On properties of the Taylor series coefficients of the Riemann xi function at $s=\frac{1}{2}$

TL;DR

This paper investigates the properties of Taylor series coefficients of the Riemann xi function at the critical point, providing integral formulas, positivity, and monotonicity results for these coefficients.

Contribution

It introduces new integral formulas involving Gaussian functions for the coefficients and proves their positivity and decreasing nature, advancing understanding of the xi function's local behavior.

Findings

Coefficients $a_k$ are positive.

Coefficients $a_k$ are decreasing.

Integral formulas involving Gaussian functions and polynomials.

Abstract

We prove some properties about the non-zero Taylor series coefficients $a_k$ of the Riemann xi function $\xi(s)$ at $s=\frac{1}{2}$. In particular, we present integral formulas that evaluate $a_k$ whose integrands involve a Gaussian function and a function we call $L(x;k)$. We use these formulas to show that $a_k$ is positive. We also define a sequence of polynomials $p(x;n)$ which arise naturally from the integral formulas and use them to prove that the coefficients $a_k$ are decreasing.

Equations

$$\xi (s)=-s(1-s){\pi }^{-\frac{s}{2}}\Gamma (\frac{s}{2})\zeta (s)$$

$$\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^s}$$

$$\Gamma (s)={\int }_0^{\infty }{e}^{-x}{x}^{s-1}dx.$$

$$\xi (s)=\xi (1-s)$$

$$\frac{e^{-\pi x^2}}{\sqrt{x}}$$

$$a_k\ge a_{k+1}\ge 0.$$

$${\pi }^{-\frac{s}{2}}\Gamma (\frac{s}{2})\zeta (s)={\int }_1^{\infty }(\sum _{n=1}^{\infty }{e}^{-\pi n^{2}x})(u^{\frac{1-s}{2}-1}+u^{\frac{s}{2}-1})dx-(\frac{1}{1-s}+\frac{1}{s}).$$

$$\xi (s)=-s(1-s){\pi }^{-\frac{s}{2}}\Gamma (\frac{s}{2})\zeta (s)=1-s(1-s){\int }_1^{\infty }(\sum _{n=1}^{\infty }{e}^{-\pi n^{2}x})(u^{\frac{1-s}{2}-1}+u^{\frac{s}{2}-1})dx$$

$$\xi (\frac{1}{2}+it)=1-(\frac{1}{4}+t^2){\int }_1^{\infty }(\sum _{n=1}^{\infty }{e}^{-\pi n^{2}x})x^{-\frac{3}{4}}(x^{-\frac{it}{2}}+x^{\frac{it}{2}})dx.$$

$$\xi (\frac{1}{2}+it)=\sum _{k=0}^{\infty }(-1{)}^k{a}_k{t}^{2k},$$

$$(-1{)}^k\frac{(\frac{d}{dt}{)}^{2k}}{(2k)!}\xi (\frac{1}{2}+it){|}_{t=0}=a_k.$$

$$a_{k+1}=4\sum _{M=1}^{\infty }{\int }_M^{M+1}(\sum _{n=1}^M\frac{e^{-\pi x^2}}{\sqrt{x}}(\frac{\log (\frac{n}{x}{)}^{2k})}{(2k)!\sqrt{n}}-\frac{\log (\frac{n}{x}{)}^{2k+2})}{(2k+2)!4\sqrt{n}})dx$$

$$a_0={\int }_{-1}^1{e}^{-\pi x^2}dx+\sum _{M=1}^{\infty }{\int }_M^{M+1}\frac{e^{-\pi x^2}}{\sqrt{x}}(2\sqrt{x}-\sum _{n=1}^M\frac{1}{\sqrt{n}})dx.$$

$$(\frac{d}{dt}{)}^{2k}(x^{-\frac{it}{2}}+x^{\frac{it}{2}}){|}_{t=0}=(-1{)}^{k}2(\frac{\log (x)}{2}{)}^{2k}$$

$$a_k=2{\int }_1^{\infty }{x}^{-\frac{3}{4}}(\frac{(\frac{\log (x)}{2}{)}^{2k-2}}{(2k-2)!}-\frac{(\frac{\log (x)}{2}{)}^{2k}}{(2k)!4})\sum _{n=1}^{\infty }{e}^{-\pi n^{2}x}dx$$

$$a_0=1-2{\int }_1^{\infty }\frac{x^{-\frac{3}{4}}}{4}\sum _{n=1}^{\infty }{e}^{-\pi n^{2}x}dx.$$

$${\int }_1^{\infty }{x}^{-\frac{3}{4}}(\frac{\log (x)}{2}{)}^{2k}{e}^{-\pi n^{2}x}dx=2{\int }_n^{\infty }\frac{\log (\frac{n}{x}{)}^{2k}}{\sqrt{n}}\frac{e^{-\pi x^2}}{\sqrt{x}}dx$$

$$a_{k+1}=4\sum _{M=1}^{\infty }{\int }_M^{M+1}(\sum _{n=1}^M\frac{e^{-\pi x^2}}{\sqrt{x}}(\frac{\log (\frac{n}{x}{)}^{2k})}{(2k)!\sqrt{n}}-\frac{\log (\frac{n}{x}{)}^{2k+2})}{(2k+2)!4\sqrt{n}})dx$$

$$a_0=1-\sum _{M=1}^{\infty }{\int }_M^{M+1}\frac{e^{-\pi x^2}}{\sqrt{x}}\sum _{n=1}^M\frac{1}{\sqrt{n}}dx.$$

$$1={\int }_{-\infty }^{\infty }{e}^{-\pi x^2}dx$$

$$a_0={\int }_{-1}^1{e}^{-\pi x^2}dx+\sum _{M=1}^{\infty }{\int }_M^{M+1}\frac{e^{-\pi x^2}}{\sqrt{x}}(2\sqrt{x}-\sum _{n=1}^M\frac{1}{\sqrt{n}})dx.$$

$$a_{k+1}=4{\int }_1^{\infty }\frac{e^{-\pi x^2}}{\sqrt{x}}L(x;k)dx$$

$$a_0={\int }_{-1}^1{e}^{-\pi x^2}dx+{\int }_1^{\infty }\frac{e^{-\pi x^2}}{\sqrt{x}}B(x)dx$$

$$L(x;k)=\sum _{n=1}^{\lfloor x\rfloor }(\frac{\log (\frac{n}{x}{)}^{2k}}{(2k)!\sqrt{n}}-\frac{\log (\frac{n}{x}{)}^{2k+2}}{(2k+2)!4\sqrt{n}})$$

$$B(x)=2\sqrt{x}-\sum _{n=1}^{\lfloor x\rfloor }\frac{1}{\sqrt{n}}.$$

$$|L(x;k)|\le 2x\log (x{)}^{2k}.$$

$$f_1(x)=\sum _{n=1}^M(\frac{\log (\frac{n}{x}{)}^{2k}}{(2k)!\sqrt{n}}-\frac{\log (\frac{n}{x}{)}^{2k+2}}{(2k+2)!4\sqrt{x}})$$

$$f_2(x)=\sum _{n=1}^{M-1}(\frac{\log (\frac{n}{x}{)}^{2k}}{(2k)!\sqrt{n}}-\frac{\log (\frac{n}{x}{)}^{2k+2}}{(2k+2)!4\sqrt{n}})$$

$$f_1(x)-f_2(x)=\frac{\log (\frac{M}{x}{)}^{2k}}{(2k)!\sqrt{M}}-\frac{\log (\frac{M}{x}{)}^{2k+2}}{(2k+2)!4\sqrt{M}}$$

$$U(x;n,k)$$

Full text

1 Introduction

The Riemann xi function $\xi(s)$ is an entire function that has a significant role in number theory: it is the “completed” form of the Riemann zeta function $\zeta(s)$

[TABLE]

where

[TABLE]

and $\Gamma(s)$ is the Gamma function

[TABLE]

This completion has many desirable properties which $\zeta(s)$ by itself does not posses; there is the functional equation

[TABLE]

and the fact that, for real $t$, $\xi(\frac{1}{2}+it)$ is real and even. These properties are manifest in the defining integral formula (1) for $\xi(s)$. Furthermore, it was conjectured by B. Riemann [1] that all the zeros of $\xi(s)$ lie on the line $s=\frac{1}{2}+it$. Thus it is natural to investigate the Taylor series coefficients of $\xi(s)$ at $s=\frac{1}{2}$.

We prove some properties about the non-zero Taylor series coefficients $a_{k}$ (see the definition of the $a_{k}$ below). In particular, we present an integral formula that evaluates $a_{k}$ whose integrand is a Gaussian function over the square root of $x$

[TABLE]

times a certain function $L(x;k)$ defined in Section 2. We use this integral formula to show that each $a_{k}\geq 0$. We also define a sequence of polynomials $p(x;n)$ which arise naturally from the integral formula. Using a positivity property of one such polynomial $p(x;2)$ we also prove that

[TABLE]

The numbers $a_{k}$ are elementary-symmetric functions evaluated at the zeroes of $\xi(\frac{1}{2}+i\sqrt{t})$. Therefore information about the positivity of the $a_{k}$ could be related to information about the zeros of $\xi(s)$. Also, the generalized Turán inequalities are positivity conditions involving the Taylor series coefficients of a function. The integral formulas we present could be applied to proving the generalized Turán inequalities for $\xi(s)$. That the growth of $L(x;k)$ is slow compared to the decay of Gaussian could be useful in estimating these quantities.

We now review the defining integral formula for $\xi(s)$ and the definition of $a_{k}$. We have for $s\in\mathbb{C}$

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See [2], chapter 6 for the derivation of this formula. Therefore we let

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and let $s=\frac{1}{2}+it$ to obtain

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Define the coefficients $a_{k}$ by

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[TABLE]

2 Integral formula using $L(x;k)$

We show how the function $L(x;k)$ arises from the integral definition of $\xi(s)$.

Lemma 1**.**

We have the integral formulas

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and

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Proof.

We differentiate (1) under the integral sign and use

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to get for integer $k\geq 1$

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and

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To the above equations for each $n$ we apply

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where we have used the change of variables $x\mapsto\frac{x^{2}}{n^{2}}$. Thus for $k\geq 0$ formula (2) becomes

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and formula (3) becomes

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Now to (5) we apply

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and obtain

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This completes the proof. ∎

We rewrite the formulas in Lemma 1 as

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and

[TABLE]

for the functions $L(x;k)$ and $B(x)$ defined next.

Definition 1**.**

For $x,k\in\mathbb{R}$, define

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and

[TABLE]

Lemma 2**.**

For integer $k\geq 1$, the function $L(x;k)$ has the bound for $x\geq e$

[TABLE]

For integer $k\geq 1$, $L(x;k)$ as a function of $x$ is $(2k-1)$-differentiable on $(0,\infty)$.

Proof.

The bound on $L(x;k)$ follows immediately from the definition. It is also follows from the definition that $L(x;k)$ is piecewise-smooth for positive $x$ except possibly at the integers. We prove that $L(x;k)$ is $(2k-1)$-differentiable at integer $x=M>0$. Now the function

[TABLE]

is equal to $L(x;k)$ for $x\in[M,M+1)$ and the function

[TABLE]

is equal to $L(x;k)$ for $x\in[M-1,M)$. Then the first $2k-1$ derivatives of the difference

[TABLE]

are each equal to 0 at $x=M$, including $f_{1}(M)-f_{2}(M)=0$. This completes the proof. ∎

The functions defined next and Lemma 3 will be used to prove the positivity of $L(x;k)$ in Theorem 1.

Definition 2**.**

For $x,k$ and $n\in\mathbb{R}$ with $x$ and $n\geq 0$, we define the functions

[TABLE]

Lemma 3**.**

For integer $k\geq 0$, we have the anti-derivatives

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and

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Proof.

This follows from successive integration by parts

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using $u=\sqrt{z}$ each time. ∎

Theorem 1**.**

We have the equality

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For $x\geq 1$ and $k\geq 0$,

[TABLE]

Proof.

We first prove $L(x;k)\geq 0$. Suppose $M\leq x<M+1$ where $M\geq 1$ is an integer. By definition

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From the definitions of $U(x;n,k)$ and $V(x;n,k)$ we have for $n\geq 1$

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This proves equation (6). We also have for $n\geq 1$

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The above equation follows from Lemma 3. And each expression in the parentheses of (7) is positive because the functions of $z$

[TABLE]

are positive decreasing and negative increasing, respectively, for $z\in(1,M)$ and $1\leq n\leq M-1$. This shows that for $k\geq 0$ and $1\leq n\leq M-1$

[TABLE]

Now

[TABLE]

for $x\geq 1$, and we prove that $U(x;M,k)$ is positive in Lemma 4 for $M\leq x<M+1$. This proves the positivity of $L(x;k)$.

For $B(x)$, we compare the integral with $M\leq x<M+1$

[TABLE]

which completes the proof. ∎

This is the Lemma used in Theorem 1.

Lemma 4**.**

For integer $M\geq 1$ and $k\geq 0$, and $M\leq x\leq M+1$,

[TABLE]

Proof.

We factor the left side of inequality in the lemma statement to obtain

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Now using the bound on $x$ we get

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Now let $M=\frac{1}{y}$ where $0\leq y\leq 1$. For $k\geq 0$ we have

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We then show that

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The above inequality is true when $y=0$, and we show that the left side is increasing in $y$. We differentiate with respect to $y$ and see that we next must show that

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Multiplying by $(1+y)$ gives

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and this is true for because for $y\geq 0$ we have

[TABLE]

∎

Corollary 1**.**

For $k\geq 0$

[TABLE]

and

[TABLE]

Proof.

These statements follow from Lemma 1 and Theorem 1.

∎

3 The polynomials $p(x;n)$

The following definition of the polynomials $p(x;n)$ is motivated by Lemma 6.

Definition 3**.**

For integer $n\geq 0$ define the polynomial $p(x;n)$ by

[TABLE]

[TABLE]

The first few $p(x;n)$ are

[TABLE]

Lemma 5**.**

For integer $n\geq 0$ and $x\in\mathbb{C}$,

[TABLE]

[TABLE]

Proof.

We denote

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Thus $c_{0,0}=1$ and $c_{m,0}=0$ for $m\geq 1$. The definition of $p(x;n)$ implies that

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We prove by induction on $n$ that

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It is true for $c_{m,0}$ for all $m$. Assume it is true for $c_{m,n}$ for all $m$ for some $n\geq 0$. Then we substitute (11) into (10), compare the coefficients of $(4k+1)^{n}$, and apply the identity

[TABLE]

for $0\leq k\leq m-1$. This proves the induction step.

Now the series on the right side of (9) is convergent for any $x\in\mathbb{C}$. We calculate its Taylor coefficients of the whole function in (9) at $x=0$ and see that they match (8). This completes the proof. ∎

Lemma 6**.**

For $c>0$ and integers $m\geq 0$ and $n\geq 0$, we have

[TABLE]

Proof.

This follows from integration by parts

[TABLE]

applied $n$ times to integral on the left side; use $\displaystyle v=\frac{\log(\frac{x}{c})^{m+h}}{(m+h)!}$ for the $h$-th application of integration by parts. ∎

Theorem 2**.**

For integers $k\geq 0$ and $n\geq 0$,

[TABLE]

Proof.

Recall

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We re-arrange this as

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We apply Lemma 6 to get

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We re-arrange in the reverse manner to obtain

[TABLE]

∎

We define $\mathrm{Wallis}(N)$ for estimates in the next results.

Definition 4**.**

The Wallis product [3] is

[TABLE]

with

[TABLE]

for $n\geq 1$. We denote

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so

[TABLE]

for all $N\geq 1$.

Theorem 3**.**

For integer $k\geq 0$,

[TABLE]

Proof.

For $k\geq 0$, we use Theorem 2 with $n=2$ to obtain

[TABLE]

Since $\displaystyle\frac{p(-\pi x^{2};2)}{4}-1\geq 0$ by Lemma 7 and $L_{k+1}(x)\geq 0$ by Lemma 1 for $x\geq 1$, this completes the proof for $k\geq 1$.

We next prove $a_{0}>a_{1}$. Recall

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and

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We apply the change of variables $x\mapsto x-1$ to the left integral of (12) and obtain

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Thus

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We have seen in Theorem 1 that

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And for $x\geq 1$

[TABLE]

This completes the proof. ∎

This is the lemma used in Theorem 3.

Lemma 7**.**

For $x\geq 1$,

[TABLE]

Proof.

We have

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Then

[TABLE]

where we bound $\frac{\pi}{2}$ from below with $\mathrm{Wallis}(0)=1$.

And

[TABLE]

where we have again bounded $\frac{\pi}{2}$ with $\mathrm{Wallis}(0)=1$. This proves the lemma. ∎

4 Further Work

• •

Apply the integral formulas to proving

[TABLE]

and the generalized Turán inequalities.

• •

Relate the polynomials $p(x;n)$ to known polynomial families.

• •

See if similar results hold for Dirichlet $L$-functions.