Zeros of the Derivatives of the Riemann Zeta-function ∗

Zeros of the Derivatives of the Riemann
Zeta-function ∗
Haseo Ki and Yoonbok Lee
Department of Mathematics, Yonsei University, Seoul 120–749, Korea
1 Introduction
haseo@yonsei.ac.kr
leeyb@yonsei.ac.kr
September 15, 2009
We study properties of zeros of the derivatives of the Riemann zeta function ζ(s). Levinson
and Montgomery [8] achieved several important theorems for the behavior of zeros of ζ (m) (s)
(m = 1, 2, 3, · · · ). If we assume the Riemann hypothesis, ζ ′ (s) has no non-real zero in Re s <
1
2 and ζ(m) (s) (m > 1) has at most finitely many zeros in Re s < 1.
Unconditionally, we are
2
able to deduce the following quantitative results by similar methods in [8].
Theorem 1. We denote ρ (m) = β (m) + iγ (m) as zeros of ζ (m) (s). Let 0 < U � T . Then, we
have
�
� �
1
− β(m) �
2 �
�
β − 1
�
+ O(U).
2
T

Theorem 3. For T a � U � T , a > 1,
we have
2
2π
�
�
β (m) − 1
�
= mU log log T + O(U).
2
T 0. Then we have
� T log T
1 � µm
2π (1 + om(1)) (T → ∞),
β (m) > 1
2
0 0.70.
We note that this improves D. W. Farmer’s result κd � 0.63952 in [5].
2

2 Lemmas
We start with the following.
Lemma 1. Let m = 1, 2, 3, . . ., χ(s) = 2sπs−1 sin πsΓ(1
− s) and s = σ + it (σ, t ∈ R).
2
Then, we have
χ (m)
χ (s) = (− log |t|)m + O(log m−1 |t|)
for |t| � t0 on any fixed vertical strip a � σ � b.
Proof of Lemma 1. From the Sterling formula, we have
Γ ′
(s) = log |t| + O(1);
Γ
dm dsm � ′ Γ
Γ (s)
�
= O(t −m ) (m = 1, 2, 3, . . .).
Thus we have
χ ′
χ
(s) = −Γ′
Γ
πs sin 2
(1 − s) + log 2π +
cos πs
2
Suppose Lemma 1 is true for m � k. Then, we have
χ (k+1) � (k) χ
(s) =
χ χ (s)
�′
+ χ(k)
χ (s)χ′
χ (s)
= − log |t| + O(1).
=O(log k |t|) + ((− log |t|) k + O(log k−1 |t|))(− log |t| + O(1))
=(− log |t|) k+1 + O(log k |t|).
By induction, we have proved the lemma.
Lemma 2. Fix a nonnegative integer m. There is t1 > 0 such that χ (m) (s) has no zero or pole
in |Im s| � t1, a � Re s � b.
Proof of Lemma 2. By the definition of χ(s) = 2sπs−1 sin πsΓ(1
− s), we know that χ(s)
2
is meromorphic on the whole complex plane with poles at s = 1, 3, 5, 7, · · · , and zeros at
s = 0, −2, −4, · · · . Thus, the Lemma 2 is true for m = 0. For the case m > 0, we use the
Lemma 1
χ (m) (s) = χ(s) χ(m)
χ (s) = χ(s)(− logm |t|) � 1 + O � log −1 |t| ��
for |Im s| = |t| � t0. Thus, we prove the lemma.
Lemma 3. Let −2 � aj � 2, bj, cj � 0, 0 < p < 1. Then, we have
� 2 �
�
�
�
cj �
�
�
x − aj + ibj
p
dx � 8
�
�
� �
�
� cj�
1 − p
p
.
−2
For Lemma 3, see [8, Lemma 4.1] or [6, Chap. 4].
j
j
3

3 Proofs of Theorems 1, 2, and 3
Proof of Theorem 1. We begin with the functional equation of the Riemann zeta function
ζ(1 − s) = χ(1 − s)ζ(s). By differentiating m times, we have
Let Jm(s) be
ζ (m) (1 − s) = χ (m) (1 − s)ζ(s) +
Jm(s) = ζ(s) +
m−1 �
j=0
m−1 �
j=0
� �
m
(−1)
j
m−j χ (j) (1 − s)ζ (m−j) (s).
� �
m m−j χ(j)
(−1)
j χ (m) (1 − s)ζ(m−j) (s). (3.1)
We know that there is Am > 1
2 such that ζ(m) (s) has no zero on Re s � Am. Consider the
rectangle with vertices 1
1
+ i(T + U), 2 2 + iT , Am + iT , Am + i(T + U). Since ζ (m) (1 −
s) = χ (m) (1 − s)Jm(s), all the zeros of Jm(s) in the rectangle are the same as the zeros of
ζ (m) (1 − s), and no poles there by Lemma 2. Now, we can apply the Littlewood Lemma [10,
Chap. 9.9] to get
� T +U
1
2π T
�
�
log �
�
Jm( 1
2
ζ( 1
2
�
+ it) �
�
+ it) � dt =
�
T

Proof of Claim. We recall that in [1], the number of zeros of ζ (k) (s) with 0 < γ (k) < T is
T T
log
2π 4πe + Ok(log T ). (3.4)
Let n be a large positive integer. For |t − n| � 1, 0 < σ < 1, k = 0, 1, 2, · · · , we have
Then, by this, (3.4) and Lemma 3, we have
� n+ 1
2
n− 1
2
�
�
�
�
ζ (k+1)
ζ (k)
From this, we have
�
1
+ it
2
ζ (k+1) �
(s) =
ζ (k)
|γ (k) 1
+ O(log n).
s − ρ (k)
−n|

4 Proofs of Theorems 4 and 5
Proof of Theorem 4. Apply the Littlewood Lemma to deduce
� � T
1 �
log �
2π �
2
JkB
� ��
1 R ���
− + it dt =
2 L �
β (k) � 1
2
0

where the infimum takes over all polynomials P satisfying P (0) = 0, P (1) = 1. Since we are
taking infimum over certain polynomial, it is equivalent to take any continuously differentiable
sinh λx
function with P (0) = 0, P (1) = 1. Thus, let P (x) = sinh λ . Then we have � 1
0 P (x)2dx =
� �
1
−2λ 1 + O λe 2λ
�� , and � 1
0 P ′ (x) 2dx = λ
� �
−2λ 1 + O λe 2
�� . Thus, we have
By taking
inf
P c(P, Qk, R) �1 +
� 1
e
0
2Ry Q(y)(Q ′ (y) + RQ(y))dy
+ λ � � −2λ
1 + O λe
2θ
�� � 1
0
+ θ � � −2λ
1 + O λe
2λ
�� � 1
�
�
�
λ = θ�
0
e 2Ry Q(y) 2 dy
� 1
0 e2Ry (Q ′ (y) + RQ(y)) 2dy � 1
0 e2RyQ(y) 2 ,
dy
e 2Ry (Q ′ (y) + RQ(y)) 2 dy.
we get the minimal value of the right hand side in the previous inequality. Since
� 1
0
e 2Ry Qk(y) 2 dy = 1 � � −1
1 + O k
2k
�� ,
� 1
we have λ = kθ (1 + O (k −1 )), and
0
� 1
e 2Ry Q ′ k(y) 2 dy = k � � −1
1 + O k
2
�� ,
inf
P c(P, Qk, R) � 2 + O � k −1�
0
e 2Ry Q ′ k(y)Qk(y)dy = 1 � � −1
1 + O k
2
�� ,
as k → ∞. By this together with (4.3) and (3.4), we conclude that for any fixed R > 0
where µk = 1 −
log 2
2R
�
β (k) > 1
2
0

with the mollifier B(s) introduced in Theorem A. For the error term O(T/L), we need some
condition of gj that will be discussed at the end of the proof. We note that simple zeros of ξ(s)
are not zeros of f(s), besides multiple zeros of ξ(s) are zeros whose multiplicities decrease
by one. From symmetry of zeros of ξ(s) to 1/2, the left hand side of (4.4) is not less than
R
L (N(T ) − Nd(T )). By Jensen’s inequality, we can deduce that
or
N(T ) − Nd(T ) �
T L
4πR log
�
1
T
� T
κd � 1 − 1
2R log
�
1
lim sup
T →∞ T
2
�
�
�
�fB � T
2
� �� �
2
1 R ���
− + it dt
2 L
�
�
�
�fB � �� �
2
1 R ���
− + it dt .
2 L
+ O(T )
All we need is to get an asymptotic formula for the mean square of fB. We have
f(s) = (g + g0)ζ(s) + g1
� ′ H
L
H (s)ζ(s) + ζ′ �
(s) =
� � s � �
log 2π 1 d �1 � −1
Q1 + ζ(s) + O |t|
2L L ds
�� ,
where Q1(x) = g + g0 + g1x. Using the last two equations, integration by parts leads us to
κd � 1 − 1
2R log
� � � T � �� �
2
1 �
�
1 R ���
T �V B − + it dt ,
2 L
lim sup
T →∞
where V (s) = Q � − 1
� �
d
1
ζ(s), and Q(x) = Q1
L ds
2 − x� = g + g0 + 1
2g1 − g1x. When
we applied the Littlewood lemma in (4.4), we needed the condition Q(0) = 1. This yields
Q(x) = 1 − g1x. In [2, p.10], Conrey made an optimal choice of this case. If we select g1 =
1.02, R = 1.2, θ = 4
7
References
2
, we have log c
R = 0.598.... Therefore, we conclude that κd > 0.70.
[1] B. C. Berndt, The number of zeros for ζ (k) (s), J. London Math. Soc.(2) 2 (1970), 577–
580
[2] J. B. Conrey, More than two fifths of the zeros of the Riemann zeta function are on the
critical line, J. reine angew. Math. 399 (1989), 1–26
[3] J. B. Conrey, A. Ghosh, and S. M. Gonek, Simple zeros of the Riemann zeta-function,
Proc. London Math. Soc. (3) 76 (1998), 497–522
[4] J. M. Deshouillers and H. Iwaniec, Kloosterman sums and Fourier coefficients of cusp
forms, Invent. Math. 70 (1982), 219–288
8

[5] D. W. Farmer, Counting distinct zeros of the Riemann-zeta function, Electron. J. Combin.
2 (1995), Research Paper 1, approx. 5 pp.
[6] N. Levinson, Gap and Density Theorems, Amer. Math. Soc. Colloq. Publ. Vol. 26, 1940.
[7] N. Levinson, More than One Third of Zeros of Riemann’s Zeta-Function are on σ = 1/2,
Adv. Math. 13 (1974), 383–436
[8] N. Levinson and H. L. Montgomery, Zeros of the derivatives of the Riemann zetafunction,
Acta Math. 133 (1974), 49–65
[9] A. Selberg, On the zeros of Riemann’s zeta function. Skr. Norske Vid. Akad. Oslo,
(1942) no. 10.
[10] E. C. Titchmarsh, The Theory of the Riemann Zeta-function, 2nd ed., revised by D. R.
Heath-Brown, Oxford University Press, Oxford, 1986
9