Number Theory: An Introduction to Mathematics

6 Alternative Formulations 391

M(x)=O(xα

∗+ε

) for everyε> 0.

It follows thatthe Riemann hypothesis holds if and only if M(x)=O(xα)for every
α> 1 /2.
It has already been mentioned that the first 1. 5 × 109 zeros ofζ(s)on the line
σ = 1 /2 are all simple. It is likely that the Riemann hypothesis does not tell the
whole story and that all zeros ofζ(s)on the lineσ = 1 /2 are simple. Thus it is
of interest that this is guaranteed by a sufficiently sharp bound forM(x). We will
show thatif

M(x)=O(x^1 /^2 logαx) as x→∞,

for someα<1,then not only do all nontrivial zeros ofζ(s)lie on the lineσ= 1 / 2
but they are all simple.

Letρ= 1 / 2 +iγbe a zero ofζ(s)of multiplicitym≥1andtakes=ρ+h,
whereh>0. Thenσ= 1 / 2 +hand, since

1 /ζ(s)=s

∫∞

1

x−s−^1 M(x)dxforσ> 1 / 2 ,

we have

| 1 /ζ(s)|≤|s|

∫∞

1

x−σ−^1 |M(x)|dx=O(|s|)

∫∞

1

x−h−^1 logαxdx

=O(|s|)

∫∞

0

e−huuαdu=O(|s|)Γ(α+ 1 )/hα+^1.

Thushα+^1 | 1 /ζ(s)|is bounded forh→+0 and hencem≤α+1. Sincemis an integer
andα<1, this impliesm=1andα≥0.

The prime number theorem, in the formM(x)=o(x), says that asymptotically
μ(n)takes the values+1and−1 with equal probability. By assuming that actually
the valuesμ(n)asymptotically behave like independent random variables Good and
Churchhouse (1968) have been led to two striking conjectures, analogous to the central
limit theorem and the law of the iterated logarithm in the theory of probability:

Conjecture AIf N(n)→∞andlogN/logn→ 0 ,then

Pn

{

M(m+N)−M(m)

( 6 N/π^2 )^1 /^2

<t

}

→( 2 π)−^1 /^2

∫t

−∞

e−u

(^2) / 2
du,
where
Pn{f(m)<t}=#{m≤n:f(m)<t}/n.