bibkey: trudgian2014pnt authors: Tim Trudgian year: 2014 title: Updating the error term in the prime number theorem doi: null url: https://arxiv.org/abs/1401.2689v2 claim: The unconditional all-prime-power Chebyshev error bound gives an explicit exterior deficit modulus for the original theta prime diagonal. Original-series majorants evaluate its scalar inputs; no signed discrepancy or spectral exclusion follows. strata_touched: [] license: bibliographic-reference-only triage: anchor

An explicit exterior modulus for the original prime diagonal

Source and preserved arithmetic quantity

The inspected source is arXiv:1401.2689v2, 13 pages, arXiv stamp 17 October 2014 and title-page date 20 October 2014, SHA-256 72594a54fe04ad73eb1e142e0365d4b3d7e576de2fad234dc2eff5b5e5ea8d79. The following locator refers to that preprint; a journal edition is not claimed inspected.

Theorem 1, printed p.3, states unconditionally

$$|\Psi (y)-y|\le yb(\log y)(y\ge 23),b(t)=\sqrt{\frac{8}{17\pi }}{\left(\frac{t}{6.455}\right)}^{1/4}\exp \left(-\sqrt{\frac{t}{6.455}}\right),$$

where $\Psi (y)={\sum }_{n\le y}\Lambda (n)$ includes all prime powers. Its separate estimate for $\theta (y)$ starts at $149$ and is not used here. The proof’s finite verified zero range is not a global RH premise. This PNT theorem is reused, rather than proved anew.

Exact Abel and continuum comparison

Retain the original positive even theta kernel, its normalization $\int \Phi (u)\cosh (u/2)du=1/2$, and the complete prime diagonal $a_p$ from the mixed spectral realization. Write $d(x)=2\cosh (x/2)$, $E(y)=\Psi (y)-y$ and

$$\begin{array}{rl}{F}_x(y)& =y^{-1/2}(\Phi (x+\log y)+\Phi (x-\log y)),\\ a_p(x)& =\frac{1}{d(x)}{\int }_{(1,\infty )}{F}_x(y)d\Psi (y),\\ a_0(x)& =\frac{1}{d(x)}{\int }_1^{\infty }{F}_x(y)dy.\end{array}$$

Abel integration gives

$$a_p(x)-a_0(x)=\frac{2\Phi (x)-{\int }_1^{\infty }E(y)F_x^{\prime }(y)dy}{d(x)},\text{\hspace{1em}(AB)}$$

with

$$F_x^{\prime }(y)=y^{-3/2}\left[-\frac{\Phi (x+\log y)+\Phi (x-\log y)}{2}+{\Phi }^{\prime }(x+\log y)-{\Phi }^{\prime }(x-\log y)\right].$$

The endpoint is positive because $E(1)=-1$ and $F_x(1)=2\Phi (x)$. The upper boundary vanishes by the original theta tail and the linear-size PNT bound. No prime power or crossing interaction is removed. The change $y=e^t$ also gives

$$a_0(x)=\frac{1}{d(x)}\int \Phi (u)e^{|x-u|/2}du.$$

Evenness and the normalization yield

$$\frac{1}{2}-a_0(x)=\frac{1}{d(x)}\int \Phi (u)e^{-|x-u|/2}du,0<\frac{1}{2}-a_0(x)\le e^{-|x|}.\text{\hspace{1em}(CT)}$$

For the identity, add the two exponentials and integrate $2\cosh ((x-u)/2)$. For the bound at $x\ge 0$, use $e^{-|x-u|/2}\le e^{-x/2}{e}^{u/2}$, $\int \Phi e^{u/2}=1/2$ and $d(x)\ge e^{x/2}$; evenness handles $x<0$. The displayed bound is deliberately loose. The comparison $a_0$ imposes no favorable sign on $a_p-1/2$.

An explicit deficit bound with the same theta constants

Define

$$H(u)=\frac{1}{2}\Phi (u)+|{\Phi }^{\prime }(u)|,H_{\infty }=\sup H,M_H=\int H(u)e^{u/2}du.$$

The original-series majorants give $H_{\infty }<60$ and $M_H\le 254/3$; $H$ is even. Let $B_0=1+\log 23$. Below $23$, $\Lambda (n)\le \log n$ gives $|E(y)|/y\le B_0$ for $1\le y<23$. Above $23$, $b(\log y)\le 1<B_0$. Indeed, writing $X=\sqrt{t/6.455}$ gives $\sqrt{X}{e}^{-X}\le 1$. Also $b^{\prime }(t)/b(t)=1/(4t)-1/(2\sqrt{6.455t})$, so $b$ decreases for $t\ge \log 23$.

After $y=e^t$, the absolute error integral in (AB), divided by $d(x)$, is at most

$$\frac{1}{d(x)}{\int }_0^{\infty }{e}^{t/2}\frac{|E(e^t)|}{e^t}(H(x+t)+H(x-t))dt.$$

For $x\ge R\ge 2\log 23$, split at $t=x/2$. The short part is at most $4B_0{H}_{\infty }{e}^{-x/4}$. On the long part use $b(t)\le b(x/2)$ and

$$\begin{array}{rl}{\int }_0^{\infty }{e}^{t/2}(H(x+t)+H(x-t))dt& =\int H(u)e^{|x-u|/2}du\\ & \le \int H(u)(e^{(x-u)/2}+e^{(u-x)/2})du\\ & =d(x)M_H.\end{array}$$

Discard only the favorable positive endpoint in (AB) and use (CT). All right-hand terms decrease in the stated range, giving

$$\boxed{{\delta }_R:=\underset{|x|\ge R}{\sup }(\frac{1}{2}-a_p(x){)}_{+}\le e^{-R}+4B_0{H}_{\infty }{e}^{-R/4}+M_{H}b(R/2),R\ge 2\log 23.}\text{\hspace{1em}(PM)}$$

This supplies the exterior-deficit input to the existing fixed-gap eigenfunction tail estimates. With the original-series constants it becomes

$${\delta }_R\le e^{-R}+240B_0{e}^{-R/4}+\frac{254}{3}b(R/2).\text{\hspace{1em}(EC)}$$

The same Abel bound, now retaining the positive endpoint, gives the global upper input

$$\Vert a_p{\Vert }_{\infty }\le \frac{1}{2}+\frac{18}{5}+B_0\frac{254}{3}<432.\text{\hspace{1em}(AC)}$$

Here $2\Phi (x)/d(x)\le \Phi (x)$, the absolute integral in (AB) divided by $d(x)$ is at most $B_0{M}_H$, and $B_0<5$ follows from $e^4>23$. No favorable error sign is imposed.

For $0<\epsilon \le 1/2$, define the completely specified radius

$$R_{\mathrm{PNT}}(\epsilon )=\max \left\{4,2\log 23,\log \frac{6}{\epsilon },4\log \frac{1440B_0}{\epsilon },\frac{1291}{25}{\log }^2\frac{1016}{5\epsilon }\right\}.\text{\hspace{1em}(PR)}$$

Then $R\ge R_{\mathrm{PNT}}(\epsilon )$ ensures ${\delta }_R\le \epsilon /2$. To check the last term, write $X=\sqrt{t/6.455}$. The elementary maximum $\sqrt{X}{e}^{-X/2}\le 1$ and $\sqrt{8/(17\pi )}<2/5$ give $b(t)\le (2/5)e^{-X/2}$. At $t=R/2$ one has $X=\sqrt{100R/1291}$. Each of the three terms in (EC) is therefore at most $\epsilon /6$ at the stated radius.

The effective cutoff estimates supply the full $B_p$ and Gamma inputs for the same fixed-gap eigenfunction task. Interior discretization and complete spectral exclusion remain separate. The bound supplies no favorable signed discrepancy, uniform $\epsilon \to 0$ exclusion, numerical Poincare constant or RH/Robin proof.

Trudgian supplies the PNT theorem. The Abel identity, continuum comparison and deficit modulus are paper-level model transfers, independently reviewed, without new Lean certification or an originality claim.