Zeros of the Derivatives of the Riemann Zeta-function ∗

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