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Math 259: Introduction to Analytic Number Theory Proof of the Prime Number Theorem; the Riemann Hypothesis
We finally have all the ingredients that we need to assemble a proof of the Prime Number Theorem with an explicit error bound. We shall give an upper bound on |(ψ(x)/x) − 1 | that decreases faster than any power of 1/ log x as x→∞, though slower than any positive power of 1/x. Specifically, we show:
Theorem. There exists an effective constant C > 0 such that
ψ(x) = x + O(x exp(−C
log x)) (1)
for all x ≥ 1.
Proof : There is no difficulty with small x, so we may and shall assume that x ≥ e, so log x ≥ 1. We use our integral approximation
ψ(x) =
2 πi
∫ (^) 1+ (^) log (^1) x +iT
1+ (^) log^1 x −iT
ζ′ ζ (s) xs^
ds s
+ O
x log^2 x T
(T ∈ [1, x]) (2)
to ψ(x). Assume that T ≥ e, and that T does not coincide with the imaginary part of any ρ. Shifting the line of integration leftwards, say to real part −1, yields
ψ(x) −
x −
| Im(ρ)|<T
xρ ρ
= I 1 + I 2 −
ζ′ ζ
(0) + O
x log^2 x T
in which I 1 , I 2 are the integrals of −(ζ′(s)/ζ(s))xs^ ds/s over the vertical line σ = − 1 , |t| < T and the horizontal lines σ ∈ [− 1 , 1 + 1/ log x], t = ±T respec- tively. We next show that I 1 is small, and that I 2 can be made small by adding O(1) to T. The vertical integral I 1 is clearly
log T x sup |t|<T
ζ′ ζ (−1 + it)
∣ ^
log^2 T x
The horizontal integrals in I 2 are
T
∫ (^) 1+ (^) log (^1) x
− 1
xσ^ dσ · sup σ∈[− 1 ,2]
ζ′ ζ (σ + iT )
The σ integral is x/ log x. We have seen already that for s = σ + iT and − 1 ≤ σ ≤ 2 we have
ζ′(s)/ζ(s) =
|T −Im ρ|< 1
s − ρ
in which the sum has O(log T ) terms. Since the number of Im ρ in the interval [T − 1 , T + 1] is log T , some point in the middle half of that interval is at distance 1 / log T from all of them; choosing that as our new value of T , we see that each term is log T , and thus that the sum is log^2 T. In conclusion, then, I 2 x log^2 T /T log x.
Better estimates can be obtained (we could save a factor of log T by averaging over [T − 12 , T + 12 ]), but are not necessary because x log^2 T /T log x is already
less than the error (x log^2 x)/T in (2).
Thus the RHS of (3) may be absorbed into the O((x log^2 x)/T ) error. In the LHS, we use our zero-free region, that is, the lower bound
1 − σ > c/ log |t|, (4)
to find that
|xρ| = xRe(ρ)^ x^1 −^ logc T = x exp
−c log x log T
Since^1 ∑
| Im(ρ)|<T
|ρ|
| Im(ρ)|<T
| Im ρ|
∫ T
1
dN (t) t
2 N (T )
T
∫ T
1
N (t) dt t^2
log T +
∫ T
1
log t dt t
log^2 T,
we thus have (^) ∑
|ρ|<T
xρ ρ
x log^2 T exp
−c
log x log T
Therefore (^) ∣ ∣ ∣ ∣
ψ(x) x
T
−c log x log T
log^2 x.
We choose T so that the logarithms − log T , − log x/ log T of the two terms 1 /T , exp(−c log x/ log T ) are equal. That is, we take T = exp
log x. Then both terms are O(exp(−C 1 log^1 /^2 x)) for some C 1 > 0. We then absorb the factor log^2 x into this estimate by changing C 1 to any positive C < C 1 , and at last complete the proof of (1).
The equivalent result for π(x) follows by partial summation:
Corollary. There exists an effective constant C > 0 such that
π(x) = li(x) + O(x exp(−C
log x)).
for all x ≥ 1.
(^1) We can use ∫^1 T because we have shown that there are no complex zeros ρ with | Im(ρ)| ≤ 1. If there were such zeros, we could absorb their terms xρ/ρ into the error estimate. We shall do this in the proof of the corresponding estimates on ψ(x, χ).
xθ^ /|ρ| < xθ^ /| Im ρ|. Hence ∣∣ ∣ ∣ ∣
| Im(ρ)|<T
xρ ρ
< 2 xθ^
0 <Im(ρ)<T
Im ρ
We have seen already that the last sum is O(log^2 T ); here T = x + O(1), so we conclude that
ψ(x) − x = O(xθ^ log^2 T ) + O(log^2 x) = O(xθ^ log^2 x),
as claimed. The corresponding estimate on π(x) − li(x) then follows from (5), since θ ≥ 1 /2.
A converse implication also holds:
Proposition. Suppose there exists θ with 1 / 2 ≤ θ < 1 such that ψ(x) = x + O(xθ+) for all > 0. Then ζ(s) has no zeros of real part > θ. The same conclusion holds if π(x) = li(x) + O(xθ+).
(So, for instance, RH is equivalent to the assertion that π(x) = li x+O(x^1 /^2 log x). The hypotheses on π(x) and ψ(x) are equivalent, again by (5).)
Proof : Write −ζ′(s)/ζ(s) =
n Λ(n)n
−s (^) as a Stieltjes integral and integrate
by parts to find
−
ζ′ ζ
(s) = s
1
ψ(x)x−s−^1 dx =
s s − 1
1
(ψ(x) − x) x−s−^1 dx (σ > 1).
If ψ(x) − x xθ+^ then the resulting integral for s/(s − 1) + ζ′(s)/ζ(s) extends to an analytic function on σ > θ, whence that half-plane contains no zeros of ζ(s).
Note the amusing consequence that an estimate ψ(x) = x + O(xθ+) would automatically improve to ψ(x) = x + O(xθ^ log^2 x), and similarly for π(x).
Remarks
One may naturally ask whether ψ(x) tends to be larger or smaller than its approximation x, and likewise whether π(x) tends to be larger or smaller than li(x). For the former question, our formula (3) suggests that ψ(x) can as easily be larger or smaller than x: the terms xρ/ρ in the formula (3) for x − ψ(x) oscillate as x increases, and if we choose log x uniformly from [1, U ] then the phase of each term tends to uniform distribution on the circle as U →∞. It may be surprising then that π(x) behaves quite differently: it is very hard to find any x such that π(x) > li(x). This is because π(x) is expressed as a Stieltjes integral involving not ψ(x) but
p<x log^ p, and
ψ(x) −
p<x
log p ∼ ψ(x^1 /^2 ) ∼ x^1 /^2.
Under the Riemann Hypothesis, x^1 /^2 is exactly of the same asymptotic order as each of the terms xρ/ρ in (3), and much larger than each single term because
|ρ|−^1 < 1 /14. For large x, we can imagine the terms xρ/ρ (Im ρ > 0) as random complex numbers∑ zρ drawn independently from the circle |z| = x^1 /^2 /ρ.^2 Then
ρ x
ρ/ρ = 2 Re ∑ (Im ρ)> 0 zρ.^ Since^
ρ 1 /|ρ|
(^2) < ∞, this heuristic suggests
that for “random large x” the scaled error x−^1 /^2 (ψ(x) − x) is drawn from a distribution symmetric about the origin, and thus that x−^1 /^2 (
p<x log^ p^ −^ x) is drawn from a distribution symmetric about −1. Since
ρ 1 /|ρ|^ = +∞, it is possible for −2 Re
(Im ρ)> 0 zρ^ to exceed^ x, and thus for^
p<x log^ p^ to exceed x and likewise for π(x) to exceed li(x). But this does not happen routinely, and indeed it was once thought that li(x) might always exceed π(x).
Littlewood first showed that the difference changes sign infinitely often. In particular, there exist x such that π(x) > li(x). But none has been found yet. The earliest explicit upper bound on the smallest such x was the (in)famously astronomical “Skewes’ number” [Skewes 1933]. That bound has since fallen, but still stands at several hundred digits, too large to reach directly even with the best algorithms known for computing π(x) — algorithms that themselves depend on the analytical formulas such as (2); see [LO 1982].
Exercises
- Use the partial-fraction decomposition of ζ′/ζ to get the following exact formula:
ψ(x) = x −
ρ
xρ ρ
ζ′ ζ
log(1 − x−^2 ).
Here
ρ is taken to mean limT^ →∞
|ρ|<T ; and if^ x^ =^ p
k, so that ψ(x) is
discontinuous at x, then we interpret ψ(x) as (ψ(x − ) + ψ(x + ))/2. Note that − 12 log(1 − x−^2 ) is the sum of −xr^ /r over the trivial zeros r = − 2 , − 4 , − 6 ,... See [Davenport 1967, Chapter 17].
- Show that the improvement 1−σ > c/ log(2/3)+^ |t| on (4) yields an estimate O
x exp(−C log(3/5)−^ x)
on the error in the Prime Number Theorem.
- Prove that
lim x→∞
log x −
∑^ x
n=
Λ(n) n
= γ,
and give an error bound both unconditionally and under the Riemann Hypoth- esis. Deduce that log x −
p<x log^ p/p^ and log log^ x^ −^
p<x 1 /p^ approach finite limits as x→∞. (The last of these refines Euler’s theorem that
p 1 /p^ diverges.)
- [A theorem of Mertens; see for instance [Titchmarsh 1951], pages 38–39.] Prove that
lim x→∞
log log x −
∑^ x
n=
Λ(n) n log n
= −γ,
(^2) We shall later make this heuristic more precise, and show that it is equivalent to the conjecture that the numbers γ > 0 such that ζ( 12 + iγ) = 0 are Q-linearly independent. This conjecture is almost certainly true and extremely difficult to prove. See [RS 1994] and [BFHR 2001] for more information.
[Hasse 1936] Hasse, H.: Zur Theorie der abstrakten elliptischen Funktionk¨orper I, II, III, J. reine angew. Math. 175 (1936), 55–62, 69–88, and 193–208.
[EHOPR 1987] Erd˝os, P., Hildebrand, A., Odlyzko, A., Pudaite, P., Reznick, B.: The asymptotic behavior of a family of sequence, Pacific J. Math. 126 (1987), 227–241.
[LO 1982] Lagarias, J.C., Odlyzko, A.M.: New algorithms for computing π(x). Pages 176–193 in Number Theory: New York 1982 (D.V. and G.V. Chudnovsky, H. Cohn, M.B. Nathanson, eds.; Berlin: Springer 1984, LNM 1052 ).
[Rawsthorne 1984] Rawsthorne, D.A.: Problem 1185, Math. Magazine 57 (1984), p.42.
[RS 1994] Rubinstein, M. Sarnak, P.: Chebyshev’s Bias. Exp. Math. 3 (1994) #3, 173–197.
[Skewes 1933] Skewes, S.: On the difference π(x) − li(x) (I). J. London Math. Soc. (1st ser.) 8 (1933), 277–283.
[Weil 1940] Weil, A.: Sur les fonctions alg´ebriques `a corps de constantes fini, C.R. Acad. Sci. Paris 210 (1940), 592–594.
[Weil 1941] Weil, A.: On the Riemann hypothesis in functionfields, Proc. Nat. Acad. Sci. USA 27 (1941), 345–347.
[Weil 1948] Weil, A.: Vari´et´es ab´eliennes et courbes alg´ebriques, Paris: Hermann
