφ(eu),

it follows from Ikehara’s Theorem 9 thatφ(x)∼x.

Proposition 12 is equivalent to the prime number theorem in the sense that either of
the relationsM(x)=o(x),ψ(x)∼xmay be deduced from the other by elementary
(but not trivial) arguments.
The Riemann hypothesis also has an equivalent formulation in terms of the func-
tionM(x). Suppose

M(x)=O(xα) asx→∞,

for someαsuch that 0<α<1. Forσ>1wehave

1 /ζ(s)=

∫∞

1 −

x−sdM(x)=s

∫∞

1

x−s−^1 M(x)dx.

But forσ>αthe integral on the right is convergent and defines a holomorphic func-
tion. Consequently it is the analytic continuation of 1/ζ(s). Thus ifα∗again denotes
the least upper bound of all zeros ofζ(s),thenα≥α∗≥ 1 /2. On the other hand,
Littlewood (1912) showed that