(August 29, 2013)
Counting zeros of ζ(s)
Paul Garrett [email protected]
http://www.math.umn.edu/egarrett/
[This document is http://www.math.umn.edu/˜garrett/m/mfms/notes 2013-14/counting zeros of zeta.pdf]
Zeros of ζ(s) in the critical strip 0 ≤ Re (s) ≤ 1 are counted [1] using the argument principle and the
Laplace-Stirling asymptotic
Γ(s) = (s − 12 ) log s − s + O(1)
(in Re (s) ≥ δ > 0 as |s| → ∞)
The counting function of interest is [2]
N (T ) = number of zeros of ζ(s) in 0 ≤ Im (s) ≤ T and 0 ≤ Re (s) ≤ 1
By the argument principle [3]
N (T ) =
1
2
1 Z ξ 0 (s) ·
ds + O(1)
2πi RT ξ(s)
where RT is the rectangle connecting 2 ± iT and −1 ± iT , traversed counter-clockwise, deformed slightly
to skirting any zeros of ζ(s). The division by 2 takes into account the double-counting of zeros off the
real interval [0, 1], and the O(1) term accommodates miscounting poles at s = 0, 1 and any zeros on [0, 1].
The proof of the following is simply an estimate of this integral, specifically, giving the leading terms in an
asymptotic in T .
[0.0.1] Theorem:
N (T ) =
log 2πe
1
T
1
· T log T −
· T + O(log T ) =
T log
+ O(log T )
2π
2π
2π
2πe
[0.0.2] Remark: In particular, there are infinitely-many zeros of ζ(s) in the critical strip.
[0.0.3] Remark: The vertical asymptotics of Γ(s) dominate and completely determine the leading terms of
the asymptotic expansion, by a direct computation which determines the relevant constants.
Proof: Using the functional equation ξ(1 − s) = ξ(s), and the symmetry ξ(s) = ξ(s), we integrate only
upward from 2 to 2 + iT , and then left from 2 + iT to 21 + iT .
The argument-principle integral computes 1/2π times the net change in the imaginary part of log ξ(s) over the
given paths, requiring continuity of the logarithm. We compute separately the net changes of the imaginary
parts of the summands in
s
s
log ξ(s) = − log π + log Γ( ) + log ζ(s)
2
2
[1] From [Backlund 1914, 1916, 1918]. See also [Titchmarsh/Heath-Brown 1951/1989] pages 212-213. [Backlund 1916]
was a thesis done under E. Lindelöf’s supervision.
[2] The convergent Euler product shows that there are no zeros in Re (s) > 1. The functional equation π − 2s Γ( s )ζ(s) =
2
1−s
s
π − 2 Γ( 1−s
2 )ζ(1 − s) shows that the only zeros of ζ(s) in Re (s) < 0 are where Γ( 2 ) has poles, namely, negative even
integers. These are the trivial zeros of ζ(s).
[3] As usual, ξ(s) is the completed zeta function ξ(s) = π −s/2 Γ( s ) ζ(s). The usual notation is S(T ) = 1 arg ζ(s),
2
π
required to be 0 at s = 2, and continuous along the vertical line from 2 to 2 + iT and then to
a zero along (0, 1) + iT , compute S(T ) slightly above T .
1
1
2
+ iT . When there is
Paul Garrett: Counting zeros of ζ(s) (August 29, 2013)
Obviously the net change of imaginary part of the logarithm of π −s/2 is
1
Im log π −( 2 +iT )/2 − log π −2/2
= Im −
1
2
+ iT
T
log π = − log π
2
2
From
log Γ(s) = (s − 21 ) log s − s + O(1)
we have
log Γ( 2s ) = ( 2s − 21 ) log
Thus, the net change from 2 to
1
2
s
2
−
s
2
+ O(1)
+ iT is
1
+ iT
2 Im log Γ( 2
) − log Γ( ) = Im
2
2
1
2
+ iT
− 12 log
2
1
2
+ iT
−
2
1
2
+ iT + O(1)
2
1 iT πi
T
T
T
T
T
= Im (− +
)
+ log + O( T1 ) − + O(1) =
log − + O(1)
4
2
2
2
2
2
2
2
Since s = 2 + iR is nicely within the region of absolute convergence of the Euler product, log ζ(2 + it) is
bounded on that line, so the net change in the imaginary part of the argument of ζ(s) from 2 to 2 + iT is
O(1).
The subtle computation concerns the net change in the argument of ζ(s) from 2 + iT to 12 + iT . We recall
a version of part of a relevant lemma from [Titchmarsh/Heath-Brown 1951/1989], page 213, which uses
Jensen’s Lemma to approximate the number of zeros, hence, the change in argument, in terms of the growth,
of a function:
[0.0.4] Lemma: Let f be a holomorphic function on a vertical strip −1 ≤ σ ≤ 4, except possibly for a
simple pole at s = 1. Suppose that f (s) = f (s). Assume a lower bound |f (2 + it)| ≥ m > 0, and a family of
upper bounds
|f (σ + it)| ≤ M (T )
(for 14 ≤ σ ≤ 4 and 1 ≤ t ≤ T )
Then, for T not the vertical coordinate of a zero of f , there is the upper bound for change in argument from
2 + iT to 21 + iT
| arg f ( 21 + iT ) − arg f (2 + iT )| ≤
π
log((2 −
1
4 )/(2
−
1
2)
1
· log M (T + 2) + log
+π
m
[0.0.5] Remark: Naturally, some of the details are insignificant, being mere artifacts of the proof. At the
same time, we give a more specific version of the result than [Titchmarsh/Heath-Brown 1951/1989].
Proof: Let q be the number of vanishings of Ref (σ + iT ) between 2 + iT and 12 + iT . The vanishings divide
the interval into q + 1 subintervals on each of which either Re f ≥ 0 or Re f ≤ 0. In particular, the value
of f stays in either the right or left half-plane, so the arg f cannot change more than π in each subinterval.
Thus,
| arg f ( 12 + iT ) − arg f (2 + iT )| ≤ (q + 1) · π
Using f (s) = f (s), the count q is the number of zeros of g(z) = 21 [f (z + iT ) + f (z − iT )] on the real interval
1
1
2 ≤ z ≤ 2. Certainly this count is bounded by the number of zeros of g(z) in the disk |z − 2| ≤ 2 − 2 .
2
Paul Garrett: Counting zeros of ζ(s) (August 29, 2013)
Let ν(r) be the number of zeros in |z − 2| ≤ r. Setting up application of Jensen’s lemma, [4] we have an
upper bound for q:
2− 41
Z
0
ν(r)
dr ≥
r
Z
2− 14
1
2− 2
2−
ν(r)
dr ≥ ν(2 − 12 ) ·
r
2−
1
4
1
2
≥ q·
2−
2−
1
4
1
2
Jensen’s lemma leads to an upper bound for the integral:
2− 14
Z
0
ν(r)
1
dr =
r
2π
Z
2π
0
1
1
log |g(2 + (2 − )eiθ )| dθ − log |g(2)| ≤ log M (T + 2) + log
4
m
giving the lemma.
///
The lemma applies to f (s) = ζ(s) since the convergent Euler product is bounded away from 0 on 2 + iR,
with bound M (t) = tN on a given vertical strip. Thus, the net change in the argument of ζ(s) from 2 + iT
to 21 + iT is O(log T ).
Altogether, the argument principle gives
T
1
T
T
·4·
log − −
2π
2
2
2
1 T
T
T
T log T
=
·
log − (1 + log π) + O(log T ) =
π
2
2
2
2π
log 2πe
1
· T log T −
·T
=
2π
2π
which is the asserted asymptotic.
N (T ) =
1
2
·
T
log π + O(log T )
2
−
T
T
log 2 −
(1 + log π) + O(log T )
2π
2π
+ O(log T )
///
Bibliography:
[Backlund 1914] R.J. Backlund, Sur les zéros de la fonction ζ(s) de Riemann, C.R. 158 (1914), 1979-81.
[Backlund 1916] R.J. Backlund, Über die Nullstellen der Riemannschen Zetafunktion, Dissertation,
Helsingfors, 1916.
[Backlund 1918] R.J. Backlund, Über die Nullstellen der Riemannschen Zetafunktion, Acta Math. 41 (1918),
345-75.
[Titchmarsh/Heath-Brown 1951/1989] E.C. Titchmarsh, The theory of the Riemann zeta function, Oxford
University Press, 1951. Second edition, revised by D.R. Heath-Brown, 1986.
[4] Jensen’s Lemma usually appears as follows: for holomorphic f on |z| ≤ r, no zeros on |z| = r, and f (0) 6= 0,
log |f (0)| −
X
ρ
Z 2π
ρ
1
log =
log |f (reiθ )| dθ
r
2π 0
(summed over zeros |ρ| < r of f )
Letting ν(t) be the number of zeros of size less than t,
−
X
ρ
Z r
X
X Z r dt
ρ
dt
log =
(log r − log |ρ|) =
=
ν(t)
r
t
t
|ρ|
0
ρ
ρ
3