Uniform Beta Smoothing ofWidder Sums for the Riemann ξ-Function

1. Introduction

Let G be the entire function determined by $G\left(y^2\right)=\xi (1/2+y)$, and put $F_{\xi }=2G^{\prime }/G$. For $m\ge 1$ define

$${\mathcal{W}}_m\left[F\right]\left(x\right)={(-1)}^{m-1}{\left(x^{m}F\left(x\right)\right)}^{(2m-1)},A_m\left(x\right)={\displaystyle \frac{{\int }_0^x{u}^{m-1}{\mathcal{W}}_m\left[F_{\xi }\right]\left(u\right)du}{2(2m-1)!B(m,m)}}.$$

The pointwise Widder expressions have zero weights concentrated around ordinate $\sqrt{x}$. Their normalized integrals instead have decreasing beta weights and can therefore be compared with cumulative zero counts.

Our main result, Theorem 1, is the uniform formula

$$A_m\left(x\right)={\displaystyle \frac{c_m\sqrt{x}}{2\pi }}\left(log{\displaystyle \frac{\sqrt{x}}{2\pi }}-1+{\displaystyle \frac{1}{2m-1}}\right)+O_L(logx),1\le m\le Lx,$$

where $c_m=\Gamma (m-1/2)\Gamma (m+1/2)/\Gamma {\left(m\right)}^2$. The main term follows by integrating the Riemann–von Mangoldt formula against the real beta kernel. The genuinely uniform step is the replacement of each complex reflected zero by its ordinate: the vertical displacement is only $O\left(x^{-1/2}\right)$, but the derivative of the beta kernel may have size $O\left(\sqrt{m}\right)$ when m grows with x. Lemma 3 proves that the total replacement error is nevertheless only $O_L(logx)$. Corollary 1 then gives a direct approximation to the ordinary zero count. Proposition 1 isolates the elementary positive-kernel argument used to transfer cumulative density bounds.

Contribution and scope.

The principal contribution of the paper is the uniform beta-smoothed zero-count asymptotic in the full range $m\le Lx$, together with its count-recovery and density-transfer consequences. The finite-height Stieltjes calculation in Section 4 is a separate application of known finite Stieltjes theory, a published rigorous zero verification, and an explicit zero-free region. It is included because it gives a concrete global finite-order consequence, but it is not presented as the source of the paper’s main novelty.

The analytic smoothing theorem uses the critical strip, the genus-zero product, and the classical zero-count remainder. It does not require a lower bound on the number of critical-line zeros. Such a bound enters only the density application: Theorem 1.1 of Lamzouri [7] gives the Montgomery–Taylor value $3/2-2^{-1/2}cot\left(2^{-1/2}\right)$.

For context, Bondesson–Simon ([1], Definition 1, Theorem 1 and Section 4.1) establish the finite Stieltjes classes, their integral representations, and the quadratic angular threshold used below. Our finite-height application combines these ingredients with the zero verification of Platt–Trudgian [4] and the zero-free region of Bellotti–Trudgian–Yang [2]. Its contribution is an explicit height-to-order conversion. The quadratic examples and finite Stieltjes representation are recalled consequences of the existing theory, not separate originality claims.

Section 2 establishes the zero-sum identities. Section 3 contains the smoothing theorem and its applications. Section 4 treats finite-height positivity. Section 5 explains the relation with existing criteria. The precise contribution investigated here is the uniform beta-smoothed zero-count statement, rather than a new RH criterion or an improved zero-density constant; no exhaustive priority claim is made.

2. Zero Sums and Widder Expressions

We use the standard completed zeta function

$$\xi \left(s\right)={\displaystyle \frac{1}{2}}s(s-1){\pi }^{-s/2}\Gamma \left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right),$$

which is entire and satisfies

$$\xi \left(s\right)=\xi (1-s),\overline{\xi \left(s\right)}=\xi \left(\overline{s}\right).$$

Put

$$H\left(y\right)=\xi \left({\displaystyle \frac{1}{2}}+y\right).$$

Then H is even. Consequently there is an entire function G such that

$$H\left(y\right)=G\left(y^2\right),G\left(x\right)=\xi \left({\displaystyle \frac{1}{2}}+\sqrt{x}\right).$$

(1)

Since H has order one, G has order $1/2$. In particular G has a genus-zero canonical product.

Define, for $x>0$,

$$F_{\xi }\left(x\right):=2{\displaystyle \frac{G^{\prime }\left(x\right)}{G\left(x\right)}}={\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{{\xi }^{\prime }}{\xi }}\left({\displaystyle \frac{1}{2}}+\sqrt{x}\right).$$

(2)

The function $F_{\xi }$ is positive on $(0,\infty )$. For completeness, we record an explicit form of the classical Riemann kernel. Put

$$\Phi \left(u\right)={\displaystyle \sum _{n=1}^{\infty }}\left(4{\pi }^2{n}^4{e}^{9u/2}-6\pi n^2{e}^{5u/2}\right)e^{-\pi n^2{e}^{2u}},u\ge 0.$$

(3)

Then the standard Fourier representation of the Riemann $\Xi$-function gives, after the substitution $t=-iy$,

$$H\left(y\right)=\xi \left({\displaystyle \frac{1}{2}}+y\right)=2{\int }_0^{\infty }\Phi \left(u\right)cosh\left(yu\right)du,y\in \mathbb{R},$$

(4)

see Titchmarsh ([8], Ch. II). Every summand in (3) is strictly positive for $u\ge 0$, since

$$4{\pi }^2{n}^4{e}^{9u/2}-6\pi n^2{e}^{5u/2}=2\pi n^2{e}^{5u/2}\left(2\pi n^2{e}^{2u}-3\right)>0.$$

The series and all differentiated integrands have the standard rapid decay needed for local uniform convergence. Hence differentiation under the integral gives

$$H^{\prime }\left(y\right)=2{\int }_0^{\infty }u\Phi \left(u\right)sinh\left(yu\right)du>0(y>0),$$

while $H\left(y\right)>0$ for $y\ge 0$. Since

$$F_{\xi }\left(x\right)={\displaystyle \frac{H^{\prime }\left(\sqrt{x}\right)}{\sqrt{x}H\left(\sqrt{x}\right)}},$$

we obtain $F_{\xi }\left(x\right)>0$ for every $x>0$.

Let ${\mathcal{Z}}_{+}$ denote the multiset of nontrivial zeros $\rho =\beta +i\gamma$ with $\gamma >0$, counted with multiplicity. The functional equation pairs the zero $\rho$ with $1-\rho$ in the lower half-plane, so the zeros of the genus-zero function G in (1) are precisely

$$z_{\rho }={(\rho -1/2)}^2,\rho \in {\mathcal{Z}}_{+}.$$

It is convenient to reflect them:

$$w_{\rho }:=-z_{\rho }=-{(\rho -1/2)}^2.$$

(5)

Writing

$$\delta =\beta -{\displaystyle \frac{1}{2}},$$

we obtain the exact identity

$$w_{\rho }=-{(\delta +i\gamma )}^2={(\gamma -i\delta )}^2.$$

(6)

Thus

$$|arg{w}_{\rho }|=2arctan{\displaystyle \frac{\left|\delta \right|}{\gamma }}.$$

(7)

For a critical-line zero, $\delta =0$ and $w_{\rho }={\gamma }^2>0$.

Since the number of zeta zeros up to height T is $O(TlogT)$,

$$\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{1}{|w_{\rho }|}}<\infty .$$

Therefore

$$G\left(x\right)=G\left(0\right)\prod _{\rho \in {\mathcal{Z}}_{+}}\left(1+{\displaystyle \frac{x}{w_{\rho }}}\right)$$

(8)

and

$$F_{\xi }\left(x\right)=2\sum _{\rho \in {\mathcal{Z}}_{+}}{\displaystyle \frac{1}{x+w_{\rho }}}.$$

(9)

For an off-line zero $\rho =1/2+\delta +i\gamma$, the symmetry $1-\overline{\rho }$ is also an upper-half-plane zero and gives $w_{1-\overline{\rho }}=\overline{w_{\rho }}$. Hence the nonreal terms in (9) occur in conjugate pairs, as used in the sector argument below.

For $m\ge 1$, define

$${\mathcal{W}}_m\left[F\right]\left(x\right):={(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}F\left(x\right)\right).$$

(10)

Lemma 1 

(Elementary derivative identity). For $m\ge 1$, $w\ne 0$, and $x\ne -w$,

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left({\displaystyle \frac{x^m}{x+w}}\right)=(2m-1)!{\displaystyle \frac{w^m}{{(x+w)}^{2m}}}.$$

(11)

Proof. 

Divide $x^m$ by $x+w$. The quotient has degree $m-1$, hence its $(2m-1)$st derivative vanishes. The remainder is ${(-w)}^m/(x+w)$. Differentiating the latter $2m-1$ times gives (11).    □

For a genus-zero product $G\left(x\right)=G\left(0\right){\prod }_j(1+x/w_j)$ with ${\sum }_j{\left|w_j\right|}^{-1}<\infty$, put $F=2G^{\prime }/G$. It follows that

$${\mathcal{W}}_m\left[F\right]\left(x\right)=2(2m-1)!\sum _j{\displaystyle \frac{w_j^m}{{(x+w_j)}^{2m}}}.$$

(12)

For later use we record the convergence explicitly. Fix a compact interval $I⋐(0,\infty )$. Under the sector bound $|arg{w}_j|\le \vartheta <\pi /2$ there is a constant $c_I>0$ such that $|x+w_j|\ge c_I(1+|w_j\left|\right)$ for $x\in I$. Hence

$$\left|{\displaystyle \frac{w_j^m}{{(x+w_j)}^{2m}}}\right|\le C_{I,m}min\{1,|w_j{|}^{-m}\}.$$

Only finitely many $w_j$ lie in a bounded set, while ${\sum }_j{\left|w_j\right|}^{-1}<\infty$; consequently the series in (12) converges absolutely and locally uniformly, for every $m\ge 1$. The same estimates, applied before differentiation to finitely many derivative orders, justify term-by-term differentiation of the logarithmic-derivative series.

3. Uniform Beta Smoothing and Density Transfer

A sector argument for pointwise Widder positivity uses a bound on every reflected zero. A cumulative lower density of critical-line zeros is a different input: it does not directly control the weight

$${\displaystyle \frac{t^{2m}}{{(x+t^2)}^{2m}}},$$

which increases up to $t=\sqrt{x}$ and then decreases. We introduce an integrated Widder expression whose zero weight is decreasing. This permits a rigorous transfer of cumulative density information. The uniformity in the order is recorded explicitly.

For $m\ge 1$ put $B_m=B(m,m)=\Gamma {\left(m\right)}^2/\Gamma \left(2m\right)$ and define

$$A_m\left(x\right)={\displaystyle \frac{1}{2(2m-1)!B_m}}{\int }_0^x{u}^{m-1}{\mathcal{W}}_m\left[F_{\xi }\right]\left(u\right)du.$$

(13)

The expression at zero is understood by analytic continuation of $F_{\xi }$ to a neighborhood of zero. Introduce the rational function

$$P_m(x,w)={\displaystyle \sum _{j=m}^{2m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m-1}{j}}\right){\displaystyle \frac{x^j{w}^{2m-1-j}}{{(x+w)}^{2m-1}}}.$$

(14)

For positive real w, this is the regularized incomplete beta function $I_{x/(x+w)}(m,m)$. Formula (14), rather than a choice of branch for the incomplete beta function, defines its complex extension here. Differentiating gives

$${\partial }_x{P}_m(x,w)={\displaystyle \frac{x^{m-1}{w}^m}{B_m{(x+w)}^{2m}}},P_m(0,w)=0.$$

Consequently, by (12) and local uniform convergence,

$$A_m\left(x\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{P}_m(x,w_{\rho }).$$

(15)

Indeed, for fixed m and bounded x, the summands are $O_m\left(\right|w_{\rho }{|}^{-m})$. The sum is real by conjugation symmetry. In particular,

$$A_1\left(x\right)={\displaystyle \frac{x{F}_{\xi }\left(x\right)}{2}},P_2(x,w)={\displaystyle \frac{x^2(x+3w)}{{(x+w)}^3}}.$$

Let $N_0^{\mathrm{s}}\left(t\right)$ count simple critical-line zeros with positive ordinate at most t. Define the positive simple-zero contribution

$$S_m\left(x\right)=\sum _{\begin{array}{c}\rho \in {\mathcal{Z}}_{+}\\ \beta =1/2\\ \rho \mathrm{simple}\end{array}}{P}_m(x,{\gamma }^2).$$

(16)

Here and below a simple zero is counted once, whereas $N\left(t\right)$ and unrestricted zero sums count multiplicities.

Theorem 1 

(Uniform smoothed asymptotic and density transfer). Set

$$c_m={\displaystyle \frac{\Gamma (m-1/2)\Gamma (m+1/2)}{\Gamma {\left(m\right)}^2}}.$$

For every fixed $L>0$, as $x\to \infty$, uniformly over integers $1\le m\le Lx$,

$$A_m\left(x\right)={\displaystyle \frac{c_m\sqrt{x}}{2\pi }}\left(log{\displaystyle \frac{\sqrt{x}}{2\pi }}-1+{\displaystyle \frac{1}{2m-1}}\right)+O_L(logx).$$

(17)

In particular $A_m\left(x\right)>0$ throughout this range once x is sufficiently large (depending on L).

Suppose, for some $p\in [0,1]$, that

$$\underset{t\to \infty }{lim\; inf}{\displaystyle \frac{N_0^{\mathrm{s}}\left(t\right)}{N\left(t\right)}}\ge p.$$

(18)

Then

$$\underset{x\to \infty }{lim\; inf}\underset{\begin{array}{c}m\in \mathbb{N}\\ 1\le m\le Lx\end{array}}{inf}{\displaystyle \frac{S_m\left(x\right)}{A_m\left(x\right)}}\ge p\mathrm{for}\mathrm{every}\mathrm{fixed}L>0.$$

(19)

We give details of the uniform estimates; a fixed-order asymptotic by itself would not establish this theorem. The homogeneity of (14) will be used repeatedly: for $T>0$,

$$P_m(T^2,{(\gamma -i\delta )}^2)=P_m\left(1,{\left({\displaystyle \frac{\gamma -i\delta }{T}}\right)}^2\right)=k_m\left({\displaystyle \frac{\gamma -i\delta }{T}}\right),$$

(20)

where on the right $k_m$ denotes the rational continuation defined by $P_m(1,z^2)$.

Lemma 2 

(Uniform real-kernel estimates). Put

$$k_m\left(y\right)=P_m(1,y^2)=I_{1/(1+y^2)}(m,m),y\ge 0.$$

Then $k_m\left(0\right)=1$, $k_m(\infty )=0$, and

$$h_m\left(y\right):=-k_m^{\prime }\left(y\right)={\displaystyle \frac{2y^{2m-1}}{B_m{(1+y^2)}^{2m}}}>0(y>0).$$

(21)

Uniformly in $m\ge 1$,

$${\int }_0^{\infty }{h}_m\left(y\right)dy=1,\underset{y\ge 0}{sup}{h}_m\left(y\right)\ll \sqrt{m},{\int }_0^{\infty }{h}_m\left(y\right)log(2+y)dy\ll 1,$$

and $h_m\left(y\right)\ll y^{-3}$ for $y\ge 2$. Moreover,

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{k}_m\left(y\right)dy& =c_m,\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{k}_m\left(y\right)logydy& =c_m\left(-1+{\displaystyle \frac{1}{2m-1}}\right).\hfill \end{array}$$

(23)

One has $1\le c_m\le \pi /2$ and $c_m=1+O\left(m^{-1}\right)$.

Proof. 

The derivative follows either from the incomplete beta integral or directly from (14). The beta integral and Stirling’s formula give $B_m^{-1}\ll 4^m\sqrt{m}$. The function $h_m$ increases to its unique maximum at $y^2=(2m-1)/(2m+1)$ and decreases thereafter. At that maximum $y\ge 1/\sqrt{3}$ and $4y^2/{(1+y^2)}^2\le 1$, which proves the supremum estimate. For $y\ge 2$, set $r\left(y\right)=4y^2/{(1+y^2)}^2\le 16/25$. Then

$$h_m\left(y\right)\ll {\displaystyle \frac{\sqrt{m}r{\left(y\right)}^m}{y}}\ll {\displaystyle \frac{r\left(y\right)}{y}}\ll y^{-3}.$$

The asserted logarithmic moment of $h_m$ follows from this tail bound and its total mass one.

For a in a neighborhood of zero, Tonelli’s theorem applied to the beta integral gives

$${\int }_0^{\infty }{y}^a{k}_m\left(y\right)dy={\displaystyle \frac{B(m-(a+1)/2,m+(a+1)/2)}{(a+1)B_m}}.$$

Evaluation and differentiation at $a=0$ prove (22), using $\psi (m+1/2)-\psi (m-1/2)=1/(m-1/2)$. Finally $c_1=\pi /2$, $c_{m+1}/c_m=1-1/\left(4m^2\right)$, and $c_m\to 1$ by Stirling’s formula. The same recurrence gives

$$0\le log{c}_m=-{\displaystyle \sum _{j=m}^{\infty }}log\left(1-{\displaystyle \frac{1}{4j^2}}\right)\ll {\displaystyle \sum _{j=m}^{\infty }}{j}^{-2}\ll m^{-1},$$

which also proves $c_m=1+O\left(m^{-1}\right)$.    □

Lemma 3 

(Replacing reflected zeros by their ordinates). For $T\to \infty$ and each fixed $L>0$,

$$\sum _{\rho \in {\mathcal{Z}}_{+}}\left|P_m(T^2,{(\gamma -i\delta )}^2)-k_m(\gamma /T)\right|{\ll }_{L}logT(1\le m\le L{T}^2).$$

(24)

Proof. 

By (20), the summand is the change of the rational function $k_m$ along the vertical segment joining $y=\gamma /T$ to $y-i\delta /T$. Since $\left|\delta \right|\le 1/2$, every such segment has length at most $1/\left(2T\right)$. The only poles of the rational continuation of $k_m$ are at $\pm i$, and differentiation of (14) gives

$$k_m^{\prime }\left(z\right)=-{\displaystyle \frac{2z^{2m-1}}{B_m{(1+z^2)}^{2m}}}.$$

(25)

We first treat $y\ge 1/2$. Put $z=y-iϵ$, with $\left|ϵ\right|\le 1/\left(2T\right)$. From (25) and (21),

$${\displaystyle \frac{|k_m^{\prime }\left(z\right)|}{h_m\left(y\right)}}={\left({\displaystyle \frac{\left|z\right|}{y}}\right)}^{2m-1}{\left({\displaystyle \frac{1+y^2}{|1+z^2|}}\right)}^{2m}.$$

(26)

Now

$${\displaystyle \frac{\left|z\right|}{y}}={\left(1+{\displaystyle \frac{{ϵ}^2}{y^2}}\right)}^{1/2}\le exp\left({\displaystyle \frac{{ϵ}^2}{2y^2}}\right),$$

while, for $T\ge 4$,

$$|1+z^2|\ge \Re (1+z^2)=1+y^2-{ϵ}^2>0.$$

Hence

$${\displaystyle \frac{1+y^2}{|1+z^2|}}\le {\left(1-{\displaystyle \frac{{ϵ}^2}{1+y^2}}\right)}^{-1}.$$

Since $y\ge 1/2$ and $\left|ϵ\right|\le 1/8$, the quantity $u={ϵ}^2/(1+y^2)$ satisfies $0\le u\le 1/80$; thus $-log(1-u)\le (80/79)u$. Substituting these two estimates into (26) yields, with room to spare,

$$|k_m^{\prime }(y-iϵ)|\le e^{8m{ϵ}^2}{h}_m\left(y\right)\le e^{2L}{h}_m\left(y\right),y\ge {\displaystyle \frac{1}{2}},$$

(27)

for $1\le m\le L{T}^2$.

It remains to control the low-y rectangle. If $0\le y<1/2$ and $\left|ϵ\right|\le 1/8$, a direct calculation gives

$$r\left(z\right):={\displaystyle \frac{{4\left|z\right|}^2}{|1+z^2{|}^2}}\le {\displaystyle \frac{4352}{6241}}<{\displaystyle \frac{3}{4}}.$$

The point that is useful near $z=0$ is to factor one power of z before using the beta normalization. From (25),

$$\begin{array}{cc}\hfill |k_m^{\prime }\left(z\right)|& ={\displaystyle \frac{2\left|z\right|}{B_m{|1+z^2|}^2}}{\left({\displaystyle \frac{{\left|z\right|}^2}{|1+z^2{|}^2}}\right)}^{m-1}\hfill \\ \hfill & \ll \sqrt{m}{\displaystyle \frac{\left|z\right|}{|1+z^2{|}^2}}r{\left(z\right)}^{m-1}.\hfill \end{array}$$

Here we used $B_m^{-1}\ll 4^m\sqrt{m}$, absorbing the remaining fixed factor 4. On the closed rectangle $0\le y\le 1/2$, $\left|ϵ\right|\le 1/8$, the factor $\left|z\right|/|1+z^2{|}^2$ is uniformly bounded. Therefore

$$|k_m^{\prime }(y-iϵ)|\ll \sqrt{m}{(3/4)}^{m-1}\ll 1,0\le y<{\displaystyle \frac{1}{2}}.$$

(28)

This explicit factorization avoids any spurious singular factor $1/\left|z\right|$ at $z=0$.

Integrating along the vertical segments and splitting at $\gamma =T/2$ now bounds the left side of (24) by

$${\displaystyle \frac{C}{T}}N(T/2)+{\displaystyle \frac{C_L}{T}}\sum _{\gamma \ge T/2}{h}_m(\gamma /T).$$

(29)

Write the classical Riemann–von Mangoldt formula as

$$N\left(t\right)=M\left(t\right)+E\left(t\right),M\left(t\right)={\displaystyle \frac{t}{2\pi }}log{\displaystyle \frac{t}{2\pi }}-{\displaystyle \frac{t}{2\pi }},E\left(t\right)=O(log(t+2)).$$

(30)

Here $M\left(0\right)=0$ by continuity; the bound for E extends to all $t\ge 0$ by adjusting the constant (see ([8], Ch. IX)). Let $a=T/2$. For the sum over ordinates $\gamma >a$, Stieltjes integration gives

$$\sum _{\gamma >a}{h}_m(\gamma /T)={\int }_{(a,\infty )}{h}_m(t/T)dM\left(t\right)+{\int }_{(a,\infty )}{h}_m(t/T)dE\left(t\right).$$

Since $M^{\prime }\left(t\right)={\left(2\pi \right)}^{-1}log(t/\left(2\pi \right))$ for $t>0$,

$$\left|{\int }_a^{\infty }{h}_m(t/T)dM\left(t\right)\right|\ll T{\int }_{1/2}^{\infty }{h}_m\left(y\right)\left(logT+log(2+y)\right)dy\ll TlogT.$$

Integration by parts against $dE$ bounds the second contribution by

$$C{h}_m(1/2)log(T+2)+C{\int }_{1/2}^{\infty }\left|h_m^{\prime }\left(y\right)\right|log(Ty+2)dy.$$

(31)

The boundary term at infinity vanishes because $h_m\left(y\right)\ll y^{-3}$ there. On $[1/2,2]$, unimodality gives

$${\int }_{1/2}^2\left|h_m^{\prime }\left(y\right)\right|dy\le 2sup{h}_m\ll \sqrt{m}.$$

On $[2,\infty )$, $h_m$ is decreasing, and another integration by parts gives

$$\begin{array}{cc}\hfill {\int }_2^{\infty }\left|h_m^{\prime }\left(y\right)\right|log(Ty+2)dy& =h_m\left(2\right)log(2T+2)+{\int }_2^{\infty }{\displaystyle \frac{T{h}_m\left(y\right)}{Ty+2}}dy\hfill \\ \hfill & \ll logT+{\int }_2^{\infty }{y}^{-4}dy\ll logT.\hfill \end{array}$$

Thus (31) is $O(\sqrt{m}logT)$. If a itself is a zero ordinate, its multiplicity is $O(log(T+2\left)\right)$, obtained by comparing (30) at $a-1$ and $a+1$. Restoring that possible boundary atom therefore costs at most $O(\sqrt{m}logT)$. We have proved

$$\sum _{\gamma \ge T/2}{h}_m(\gamma /T)\ll (T+\sqrt{m})logT.$$

(32)

Substituting this into (29), using $N(T/2)\ll TlogT$ and $\sqrt{m}/T\le \sqrt{L}$, proves (24).    □

Proposition 1 

(Transfer by a decreasing kernel). Let N and Q be nonnegative, nondecreasing counting functions on $[0,\infty )$ with $N\left(0\right)=Q\left(0\right)=0$. Suppose

$$Q\left(t\right)\ge qN\left(t\right)-C(t\ge 0)$$

for constants $q,C\ge 0$. Let k be an absolutely continuous decreasing function with $k\left(0\right)=1$, $k(\infty )=0$, and $h=-k^{\prime }\ge 0$. For any $T>0$ for which the integrals below are finite and the boundary products vanish,

$${\int }_0^{\infty }k(t/T)dQ\left(t\right)\ge q{\int }_0^{\infty }k(t/T)dN\left(t\right)-C.$$

Proof. 

Integration by parts writes the two integrals as ${\int }_0^{\infty }Q\left(Ty\right)h\left(y\right)dy$ and ${\int }_0^{\infty }N\left(Ty\right)h\left(y\right)dy$. Integrate the assumed inequality and use ${\int }_0^{\infty }h\left(y\right)dy=1$.    □

Proof 

(Proof of Theorem 1). Write

$$R_m\left(T\right)=\sum _{\rho \in {\mathcal{Z}}_{+}}{k}_m(\gamma /T).$$

Partial summation in (30) gives

$$R_m\left(T\right)={\displaystyle \frac{T}{2\pi }}{\int }_0^{\infty }{k}_m\left(y\right)log{\displaystyle \frac{Ty}{2\pi }}dy+O(logT),$$

uniformly in all $m\ge 1$. More explicitly, $R_m\left(T\right)={\int }_0^{\infty }{k}_m(t/T)dN\left(t\right)$; the $dM$ part becomes the displayed integral after $t=Ty$, while integration by parts in the $dE$ part gives an error bounded by a constant multiple of

$${\int }_0^{\infty }{h}_m\left(y\right)log(Ty+2)dy=O(logT)$$

by Lemma 2. The boundary terms vanish: at zero one uses $M\left(0\right)=E\left(0\right)=N\left(0\right)=0$, and at infinity $k_m(t/T)=O_m\left(t^{-2m}\right)$ for each fixed m occurring in the Stieltjes integral. The resulting error bound is uniform because only the normalized mass and logarithmic moment of $h_m$ enter the estimate. Lemma 2 evaluates the integral, while Lemma 3 gives

$$A_m\left(T^2\right)=R_m\left(T\right)+O_L(logT)(1\le m\le L{T}^2).$$

(33)

This proves (17). Since $1\le c_m\le \pi /2$, the main term is bounded below by a positive constant times $TlogT$ for sufficiently large T, uniformly in this range.

For the density assertion, first take $p>0$ and fix $0<\epsilon <p$. Assumption (18) implies

$$N_0^{\mathrm{s}}\left(t\right)\ge (p-\epsilon )N\left(t\right)-C_{\epsilon }(t\ge 0)$$

for some finite $C_{\epsilon }$. Proposition 1, with $q=p-\epsilon$, gives

$$\begin{array}{cc}\hfill S_m\left(T^2\right)& ={\int }_0^{\infty }{N}_0^{\mathrm{s}}\left(Ty\right)h_m\left(y\right)dy\hfill \\ \hfill & \ge (p-\epsilon )R_m\left(T\right)-C_{\epsilon }.\hfill \end{array}$$

Combine this with (33) and the uniform lower bound $A_m\left(T^2\right)\gg TlogT$. Let $T\to \infty$ and then $\epsilon \downarrow 0$. If $p=0$, the conclusion follows directly from $S_m\ge 0$ and eventual positivity of $A_m$.    □

Corollary 1 

(Recovery of the ordinary zero count). For fixed $a,L>0$, uniformly for integers $a\sqrt{x}\le m\le Lx$ as $x\to \infty$,

$$A_m\left(x\right)=N\left(\sqrt{x}\right)+O_{a,L}(logx).$$

Proof. 

Put $T=\sqrt{x}$. Theorem 1 and $c_m=1+O\left(m^{-1}\right)$ give

$$A_m\left(T^2\right)=M\left(T\right)+O_L\left(logT+{\displaystyle \frac{TlogT}{m}}\right).$$

If $m\ge aT$, the second error is $O_a(logT)$. Apply (30) to replace $M\left(T\right)$ by $N\left(T\right)$.    □

Remark 1 

(Meaning of the density ratio). Although $S_m$ is a positive sum, individual off-line blocks in $A_m$ need not be nonnegative in the range of Theorem 1. The ratio $S_m/A_m$ is therefore not a probability or a literal partition of positive mass. Nevertheless, $0\le S_m\left(T^2\right)\le R_m\left(T\right)$ and (33) imply

$$0\le {\displaystyle \frac{S_m\left(T^2\right)}{A_m\left(T^2\right)}}\le 1+O_L\left(T^{-1}\right)$$

uniformly for $1\le m\le L{T}^2$, once T is sufficiently large. The density theorem is an asymptotic comparison with an eventually positive normalizing sum.

Corollary 2 

(Application of Lamzouri’s stated proportion theorem). Using Theorem 1.1 of Lamzouri [7], Theorem 1 applies with the Montgomery–Taylor constant

$$C_0={\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{\sqrt{2}}}cot{\displaystyle \frac{1}{\sqrt{2}}}=0.6725007036794116457\dots .$$

Thus for every fixed $L>0$,

$$\underset{x\to \infty }{lim\; inf}\underset{\begin{array}{c}m\in \mathbb{N}\\ 1\le m\le Lx\end{array}}{inf}{\displaystyle \frac{S_m\left(x\right)}{A_m\left(x\right)}}\ge C_0.$$

The dependence on this external theorem is confined to the numerical value of p; the smoothed asymptotic and the general transfer theorem do not use it.

Remark 2 

(What this extension establishes). The construction links the Widder expressions to cumulative zero counts through an exact beta-kernel identity. The estimates are uniform for $m\le Lx$, rather than only for fixed m. The density transfer itself is an Abel-summation consequence of (18); it does not improve $C_0$. Positivity of $A_m\left(x\right)$ is positivity of an integral of a Widder expression and does not imply pointwise positivity of that expression. The theorem therefore gives no increase in the global finite Stieltjes cutoff and no implication to RH. No explicit onset $x_0\left(L\right)$ is asserted.

4. Finite-Height Positivity in the Existing Stieltjes Framework

For a nonnegative smooth function f, membership in the finite Stieltjes class ${\mathcal{S}}_k$ means ${\mathcal{W}}_m\left[f\right]\ge 0$ for $1\le m\le k$ ([1], Definition 1). Consider a real genus-zero product

$$G\left(x\right)=G\left(0\right)\prod _j(1+x/w_j),G\left(0\right)\ne 0,\sum _j{\left|w_j\right|}^{-1}<\infty .$$

(34)

Nonreal parameters occur in conjugate pairs. With $F=2G^{\prime }/G$, formula (12) applies. The following argument extends the quadratic calculation of Bondesson–Simon ([1], Section 4.1) by summation.

Lemma 4 

(Sector geometry). Let $w=r{e}^{i\varphi }$ with $r>0$ and $\left|\varphi \right|<\pi /2$, and let $x\ge 0$. Then

$$\left|arg\left({\displaystyle \frac{w}{{(x+w)}^2}}\right)\right|\le \left|\varphi \right|.$$

(35)

If $x>0$ and $\varphi \ne 0$, the inequality is strict.

Proof. 

Assume first $\varphi \ge 0$ and put $\psi =arg(x+w)$. Adding the nonnegative real number x moves the argument toward the positive real axis, so $0\le \psi \le \varphi$. Hence

$$-\varphi \le \varphi -2\psi \le \varphi .$$

This is (35). If $x>0$ and $\varphi >0$, then $0<\psi <\varphi$, which gives strictness. The case $\varphi \le 0$ follows by conjugation.    □

Theorem 2 

(Sector-to-Widder positivity). Assume in (34) that

$$|arg{w}_j|\le \vartheta <{\displaystyle \frac{\pi }{2}}\mathrm{for}\mathrm{every}j.$$

(36)

Then for every integer $m\ge 1$ satisfying

$$m\vartheta \le {\displaystyle \frac{\pi }{2}},$$

(37)

one has

$${\mathcal{W}}_m\left[F\right]\left(x\right)>0\mathrm{for}\mathrm{every}x>0,$$

(38)

provided G has at least one zero. More quantitatively,

$${\mathcal{W}}_m\left[F\right]\left(x\right)\ge 2(2m-1)!cos\left(m\vartheta \right)\sum _j{\displaystyle \frac{|w_j{|}^m}{|x+w_j{|}^{2m}}}.$$

(39)

Proof. 

A positive real $w_j$ contributes positively to (12). For a nonreal conjugate pair put $q=w/{(x+w)}^2$. Lemma 4 gives $|argq|<|argw|\le \vartheta$ for $x>0$. Hence

$$q^m+{\overline{q}}^m=2{\left|q\right|}^{m}cos(margq)>0$$

whenever $m\vartheta \le \pi /2$, including equality. Also

$$cos(margq)\ge cos\left(m\vartheta \right)\ge 0.$$

Absolute convergence permits summation over the real zeros and conjugate pairs. This proves both conclusions. At $m\vartheta =\pi /2$ the displayed lower bound is zero, but the individual blocks still establish strict positivity. If $\vartheta =0$, all reflected zeros are positive real and every order is positive.    □

Attribution of the quadratic examples.

Bondesson–Simon ([1], Section 4.1, Proposition 3 and its proof) use $f_{\alpha }\left(x\right)={(1+2cos\left(\pi \alpha \right)x+x^2)}^{-1}$ and $g_{\alpha }=-{(log{f}_{\alpha })}^{\prime }$. Their calculation gives the finite threshold $2\alpha k\le 1$. Below $F_{\varphi }=2g_{\alpha }$ with $\varphi =\pi \alpha$. The examples and threshold are therefore recorded as known building blocks; the extension to products uses dilation, summation and convergence.

Proposition 2 

(Exact universal sector threshold). Let $0\le \vartheta <\pi /2$ and $m\ge 1$. The inequality ${\mathcal{W}}_m[2G^{\prime }/G]\left(x\right)>0$ for all $x>0$ holds for every nonconstant real genus-zero product (34) with $|arg{w}_j|\le \vartheta$ if and only if $m\vartheta \le \pi /2$.

Proof. 

Sufficiency is Theorem 2. If $m\vartheta >\pi /2$, choose

$${\displaystyle \frac{\pi }{2m}}<\varphi <min\{\vartheta ,\pi /m\}.$$

The interval is nonempty. The quadratic

$$G_{\varphi }\left(x\right)=(1+x{e}^{i\varphi })(1+x{e}^{-i\varphi })=1+2xcos\varphi +x^2$$

has reflected zeros $e^{i\varphi },e^{-i\varphi }$ in the sector. Its logarithmic derivative $F_{\varphi }=2G_{\varphi }^{\prime }/G_{\varphi }$ is positive on $(0,\infty )$. Formula (12), which here extends continuously to $x=0$, gives

$${\mathcal{W}}_m\left[F_{\varphi }\right]\left(0\right)=4(2m-1)!cos\left(m\varphi \right)<0.$$

Continuity gives a negative value for all sufficiently small positive x.    □

Proposition 3 

(Finite positivity with nonreal poles). For every integer $N\ge 1$, there is a positive rational function $F_{\varphi }=2G_{\varphi }^{\prime }/G_{\varphi }$ in ${\mathcal{S}}_N\setminus {\mathcal{S}}_{N+1}$ whose poles are nonreal.

Proof. 

Choose

$${\displaystyle \frac{\pi }{2(N+1)}}<\varphi <{\displaystyle \frac{\pi }{2N}}$$

and use the preceding quadratic $G_{\varphi }$. It has nonreal zeros, and

$$F_{\varphi }\left(x\right)={\displaystyle \frac{4(x+cos\varphi )}{x^2+2xcos\varphi +1}}>0.$$

Since $N\varphi <\pi /2$, Theorem 2 proves all the first N Widder inequalities. On the other hand,

$${\displaystyle \frac{\pi }{2}}<(N+1)\varphi <{\displaystyle \frac{(N+1)\pi }{2N}}\le \pi ,$$

so ${\mathcal{W}}_{N+1}\left[F_{\varphi }\right]\left(0\right)<0$ and the same holds for small positive x. Thus $F_{\varphi }\notin {\mathcal{S}}_{N+1}$. Its nonreal poles also exclude an ordinary Stieltjes representation.    □

These examples establish a limitation of the general finite hierarchy. They do not assert the existence of an off-line zero of $\xi$, or optimality of the numerical cutoff for $F_{\xi }$ itself.

4.1. Transfer from a Verified Height

We now combine a finite verification of RH with any explicit two-sided zero-free strip above the verified height.

Theorem 3 

(Finite-height transfer principle). Let $H>1$ and let $d:[H,\infty )\to [0,\infty )$ be finite-valued with $d\left(t\right)/t$ nonincreasing. Assume that every nontrivial zero with $0<\gamma \le H$ lies on the critical line, and that every zero with $\gamma >H$ satisfies

$$|\beta -1/2|\le d\left(\gamma \right).$$

(40)

Set $b\left(t\right)=min\left\{d\right(t),1/2\}$ and

$${\vartheta }_H=2arctan{\displaystyle \frac{b\left(H\right)}{H}},N_H=\left\{\begin{array}{cc}⌊{\displaystyle \frac{\pi }{2{\vartheta }_H}}⌋,\hfill & b\left(H\right)>0,\hfill \\ +\infty ,\hfill & b\left(H\right)=0.\hfill \end{array}\right.$$

(41)

Then ${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)>0$ for every $x>0$ and every integer $1\le m\le N_H$. When $N_H=+\infty$, this means all positive integer orders; in that case the assumptions themselves imply RH.

Proof. 

The critical strip and (40) give $|\beta -1/2|\le b\left(\gamma \right)$ above H. Moreover,

$${\displaystyle \frac{b\left(t\right)}{t}}=min\left\{{\displaystyle \frac{d\left(t\right)}{t}},{\displaystyle \frac{1}{2t}}\right\}$$

is nonincreasing, since both entries are nonincreasing. For zeros above H,

$$|arg{w}_{\rho }|\le 2arctan{\displaystyle \frac{b\left(\gamma \right)}{\gamma }}\le 2arctan{\displaystyle \frac{b\left(H\right)}{H}}={\vartheta }_H.$$

Zeros below H give positive real reflected parameters. Since $H>1$ and $b\left(H\right)\le 1/2$, one has $0\le {\vartheta }_H<\pi /2$. If $b\left(H\right)>0$, Theorem 2 applies for $m\le \lfloor \pi /\left(2{\vartheta }_H\right)\rfloor$, including a possible integer endpoint. If $b\left(H\right)=0$, monotonicity and nonnegativity force $b\left(t\right)=0$ for all $t\ge H$. All remaining zeros then lie on the critical line, and every Widder order is strictly positive.    □

Even without a zero-free region, the critical strip gives $|\beta -1/2|<1/2$, so finite verification to height H alone yields the simpler sector

$${\vartheta }_H^{\left(0\right)}=2arctan{\displaystyle \frac{1}{2H}}.$$

The explicit zero-free region sharpens this slightly but significantly at the trillion-scale order considered here.

4.2. An Explicit Numerical Application

Platt and Trudgian [4] rigorously verified, using interval arithmetic and Turing’s method, that every nontrivial zero with

$$0<\gamma \le H,H=3·{10}^{12},$$

(42)

lies on the critical line, and their published result also states that all such zeros are simple. Only the critical-line location is needed for the finite-order positivity theorem below; simplicity is not used in its proof.

Bellotti, Trudgian and Yang ([2], Theorem 1) establish the explicit classical zero-free region

$$\zeta (\sigma +it)\ne 0\mathrm{if}t\ge 3,\sigma \ge 1-{\displaystyle \frac{1}{Rlogt}},R=4.896.$$

(43)

By conjugation and the functional equation, a nontrivial zero at height $t>H$ obeys

$${\displaystyle \frac{1}{Rlogt}}\le \beta \le 1-{\displaystyle \frac{1}{Rlogt}}.$$

(44)

Thus

$$\left|\beta -{\displaystyle \frac{1}{2}}\right|\le d\left(t\right):={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{Rlogt}}.$$

(45)

Moreover,

$${\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{d\left(t\right)}{t}}\right)={\displaystyle \frac{1}{t^2}}\left[{\displaystyle \frac{1}{R{(logt)}^2}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{Rlogt}}\right]<0$$

for $t\ge H$, since $logt>28$ and $R>4$. Here $0<d\left(t\right)<1/2$, so $b=d$ in Theorem 3. Although (45) is weak, $\gamma >H$ and strict decrease of $d\left(t\right)/t$ give a strict angular bound above H. Consequently all reflected zero parameters lie in the sector

$$|arg{w}_{\rho }|<{\vartheta }_H,{\vartheta }_H=2arctan\left[{\displaystyle \frac{1}{H}}\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{RlogH}}\right)\right].$$

(46)

4.3. Arithmetic Verification

The source package contains the script verify_cutoff.py, which evaluates (46) using the interval context of mpmath at 80 decimal digits. The endpoints are outward rounded. It gives the enclosures

$$\begin{array}{cc}\hfill 0.4928906727372165928820573& <d\left(H\right)<0.4928906727372165928820574,\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill 3.2859378182481106192137157\times {10}^{-13}& <{\vartheta }_H<3.2859378182481106192137158\times {10}^{-13},\hfill \end{array}$$

(48)

and

$$\begin{array}{cc}\hfill 4,780,359,257,170.4922923019926& <{\displaystyle \frac{\pi }{2{\vartheta }_H}}<4,780,359,257,170.4922923019927.\hfill \end{array}$$

(49)

The script also checks these displayed enclosures directly, starting from the exact decimal inputs H and R. Therefore

$$N_H=4,780,359,257,170,N_H{\vartheta }_H<{\displaystyle \frac{\pi }{2}}<(N_H+1){\vartheta }_H.$$

(50)

Theorem 4 

(Explicit finite-order consequence). Let $F_{\xi }$ be defined by (2). Then, unconditionally,

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left[x^m{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{{\xi }^{\prime }}{\xi }}\left({\displaystyle \frac{1}{2}}+\sqrt{x}\right)\right]>0$$

(51)

for every $x>0$ and every integer

$$1\le m\le \begin{array}{|c|}\hline 4,780,359,257,170\\ \hline\end{array}.$$

(52)

In particular, $F_{\xi }\in {\mathcal{S}}_{4,780,359,257,170}$ in the sense of Bondesson–Simon. Membership requires nonnegative inequalities; the displayed strict inequalities give a stronger conclusion.

Proof. 

Apply Theorem 3 with (42), (45), and (46). The certified enclosure (49) gives (50), so Theorem 2 yields strict positivity for all orders in (52).    □

Remark 3 

(Published 2024 zero-free region versus the 2026 update). Using the published Mossinghoff–Trudgian–Yang constant $R=5.558691$ [3] gives the smaller cutoff $4,772,153,270,068$. The present numerical theorem uses the stronger 2026 Bellotti–Trudgian–Yang region (43). The former cutoff is a consequence of that published region; the explicit cutoff uses the stated 2026 preprint theorem. Neither integer is claimed to be optimal for $F_{\xi }$.

4.4. A Prescribed Order and a Positive Margin

The transfer principle can be inverted. If one uses only the critical-strip bound $|\beta -1/2|<1/2$, a sufficient condition for the mth Widder inequality is

$$2marctan{\displaystyle \frac{1}{2H}}\le {\displaystyle \frac{\pi }{2}}.$$

Equivalently,

$$H\ge {\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}.$$

(53)

Since

$${\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}={\displaystyle \frac{2m}{\pi }}+O\left(m^{-1}\right),$$

we obtain the following simple rule.

Corollary 3 

(Linear height-order law). For $H>1$, a rigorous verification of RH through height

$$H\ge {\displaystyle \frac{1}{2}}cot{\displaystyle \frac{\pi }{4m}}$$

is sufficient, even without an additional zero-free region, to prove the mth Widder inequality globally on $(0,\infty )$. Asymptotically, height H certifies all orders

$$m\le {\displaystyle \frac{\pi }{4arctan(1/(2H\left)\right)}}={\displaystyle \frac{\pi }{2}}H+{\displaystyle \frac{\pi }{24H}}+O\left(H^{-3}\right).$$

Thus a successful verification of RH through the indicated finite height suffices for each prescribed order. This is a sufficient implication, not a guarantee that arbitrarily high successful verifications can be obtained. The zero-free region improves the constant at a fixed verified height by reducing the maximum possible horizontal displacement of all as-yet-unverified zeros.

For comparison, if one ignored the explicit zero-free region and used only $|\beta -1/2|<1/2$ above $H=3·{10}^{12}$, one would still obtain

$$1\le m\le 4,712,388,980,384.$$

The 2026 zero-free region increases the certified order by exactly $67,970,276,786$ relative to the critical-strip-only cutoff above.

In the range of Theorem 4, retaining the first critical-line zero ${\gamma }_1=14.1347251417\dots$ in (12) gives

$${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)\ge 2(2m-1)!{\displaystyle \frac{{\gamma }_1^{2m}}{{(x+{\gamma }_1^2)}^{2m}}}>0.$$

This is a pointwise margin, not a uniform lower bound on the entire half-line. Only the previously published zero verification is computationally substantial; the accompanying script checks the elementary cutoff arithmetic.

5. Relation with Existing Positivity Hierarchies

The finite Widder inequalities are exactly the defining conditions of the Bondesson–Simon classes ${\mathcal{S}}_k$ [1]. There is also a precise algebraic relation with the genus-zero operators studied by Zhang, which is useful for positioning the present result correctly.

For a sufficiently differentiable function h, introduce

$${\mathcal{Z}}_{p,k}\left[h\right]\left(x\right):={\left(-{\displaystyle \frac{1}{x}}\right)}^p{\displaystyle \frac{d^k}{d{x}^k}}\left(x^{k+p}{\displaystyle \frac{d^{p}h}{d{x}^p}}\left(x\right)\right),p,k\in {\mathbb{N}}_0.$$

(54)

Operators of this form occur in Zhang’s complete-monotonicity criteria for genus-zero entire functions ([6], (1.8)). The following elementary identity shows that the Widder family is a diagonal slice.

Lemma 5 

(Widder–Zhang operator identity). For every integer $m\ge 1$ and every $h\in C^{2m-1}(0,\infty )$,

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}h\left(x\right)\right)={\mathcal{Z}}_{m-1,m}\left[h\right]\left(x\right).$$

(55)

Consequently, with $h=G^{\prime }/G$,

$${\mathcal{W}}_m\left[F_{\xi }\right]\left(x\right)=2{\mathcal{Z}}_{m-1,m}[G^{\prime }/G]\left(x\right).$$

Proof. 

It is enough to prove

$${\displaystyle \frac{d^m}{d{x}^m}}\left(x^{2m-1}{h}^{(m-1)}\left(x\right)\right)=x^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}h\left(x\right)\right).$$

Expanding both sides by Leibniz’ rule, the coefficient of $x^{m-1+j}{h}^{(m-1+j)}\left(x\right)$, $0\le j\le m$, is in both cases

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right){\displaystyle \frac{(2m-1)!}{(m-1+j)!}}.$$

Dividing by $x^{m-1}$ and multiplying by ${(-1)}^{m-1}$ gives (55).    □

The identity is algebraic. It does not identify a finite set of inequalities with complete monotonicity. Zhang’s Theorem 1 assumes positive Taylor coefficients, order less than one, and reflected zeros satisfying $\Re w_j\ge {\beta }_0\left|w_j\right|>0$. Under those hypotheses, negative real zeros are characterized by the existence of a fixed $p\ge 0$ for which ${\mathcal{Z}}_{p,k}[G^{\prime }/G]$ is completely monotone for every $k\ge 0$. Here $p=m-1$ varies with the diagonal index m, and only positivity of finitely many functions is concluded.

Related reciprocal and generalized-gamma-convolution formulations are studied by Polson [5]. We use no theorem from that work. The direct identity

$$L\left(\theta \right)={\displaystyle \frac{G\left(0\right)}{G\left(2\theta \right)}},-{\displaystyle \frac{d}{d\theta }}logL\left(\theta \right)=F_{\xi }\left(2\theta \right)$$

explains the relation of notation, but does not identify positivity conditions on L with those on its logarithmic derivative.

5.1. The Infinite Hierarchy and Finite Representations

Recall that a Stieltjes function has the form

$$f\left(x\right)=a+{\int }_{[0,\infty )}{\displaystyle \frac{d\mu \left(t\right)}{x+t}},a\ge 0,\mu \ge 0,$$

(56)

with ${\int }_{[0,\infty )}{(1+t)}^{-1}d\mu \left(t\right)<\infty$. Widder’s real-variable criterion, in the form recalled in ([1], (4)), states that a nonnegative smooth function is Stieltjes if and only if

$${(-1)}^{m-1}{\displaystyle \frac{d^{2m-1}}{d{x}^{2m-1}}}\left(x^{m}f\left(x\right)\right)\ge 0(m\ge 1,x>0).$$

(57)

Bondesson and Simon call a nonnegative function satisfying (57) for $m=1,\dots ,k$ a k-Stieltjes function.

Proposition 4 

(Stieltjes formulation). The Riemann hypothesis is equivalent to $F_{\xi }$ being a Stieltjes function on $(0,\infty )$.

Proof. 

Assume RH. Write the upper-half-plane nontrivial zeros as $\rho =1/2+i\gamma$, $\gamma >0$, with multiplicity. Since G has order $1/2$ and genus zero,

$$G\left(x\right)=G\left(0\right)\prod _{\gamma >0}\left(1+{\displaystyle \frac{x}{{\gamma }^2}}\right),$$

with locally uniform convergence. Hence

$$F_{\xi }\left(x\right)=2\sum _{\gamma >0}{\displaystyle \frac{1}{x+{\gamma }^2}}.$$

Because ${\sum }_{\gamma >0}{\gamma }^{-2}<\infty$, this is a Stieltjes representation with positive discrete measure $2{\sum }_{\gamma >0}{\delta }_{{\gamma }^2}$.

Conversely, suppose $F_{\xi }$ is Stieltjes. Its representation extends to a holomorphic function $S\left(z\right)$ on the slit plane

$$\Omega =\mathbb{C}\setminus (-\infty ,0].$$

On $(0,\infty )$ it agrees with the meromorphic function $M\left(z\right)=2G^{\prime }\left(z\right)/G\left(z\right)$. Suppose, toward a contradiction, that G has a zero $z_0\in \Omega$. Choose a small disk D centered at $z_0$ containing no other zero of G, and join a point of $(0,\infty )$ to a point of the punctured disk $D\setminus \left\{z_0\right\}$ by a path in $\Omega$ that avoids the discrete zero set of G. Analytic continuation of the identity $S=M$ along this path shows that $S=M$ on $D\setminus \left\{z_0\right\}$. But if $z_0$ has multiplicity $r\ge 1$, then

$$M\left(z\right)={\displaystyle \frac{2r}{z-z_0}}+O\left(1\right)(z\to z_0),$$

whereas S is holomorphic at $z_0$. This contradiction shows that every zero of G lies on $(-\infty ,0)$.

If $\rho$ is a nontrivial zero of $\zeta$, then ${(\rho -1/2)}^2$ is a zero of G, so

$${(\rho -1/2)}^2<0.$$

Therefore $\rho -1/2$ is purely imaginary, and RH follows.    □

Thus RH asks for the entire infinite Widder hierarchy (57). A single finite level does not suffice, as Proposition 3 shows.

Bondesson and Simon ([1], Theorem 1) define ${\mathcal{S}}_k$ by nonnegativity together with the first k Widder conditions. Their Theorem 1 states that, for every $k\ge 2$, $f\in {\mathcal{S}}_k$ if and only if there exist $a_k\ge 0$ and a nonnegative measure ${\mu }_k$ on $[0,\infty )$, integrating ${(1+t)}^{-1}$, such that

$$f\left(x\right)=a_k+{\int }_0^{\infty }{\Phi }_{k-1}(x,t)d{\mu }_k\left(t\right).$$

The kernels ${\Phi }_{k-1}$ are their finite-type approximants to the ordinary Stieltjes kernel.

In particular, Theorem 4 yields this representation for $F_{\xi }$ at $k=4,780,359,257,170$. This is an immediate application of their representation theorem. Its kernel is not the ordinary Stieltjes kernel, so this conclusion does not imply RH.