Series representation of a cotangent sum related to the Estermann zeta function

TL;DR

This paper derives an explicit series expansion for a cotangent sum related to the Estermann zeta function, improving upon previous identities and providing a clearer analytical representation.

Contribution

It introduces a new explicit series formula for the cotangent sum c0(q/p) when q=1, enhancing the understanding of its relation to the Estermann zeta function.

Findings

Derived an explicit series expansion for c0(1/p).

Improved upon previous identities related to the cotangent sum.

Provides analytical tools for further study of the Estermann zeta function.

Abstract

In this paper, we are interested by the cotangent sum c0(q/p) related to the Estermann zeta function for the special case when q = 1 and get explicit formula for its series expansion, which represents an improvement of the identity (2:1) Theorem (2:1) in the recent work [11].

Equations

$$c_0\left(\frac{q}{p}\right)=-\sum _{k=1}^{p-1}\frac{k}{p}\cot \frac{\pi kq}{p}.$$

$$E_0\left(0,\frac{q}{p}\right)=\frac{1}{4}+\frac{i}{2}{c}_0\left(\frac{q}{p}\right)$$

$$E_0\left(s,\frac{q}{p}\right)=\sum _{k⩾1}\frac{d(k)}{k^s}\exp \left(\frac{2\pi ikq}{p}\right).$$

$$V\left(\frac{q}{p}\right)=\sum _{r=1}^{p-1}\left\{\frac{rq}{p}\right\}\cot \left(\frac{\pi r}{p}\right)=-c_0\left(\frac{\overline{q}}{p}\right)$$

$$\begin{array}{rl}& \frac{1}{2\pi \sqrt{pq}}{\int }_{-\infty }^{+\infty }{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^2{\left(\frac{q}{p}\right)}^{it}\frac{dt}{\frac{1}{4}+t^2}=\\ & \frac{\log 2\pi -\gamma }{2}\left(\frac{1}{p}+\frac{1}{q}\right)+\frac{p-q}{2pq}\log \frac{q}{p}-\frac{\pi }{2pq}\left(V\left(\frac{p}{q}\right)+V\left(\frac{q}{p}\right)\right).\end{array}$$

$$d_N^2=\underset{A_N}{\inf }\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{\left|1-\zeta A\left(\frac{1}{2}+it\right)\right|}^2\frac{dt}{\frac{1}{4}+t^2}$$

$$A_N(s)=\sum _{n=1}^N\frac{a_n}{n^s}.$$

$$c_0\left(\frac{1}{p}\right)=\frac{p\left(p-1\right)\left(p-2\right)}{\pi }\sum _{k⩾0}\frac{b_k}{\left(k+1\right)\left(k+p+1\right)\left(k+2\right)\left(k+p\right)}$$

$$b_k-2b_{k-1}+b_{k-2}=0,2⩽k⩽p-1,k=p+1,$$

$$b_p-2b_{p-1}+b_{p-2}=1$$

$$b_k-2b_{k-1}+b_{k-2}-b_{k-p}+2b_{k-p-1}-b_{k-p-2}=0,k⩾p+2.$$

$$c_0\left(\frac{1}{p}\right)=\frac{p\left(p-1\right)\left(p-2\right)}{\pi }\sum _{k⩾0}\frac{\left(k+1-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \right)\left(\lfloor \frac{k}{p}\rfloor +1\right)}{\left(k+1\right)\left(k+p+1\right)\left(k+2\right)\left(k+p\right)}.$$

$$s_n(p)=p\left(p-1\right)\left(p-2\right)\sum _{k=0}^n\frac{\left(k+1-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \right)\left(\lfloor \frac{k}{p}\rfloor +1\right)}{\left(k+1\right)\left(k+p+1\right)\left(k+2\right)\left(k+p\right)}.$$

$$\begin{array}{rl}& \frac{1}{2\pi \sqrt{p}}{\int }_{-\infty }^{+\infty }{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^2{\left(\frac{1}{p}\right)}^{it}\frac{dt}{\frac{1}{4}+t^2}=\\ & \frac{p-1}{2p}\left[2\left(\log 2\pi -\gamma \right)-\log p+p\left(p-2\right)\sum _{k⩾0}\frac{\left(k+1-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \right)\left(\lfloor \frac{k}{p}\rfloor +1\right)}{\left(k+1\right)\left(k+p+1\right)\left(k+2\right)\left(k+p\right)}\right].\end{array}$$

$$\theta (i,j)=\frac{p{i}^2+\left(p+2r+2\right)i+2r+2}{\left(ip+r+1\right)\left(ip+p+r+1\right)\left(ip+r+2\right)\left(ip+p+r\right)}$$

$$c_0\left(\frac{1}{p}\right)=\frac{p\left(p-1\right)\left(p-2\right)}{2\pi }\sum _{i⩾0}\sum _{r=0}^{r-1}\theta (i,j)$$

$$\frac{1}{{\left(1-x\right)}^2\left(1-x^p\right)}=\sum _{k⩾0}{b}_k{x}^k,|x|<1$$

$$b_k=\left(k+1-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \right)\left(\lfloor \frac{k}{p}\rfloor +1\right)$$

$$\frac{1}{1-x}=\sum _{k⩾0}{x}^k,|x|<1$$

$$\frac{1}{1-x^p}=\sum _{k⩾0}{b}_p(k)x^k,|x|<1.$$

$$\frac{1}{{\left(1-x\right)}^2}=\sum _{k⩾0}\left(k+1\right)x^k,|x|<1$$

$$\frac{1}{{\left(1-x\right)}^2\left(1-x^p\right)}=\sum _{k⩾0}\sum _{j=0}^k\left(k-j+1\right)b_p(j)x^k,|x|<1.$$

$$\lfloor \frac{k}{p}\rfloor p⩽k<\left(\lfloor \frac{k}{p}\rfloor +1\right)p$$

$$\sum _{j=0}^k\left(k-j+1\right)b_p(j)=\sum _{j=0}^{\lfloor \frac{k}{p}\rfloor }\left(k-jp+1\right)$$

$$\sum _{j=0}^{\lfloor \frac{k}{p}\rfloor }\left(k-jp+1\right)=\left(k+1\right)\left(\lfloor \frac{k}{p}\rfloor +1\right)-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \left(\lfloor \frac{k}{p}\rfloor +1\right).$$

$$b_k=\left(k+1\right)\left(\lfloor \frac{k}{p}\rfloor +1\right)-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \left(\lfloor \frac{k}{p}\rfloor +1\right).$$

$$b_k=\left(k+1-\frac{p}{2}\lfloor \frac{k}{p}\rfloor \right)\left(\lfloor \frac{k}{p}\rfloor +1\right)$$

$$\displaystyle b_{k}=\left\{\begin{array}[]{lll}k+1&\quad\textrm{if}\ k<p,\\ b_{k-p}+k+1&\quad\textrm{if}\ k\geqslant p.\end{array}\right.$$

$$\frac{d^k(1-x^p)f(x)}{d{x}^k}=\frac{(k+1)!}{(1-x{)}^{k+2}}.$$

$$\frac{d^k(1-x^p)f(x)}{d{x}^k}=\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)(1-x^p{)}^{(j)}{f}^{k-j}(x).$$

Full text

1. Introduction and main results

For $p$ a positive integer and $q=1,2,\cdots,p-1$ such that $\left(p,q\right)=1$, let the cotangent sum [15]

[TABLE]

$c_{0}\left(\frac{q}{p}\right)$ is the value at $s=0$,

[TABLE]

of the Estermann zeta function

[TABLE]

This sum is related directly to Vasyunin cotangent sum [16];

[TABLE]

which arises in the study of the Riemann zeta function by virtue of the formula [4, 14].

[TABLE]

This formula is connected to the approach of Nyman, Beurling and Báez-Duarte to the Riemann hypothesis [13, 12]. Which states that the Riemann hypothesis is true if and only if $\displaystyle\lim_{n\to\infty}d_{N}=0$, where

[TABLE]

and the infimum is taken over all Dirichlet polynomials

[TABLE]

In the literature; different results about $V(\frac{q}{p})$ and $c_{0}(\frac{q}{p})$ are obtained. For more details we refer to [10, 15, 14, 5, 8] and reference therein.

Exactly our interest in this work is the case $q=1$ in order to compute explicitly the sequence $b_{k}$ in the series expansion $(2.1)$ Theorem $(2.1)$ [11] of $c_{0}(\frac{1}{p})$:

[TABLE]

where $b_{k}$ is generated by the function $f\left(x\right)=\frac{1}{\left(1-x\right)^{2}\left(1-x^{p}\right)}=\sum_{k\geqslant 0}b_{k}x^{k}$. The first terms are $b_{0}=1,b_{1}=2$ and the others are given by the recursive formulae:

[TABLE]

[TABLE]

and

[TABLE]

2. Statement of main results

In the following theorem, we prove that $b_{k}=\left(k+1-\frac{p}{2}\left\lfloor\frac{k}{p}\right\rfloor\right)\left(\left\lfloor\frac{k}{p}\right\rfloor+1\right)$, where $\left\lfloor.\right\rfloor$ is the well known floor function.

Theorem 2.1**.**

[TABLE]

Let the integer sequence $s_{n}(p)$ defined by

[TABLE]

Then $\pi c_{0}\left(\frac{1}{p}\right)=\lim_{n\to\infty}s_{n}(p)$ which explains that $\pi c_{0}\left(\frac{1}{p}\right)$ lies to $\overline{\mathbb{Q}}.$ Only in means of the Theorem 2.1, for $q=1$ the identity (1.3) becomes

[TABLE]

Let the arithmetical function of two variables

[TABLE]

then the following proposition is obtained

Proposition 2.1**.**

[TABLE]

3. Proof of main results

3.1. Proof of Theorem 2.1

We take inspiration from the theory of generating functions [7, 9], and prove that the sequence $b_{k}$ generated by the rational function $f(x)$ is explicitly done in the following lemma. And the identity (2.1), Theorem 2.1 is deduced.

Lemma 3.1**.**

[TABLE]

with

[TABLE]

Proof.

It is well known that

[TABLE]

and

[TABLE]

with $b_{p}(k)=1$ if $p$ divides $k$ and zero otherwise. Then

[TABLE]

and

[TABLE]

But

[TABLE]

then

[TABLE]

Since

[TABLE]

Then

[TABLE]

Finally

[TABLE]

∎

3.2. Proof of Proposition 2.1.

In this subsection, we expose two methods to prove the Proposition 2.1.

Analytic method. We began by the following interesting lemma

Lemma 3.2**.**

[TABLE]

Proof.

We remember for $|x|<1$ that $f(x)=\frac{1}{\left(1-x\right)^{2}\left(1-x^{p}\right)}=\sum_{k\geqslant 0}b_{k}x^{k}$. It is well known that $f^{(k)}(0)=\frac{d^{k}f(x)}{dx^{k}}|_{x=0}=k!b_{k}$. Taking $g(x)=\left(1-x\right)^{-2}$, it is easy to show that $g^{(k)}(x)=\frac{(k+1)!}{(1-x)^{k+2}}$.

Using Leibnitz formula for successive derivatives of any infinitely derivable function explained in the work [9], and the identity $(1-x^{p})f(x)=(1-x)^{2}$ we deduce that

[TABLE]

But

[TABLE]

Since $(1-x^{p})^{(j)}=-(p)_{j}\sigma_{p}(j)x^{p-j}$, where $\sigma_{p}(j)=1$ if $j\leqslant p$ and zero otherwise, then

[TABLE]

Furthermore $f^{(k)}(0)=(k+1)!$ for $k<p$ and for $k\geqslant p$ we have

[TABLE]

Which means that $b_{k}=k+1$ for $k<p$ and $b_{k}=b_{k-p}+k+1$ for $k\geqslant p.$ ∎

Corollary 3.1**.**

For $0\leqslant r\leqslant p-1$ and $i\geqslant 0$ we have

[TABLE]

Proof.

First in means of the formula (3.7) we have $b_{r}=r+1$, $b_{p+r}=p+2r+2$, $b_{2p+r}=3p+3r+3$, $b_{3p+r}=6p+4r+4$ and $b_{4p+r}=10p+5r+5.$ Then for $i\geqslant 0$ suppose that $b_{ip+r}=\frac{1}{2}i(i+1)p+(i+1)(r+1)$ then

[TABLE]

Finally

[TABLE]

and the result follows. ∎

Combining the identities in Lemma 3.2 and Corollary 3.1 we get the desired result (2.3) in Proposition 2.1.

Arithmetic method. This is not expensive, just writing $k=ip+r$ which is the Euclidean division of $k$ over $p$ then $\left\lfloor\frac{k}{p}\right\rfloor=i$. Substituting this value in the expression of $b_{k}$ in Lemma 3.1, we get

[TABLE]

After development we get the result (3.8) of Corollary 3.1, which implies the result (2.3) Proposition 2.1.

4. Consequences

4.1. Limit at infinity of $c_{0}(\frac{1}{p})$

Theorem 4.1**.**

[TABLE]

An interesting open question is how about the exact value of this limit. Since $c_{0}\left(\frac{1}{p}\right)$ is related directly to Riemann hypothesis where the real part of the zeros of the Riemann zeta function is $\frac{1}{2}$, one can conjecture that $\lim_{p\to+\infty}\frac{1}{p^{3}}c_{0}\left(\frac{1}{p}\right)=\frac{1}{2}$.

4.2. Proof of Theorem 4.1

We also use notational convention $0^{0}=1$ and the functions

[TABLE]

It’s clear that $\varphi_{m}(r,p)$ converge for $i\in\left\{0,1,2\right\}$ and diverge for the others. In means of these functions another expression of $c_{0}\left(\frac{1}{p}\right)$ is deduced by using the identity (3.8) of the last Corollary 3.1:

[TABLE]

In the following lemma, we develop some inequalities satisfied by the functions $\varphi_{i}(r,p)$ for $i\in\left\{0,1,2\right\}.$

Lemma 4.1**.**

[TABLE]

[TABLE]

[TABLE]

Proof.

To get the proof of Lemma 4.1, just remark that for $i\geqslant 1$ we have

[TABLE]

and for $0\leqslant r\leqslant p-1$

[TABLE]

furthermore the last inequalities are deduced. ∎

For simplifying calculus, let us denoting the function $\phi(r,p)$ to be

[TABLE]

Then in one hand we have

[TABLE]

and

[TABLE]

Which inducts that

[TABLE]

In other hand

[TABLE]

and

[TABLE]

Furthermore

[TABLE]

The passage to the limit in two last inequalities (4.6) and (4.7) states that

[TABLE]

4.3. New property of the floor function

Returning back to own recursive formula of $b_{k}$ and using its expression (3.2) in Lemma 3.1 we obtain

Theorem 4.2**.**

For $2\leqslant k\leqslant p-1$ and $k=p+1$ we have

[TABLE]

at $k=p$

[TABLE]

and for $k\geqslant p+2$;

[TABLE]

Proof.

We can write for $2\leqslant k\leqslant p+1$

[TABLE]

and for $k\geqslant p+2$;

[TABLE]

But from the recursive formulae (1.5), (1.6) and (1.7) of the sequence $b_{k}$ we deduce respectively the formulae (4.8), (4.9) and (4.10). ∎

4.4. sum of some numerical series

Inspired from the results $c_{0}(\frac{1}{3})=\frac{1}{3\sqrt{3}}$, $c_{0}(\frac{1}{4})=\frac{1}{2}$ and $c_{0}(\frac{1}{6})=\frac{7}{3\sqrt{3}}$ in [8] and $c_{0}(\frac{1}{5})=\frac{\left(\sqrt{5}-1\right)\sqrt{5-\sqrt{5}}+3\left(\sqrt{5}+1\right)\sqrt{5+\sqrt{5}}}{10\sqrt{10}}$ in [6], we conclude that

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]