33

数理解析研究所講究録
第2013巻 2016年 33-43

On the zeros of the derivatives of the Riemann zeta function
under the Riemann hypothesis

Ade Irma Suriajaya
Graduate School of Mathematics, Nagoya University
Abstract
The number of zeros and the distribution of the real part of non‐real zeros of thederivatives of the Riemann zeta function have been investigated by Berndt, Levin‐son,

Montgomery, and Akatsuka. Berndt, Levinson, and Montgomery investigated

the general case, meanwhile Akatsuka gave sharper estimates for the first derivativeof the Riemann zeta function under the truth of the Riemann hypothesis. In this re‐port, we introduce a generalization of the results of Akatsuka to the k‐th derivative(for positive integer k) of the Riemann zeta function.

1

Introduction
Zeros of the derivatives of the Riemann zeta function$\zeta$(s) have been studied forabout 80 years. In 1935, Speiser [Spe] showed that the Riemann hypothesis is equivalent
to the first derivative of the Riemann zeta function{\rm Re}(s)<1/2
function.

.

$\zeta$'(s) having no non‐real zeros inThis result is a breakthrough in the study of zeros of the Riemann zeta

Following the work of Speiser, Spira [Spi65, Spi70] studied the zero‐free regionsof higher order derivatives of the Riemann zeta function, we write $\zeta$^{(k)}(s) to denote the k‐th derivative of the Riemann zeta function for positive integers kfurther study in the zeros of $\zeta$^{(k)}(s)number of zeros of$\zeta$^{(k)}(s)

.

.

.

These results encourage

For example, in 1970, Berndt[Ber] investigated theHe [Ber, Theorem] proved that for any positive integer k,

N_{k}(T)=\displaystyle \frac{T}{2 $\pi$}\log\frac{T}{4 $\pi$}-\frac{T}{2 $\pi$}+O(\log T)(1.1)

holds, where N_{k}(T) denotes the number of zeros of $\zeta$^{(k)}(s) with 0<{\rm Im}(s)\leq T counted
Furthermore, in 1973, Spira [Spi73] also studied the relation between
the zeros of $\zeta$'(s) and the Riemann hypothesis. In 1974, Levinson and Montgomery[LM] studied many properties related to the distribution of zeros of $\zeta$^{(k)}(s) including thelocation of zeros. They [LM, Theorem 10] also showed that for any positive integer k,,with multiplicity.

,$\zeta$^{(k)}($\rho$^{(k)}=0, <$\gam a$^{(k)\prime}\displayst le\leqT\sum_{$\rho$^{(k)}-$\beta$^{(k)}+i$\gam a$^{(k)}($\beta$^{(k)}-\frac{1}2)=\frac{kT}{2$\pi$}\log\log\frac{T}2$\pi$}+\frac{1}2$\pi$}(\frac{1}2\log2-k\log\log2)T-k\displaystyle \mathrm{L}\mathrm{i}(\frac{T}{2 $\pi$})+(1.2)
O(\log T)2010 Mathematics Sub_{2}ect Classification:

Primary llM06.
Keywords and phrases: Riemann zeta function, derivative, zeros.
The original paper [Sur] is to appear in Functiones et Approximatio Commentarii Mathematici.
This work was partly supported by Nitori International Scholarship Foundation and Iwatani Naoji
Foundation.

34

holds, where the sum is counted with multiplicity and\displaystyle\mathrm{L}\mathrm{i}(x):=\int_{2}^{x}\frac{dt}{\logt}.
The above estimate shows how the real parts of non‐real zeros of $\zeta$^{(k)}(s)around the critical line {\rm Re}(s)=1/2.

The zeros of$\zeta$^{(k)}(s)

near

are

distributed

the critical line were thenstudied further by Conrey and Ghosh [CG] in 1989.

[Yi196, YilOO] investigated non‐real zeros of $\zeta$''(s) and $\zeta$'' (s) in the{\rm Re}(s)<1/2 He succeeded in
and
that
that
the
Riemannhypothesis implies$\zeta$'' (s) each has only one pairshowing$\zeta$''(s)of non‐real zeros in {\rm Re}(s)<1/2 Unfortunately, results analogous to Speisers [Spe] were
not obtained. Currently, no results similar to Speisers [Spe] are known for higher order
In 1996, Yildirim

region to the left of the critical line, that is in the region.

.

derivatives.
In 2012, Akatsuka[Aka, Theorems 1 and 3] improved each of the error term of the(1.1)

results obtained by Berndt and by Levinson and Montgomery mentioned above (eq.

and (1.2)) for the case k=1 under the assumption of the truth of the Riemann hypothesis.
He showed that$\zeta$'($\rho$,)=0, <$\gam a$^{i_\leq}T\displaystyle\sum_{$\rho$'=$\beta$'+i$\gam a$'}($\beta$'-\frac{1}{2})=\frac{T}{2$\pi$}\log\log\frac{T}{2$\pi$}+\frac{1}{2$\pi$}(\frac{1}{2}\log2-\log\log2)T-\displaystyle \mathrm{L}\mathrm{i}(\frac{T}{2 $\pi$})+O( \log\log T)^{2})

and

N_{1}(T)=\displaystyle \frac{T}{2 $\pi$}\log\frac{T}{4 $\pi$}-\frac{T}{2 $\pi$}+O(\frac{\log T}{(\log\log T)^{1/2} )hold if the Riemann hypothesis is true. In this report, we are interested in investigating
Akatsukas method in the case when k\geq 2.

Some notation and main results2

Before wc introduce our results, we define some notation.
In this report we denote by \mathbb{R} and \mathb {C} the set of all real numbers and the set of allcomplex numbers, respectively. Throughout this report, the letter k is used as a fixed
positive integer, unless otherwise specified. For convenience, we let $\rho$^{(k)}=$\beta$^{(k)}+i$\gamma$^{(k)}represent non‐real zeros of $\zeta$^{(k)}(s).

Each of the following results introduced in this report is a generalization of Theorem 1,

Corollary 2, and Theorem 3 of [Aka], respectively. Note that each sum counts the non‐realof $\zeta$^{(k)}(s) with multiplicity and that O_{k} denotes the error terms which depend only

zeros

on

k.

Theorem 1. Assume that the Riemann hypothesis is true.

Then for any T>4 $\pi$ ,have$\rho$^{(k)}-$\beta$^{(k)}+i$\gam $^{(k)}\displayte\sum_{0<$\gam $^{(k)}\leqT},($\beta$^{(k)}-\displaystyle \frac{1}{2})=\frac{kT}{2 $\pi$}\log\log\frac{T}{2 $\pi$}+\frac{1}{2 $\pi$}(\frac{1}{2}\log 2-k\log\log 2)T

we

35

-k\displaystyle \mathrm{L}\mathrm{i}(\frac{T}{2 $\pi$})+O_{k}( \log\log T)^{2})
Corollary 2. (Cf. [LM,.

Theorem 3].) Assume that the Riemann hypothesis is true. Thenfor 0<U<T (where T is restricted to satisfy T>4 $\pi$ ),

we

have

$\rho^{(k)}=$\beta$^{(k)}+i$\gam $^{(k)}\displayte\sum_{T<$\gam $^{(k)}\leqT+U}, ($\beta$^{(k)}-\displaystyle \frac{1}{2})=\frac{kU}{2 $\pi$}\log\log\frac{T}{2 $\pi$}+\frac{1}{2 $\pi$}(\frac{1}{2}\log 2-k\log\log 2)U+O(\displaystyle \frac{U^{2} {T\log T})+O_{k}( \log\log T)^{2})

Here the

error

term

any parameters.

o(\displaystyle \frac{U^{2} {T\log T})

holds

.

uniformly, in other words, it does not depend on

Theorem 3. Assume that the Riemann hypothesis is true. Then for T\geq 2_{f}N_{k}(T)=\displaystyle \frac{T}{2 $\pi$}\log\frac{T}{4 $\pi$}-\frac{T}{2 $\pi$}+O_{k}(\frac{\log T}{(\log\log T)^{1/2} )

we

have

,

where N_{k}(T) is as defined in equation (1.1).

We write {\rm Re}(s) and {\rm Im}(s)(for any s\in \mathbb{C} ) as $\sigma$ and t respectively.

We abbreviate

the Riemann hypothesis as RH, and finally, we define two functions F(s) and G_{k}(s),asfollows:

F(s):=2^{s}$\pi$^{s-1}\displaystyle \sin(\frac{ $\pi$ s}{2}) $\Gamma$(1-s) , G_{k}(s):=(-1)^{k}\frac{2^{s} {(\log 2)^{k} $\zeta$^{(k)}(s)
By the above definition of F(s),we can.

check easily that the functional equation for $\zeta$(s)states$\zeta$(s)=F(s) $\zeta$(1-s).

Sketch of proofs

3

In this report, we mainly give only sketch of the proofs. Refer to the original paper[Sur] for the details.
3.1

Key lemmas

We first introduce a few lemmas and propositions which are analogues of those in[Aka].
Lemma 3.1. There exists an a_{k}\geq 10 such that

|G_{k}(s)-1|\displaystyle \leq\frac{1}{2}(\frac{2}{3})^{ $\sigma$/2}
holds for any $\sigma$\geq a_{k}.

36\squareProof. See [LM, inequality (3.2) (p. 54
Lemma 3.2. There exists a $\sigma$_{k}\leq-1 such that|\displayst le\sum_{j=1}^{k}\left(\begin{ar y}{l k\ j \end{ar y}\right)(-1)^{j-1}\frac{1}\frac{F(k)}{F(k-,)}(s)}\frac{$\zeta$^{(j)}{$\zeta$}(1-s)|\leq2^{$\sigma$}holds in the region $\sigma$\leq$\sigma$_{k}, t\geq 2.

Proof. (Sketch)
We begin by estimatingin the region\displaystyle \frac{F^{(k)} {F(k-j)}(s) (j=1,2, \cdots, k)$\sigma$<1, t\geq 2 Using methods similar to [LM, pp. 54−55], we can show that.

for any positive integer k , we can take $\sigma$_{k_{1}}\leq-1 sufficiently small (i.e.

sufficiently large

in the negative direction) so that for any s with $\sigma$\leq$\sigma$_{k_{1} and t\geq 2 , we have|\displaystyle \frac{F^{(k)} {F(k-j)}(s)|\geq\frac{1}{2k}(\log(1- $\sigma$) ^{j}

(3.1)

.

Next we estimate\displaystyle \frac{$\zeta$^{(j)} { $\zeta$}(1-s) (j=1,2, \cdots, k).

In the region $\sigma$\leq-1, t\geq 2 , we can make use of the Dirichlet series of $\zeta$(s) and$\zeta$^{(j)}(s) toobtain|\displaystyle\frac{$\zeta$^{(j)}{$\zeta$}(1-s)|\leq\frac{2^{$\sigma$}{2-\frac{$\pi$^{2}{6}(\frac{1}{2}(\log2)^{j}+\sum_{l=0}^{j}\frac{(\log2)^{j-l}\frac{j^1}{(j-l)!}{(-$\sigma$)^{l+1}\mathrm{I}(3.2)

Now combining inequalities (3.1) and (3.2), for $\sigma$\leq$\sigma$_{k_{1} and t\geq 2 , and noting that forany positive integer k,\displayst le\lim_{$\sigma$\rightar ow-\infty}\frac{2k}{2-\frac{$\pi$^{2}{6}\sum_{j=1}^{k}\left(\begin{ar ay}{l k\ j \end{ar ay}\right)\frac{1}(\log(1-$\sigma$)^{j}(\frac{1}2(\log2)^{j}+\sum_{l=0}^{j}\frac{(\log2)^{j-l}\frac{j!}(j-l)^{1} {(-$\sigma$)^{l+1}\mathrm{I}=0,
we canfind $\sigma$_{k}\leq$\sigma$_{k_{1}}(\leq-1) such that\displaystyle\frac{2k}{2-\frac{$\pi$^{2}{6}\sum_{j=1}^{k}\left(\begin{ar ay}{l k\ j \end{ar ay}\right)\frac{1}{(\log(1-$\sigma$)^{j}(\frac{1}{2}(\log2)^{j}+\sum_{l=0}^{j}\frac{(\log2)^{j-l}\frac{j!}(j-l)^{1} {(-$\sigma$)^{l+1})\leq1
holds for any $\sigma$\leq$\sigma$_{k}.

This implies our lemma.

\squareNow we fix the above a_{k} and $\sigma$_{k} to show the following lemma.
Lemma 3.3. Assume RH Then there exists a t_{k}\displaystyle \geq\max\{a_{k}^{2}, -$\sigma$_{k}\} such that the following.

conditions are satisfied:37

1. For any ssatisfying $\sigma$_{k}\leq $\sigma$\leq 1/2 and t\geq t_{k}-1,|\displaystyle \frac{F^{(k)} {F}(s)|\geq 1
holds.

Furthermore, we can take a branch of \log(F^{(k)}/F)(s) in that region such that

it is holomorphic there and\displaystyle \frac{$\alpha$_{k} $\pi$}{6}<\arg\frac{F^{(k)} {F}(s)<\frac{$\beta$_{k} $\pi$}{6}holds, where($\alpha$_{k},$\beta$_{k})=\left\{ begin{ar ay}{l (5,7)ifkisod ,\ (-1, )ifkisev n. \end{ar ay}\right.
2. For any s satisfying $\sigma$_{k}\leq $\sigma$<1/2 and t\geq t_{k}-1_{f}\displayst le\frac{$\zeta$^{(k)}{$\zeta$}(s)\neq0
holds.

Furthermore, we can take a branch of \log($\zeta$^{(k)}/ $\zeta$)(s) in that region such that

it is holomorphic there and\displaystyle\frac{k$\pi$}{2}<\arg\frac{$\zeta$^{(k)} {$\zeta$}(s)<\frac{3k$\pi$}{2}holds.
3. For all $\sigma$\in \mathbb{R} ,

we

have

$\zeta$( $\sigma$+it_{k})\neq 0, $\zeta$^{(k)}( $\sigma$+it_{k})\neq 0.
Proof. (Sketch)
1, we apply Stirlings formula and methods similar to the proof of
inequality (3.1). We can show that
To prove conditionF^{(k)}(s)=F(s)(- \displaystyle \log(1-s)+O(1))^{k}(1+O(\frac{1}{|\log s|^{2} ))holds in the region $\sigma$_{k}\leq $\sigma$\leq 1/2, t\geq 99

.

(3.3)

Thus for any integer k\geq 1 , we can take somet_{k_{1}}\geq 100 such that|\displaystyle \frac{F^{(k)} {F}(s)|\geq 1(3.4)

holds for $\sigma$_{k}\leq $\sigma$\leq 1/2 and t\geq t_{k_{1}}-1.
We note from equation(3.3) that (F^{(k)}/F)(s)=(-1)^{k}(\log t)^{k}+O((\log t)^{k-1}) when$\sigma$_{k}\leq $\sigma$\leq 1/2 and t\geq 99 Consequently, for odd integer k\geq 1.

sufficiently large such that\displaystyle \frac{5 $\pi$}{6}<\arg\frac{F^{(k)} {F}(s)<\frac{7 $\pi$}{6},

we canfind t_{k_{2}}'\geq 100

38

holds for $\sigma$_{k}\leq $\sigma$\leq 1/2 and t\geq t_{k_{2}}'-1t_{k_{2}}''\geq 100.

Similarly, when k is even, we can also findlarge enough such that-\displaystyle \frac{ $\pi$}{6}<\arg\frac{F^{(k)} {F}(s)<\frac{ $\pi$}{6}holds for$\sigma$_{k}\leq $\sigma$\leq 1/2 and t\geq t_{k_{2}}''-1.

Since allpoles of F(s) lie on
inequality (3.4) implies that

zeros

and

poles
(F^{(k)}/F)(s)\log(F^{(k)}/F)(s) is holomorphic in the region with this branch. We set
has\mathbb{R},noin t>0.

This

along with($\alph$_{k},$\beta$_{k}):=\left{\begin{ar y}{l (5,7)&\mathrm{i}\ athrm{f}k\mathrm{i}\ athrm{s}\mathrm{o}\mathrm{d}\mathrm{d},\ (-1,)&\mathrm{i}\ athrm{f}\ athrm{k}\mathrm{i}\ athrm{s}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}; \end{ar y}\right.
and

t_{k2}:=\left{begin{ary}l t_{k2}',&\mathr{i}\mathr{f}k\mathr{i}\mathr{s}\mathr{o}\mathr{d}\mathr{d},\ t_{k2}',&\mathr{i}\mathr{f}k\mathr{i}\mathr{s}\mathr{e}\mathr{v}\mathr{e}\mathr{n}. \end{ary}\ight.
From the above calculations, we find that \displaystyle \max\{t_{k_{1} , t_{k_{2} , a_{k}^{2}, -$\sigma$_{k}\} is a candidate for t_{k}.

Thus we have proven that t_{k}\displaystyle \geq\max\{a_{k}^{2}, -$\sigma$_{k}\} for which condition 1 holds exists. Since
we

want

t_{k} to also satisfy conditions 2 and 3, we need to examine those conditions tocompletely prove the existence of t_{k}.
To prove condition 2, we first make use of the finiteness of the number of non‐real zeros

of $\zeta$^{(j)}(s) in the regionof Theorem 7 (p. 51$\sigma$<1/2 under RH for any positive integer j (cf. [LM, Corollary
to find some t_{k_{3} such that for all j=1,2, \cdots, k , we have$\zeta$^{(j)}(s)\neq 0
in the region(3.5)$\sigma$<1/2, t\geq t_{k_{3}}-1.

Next we show that we can take a branch of \log($\zeta$^{(k)}/ $\zeta$)(s) in the region $\sigma$_{k}\leq $\sigma$<

1/2, t\geq t_{k_{4}}-1 for some t_{k_{4}}\geq 100 so that it is holomorphic there and,\displaystyle\frac{k$\pi$}{2}<\arg\frac{$\zeta$^{(k)} {$\zeta$}(s)<\frac{3k$\pi$}{2}holds there by making use of the following inequality{\rm Re}(\displaystyle \frac{$\zeta$^{(j)} {$\zeta$^{(j-1)} (s) \leq-\frac{2}{9}\log|s+O_{$\sigma$_{k} (1)which holds for any j=1 , 2, \cdots, k when $\sigma$_{k}\leq $\sigma$<1/2 and t\geq t_{k_{3}}-1(see [LM, pp.

64−65]). We omit details of the proof here.
We then have \displaystyle \max\{t_{k_{1} , t_{k_{2} , t_{k_{3} , t_{k_{4} , a_{k}^{2}, -$\sigma$_{k}\} as a candidate for t_{k}.
Now we set t_{k_{5} :=\displaystyle \max\{t_{k_{1}}, t_{k_{2}}, t_{k_{3}}, t_{k_{4}}, a_{k}^{2}, -$\sigma$_{k}\}.

Since we are assuming RH,$\zeta$( $\sigma$+it)\neq 0 for any t>0 if $\sigma$\neq 1/2.

According to [Spi65, Table 1 (p. 678) and Theorem 1], for any positive integer k,
we

have

$\zeta$^{(k)}( $\sigma$+it)\neq 0 ( $\sigma$\geq 7k/4+2, t\in \mathbb{R}).

39

Since t_{k_{5}}\geq t_{k_{3}},from (3.5), we have $\zeta$^{(k)}( $\sigma$+it)\neq 0 forHence, for any positive integer k$\zeta$(1/2+it_{k})\neq 0,

weonly need to find t_{k}\in[t_{k_{5}}+1, t_{k_{5}}+2] for which$\zeta$^{(k)}( $\sigma$+it_{k})\neq 0

and$\sigma$<1/2 and t\geq t_{k_{5}}.

for 1/2\leq $\sigma$\leq 7k/4+2

hold.

Thus, we have shown that t_{k} defined above satisfies t_{k}\displaystyle \geq\max\{a_{k}^{2}, -$\sigma$_{k}\} and also
conditions 1 to 3.

\squareWe now fix a_{k}, $\sigma$_{k} , and t_{k} which satisfy Lemmas 3.1, 3.2, and 3.3.
Now we give two bounds for— \arg $\zeta$( $\sigma$+iT)+\arg G_{k}( $\sigma$+iT).

We use methods similar

to [Aka, Lemmas 2.3, 2.4, 2.6]. We take the logarithmic branches of \log $\zeta$(s) and \log G_{k}(s)such that they tend to 0 as $\sigma$\rightarrow\infty and are holomorphic in \mathbb{C}\backslash \{ $\rho$+ $\lambda$| $\zeta$( $\rho$)=0 or \infty,0\} and \mathb {C}\backslash {$\rho$^{(k)}+ $\lambda$|$\zeta$^{(k)}($\rho$^{(k)})=0‐

or$\lambda$\leq$\lambda$\leq 0 }, respectively. We write\infty,\displaystyle \arg $\zeta$( $\sigma$+iT)+\arg G_{k}( $\sigma$+iT)=\arg\frac{G_{k} { $\zeta$}( $\sigma$+iT),

where the argument on the right hand side is determined so that \log(G_{k}/ $\zeta$)(s) tends to
0 as $\sigma$\rightarrow\infty and is holomorphic in \mathb {C}\backslash {z+ $\lambda$|($\zeta$^{(k)}/ $\zeta$)(z)=0 or \infty, $\lambda$\leq 0 }.

Lemma 3.4. Assume RH and let T\geq t_{k}(since T\geq t_{k}\geq 100, $\epsilon$_{0}<1/8)

,

we

.

Then for any $\epsilon$_{0}>0 satisfying $\epsilon$_{0}<(2\log T)^{-1}have for 1/2+$\epsilon$_{0}< $\sigma$\leq a_{k},\displaystyle\arg\frac{G_{k}{$\zeta$}($\sigma$+iT)=O_{a_{k},t_{k}(\frac{\log\frac{\logT}{$\epsilon$_{0} {$\sigma$-\frac{1}2-$\epsilon$_{0})We omit the proof of the above lemma (refer to [Sur, Lemma 2.3]).
Lemma 3.5. Assume RH and let A\geq 2 be fixed.

Then there exists a constant C_{0}>0such that|$\zeta$^{(k)}( $\sigma$+it)|\displaystyle \leq\exp(C_{0}(\frac{(\log T)^{2(1- $\sigma$)} {\log\log T}+(\log T)^{1/10})
holds for T\geq t_{k}, T/2\leq t\leq 2T,

1/2-(\log\log T)^{-1}\leq $\sigma$\leq A.

Proof Referring to [Tit, (14.14.2), (14.14.5) and the first equation on p. 384], we canshow that| $\zeta$( $\sigma$+it)|\displaystyle \leq\exp(C_{1}(\frac{(\log T)^{2(1- $\sigma$)} {\log\log T})+(\log T)^{1/10})(3.6)

holds for 1/2-2(\log\log T)^{-1}\leq $\sigma$\leq A+1, T/3\leq t\leq 3T for some constant C_{1}>0 (cf.

[Aka, pp. 2251−2252]).

Applying Cauchys integral formula, we see that$\zeta$^{(k)}(s)=\displaystyle \frac{k!}{2 $\pi$ i}\int_{|z-s|= $\epsilon$}\frac{ $\zeta$(z)}{(z-s)^{k+1} dz

for

0< $\epsilon$<1/2

holds in the region defined by 1/2-(\log\log T)^{-1}\leq $\sigma$\leq A and T/2\leq t\leq 2T

inequality (3.6) and by taking $\epsilon$=(2(\log\log T)^{1/k})^{-1}(<1/2)

,

we

.

Applying

obtain Lemma 3.5.

\square40

Lemma 3.6. Assume RH and let T\geq t_{k}.

Then for any 1/2\leq $\sigma$\leq 3/4 ,\displaystyle \mathrm{a}x\mathrm{g}G_{k}( $\sigma$+iT)=O_{a_{k} (\frac{(\log T)^{2(1- $\sigma$)} {(\log\log T)^{1/2} )
Proof

we

have

.

The proof proceeds in the same way as the proof of Lemma 2.4 of [Aka]. Refer to[Aka, pp. 2252−2253] for the detailed proof and use Lemma 3.5 above in place of Lemma
2.6 of [Aka].

\squareRemark 1. The restrictions of the lower bound of T we gave in Lemmas 3.4, 3.5, and
3.6 are not essential, but they are sufficient for our needs.

3.2

Proof of Theorem 1

The following proposition gives the main term of Theorem 1.

Proposition 3.7. Assume RH Take a_{k} and t_{k} which satisfy Lemmas 3.1 and 3.3 respec‐tively. Then for T\geq t_{k} which satisfies $\zeta$^{(k)}( $\sigma$+iT)\neq 0 and $\zeta$( $\sigma$+iT)\neq 0 for any $\sigma$\in \mathbb{R},
.

we

have

$\rho$^{(k)}-$\beta$^{(k)}+i$\gam $^{(k)}\displayt e\sum_{0<$\gam $^{(k)}\leqT}, ($\beta$^{(k)}-\displaystyle \frac{1}{2})=\frac{kT}{2 $\pi$}\log\log\frac{T}{2 $\pi$}+\frac{1}{2 $\pi$}(\frac{1}{2}\log 2-k\log\log 2)T-k\mathrm{L}\mathrm{i}(\frac{T}{2 $\pi$})+\displaystyle \frac{1}{2 $\pi$}\int_{1/2}^{a_{k} (- \arg $\zeta$( $\sigma$+iT)+\arg G_{k}( $\sigma$+iT) d $\sigma$+O_{k}(1)where the logarithmic branches are taken as in Section 3.1,(see the paragraph preceding

Lemma 3.4).
We omit the proof (refer to [Sur, Proposition 2.2]). The proof of Theorem 1 is doneasfollows.
First of all, we consider for T\geq t_{k} which satisfies

for any $\sigma$\in \mathbb{R}.

$\zeta$^{(k)}( $\sigma$+iT)\neq 0 and $\zeta$( $\sigma$+iT)\neq 0

From Lemma 3.4, we have\displaystyle\int_{1/2+ $\epsilon$_{0}^{a_{k}\arg\frac{G_{k}{$\zeta$}($\sigma$+iT)d$\sigma$\l_{a_{k},t_{k}\int_{1/2+ $\epsilon$_{0}^{a_{k}\frac{\log\frac{\logT}{$\epsilon$_{0} {$\sigma$-\frac{1}2-$\epsilon$_{0}d$\sigma$\l_{a_{k}\log\frac{\logT}{$\epsilon$_{0}\log\frac{1}$\epsilon$_{0}.
Next, from Lemma 3.6,

\displaystyle \arg G_{k}( $\sigma$+iT)=O_{a_{k} (\frac{(\log T)^{2(1- $\sigma$)} {(\log\log T)^{1/2} )and from equation (2.23) of [Aka, p.

for

1/2\leq $\sigma$\leq 3/42251] (cf. [Tit, equations (14.14.3) and (14.14.5)]),

RH implies that\displaystyle \arg $\zeta$( $\sigma$+iT)=O(\frac{(\log T)^{2(1- $\sigma$)} {\log\log T})

holds uniformly for 1/2\leq $\sigma$\leq 3/4

.

Thus,

\displaystyle\int_{1/2}^{1/2+2$\epsilon$_{0} \arg\frac{G_{k} {$\zeta$}($\sigma$+iT)d$\sigma$\l_{a_{k} \frac{\logT}{(\log\logT)^{1/2} $\epsilon$_{0}.

41Now we take $\epsilon$_{0}=(4\log T)^{-1}(<(2\log T)^{-1}) , then we have\displaystyle \int_{1/2}^{a_{k} \arg\frac{G_{k} { $\zeta$}( $\sigma$+iT)d $\sigma$\l _{a_{k},t_{k} (\log\log T)^{2}
Applying this to Proposition 3.7 and noting that a_{k} and t_{k} are fixed constants that depend
only on k

,

have

we

$\rho$^{(k)}-$\beta$^{(k)}+i$\gam $^{(k)}\displayt e\sum_{0<$\gam $^{(k)}\leqT}, ($\beta$^{(k)}-\displaystyle \frac{1}{2})=\frac{kT}{2 $\pi$}\log\log\frac{T}{2 $\pi$}+\frac{1}{2 $\pi$}(\frac{1}{2}\log 2-k\log\log 2)T-k\mathrm{L}\mathrm{i}(\frac{T}{2 $\pi$})+O_{k}((\log\log T)^{2})

.

(3.7)

For 4 $\pi$<T<t_{k} , we are adding some finite number of terms which depend on t_{k} , andthus depend only on k so this can be included in the error term.
For T\geq t_{k} such that$\zeta$^{(k)}( $\sigma$+iT)=0 or $\zeta$( $\sigma$+iT)=0 for some $\sigma$\in \mathbb{R} there is some,increment in the value ofas$\rho$^{(k)}-$\beta$^{(k)}+i$\gam a$^{(k)}\displaystle\sum_{0<$\gam a$^{(k)}\leqT},($\beta$^{(k)}-\frac{1}2)$\rho$^{(k)}-$\beta$^{(k)}+i$\gam a$^{(k)}\displaystle\sum_{$\gam a$^{(k)}=T,($\beta$^{(k)}-\frac{1}2)much as.

We estimate this and show that this can be included in the error term of equation (3.7).

We start by taking a small 0< $\epsilon$<1 such that $\zeta$^{(k)}( $\sigma$+i(T\pm $\epsilon$))\neq 0 and $\zeta$( $\sigma$+i(T\pm $\epsilon$))\neq 0
for any $\sigma$\in \mathbb{R}$\rho^{(k)}=$\beta$^{(k)}+i$\gam $^{(k)}\displayte\sum_{0<$\gam $^{(k)}\leqTpm$\epsilon$},.

According to equation (3.7),($\beta$^{(k)}-\displaystyle \frac{1}{2})=\frac{k(T\pm $\epsilon$)}{2 $\pi$}\log\log\frac{T\pm $\epsilon$}{2 $\pi$}+\frac{1}{2 $\pi$}(\frac{1}{2}\log 2-k\log\log 2)(T\pm $\epsilon$)-k\displaystyle \mathrm{L}\mathrm{i}(\frac{T\pm $\epsilon$}{2 $\pi$})+O_{k}( \log\log T)^{2})

.

Thus,

$\rho$^{(k)}-$\beta$^{(k)}+i$\gam a$^{(k)}\displayst le\sum_{T-$\epsilon$< \gam a$^{(k)}\leqT+$\epsilon$'}($\beta$^{(k)}-\frac{1}2)=\frac{k$\epsilon$}{$\pi$}\log\log\frac{T}2$\pi$}+\frac{$\epsilon$}{$\pi$}(\frac{1}2\log2-k\log\log2)+O_{k}(\log\logT)^{2})=O_{k}((\log\log T)^{2}).

This implies$\rho$^{(k)}-$\beta$^{(k)}+i$\gam a$^{(k)}\displaystyle\sum_{$\gam a$^{(k)}=T},($\beta$^{(k)}-\frac{1}2)=O_{k}(\log\logT)^{2}).

Therefore, this increment can also be included in the error term and the proof is complete.
\square42

Proof of Corollary 2

3.3(the ending part ofThis is an immediate consequence of Theorem 1. See [LM, p. 58

Section 3\squareProof of Theorem 3

3.4

Finally, we give the proof of Theorem 3. We first introduce the following proposition
which give the main term of our estimate.

Proposition 3.8. Assume RH Take t_{k} which satisfies all conditions of Lemma 3.3. Thenfor T\geq 2 which satisfies $\zeta$( $\sigma$+iT)\neq 0 and $\zeta$^{(k)}( $\sigma$+iT)\neq 0 for all $\sigma$\in \mathbb{R} we have.

,N_{k}(T)=\displaystyle \frac{T}{2 $\pi$}\log\frac{T}{4 $\pi$}-\frac{T}{2 $\pi$}+\frac{1}{2 $\pi$}\arg G_{k}(\frac{1}{2}+iT)+\frac{1}{2 $\pi$}\arg $\zeta$(\frac{1}{2}+iT)+O_{k}(1),where the arguments are determined as in Proposition 3. 7.
We omit the proof (refer to[Sur, Proposition 3.1]).

The proof of Theorem 3 isasfollows.

Firstly we consider for T\geq 2 which satisfies $\zeta$^{(k)}( $\sigma$+iT)\neq 0 and $\zeta$( $\sigma$+iT)\neq 0 for
By Lemma 3.6,

any $\sigma$\in \mathbb{R}.

\displaystyle \arg G_{k}(\frac{1}{2}+iT)=O_{a_{k} (\frac{\log T}{(\log\log T)^{1/2} )and again from equation (2.23) of [Aka, p.

2251], we have\displaystyle \arg $\zeta$(\frac{1}{2}+iT)=O(\frac{\log T}{\log\log T}).

Substituting these into Proposition 3.8, we obtain

N_{k}(T)=\displaystyle \frac{T}{2 $\pi$}\log\frac{T}{4 $\pi$}-\frac{T}{2 $\pi$}+O_{k}(\frac{\log T}{(\log\log T)^{1/2} )Next, if $\zeta$( $\sigma$+iT)=0 or $\zeta$^{(k)}( $\sigma$+iT)=0 for some $\sigma$\in \mathbb{R}a.

(T\geq 2) then again we take,small 0< $\epsilon$<1 such that $\zeta$^{(k)}( $\sigma$+i(T\pm $\epsilon$))\neq 0 and $\zeta$( $\sigma$+i(T\pm $\epsilon$))\neq 0 for any $\sigma$\in \mathbb{R} as

in the proof of Theorem 1. Then similarly, we can show that the increment of the valueof N_{k}(T)canbe included in the error term of the above equation which completes theproof.

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Graduate School of Mathematics

Nagoya University
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m12026a@math.nagoya‐u.ac.jp,

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