On the Generalized Li’s Criterion Equivalent to the Riemann Hypothesis and Its First Applications

Abstract

We present some new results and first applications of the recently established by us (Ukrainian Math. J. 2014, 66, pp. 371–383) generalized Li’s criterion equivalent to the Riemann Hypothesis. This criterion is the statement that the Riemann hypothesis is equivalent to the non-negativity of the sums over non-trivial Riemann zeroes $\rho$ $\sum\nolimits_{\rho } {(1 - (\frac{\rho + b}{\rho - b - 1})^{n} )}$ (generalized Li’s coefficients), which are equal to the derivatives $\frac{(2b + 1)}{(n - 1)!}\frac{{d^{m} }}{{dz^{m} }}((z + b)^{m - 1} \ln (\xi (z)))|_{z = 1 + b}$ of the Riemann $\xi$-function, for some real b > −1/2 (if so, this takes place for all such b) and all m = 1, 2, 3 … Namely, we give an expression of the generating function for the generalized Li’s coefficients, and, assuming RH, for their asymptotic for large n. As the first applications, we show that for any positive integer n there is such a value of b_n (depending on n) that for all $m \le n$ and b > b_n, the inequality $\frac{{d^{m} }}{{dz^{m} }}((z + b)^{m - 1} \ln (\xi (z)))|_{z = 1 + b} \ge 0$ does hold true.

Appendices

Appendix 1: Generating Function for the Generalized Li’s Sums

Theorem 4

For any real b, $\sigma$, $\sigma \ne - b$, the sums $\lambda_{n,b,\sigma } : = \sum\limits_{\rho } {(1 - (1 + \frac{2b + 2\sigma }{\rho - b - 2\sigma })^{n} )}$$= \sum\limits_{\rho } {(1 - (\frac{\rho + b}{\rho - b - 2\sigma })^{n} )}$ are given by the coefficients of the following Taylor expansion:

$$ \ln (\xi \left( {b + 2\sigma + \frac{(2b + 2\sigma )z}{1 - z}} \right)) = \ln (\xi (b + 2\sigma )) + \sum\limits_{n = 1}^{\infty } {\frac{{\lambda_{n,b,\sigma } }}{n}z^{n} } $$

(A1)

Proof

Proof is a straightforward generalization of the approach given in [3]. We start from the well known property concerning the representation of the Riemann $\xi$-function as an infinite Hadamard product over its zeroes [4]

$$ \xi (s) = \frac{1}{2}\prod_{\rho } (1 - \frac{s}{\rho }) $$

(A2)

and then introduce the function $\xi^{*} (z): = \xi (z + b + 2\sigma )$. We have

$$ \xi^{*}(s) = \xi (b + 2\sigma ) \cdot \prod_{\rho *} (1 - \frac{s}{{\rho^{*} }}) $$

(A3)

where $\rho^{*} = \rho - b - 2\sigma$ are zeroes of $\xi^{*} (z)$. (This, similarly to (A2), is a direct consequence of the general Hadamard product formula and the relation $\frac{\xi '}{\xi }(b + 2\sigma ) = \sum\limits_{\rho } {\frac{1}{b + 2\sigma - \rho }}$ [4]. Now we take $s = - \frac{(2b + 2\sigma )z}{z - 1}$ hence $1 - \frac{s}{\rho *} =$ $= \frac{{1 - z(1 + \frac{2b + 2\sigma }{\rho *})}}{1 - z}$. We have, by taking a logarithm and applying Taylor expansion $\ln (1 - \frac{s}{\rho *}) = \ln (1 - z(1 + \frac{2\sigma + 2b}{\rho *})) - \ln (1 - z) = \sum\nolimits_{n = 1}^{\infty } {\frac{1}{n}(1 - (1 + \frac{2b + 2\sigma }{\rho *}))^{n} )z^{n} }$, so that, according to Eq. (A3), the sums $\lambda_{n,b,\sigma } : = \sum\nolimits_{\rho } {(1 - (1 + \frac{2b + 2\sigma }{\rho - b - 2\sigma })^{n} )}$ in question are given by Taylor expansion of the function $\ln (\xi^{*} (\frac{(2b + 2\sigma )z}{1 - z}))$$= \ln (\xi (b + 2\sigma + \frac{(2b + 2\sigma )z}{1 - z}))$$= \ln (\xi (\frac{b + 2\sigma + bz}{1 - z}))$ at the point z = 0: $\ln (\xi \left( {b + 2\sigma + \frac{(2b + 2\sigma )z}{1 - z}} \right)) = \ln (\xi (b + 2\sigma )) + \sum\nolimits_{n = 1}^{\infty } {\frac{{\lambda_{n,b,\sigma } }}{n}z^{n} }$.

In particular, if we take $b = - 2\sigma \ne 0$, we have the function $\ln (\xi (\frac{2\sigma z}{z - 1}))$ which is quite similar to the original Li’s function $\ln (\xi (\frac{z}{z - 1}))$: the non-negativity of all coefficients of the expansion $\ln (\xi \left( {\frac{2\sigma z}{z - 1}} \right)) = - \ln 2 + \sum\nolimits_{n = 1}^{\infty } {\frac{{\lambda_{n, - 2\sigma ,\sigma } }}{n}z^{n} }$ for some $0 < \sigma < 1/2$ is equivalent to the statement that there are no zeroes with $\text{Re} \rho \le \sigma$.

Appendix 2: Asymptotic of Generalized Li’s Sums Assuming the Riemann Hypothesis

Let us now calculate the asymptotics of the sums $\lambda_{n,b,1/2} = \sum\limits_{\rho } {(1 - \left( {\frac{\rho + b}{\rho - 1 - b}} \right)^{n} )}$ over the non-trivial zeroes of the Riemann zeta-function for large n (and thus also an asymptotics of the derivatives related with them) assuming the Riemann hypothesis.

Theorem 5

Assume the Riemann hypothesis. Then for large enough n, for any real fixed $b \ne - 1/2$

$$ \begin{aligned} \lambda_{n,b,1/2} = \sum\limits_{\rho } {(1 - \left( {\frac{\rho + b}{\rho - 1 - b}} \right)^{n} )} = \sum\limits_{\rho } {(1 - \left( {\frac{\rho - b - 1}{\rho + b}} \right)}^{n} ) = \hfill \\ \frac{|2b + 1|}{2}n\ln n + \frac{|2b + 1|}{2}(\gamma - 1 - \ln \frac{2\pi }{|2b + 1|})n + o(n) \hfill \\ \end{aligned} $$

(A4)

Proof

The proof is a straightforward generalization of the method presented in Coffey paper [17] (see also [20]; a similar asymptotics for Li’s sums was also obtained by Voros with another method [21]). Let us first put $b > - 1/2$. Using $\rho = 1/2 + iT$, we write for an argument $\vartheta$ of the function $\frac{\rho + b}{\rho - b - 1}$:

$$ \tan \vartheta = - \frac{(2b + 1)T}{{T^{2} - 1/4 - b - b^{2} }} = - \frac{(2b + 1)T}{{T^{2} - (2b + 1)^{2} /4}} $$

(A5)

Correspondingly, $\sin \vartheta = - \frac{(2b + 1)T}{{T^{2} + (2b + 1)^{2} /4}}$ and $\cos \vartheta =$$\frac{{T^{2} - (2b + 1)^{2} /4}}{{T^{2} + (2b + 1)^{2} /4}}$; here we used $(T^{2} - 1/4 - b - b^{2} )^{2} + (2b + 1)^{2} T^{2} = (T^{2} + 1/4 + b + b^{2} )^{2}$. The derivative $d\vartheta /dT$ is found from (A5): $\frac{d\vartheta }{dT} = \frac{2b + 1}{{T^{2} + (2b + 1)^{2} /4}}$, and now we are in a position to calculate the sum in question on Riemann hypothesis: $\lambda_{n,b,1/2} = \varSigma_{\rho } (1 - (\frac{\rho + b}{\rho - b - 1})^{n} ) = 2\sum\nolimits_{\rho } {(1 - \cos (n\vartheta_{\rho } ))}$ so that, expressed as an integral over the number of non-trivial zeroes dN, $k_{n,b} = 2\int\nolimits_{0}^{\infty } {(1 - \cos (n\vartheta (T)))dN}$. Integrating by parts, we obtain

$$ k_{n,b} = 2\int\limits_{0}^{\infty } {(1 - \cos (n\vartheta (T)))dN} = - 2n\int\limits_{0}^{\infty } {\sin (n\vartheta )\frac{d\vartheta }{dT}N(T)dT} $$

(A6)

Below we will use the approximations [4]

$$ N(T) = \frac{T}{2\pi }\ln \frac{T}{2\pi } - \frac{T}{2\pi } + O(\ln T) $$

(A7)

$\vartheta = - \frac{2b + 1}{T} + O(1/T^{3} )$, $\frac{d\vartheta }{dT} = \frac{2b + 1}{{T^{2} }} + O(1/T^{4} )$ for the functions under the integral sign in (A6), and then will use the variable change $y = \frac{(2b + 1)n}{T}$. This variable change implies $\frac{1}{T} = \frac{y}{(2b + 1)n}$ hence all terms of the type $O(\frac{1}{{T^{k} }})$ with “larger than necessary” k will add only o(n) contributions to the integral value and can be neglected. Thus we get $\lambda_{n,b,1/2} = 2n\int_{{T_{1} }}^{\infty } {\frac{(2b + 1)}{{T^{2} }}\sin (\frac{(2b + 1)n}{T})N(T)dT} + o(n)$ where T₁ = 14, say (the first zero lies at ½ + i14.1347… [4]), and further $k_{n,b} = 2\int_{0}^{{(2b + 1)n/T_{1} }} {\sin (y)N(\frac{(2b + 1)n}{y})dy} = - \frac{(2b + 1)n}{\pi }\int_{0}^{\infty } {\frac{\sin y}{y}(\ln \frac{2\pi y}{(2b + 1)n} + 1)dy} + o(n)$. Using examples N3.721.1 $\int_{0}^{\infty } {\frac{\sin y}{y}dy} = \frac{\pi }{2}$ and N4.421.1 $\int_{0}^{\infty } {\ln y\frac{\sin y}{y}dy} = - \frac{\pi }{2}\gamma$ from [18], we obtain Eq. (A4) for positive 2b + 1. The case b < −1/2 is quite similar with changes of signs whenever appropriate, and in this manner we finish the proof of Theorem 5.

This asymptotics depends only on the density of zeroes (A7) and thus can be elementarily modified for numerous other zeta-functions-of course, assuming correspondingly the generalized Riemann hypothesis.

Remark 6

Following [17, 20], the sum (derivative) in question can be rewritten, using $\sin n\vartheta = \sin \vartheta \cdot U_{n - 1} (\cos \vartheta )$, where U_k is the k-th Chebyshev polynomial of the second kind [19, 22], in a rather elegant form

$$ k_{n,b} = 2n\int\limits_{0}^{\infty } {\frac{{(2b + 1)^{2} T}}{{(T^{2} + (2b + 1)^{2} /4)^{2} }}U_{n - 1} (\frac{{T^{2} - (2b + 1)^{2} /4}}{{T^{2} + (2b + 1)^{2} /4}})N(T)dT} $$

(A8)

We will not use any properties of this polynomial below, but would like to note the next logical step which is the variable change $x = \frac{{T^{2} - (2b + 1)^{2} /4}}{{T^{2} + (2b + 1)^{2} /4}} = 1 - \frac{{(2b + 1)^{2} /2}}{{T^{2} + (2b + 1)^{2} /4}}$. Clearly, $dx = \frac{{T(2b + 1)^{2} }}{{(T^{2} + (2b + 1)^{2} /4)^{2} }}dT$ so that

$$ k_{n,b} = 2n\int\limits_{ - 1}^{1} {U_{n - 1} (x)N(x)dx} $$

(A9)

Using $T = \frac{1}{2}(2b + 1)\sqrt {\frac{1 + x}{1 - x}}$ and limiting ourselves with the $N(T) = \frac{T}{2\pi }\ln \frac{T}{2\pi } - \frac{T}{2\pi } + O(\ln T)$ precision, we may write $N(x) = \frac{2b + 1}{4\pi }\sqrt {\frac{1 + x}{1 - x}} \ln (\frac{2b + 1}{4\pi }\sqrt {\frac{1 + x}{1 - x}} ) - \frac{2b + 1}{4\pi }\sqrt {\frac{1 + x}{1 - x}} + O(\ln \frac{1 + x}{1 - x})$ which is to be substituted into (A5). Note, that integrals of the type $\int_{ - 1}^{1} {U_{n} (x)(1 - x)^{\alpha } (1 + x)^{\beta } dx}$ quite naturally appear in some applications of the Chebyshev polynomials; see e.g. example N 7.347.2 of book [18].