Li coefficients and the quadrilateral zeta function

Abstract

In this note, we study the Li coefficients $\lambda _{n,a}$ for the quadrilateral zeta function. Furthermore, we give an arithmetic and asymptotic formula for these coefficients. Especially, we show that for any fixed $n \in {\mathbb {N}}$, there exists $a>0$ such that $\lambda _{2n-1,a}> 0$ and $\lambda _{2n,a} < 0$.

1 Introduction and statement of main results

1.1 Li coefficients

The Riemann hypothesis (RH) is a critical question in analytic number theory. As such, it is interesting to examine different ways to frame it, which may shed more light on its resolution. In 1997, Xian-Jin Li has discovered a new positivity criterion for the RH. In [10], he defined the Li coefficients for the Riemann zeta function as

$$ \begin{align*} \lambda_{n}=\frac{1}{(n-1)!}\frac{d^{n}}{ds^{n}}\left[s^{n-1}\log\xi(s)\right]_{s=1}, \end{align*} $$

where $\xi $ is the completed Riemann zeta function defined by

$$ \begin{align*} \xi(s)=s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s), \end{align*} $$

which satisfies $\xi (s)=\xi (1-s)$ and gave a simple equivalence criterion for the RH: RH is true if and only if these coefficients are nonnegative for every positive integer n. The Li coefficients $\lambda _{n}$ can be written as follows:

$$ \begin{align*} \lambda_{n}=\sum_{\rho}^{\ast}\left[1- \left(1-\frac{1}{\rho}\right)^{n}\right]=\lim_{T\to\infty}\sum_{\rho; \lvert {\mathrm{Im}}(\rho) \rvert \leq T}\left[1-\left(1-\frac{1}{\rho}\right)^{n}\right], \end{align*} $$

where the sum runs over the nontrivial zeros of the Riemann zeta function counted with multiplicity. This criterion is generalized by Bombieri and Lagarias [4] for any arbitrarily multiset of numbers assuming certain convergence conditions. Voros [19, Section 3.3] has proved that the RH true is equivalent to the growth of $\lambda _{n}$ as $\frac {1}{2}n \log n$ determined by its archimedean part, while the RH false is equivalent to the oscillations of $\lambda _{n}$ with exponentially growing amplitude, determined by its finite part. The Li coefficients were generalized in two ways: by generalizing these coefficients to various sets of functions (the Selberg class, the class of automorphic L-functions, zeta function on function fields,…[8, 11, 17]) and by introducing new parameter in its definition (see [12]). The Li coefficients (and its generalizations) have generated a lot of research interest due to its applicability and simplicity.

1.2 Quadrilateral zeta function

Recall the definitions of Hurwitz and periodic zeta functions. The Hurwitz zeta function $\zeta (s,a)$ is defined by the series

$$\begin{align*}\zeta (s,a) := \sum_{n=0}^\infty \frac{1}{(n+a)^s}, \qquad \sigma>1, \quad 0<a \le 1. \end{align*}$$

The function $\zeta (s,a)$ is a meromorphic function with a simple pole at $s=1$ whose residue is $1$ (see, for example, [1, Section 12]). The periodic zeta function ${\mathrm {Li}}_s (e^{2\pi ia})$ is defined by

$$\begin{align*}{\mathrm{Li}}_s (e^{2\pi ia}) := \sum_{n=1}^\infty \frac{e^{2\pi ina}}{n^s}, \qquad \sigma>1, \quad 0<a \le 1 \end{align*}$$

(see, for instance, [1, Exercise 12.2]). Note that the function ${\mathrm {Li}}_s (e^{2\pi ia})$ with $0<a<1$ is analytically continuable to the whole complex plane since ${\mathrm {Li}}_s (e^{2\pi ia})$ does not have any pole, that is shown by the fact that the Dirichlet series of ${\mathrm {Li}}_s (e^{2\pi ia})$ converges uniformly in each compact subset of the half-plane $\sigma>0$ when $0<a<1$ (see, for example, [9, p. 20]). For $0 <a \le 1/2$ , we define zeta functions

$$ \begin{align*} \begin{aligned} &Z(s,a) := \zeta (s,a) + \zeta (s,1-a), \qquad P(s,a) := {\mathrm{Li}}_s (e^{2\pi ia}) + {\mathrm{Li}}_s (e^{2\pi i(1-a)}), \\ &2Q(s,a) := Z(s,a) + P(s,a) , \qquad \xi_Q (s,a) := s(s-1) \pi^{-s/2} \Gamma (s/2) Q(s,a). \end{aligned} \end{align*} $$

We can see that $Q(s,a)$ is meromorphic functions with a simple pole at $s=1$ . In addition, we have $Q(0,a)=-1/2=\zeta (0)$ and $\xi _Q(s,a) = \xi _Q(1-s,a)$ , which is proved by

(1.1) $$ \begin{align} Q(1-s,a) = \Gamma_{\!\! {\mathrm{cos}}} (s) Q(s,a), \qquad \Gamma_{\!\! {\mathrm{cos}}} (s) : = \frac{2\Gamma (s)}{(2\pi )^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \end{align} $$

(see [13, Theorem 1.1]). Moreover, the function $Q(s,a)$ has the following properties. When $a=1/6, 1/4, 1/3$ , and $1/2$ , the RH holds true if and only if all nonreal zeros of $Q(s,a)$ are on the line ${\mathrm {Re}}(s)=1/2$ (see [14, Proposition 1.3]). Let $N_{\mathrm {Q}}^{\mathrm {CL}} (T)$ be the number of the zeros of $Q(s,a)$ on the line segment from $1/2$ to $1/2 +iT$ . In [13, Theorem 1.2], the third author proved that for any $0 < a \le 1/2$ , there exist positive constants $A(a)$ and $T_0(a)$ such that

$$\begin{align*}N_{\mathrm{Q}}^{\mathrm{CL}} (T) \ge A(a) T \quad \mbox{whenever} \quad T \ge T_0(a). \end{align*}$$

Next, let $N_{F}(T)$ count the number of nonreal zeros of a function $F(s)$ having $|{\mathrm {Im}}(s)|<T$ . Then, for any $0<a\leq 1/2$ ,

$$ \begin{align*} N_{\zeta}(T)-N_{Q}(T)=O_{a}(T), \end{align*} $$

and the third author [14, Proposition 1.8] proved that

$$ \begin{align*} N_{Q}(T)=\frac{T}{\pi}\log T-\frac{T}{\pi}\log(2e\pi a^{2})+O_{a}(\log T). \end{align*} $$

Furthermore, he [14, Theorem 1.1] proved that there is a unique absolute $a_{0}\in {(0,1/2)}$ such that

$$ \begin{align*} Q(1/2,a)>0\ \ \iff \ 0<a<a_{0}. \end{align*} $$

In addition, it is proved in [14, Corollary 1.2] that all real zeros of $Q(s,a)$ are simple and are located only at the negative even integers just like $\zeta (s)$ if and only if ${a_0 < a \le 1/2}$ . Let us note by $Z_Q$ the set of all nontrivial zeros $\rho _{a}$ of $\xi _{Q}(s,a)$ . Since it is an entire function of order 1, one has

(1.2) $$ \begin{align} \xi_{Q}(s,a)=e^{A+Bs}\prod_{\rho_{a}\in{Z_{Q}}}\left(1-\frac{s}{\rho_{a}}\right)e^{\frac{s}{\rho_{a}}} =\xi_{Q}(0,a)\prod_{\rho_{a}\in{Z_{Q}}}\left(1-\frac{s}{\rho_{a}}\right), \end{align} $$

where $e^{A}=1/2,\ B=\frac {Q'}{Q}(0,a)-1-\frac {\gamma +\log \pi }{2}$ , and $\gamma $ denotes the Euler constant. Note that $Q'(0,a)$ is given explicitly in [14, Theorem 1.5].

1.3 Main results

Recall that $\zeta (1-s) = \Gamma _{\!\! {\mathrm {cos}}} (s) \zeta (s)$ and $Q(1-s,a) = \Gamma _{\!\! {\mathrm {cos}}} (s) Q(s,a)$ by (1.1). However, the function $Q(s,a)$ does not have an Euler product except for $a=1/6, 1/4, 1/3$ , and $1/2$ . Hence, the function $Q(s, a)$ is a suitable object to consider the influence of not Riemann’s functional equation but an Euler product to zeros of zeta functions. We show a criterion for nonvanishing of $Q(s,a)$ in terms of the positivity of the Li coefficients, an arithmetic and asymptotic formula for these coefficients in Theorems 1.1, 1.2, and 1.4, respectively. It should be emphasized that $\lambda _{n,a}$ defined in (1.3) are the first Li coefficients that we can explicitly give $n \in {\mathbb {N}}$ such that $\lambda _{n,a} <0$ . There is a possibility that this fact would give an idea to find negative Li coefficients for $\zeta (s)$ if they would exist.

For $n\neq 0$ , the Li coefficients attached to $Q(s,a)$ nonvanishing at zero are defined by the sum

$$ \begin{align*} \lambda_{n,a}:=\sum_{\rho_{a}\in{Z_{Q}}}^{*}\left(1-\left(1-\frac{1}{\rho_{a}}\right)^{n}\right)= \lim_{T\longmapsto \infty}\sum_{|{\mathrm{Im}}(\rho_{a})|\leq T}^{*}\left(1-\left(1-\frac{1}{\rho_{a}}\right)^{n}\right). \end{align*} $$

The symmetry $\rho _{a}\longmapsto 1-\rho _{a}$ in the set $Z_Q$ of nontrivial zeros of $Q(s,a)$ implies that $\lambda _{-n,a}=\overline {\lambda _{n,a}}=\lambda _{n,a}$ for all $n\in {\mathbb {N}}$ . So, $\lambda _{n,a}$ are real. We have also

(1.3) $$ \begin{align} \lambda_{n,a} :=\frac{1}{(n-1)!} \frac{d^n}{d s^n} \bigl[ s^{n-1} \log \xi_Q(s,a) \bigr]_{s=1}. \end{align} $$

Moreover, from (1.2), we have (see [4, Equations (2.3) and (2.4)] or [17, Appendix A])

$$ \begin{align*} \sum_{n=0}^{\infty}\lambda_{n+1,a}s^{n}=\frac{d}{ds}\log \left[\xi_{Q}\left(\frac{1}{1-s},a\right)\right].\end{align*} $$

As an analogue of Li coefficients for the Riemann zeta function, we have the following.

Theorem 1.1 The function $Q(s,a)$ does not vanish when $\mathrm {Re} (s)>1/2$ if and only if $\lambda _{n,a} \geq 0$ for all $n \in {\mathbb {N}}$ .

An arithmetic formula for $\lambda _{n,a}$ is stated in the following theorems.

Theorem 1.2 We have

$$\begin{align*}\lambda_{n,a} = 1-\frac{n}{2}(\log(4\pi)+\gamma)+\sum_{k=2}^{n}(-1)^{k}\binom{n}{k} \left(1-2^{-k}\right)\zeta(k)+\sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1), \end{align*}$$

where $\gamma _{Q}(n)$ are defined as follows:

$$ \begin{align*} \frac{Q'}{Q}(s+1,a)+\frac{1}{s}=\sum_{n=0}^{\infty}\gamma_{Q}(n)s^{n}. \end{align*} $$

Theorem 1.3 For $a=1/2,1/3,1/4,1/6$ , under the RH, we have

$$ \begin{align*} \lambda_{n,a}=\frac{n}{2}\log n+\frac{n}{2}\left(\gamma-1-\log 2\pi\right)+O(\sqrt{n}\log n). \end{align*} $$

For a fixed $l \in {\mathbb {N}}$ , we have the following asymptotic formula of $\lambda _{l,a}$ when $a \to +0$ . We can see that there exists $n \in {\mathbb {N}}$ such that $\lambda _{n,a} <0$ by Theorem 1.1 and the fact that $Q(s,a)$ does not satisfy an analogue of the RH when $a \in {\mathbb {Q}} \cap (0,1/2) \backslash \{1/6, 1/4, 1/3\}$ (see [14, Proposition 1.4]). Clearly, this argument gives no information on the frequency of $n \in {\mathbb {N}}$ , the smallest $n \in {\mathbb {N}}$ such that $\lambda _{n,a} <0$ and so on. However, the next theorem implies that $\lambda _{2n,a} <0$ if we fix any $n \in {\mathbb {N}}$ and then we take $a>0$ sufficiently small.

Theorem 1.4 Fix $l \in {\mathbb {N}}$ . Then it holds that

$$\begin{align*}\lambda_{l,a} = \frac{(-1)^{l+1}}{(2a)^l} + O_l \bigl( a^{1-l} |\log a| \bigr), \qquad a \to +0. \end{align*}$$

Especially, for any fixed $n \in {\mathbb {N}}$ , there are $a>0$ such that

$$\begin{align*}\lambda_{2n-1,a}> 0 \quad \mbox{ and } \quad \, \lambda_{2n,a} < 0. \end{align*}$$

2 Proofs

2.1 Proof of Theorem 1.1

Since $\lambda _{-n,a}=\overline {\lambda _{n,a}}=\lambda _{n,a}$ for all $n\in {\mathbb {N}}$ , then $\mathrm {Re}(\lambda _{-n,a})=\mathrm {Re}(\lambda _{n,a}) =\lambda _{n,a}$ . Using that $\xi _{Q}(s,a)$ is an entire function of order 1, and its zeros lie in the critical strip ${0< \mathrm {Re}(s)<1}$ , we obtain that the series $\sum _{\rho \in {Z_{Q}}}\frac {1+|\mathrm {Re}(\rho )|}{(1+|\rho |)^{2}}$ is convergent. Application of [4, Theorem 1] to the multiset $Z_{Q}$ of zeros of $Q(s,a)$ gives that $\mathrm {Re}(\rho )\leq 1/2$ if and only if $\lambda _{n,a}\geq 0$ for all $n\in {\mathbb {N}}$ . Now, the application of the same theorem to the multiset $1-Z_{Q}=Z_{Q}$ gives $\mathrm {Re}(\rho )\geq 1/2$ if and only if $\lambda _{n,a}\geq 0$ . This completes the proof.

Theorem 1.1 can also be proved by the same argument used in [5, Theorem 1], which is due to Oesterlé.

2.2 Proof of Theorem 1.2

From the expression of $\xi _{Q}(s,a)$ , one has

$$ \begin{align*}\frac{\xi'_{Q}}{\xi_{Q}}(s,a)=\frac{1}{s}+\frac{1}{s-1}- \frac{1}{2}\log\pi+\frac{1}{2}\frac{\Gamma'}{\Gamma}(s/2)+\frac{Q'}{Q}(s,a),\end{align*} $$

which is rewritten as

(2.1) $$ \begin{align} \frac{\xi'_{Q}}{\xi_{Q}}(s+1,a)=\frac{1}{s+1}+\frac{1}{s}-\frac{1}{2}\log\pi+ \frac{1}{2}\frac{\Gamma'}{\Gamma}((s+1)/2)+\frac{Q'}{Q}(s+1,a). \end{align} $$

Note that $Q(s,a)$ is a meromorphic function on the whole complex plane, which is holomorphic everywhere except for a simple pole at $s = 1$ with residue 1 (see [13, Section 2.1]). Let us define the coefficients $\gamma _{Q}(n)$ and $\tau _{Q}(n)$ as follows:

(2.2) $$ \begin{align} \frac{Q'}{Q}(s+1,a)+\frac{1}{s}=\sum_{n=0}^{\infty}\gamma_{Q}(n)s^{n} \end{align} $$

and

(2.3) $$ \begin{align} -\frac{1}{2}\log\pi+\frac{1}{2}\frac{\Gamma'}{\Gamma}((s+1)/2)=\sum_{n=0}^{\infty}\tau_{Q}(n)s^{n}. \end{align} $$

By Equation (1.2), one has

$$ \begin{align*} \log\xi_{Q}(s,a)=\log\xi_{Q}(0,a)-\sum_{\rho_{a}\in{Z_{Q}}}\sum_{m=1}^{\infty}\frac{1}{m\rho^{m}}s^{m}. \end{align*} $$

From the functional equation for the function $\xi _{Q}(s,a)$ , in the neighborhood of $s=0$ , we have

(2.4) $$ \begin{align} \frac{\xi_{Q}^{\prime}}{\xi_{Q}}(s+1,a)=-\frac{\xi_{Q}^{\prime}}{\xi_{Q}} (-s,a)=\sum_{m=0}^{\infty}(-1)^{m}\sum_{\rho_{a}\in{Z_{Q}}}\frac{1}{\rho^{m+1}}s^{m}. \end{align} $$

Comparing Equations (2.1)–(2.4), we get

$$ \begin{align*} (-1)^{m}\sum_{\rho_{a}\in{Z_{Q}}}\frac{1}{\rho^{m+1}} =(-1)^{m}+\gamma_{Q}(m)+\tau_{Q}(m), \end{align*} $$

for $m\geq 0$ . Hence, the definition of $\lambda _{n,a}$ yields

$$ \begin{align*} \lambda_{n,a}=\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\sum_{\rho_{a}\in{Z_{Q}}}\frac{1}{\rho^{k}}=1+ \sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)+\sum_{k=1}^{n}\binom{n}{k}\tau_{Q}(k-1), \end{align*} $$

where

$$ \begin{align*} \tau_{Q}(0)=-\frac{1}{2}\log\pi+\frac{1}{2}\psi(1/2)\ \mbox{and}\ \tau_{Q}(k-1)=(-1)^{k} \sum_{m=0}^{\infty}\frac{1}{(2m+1)^{k}} \end{align*} $$

using that $\psi (z)=-\gamma -\frac {1}{z}+\sum _{k=1}^{\infty }\frac {z}{k(k+z)}$ . Here, $\psi (s)=\frac {\Gamma '}{\Gamma }(s)$ is the logarithmic derivative of the Gamma function. Since $\psi (1/2)=-\gamma -2\log 2$ , we obtain

$$ \begin{align*} \lambda_{n,a}&=1-\frac{n}{2}(\log(4\pi)+\gamma)+\sum_{k=2}^{n}(-1)^{k} \binom{n}{k}\sum_{m=0}^{\infty}\frac{1}{(2m+1)^{k}} +\sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)\\ &=1-\frac{n}{2}(\log(4\pi)+\gamma)+\sum_{k=2}^{n}(-1)^{k}\binom{n}{k} \left(1-2^{-k}\right)\zeta(k)+\sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1). \end{align*} $$

The equality above implies Theorem 1.2.

2.3 Proof of Theorem 1.3

Let us note that

$$ \begin{align*} \sum_{k=2}^{n}(-1)^{k}\binom{n}{k}\left(1-2^{-k}\right)\zeta(k)= \sum_{k=2}^{n}(-1)^{l}\binom{n}{k}\frac{\zeta(k,1/2)}{2^{k}}, \end{align*} $$

where $\zeta (s, a)$ is the Hurwitz zeta function defined in Section 1.2. With the notation of Flajolet and Vespas [7, Lines 2–4, p. 70], this is $A_{n}(1,2)$ and which is equal to

$$ \begin{align*} \frac{n}{2}\psi(n)+n\left(\gamma-\frac{1}{2}+\frac{1}{2}\log2\right)+o(1), \end{align*} $$

where the o(1) error term above is exponentially small and oscillating and equal to

$$ \begin{align*} \frac{1}{2}\left(\frac{n}{\pi}\right)^{1/4}\exp(-\sqrt{2\pi n})\cos\left(\sqrt{2\pi n}-\frac{5\pi}{8}\right)+O\left(n^{-1/4}e^{-\sqrt{2\pi n}}\right). \end{align*} $$

Then we have

$$ \begin{align*} \lambda_{n,a}=\frac{n}{2}\log n+\frac{n}{2}\left(\gamma-1-\log 2\pi\right)+\sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)+O(1). \end{align*} $$

It remains to prove that

(2.5) $$ \begin{align} \sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)=O(\sqrt{n}\log n). \end{align} $$

To do so, we follow very closely the lines of the proof of the corresponding result in [8, Theorem 6.1] or [16, Lemma 3.3] and it will be shortened. We use the following kernel function:

$$ \begin{align*} k_{n}(s):=\left(1+\frac{1}{s}\right)^{n}-1=\sum_{k=1}^{n}\binom{n}{k} \frac{1}{s^{k}}. \end{align*} $$

The residue theorem gives

$$ \begin{align*} \sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)=\frac{1}{2i\pi}\int_{C}k_{n}(s)\left(-\frac{Q'}{Q}(s+1,a)\right)ds, \end{align*} $$

where C is a contour enclosing the point $s = 0$ counterclockwise on a circle of small enough positive radius. The residue comes entirely from the singularity at $s = 0$ , as no other singularities lie inside the contour. Let $T=\sqrt {n}+\epsilon _{n}$ , for some $0<\epsilon _{n}<1$ . Now we follow very closely the lines in [16, pp. 1106–1107] using that the function $\frac {Q'}{Q}(s,a)$ satisfies the properties ¹

$$ \begin{align*} \frac{Q'}{Q}(s,a)=\sum_{\rho_a;\ |{\mathrm{Im}}(\rho_a -s)|<1}\frac{1}{s-\rho_{a}}+O(\log(1+|s|)), \end{align*} $$

for $-2<\mathrm {Re}(s)<2$ and

$$ \begin{align*} \left|\frac{Q'}{Q}(s+1,a)\right|=O(\log^{2}T), \end{align*} $$

for $-2\leq \mathrm {Re}(s)\leq 2$ , and we get

$$ \begin{align*} \sum_{k=1}^{n}\binom{n}{k}\gamma_{Q}(k-1)=\lambda_{-n,a,T}+O(\sqrt{n}\log n), \end{align*} $$

where

$$ \begin{align*} \lambda_{-n,a,T}=\sum_{\rho_{a}\in{Z_{Q}};|Im (\rho_{a}|\leq T}^{*} \left(1-\left(1-\frac{1}{\rho_{a}}\right)^{n}\right), \end{align*} $$

with $T=\sqrt {n}+\epsilon _{n}$ . For $a=1/2,1/3,1/4,1/6$ , under the RH, since $\left |1-\frac {1}{\rho _{a}}\right |=1$ and using formula of $N_{Q}(T)$ given in Section 1.2, we obtain $\lambda _{n,a,T}=O(T\log T+1)$ . Therefore, Equation (2.5) follows from that $\lambda _{-n,a,\sqrt {n}}=\lambda _{-n,a,\sqrt {n}}=O(\sqrt {n}\log n)$ .

Remark Since $2Q(s,a) := Z(s,a) + P(s,a)$ , from Corollary 2.3 below and [6, Equation (1.18)], we obtain

$$ \begin{align*} \gamma_{Q}(n)=\frac{1}{2}\left(\delta_{n}(a)+\frac{(-1)^{n}}{n!}(l_{n}(a)+l_{n}(1-a))\right), \end{align*} $$

where $\delta _n (a) = \frac {|\log a|^{n}}{an!} + O(1)$ and $l_{n}(a)$ are the coefficients in the expansion of ${\mathrm {Li}}_s (e^{2\pi ia})$ at $s = 1$ ; for $a\notin {\mathbb {Z}}$ , one has

$$ \begin{align*} {\mathrm{Li}}_s (e^{2\pi ia})=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}l_{n}(a)(s-1)^{n}. \end{align*} $$

2.4 Proof of Theorem 1.4

To show Theorem 1.4, we quote the following lemmas from [2, 3].

Lemma 2.1 [3, Theorem 1]

We set

$$\begin{align*}(s-1) \zeta (s,a) = 1+ \sum_{n=0}^\infty \gamma_n (a) (s-1)^{n+1}, \qquad 0 <a \le 1. \end{align*}$$

Then it holds that

$$\begin{align*}\gamma_n (a) = \frac{(-1)^n}{n!} \lim_{m \to \infty} \Biggl( \sum_{k=0}^m \frac{\log^n (k+a) }{k+a} - \frac{\log^{n+1} (m+a) }{n+1} \Biggr). \end{align*}$$

Lemma 2.2 [2, Equation (26)]

Let $0 < a \le 1$ , and let n be a nonnegative integer. Then one has

$$ \begin{align*} \zeta^{(n)} (0,a) &= \biggl( \frac{1}{2} - a \biggr) |\log a|^n - n! + n! a \sum_{m=n}^\infty \frac{|\log a|^m }{m!}\\&\quad + (-1)^n n \int_0^\infty \frac{\varphi (x) \log^{n-1} (x+a)}{(x+a)^2} dx\\&\quad - (-1)^n n(n-1) \int_0^\infty \frac{\varphi (x) \log^{n-2} (x+a)}{(x+a)^2} dx,\end{align*} $$

where $\varphi (x) = \int _0^x (y- \lfloor y \rfloor - 1/2)dy$ is periodic with period $1$ and satisfies $2\varphi (x) = x(x-1)$ if $0 \le x \le 1$ .

By using the lemmas above, we immediately obtain the following.

Corollary 2.3 When $a>0$ is sufficiently small,

$$ \begin{align*} \begin{aligned} (s-1) Z(s,a) &= 2+ \sum_{n=0}^\infty \delta_n (a) (s-1)^{n+1}, \qquad \delta_n (a) = \frac{|\log a|^n}{an!} + O(1),\\ &Z(s,a) = \sum_{n=1}^\infty \epsilon_n (a) s^n, \qquad \epsilon_n (a) = O ( |\log a|^n ). \end{aligned} \end{align*} $$

Proof The first formula and estimation are easily proved by Lemma 2.1 (see also [3, Theorem 2]). For the first integral in Lemma 2.2, one has

$$ \begin{align*} \begin{aligned} & \int_0^1 \frac{\varphi (x) \log^{n-1} (x+a)}{(x+a)^2} dx \ll \int_0^1 \frac{\log^{n-1} (x+a)}{x+a} dx = O ( |\log a|^n ), \\ & \int_1^\infty \frac{\varphi (x) \log^{n-1} (x+a)}{(x+a)^2} dx \ll \int_1^\infty \frac{\log^{n-1} (x+a)}{(x+a)^2} dx = O (1) \end{aligned} \end{align*} $$

from $x < x+a$ when $x,a>0$ . In addition, we have

$$\begin{align*}a \sum_{m=n}^\infty \frac{|\log a|^m }{m!} \le a \sum_{m=0}^\infty \frac{|\log a|^m }{m!} = a e^{|\log a|} = a e^{-\log a} =1, \qquad 0< a <1/2. \end{align*}$$

Hence, we obtain

$$\begin{align*}\zeta (s,a) = \sum_{n=0}^\infty \frac{\zeta^{(n)} (0,a)}{n!} s^n, \qquad \zeta^{(n)} (0,a) = O ( |\log a|^n ). \end{align*}$$

Therefore, we have $\epsilon _n (a) = O ( |\log a|^n )$ and the second formula in this corollary by the definition of $Z(s,a)$ and $Z(0,a) = \zeta (0,a) +\zeta (0,1-a) = 0$ (see [15, Equation (4.11)]).

Proof of Theorem 1.4

Recall the functional equation

$$\begin{align*}Z (1-s,a) = \Gamma_{\!\! {\mathrm{cos}}} (s) P(s,a) , \qquad \Gamma_{\!\! {\mathrm{cos}}} (s) := \frac{2\Gamma (s)}{(2\pi)^s} \cos \Bigl( \frac{\pi s}{2} \Bigr) \end{align*}$$

(see [15, Lemma 4.11]). By using $\Gamma _{\!\! {\mathrm {cos}}} (s) \Gamma _{\!\! {\mathrm {cos}}} (1-s) =1$ , we have

$$\begin{align*}2 Q(s,a) = Z(s,a) + P(s,a) = Z(s,a) + \Gamma_{\!\! {\mathrm{cos}}} (1-s) Z(1-s,a). \end{align*}$$

Let $|s-1|$ be sufficiently small. Then, by $\lim _{s \to 1} (s-1) Q(s,a) = 1$ , the equation above, and the definitions of $Q(s,a)$ and $\xi _Q(s,a)$ , we have

$$ \begin{align*} &\frac{d^l}{d s^l} \big[ s^{l-1} \log \xi_Q(s,a) \big]_{s=1} = \frac{d^l}{d s^l} \Big[ s^{l-1} \log \big( (s-1) Q(s,a) \big) + s^{l-1} \log \big( s \pi^{-s/2} \Gamma (s/2) \big) \Big]_{s=1} \\& \quad = \frac{d^l}{d s^l} \bigg[ s^{l-1} \log \bigg( \frac{s-1}{2} \Big( Z (s,a) + \Gamma_{\!\! {\mathrm{cos}}} (1-s) Z(1-s,a) \Big) \bigg) \bigg]_{s=1} + O_l(1) \\& \quad = \frac{d^l}{d s^l} \bigg[ s^{l-1} \log \bigg( 1 + \sum_{n=0}^\infty \Big( \delta_n^{\prime} (a) + \epsilon_n^{\prime} (a) \Big) (s-1)^{n+1} \bigg) \bigg]_{s=1} + O_l (1), \end{align*} $$

where $\delta _n^{\prime } (a)$ and $\epsilon _n^{\prime } (a)$ are defined by

$$\begin{align*}\delta_n^{\prime} (a) := \frac{\delta_n (a)}{2}, \qquad (s-1) \Gamma_{\!\! {\mathrm{cos}}} (1-s) Z(1-s,a) = 2 \sum_{n=0}^\infty \epsilon_n^{\prime} (a) (s-1)^{n+1}. \end{align*}$$

Clearly, the second estimation in Corollary 2.3 implies

$$\begin{align*}Z(1-s,a) = \sum_{n=1}^\infty \epsilon_n (a) (1-s)^n,\qquad \epsilon_n (a) = O ( |\log a|^n ). \end{align*}$$

Thus, we can see that $\epsilon _n^{\prime } (a) = O ( |\log a|^{n+1} )$ from $\lim _{s \to 1} (s-1) \Gamma _{\!\! {\mathrm {cos}}} (1-s) = -2$ and the fact that the function $(s-1) \Gamma _{\!\! {\mathrm {cos}}} (1-s)$ does not depend on a. Put $\eta _n(a) := \delta _n^{\prime } (a) + \epsilon _n^{\prime } (a)$ . Then, for $n \ge 0$ , we have

(2.6) $$ \begin{align} \eta_n (a) = \frac{1}{n!} \frac{|\log a|^n}{2a} + O ( |\log a|^{n+1} ), \qquad a \to +0 \end{align} $$

by Corollary 2.3. By virtue of

$$ \begin{align*} \begin{aligned} (a_0 x + a_1 x^2 + a_2 x^3 + \cdots)^m = a_0^m & x^m + \binom{m}{1} a_0^{m-1 }a_1 x^{m+1} + \cdots \\ (a_0 x + a_1 x^2 + a_2 x^3 + \cdots)^{m-1} = a_0^m x^{m-1} + \binom{m-1}{1} a_0^{m-2} a_1 & x^{m} + \cdots \\ & \vdots \\ (a_0 x + a_1 x^2 + a_2 x^3 + \cdots)^{1} = \cdots + a_m & x^m + \cdots, \end{aligned} \end{align*} $$

where $m \in {\mathbb {N}}$ and $a_m,x \in {\mathbb {C}}$ , the coefficient of $(s-1)^l$ in the function

$$\begin{align*}f(s,a) := \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m} \Biggl( \sum_{n=0}^\infty \eta_n(a) (s-1)^{n+1} \Biggr)^{\!\!\!m} \end{align*}$$

is expressed as

(2.7) $$ \begin{align} \frac{(-1)^{l+1}}{l} \bigl( \eta_0 (a) \bigr)^l + \frac{(-1)^{l}}{l-1} \binom{l-1}{1} \eta_0 (a)^{l-2} \eta_1 (a) + \cdots+ \frac{(-1)^{1+1}}{1} \eta_{l-1} (a). \end{align} $$

Note that the function above is estimated by

(2.8) $$ \begin{align} \frac{(-1)^{l+1}}{l} \bigl( \eta_0 (a) \bigr)^l + O_l \bigl( \eta_0 (a)^{l-2} \eta_1 (a) \bigr) = \frac{(-1)^{l+1}}{l} (2a)^{-l} + O_l \bigl(a^{1-l} |\log a| \bigr) \end{align} $$

from (2.6) when $a \to +0$ . We can find that

$$\begin{align*}(s-1) \Bigl( Z (s,a) + \Gamma_{\!\! {\mathrm{cos}}} (1-s) Z(1-s,a) \Bigr) = 1 + \sum_{n=0}^\infty \eta_n (a) (s-1)^{n+1} \end{align*}$$

is analytic when $|s-1|<1$ form the poles of $Z (s,a)$ and $\Gamma _{\!\! {\mathrm {cos}}} (1-s)$ . So we can choose $|s-1|>0$ such that

$$\begin{align*}\sum_{n=0}^\infty \bigl|\eta_n (a) \bigr| |s-1|^{n+1} < \frac{1}{2}. \end{align*}$$

Then, from (2.7), the Leibniz product rule, the definition of $\eta _n(a)$ , and the Taylor expansion of $\log (1+x)$ with $|x| <1$ , one has

(♭) $$ \begin{align} &\frac{d^l}{d s^l} \Bigl[ s^{l-1} \log \xi_Q(s,a) \Bigr]_{s=1} = \frac{d^l}{d s^l} \Bigl[ s^{l-1} f (s,a) \Bigr]_{s=1} + O_l (1) \notag\\& \quad = \binom{l}{l} \frac{(-1)^{l+1}}{l} l! \bigl( \eta_0 (a) \bigr)^l + O_l \bigl( \eta_0 (a)^{l-2} \eta_1 (a) \bigr)\end{align} $$

(♮) $$ \begin{align} & + \binom{l}{l-1} (l-1) \frac{(-1)^{l}}{l-1} (l-1)! \bigl( \eta_0 (a) \bigr)^{l-1} + O_l \bigl( \eta_0 (a)^{l-3} \eta_1 (a) \bigr) \end{align} $$

(♯) $$ \begin{align} & \quad + \cdots + \binom{l}{1} (l-1) ! \frac{(-1)^{1+1}}{1} \bigl( \eta_0 (a) \bigr)^{1} + O_l (1). \end{align} $$

Note that ( $\flat $ ) comes from $f^{(l)} (s,a)$ , ( $\natural $ ) is deduced by $f^{(l-1)} (s,a)$ , and ( $\sharp $ ) derives from $f^{(1)} (s,a)$ , $f^{(0)} (s,a)$ , and $O_l (1)$ in the left-hand side of the formula above. Therefore, by (2.8), we obtain

$$ \begin{align*} \begin{split} \frac{d^l}{d s^l} \Bigl[ s^{l-1} \log \xi_Q(s,a) \Bigr]_{s=1} &= (-1)^{l+1} (l-1)! \bigl( \eta_0 (a) \bigr)^l + O_l \bigl( \eta_0 (a)^{l-2} \eta_1 (a) \bigr) \\ &= (-1)^{l+1} \frac{(l-1)!}{(2a)^l} + O_l \bigl(a^{1-l} |\log a| \bigr), \end{split} \end{align*} $$

which implies Theorem 1.4.

At the end of the paper, we give numerical computation for $\lambda _{n,a}$ by Mathematica 13.0. Let

$$\begin{align*}\lambda_{n,a}^{[k]} := \frac{1}{(n-1)!} \frac{d^n}{d s^n} \bigl[ s^{n-1} \log \xi_Q(s,a) \bigr]_{s=1-10^{-k}}, \qquad \lambda_{n,a}^* := \frac{(-1)^{n+1}}{(2a)^n}. \end{align*}$$

Then, we have the following.

For $n=1$ , we have

$$ \begin{align*}\begin{array}{ccrll} a := 2^{-17}& \lambda_{1,a}^{[10]} = 65,537... & \lambda_{1,a}^{[10]} / \lambda_{1,a}^* = 1.00001...\\ a := 2^{-18}& \lambda_{1,a}^{[10]} = 131,074... & \lambda_{1,a}^{[10]} / \lambda_{1,a}^* = 1.00002...\\ a := 2^{-19}& \lambda_{1,a}^{[10]} = 262,151... & \lambda_{1,a}^{[10]} / \lambda_{1,a}^* = 1.00003...\\ a := 2^{-17}& \lambda_{1,a}^{[11]} = 65,536.6... & \lambda_{1,a}^{[11]} / \lambda_{1,a}^* = 1.00001...\\ a := 2^{-18}& \lambda_{1,a}^{[11]} = 131,073... & \lambda_{1,a}^{[11]} / \lambda_{1,a}^* = 1.00001...\\ a := 2^{-19}& \lambda_{1,a}^{[11]} = 262,145... & \lambda_{1,a}^{[11]} / \lambda_{1,a}^* = 1.00000...\\ a := 2^{-17}& \lambda_{1,a}^{[12]} = 655,365... & \lambda_{1,a}^{[12]} / \lambda_{1,a}^* = 1.00001...\\ a := 2^{-18}& \lambda_{1,a}^{[12]} = 131,073... & \lambda_{1,a}^{[12]} / \lambda_{1,a}^* = 1.00000...\\ a := 2^{-19}& \lambda_{1,a}^{[12]} = 262,145... & \lambda_{1,a}^{[12]} / \lambda_{1,a}^* = 1.00000... \end{array}\end{align*} $$

For $n=2$ , we have

$$ \begin{align*}\begin{array}{ccrll} a := 2^{-17}&\ \ \ \ \ \ \lambda_{2,a}^{[10]} = -4.29352... \times 10^9 & \ \ \ \ \ \ \lambda_{2,a}^{[10]} / \lambda_{2,a}^* = 0.999663...\\ a := 2^{-18}& \ \ \ \ \ \ \lambda_{2,a}^{[10]} = -1.7177... \times 10^{10}&\ \ \ \ \ \ \lambda_{2,a}^{[10]} / \lambda_{2,a}^* = 0.999836...\\ a := 2^{-19}&\ \ \ \ \ \ \lambda_{2,a}^{[10]} = -6.87162... \times 10^{10}& \ \ \ \ \ \ \lambda_{2,a}^{[10]} / \lambda_{2,a}^* = 0.999952...\\ a := 2^{-17}&\ \ \ \ \ \ \lambda_{2,a}^{[11]} = -4.29478... \times 10^9&\ \ \ \ \ \ \lambda_{2,a}^{[11]} / \lambda_{2,a}^* = 0.999956...\\ a := 2^{-18}&\ \ \ \ \ \ \lambda_{2,a}^{[11]} = -1.71753... \times 10^{10}&\ \ \ \ \ \ \lambda_{2,a}^{[11]} / \lambda_{2,a}^* = 0.999736...\\ a := 2^{-19}&\ \ \ \ \ \ \lambda_{2,a}^{[11]} = -6.87149... \times 10^{10}& \ \ \ \ \ \ \lambda_{2,a}^{[11]} / \lambda_{2,a}^* = 0.999933...\\ a := 2^{-17}&\ \ \ \ \ \ \lambda_{2,a}^{[12]} = -4.29477... \times 10^9& \ \ \ \ \ \ \lambda_{2,a}^{[12]} / \lambda_{2,a}^* = 0.999955...\\ a := 2^{-18}& \ \ \ \ \ \ \lambda_{2,a}^{[12]} = -1.6911... \times 10^{10}&\ \ \ \ \ \ \lambda_{2,a}^{[12]} / \lambda_{2,a}^* = 0.984353...\\ a := 2^{-19}&\ \ \ \ \ \ \lambda_{2,a}^{[12]} = -6.87187... \times 10^{10}& \ \ \ \ \ \ \lambda_{2,a}^{[12]} / \lambda_{2,a}^* = 0.999989... \end{array} \end{align*} $$