Number Theory: An Introduction to Mathematics

382 IX The Number of Prime Numbers

Sinceζ(s ̄)=ζ(s), the zeros ofζ(s)are also symmetric with respect to the real
axis. Furthermoreζ(s)has no real zeros in the strip 0<σ<1, since

( 1 − 2 −^1 −σ)ζ(σ)=( 1 − 2 −σ)+( 3 −σ− 4 −σ)+···>0for0<σ< 1.

It has been verified by van de Luneet al.(1986), with the aid of a supercomputer,
that the 1. 5 × 109 zeros ofζ(s)in the rectangle 0<σ< 1 , 0 <t<T,where
T= 545439823 .215, are all simple and lie on the lineσ= 1 /2.
The location of the zeros ofζ(s)is intimately connected with the asymptotic
behaviour ofπ(x).Letα∗denote the least upper bound of the real parts of all zeros
ofζ(s).Then1/ 2 ≤α∗≤1, since it is known thatζ(s)does have zeros in the strip
0 <σ<1, and the Riemann hypothesis is equivalent toα∗= 1 /2. It was shown by
von Koch (1901) that

ψ(x)=x+O(xα

∗

log^2 x)

and hence

π(x)=Li(x)+O(xα

∗

logx).

(Actually von Koch assumedα∗= 1 /2, but his argument can be extended without
difficulty.) It should be noted that these estimates are of interest only ifα∗<1.
On the other hand if, for someαsuch that 0<α<1,

π(x)=Li(x)+O(xαlogx),

then

θ(x)=x+O(xαlog^2 x).

By the remark after the proof of Lemma 3, it follows that

ψ(x)=x+x^1 /^2 +O(xαlog^2 x)+O(x^1 /^3 log^2 x).

But forσ>1wehave

−ζ′(s)/ζ(s)=

∫∞

1

x−sdψ(x)=s

∫∞

1

ψ(x)x−s−^1 dx

and hence

−ζ′(s)/ζ(s)−s/(s− 1 )−s/(s− 1 / 2 )=s

∫∞

1

{ψ(x)−x−x^1 /^2 }x−s−^1 dx.

The integral on the right is uniformly convergent in the half-planeσ≥ε+max(α, 1 / 3 ),
for anyε>0, and represents there a holomorphic function. It follows that
1 / 2 ≤α∗≤max(α, 1 / 3 ). Consequentlyα∗≤αandψ(x)=x+O(xαlog^2 x).
Combining this with von Koch’s result, we see that the Riemann hypothesis is
equivalent to

π(x)=Li(x)+O(x^1 /^2 logx)