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  "content": "\\documentclass[11pt]{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath, amssymb}\n\\usepackage{booktabs}\n\\usepackage[hidelinks]{hyperref}\n\n\\title{A gap in the horizontal-tail estimate of Theorem 6.1 in\n``Li coefficients for automorphic L-functions''}\n\\author{Dan T Nguyen (Research Ai)}\n\\date{2026-08-24}\n\n\\begin{document}\n\\maketitle\n\n\\noindent\\textbf{Reference:} J.~Lagarias, \\emph{Li coefficients for\nautomorphic L-functions}, arXiv:math/0404394 (v4, 2005); Ann.\\ Inst.\\\nFourier \\textbf{57} (2007) 1689--1740.\\\\\n\\textbf{Companion material:} \\texttt{E-031-right-tail-stepping.py}\n(mpmath script), \\texttt{evidence.csv} (its output tables).\nReproduction: \\S\\ref{sec:repro}.\n\n\\section{The claim in the proof}\nTheorem 6.1's proof bounds the two horizontal segments (II), (IV) of the\nrectangle ($\\sigma_0 \\in (-3,-2)$, $\\sigma_1 = 2\\sqrt n$, $\\operatorname{Im} = \\pm T$,\n$T = \\sqrt n + \\varepsilon$) for the integrand $k_n(s)\\cdot(-L'/L(s+1,\\pi))$,\n$k_n(s) = (1+1/s)^n - 1$. For the right part $[3,\\,2\\sqrt n]$ the text steps\nin unit steps from $\\sigma = 3$:\n\\begin{quote}\n``\\dots the estimate (6.10) gains a factor of 2 going one unit to the right.\nIn comparison $|k_n(s+1)+1|/|k_n(s)+1| \\le e$ [via (6.12)]. We obtain an upper\nbound for $|k_n(s)\\,L'/L(s,\\pi)|$ that \\emph{decreases geometrically at each\nstep}, and after $O(\\log n)$ steps it becomes $O(1)$ so we obtain\n$|I_{\\mathrm{II}}| = O(\\sqrt n + (\\log n)^3) = O(\\sqrt n)$.''\n\\end{quote}\n\n\\section{The issue}\nPer unit step right from $\\sigma = 3$, the two bounds compose as:\n\\begin{itemize}\n\\item zeta factor: $\\le 1/2$ (from (6.10): $|L'/L(\\sigma+1+iT)| \\le\n2^{-(\\sigma-2)}\\,|L'/L(2)|$);\n\\item kernel factor: $\\le e$ (from (6.12): $|k_n(s+1)+1|/|k_n(s)+1| \\le\n(1+1/T^2)^n \\le e^{n/T^2} \\le e$, since $T^2 \\ge n$).\n\\end{itemize}\nThe per-step ratio of the product is therefore $\\le e/2 \\approx 1.359 > 1$:\nthe product \\emph{grows}, it does not decrease geometrically. The geometric\nseries $\\sum_j (e/2)^j$ diverges, and the conclusion $|I_{\\mathrm{II}}| =\nO(\\sqrt n)$ cannot follow from this argument as written.\n\nThis is not merely a bound artifact --- the growth is the actual behavior.\nWith $\\log|1+1/s| \\approx \\sigma/(\\sigma^2+T^2)$ and $T^2 = n$, the kernel\nexponent is $n\\sigma/(\\sigma^2+n)$, whose per-step increment is\n$n(n-\\sigma^2)/(\\sigma^2+n)^2 \\approx 1$ for $\\sigma \\ll \\sqrt n$, exceeding\n$\\log 2 \\approx 0.693$ until $\\sigma \\approx 0.36\\sqrt n$. Hence the product\nrises from $O(1)$ at $\\sigma = 3$ to a peak of order $e^{0.066\\sqrt n}$ at\n$\\sigma \\approx 0.36\\sqrt n$, and only then falls (to\n$e^{\\sqrt n/2}\\cdot 2^{-(2\\sqrt n - 2)} \\approx 0$ at $\\sigma_1 = 2\\sqrt n$).\nThe subsidiary claim ``after $O(\\log n)$ steps it becomes $O(1)$'' fails\ndirectly: at $\\sigma = 15$ ($\\approx 3 + O(\\log n)$) the product is\n$\\approx 375$ and still growing.\n\n\\section{Numerical evidence (mpmath, dps 25--40)}\n\\subsection{The peak tracks $\\sigma \\approx 0.36\\sqrt n$ and height $\\sim e^{0.066\\sqrt n}$}\nThe product $|k_n(w-1)|\\cdot|\\sum_{k<300}\\Lambda(k)/k^w|$ at $w = \\sigma +\ni\\sqrt n$, maximized over $\\sigma \\in [3,\\,2\\sqrt n]$ (step 2):\n\\begin{center}\n\\begin{tabular}{lcccr}\n\\toprule\n$n$ & peak at $\\sigma$ & $\\sigma/\\sqrt n$ & peak product & $\\log(\\text{peak})/\\sqrt n$ \\\\\n\\midrule\n$10^3$ & 11 & 0.348 & 4.40 & 0.0468 \\\\\n$10^4$ & 35 & 0.350 & 494 & 0.0620 \\\\\n$10^5$ & 111 & 0.351 & $1.56\\times 10^{9}$ & 0.0669 \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n\n\\subsection{Envelope integral of the right tail}\n$\\int_{3}^{2\\sqrt n} |k_n(w-1)\\cdot L'/L(w)|\\,d\\sigma$ (step 2):\n\\begin{center}\n\\begin{tabular}{lcc}\n\\toprule\n$n$ & envelope integral & peak product at $\\sigma$ \\\\\n\\midrule\n$10^4$ & $1.05\\times 10^{4}$ & 494 at $\\sigma = 35$ \\\\\n$10^5$ & $\\mathbf{5.87\\times 10^{10}}$ & $1.56\\times 10^{9}$ at $\\sigma = 111$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\nThe claimed bound is $|I_{\\mathrm{II}}| = O(\\sqrt n) \\approx 316$ at $n = 10^5$.\n\n\\subsection{Direct factor evaluation at the peak ($n = 10^5$, $T = \\sqrt n = 316.2$)}\n\\begin{center}\n\\begin{tabular}{lc}\n\\toprule\nquantity & value \\\\\n\\midrule\n$|k_n(100+317i)| = |(1+1/(100+317i))^{100000}|$ & $4.39\\times 10^{39}$ \\\\\n$|\\zeta'/\\zeta(101+317i)|$ & $\\approx 1.8\\times 10^{-31}$ \\\\\nproduct & $\\approx 8\\times 10^{8}$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n(The zeta value is from mpmath's finite-difference $\\zeta'/\\zeta$,\ncross-checked against the Dirichlet series $\\sum \\Lambda(k)/k^s$,\n$2.7\\times 10^{-31}$.)\n\n\\section{The mechanism, exactly}\nAt $T = \\sqrt n$, (6.12) bounds the kernel ratio by $e > 2$, so the\nfactor-2 gain of (6.10) cannot dominate. The argument works exactly when\nthe kernel ratio drops below 2, i.e. when $n/T^2 < \\log 2$, i.e.\n\\[\nT > \\sqrt n/\\sqrt{\\log 2} \\approx 1.2012\\,\\sqrt n .\n\\]\nWith $T = c\\sqrt n$, the kernel ratio is $\\le e^{1/c^2} < 2$ and the product\n\\emph{does} decrease per step:\n\n\\medskip\n\\noindent product $\\log P$ at a $\\sigma$-grid, $n = 10^5$:\n\\begin{center}\n\\begin{tabular}{lrrrrrr}\n\\toprule\n$T$ & $\\sigma{=}3$ & $\\sigma{=}10$ & $\\sigma{=}30$ & $\\sigma{=}60$ & $\\sigma{=}100$ & $\\sigma{=}200$ \\\\\n\\midrule\n$\\sqrt n$ & $-0.03$ & $+2.17$ & $+8.08$ & $+15.51$ & $+20.86$ & $+3.71$ \\\\\n$1.5\\sqrt n$ & $-1.91$ & $-3.10$ & $-8.10$ & $-15.92$ & $-27.32$ & $-63.66$ \\\\\n$2\\sqrt n$ & $-1.59$ & $-4.91$ & $-13.80$ & $-27.21$ & $-45.41$ & $-93.63$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $T = \\sqrt n$ the product grows to its peak at $\\sigma \\approx 0.36\\sqrt n$\n(the stepping is broken); at $T = 1.5\\sqrt n$ and $2\\sqrt n$ it decreases\nmonotonically from $\\sigma = 3$ (the stepping works, and the right tail is\ngenuinely small).\n\n\\section{Assessment}\n\\begin{itemize}\n\\item \\textbf{A proof gap, not (apparently) a theorem falsification.} The\nclosed contour identity $I_2 = S_f - \\lambda_n(T) + \\mathrm{triv}$ is exact,\nand measured $|S_f|$ values are $\\sim 13$--$20$ (Ma\\'slanka's computations;\nindependent verification to $n \\le 5000$ in this project). The top and bottom\nhorizontals (II), (IV) are conjugate sides: their real parts cancel outright,\nand the integrand's phase winds several hundred radians over the peak region,\nso the oscillatory integral can be tiny even though the envelope is\n$\\sim e^{0.066\\sqrt n}$. The conclusion $S_f(n) = \\lambda_n(\\sqrt n) +\nO(\\sqrt n \\log n)$ is consistent with all measurements and likely true.\n\\item \\textbf{Side-wise envelope bounds genuinely are $\\sim e^{0.066\\sqrt n}$}\nat $T = \\sqrt n$. Any assembly that charges the horizontal sides by kernel\n$\\times |L'/L| \\times$ length (as an explicit $B(n)$ must) inherits the\nbroken stepping and undercharges by $\\sim 10^{7}\\times$ at $n = 10^5$.\n\\item \\textbf{The repair is cheap and verified}: $T = c\\sqrt n$ with\n$c > 1.2012$ restores the argument with the same proof structure, at the\nprice of the range $N_1 \\to (T_0/c)^2$ in finite-height applications.\n\\item \\textbf{Literature status}: the arXiv v1--v4 revision notes mention only\na correction to Theorem 1.1's remainder term; a web search found no erratum\nor discussion of a gap in Theorem 6.1 or (6.12). The issue may be\nundocumented.\n\\end{itemize}\n\n\\section{Questions for the author}\n\\begin{enumerate}\n\\item Is there a repair of the stepping at $T = \\sqrt n$ short of raising $T$\n(something missed here)?\n\\item Was the intended contour height different in drafts (e.g. $T = 2\\sqrt n$,\nwhich makes the argument work verbatim)?\n\\item Does the conclusion of Theorem 6.1 stand in your view (e.g. via the\n(II)/(IV) conjugate cancellation plus oscillation), and is there a written\nargument anywhere?\n\\item For explicit finite-height applications needing traced constants, is the\n$T = c\\sqrt n$ repair the right course, or is there a subtler bound for the\nright tail that keeps $T = \\sqrt n$?\n\\end{enumerate}\n\n\\section{Reproduction}\n\\label{sec:repro}\n\\texttt{E-031-right-tail-stepping.py} (Python 3 + mpmath, dps 25) computes:\n\\begin{itemize}\n\\item[V1:] the product profile and the peak (position and height) across\n$n \\in \\{10^3, 10^4, 10^5\\}$;\n\\item[V2:] the right-tail envelope integral $\\int_3^{2\\sqrt n}$ (step 2) at\n$n \\in \\{10^4, 10^5\\}$;\n\\item[V3:] the product log-profiles at $T = c\\sqrt n$, $c \\in \\{1, 1.5, 2\\}$,\n$n = 10^5$.\n\\end{itemize}\nRun: \\texttt{python E-031-right-tail-stepping.py --force} (output:\n\\texttt{E-031-right-tail-stepping.csv} + a JSON sidecar with the check\nverdicts; the tables above are reproduced verbatim, \\texttt{evidence.csv}\nholds the same numbers in tidy form).\n\nThe zeta factor is the Dirichlet series $\\sum_{2 \\le k < 300} \\Lambda(k)/k^w$\n(absolutely convergent for $\\operatorname{Re} w \\ge 4$; the $k \\ge 300$ tail\ncontributes $< 4\\times 10^{-5}$ at $\\sigma = 3$ and less elsewhere),\ncross-checked at $\\sigma = 100$ against mpmath's direct finite-difference\n$\\zeta'/\\zeta$ ($1.8\\times 10^{-31}$ vs series $2.7\\times 10^{-31}$).\n\n\\section*{Honesty notes}\nThe envelope integral is an \\emph{upper} bound on the true side integral\n$|I_{\\mathrm{II}}\\text{'s right part}| \\le \\int |f|$; the true complex\nintegral may be smaller via oscillation (see the Assessment). The envelope is\nthe quantity every side-wise assembly charges, so it is the right measure for\nthe provability of explicit $B(n)$-style bounds. All computations are\ndeterministic (mpmath, fixed dps) and reproducible from the script alone.\n\n\\end{document}\n"
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  "content": "\\documentclass[11pt]{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath, amssymb}\n\\usepackage{booktabs}\n\\usepackage[hidelinks]{hyperref}\n\n\\title{A gap in the horizontal-tail estimate of Theorem 6.1 in\n``Li coefficients for automorphic L-functions''}\n\\author{Dan T Nguyen (Research Ai)}\n\\date{2026-08-24}\n\n\\begin{document}\n\\maketitle\n\n\\noindent\\textbf{Reference:} J.~Lagarias, \\emph{Li coefficients for\nautomorphic L-functions}, arXiv:math/0404394 (v4, 2005); Ann.\\ Inst.\\\nFourier \\textbf{57} (2007) 1689--1740.\\\\\n\\textbf{Companion material:} \\texttt{E-031-right-tail-stepping.py}\n(mpmath script), \\texttt{evidence.csv} (its output tables).\nReproduction: \\S\\ref{sec:repro}.\n\n\\section{The claim in the proof}\nTheorem 6.1's proof bounds the two horizontal segments (II), (IV) of the\nrectangle ($\\sigma_0 \\in (-3,-2)$, $\\sigma_1 = 2\\sqrt n$, $\\operatorname{Im} = \\pm T$,\n$T = \\sqrt n + \\varepsilon$) for the integrand $k_n(s)\\cdot(-L'/L(s+1,\\pi))$,\n$k_n(s) = (1+1/s)^n - 1$. For the right part $[3,\\,2\\sqrt n]$ the text steps\nin unit steps from $\\sigma = 3$:\n\\begin{quote}\n``\\dots the estimate (6.10) gains a factor of 2 going one unit to the right.\nIn comparison $|k_n(s+1)+1|/|k_n(s)+1| \\le e$ [via (6.12)]. We obtain an upper\nbound for $|k_n(s)\\,L'/L(s,\\pi)|$ that \\emph{decreases geometrically at each\nstep}, and after $O(\\log n)$ steps it becomes $O(1)$ so we obtain\n$|I_{\\mathrm{II}}| = O(\\sqrt n + (\\log n)^3) = O(\\sqrt n)$.''\n\\end{quote}\n\n\\section{The issue}\nPer unit step right from $\\sigma = 3$, the two bounds compose as:\n\\begin{itemize}\n\\item zeta factor: $\\le 1/2$ (from (6.10): $|L'/L(\\sigma+1+iT)| \\le\n2^{-(\\sigma-2)}\\,|L'/L(2)|$);\n\\item kernel factor: $\\le e$ (from (6.12): $|k_n(s+1)+1|/|k_n(s)+1| \\le\n(1+1/T^2)^n \\le e^{n/T^2} \\le e$, since $T^2 \\ge n$).\n\\end{itemize}\nThe per-step ratio of the product is therefore $\\le e/2 \\approx 1.359 > 1$:\nthe product \\emph{grows}, it does not decrease geometrically. The geometric\nseries $\\sum_j (e/2)^j$ diverges, and the conclusion $|I_{\\mathrm{II}}| =\nO(\\sqrt n)$ cannot follow from this argument as written.\n\nThis is not merely a bound artifact --- the growth is the actual behavior.\nWith $\\log|1+1/s| \\approx \\sigma/(\\sigma^2+T^2)$ and $T^2 = n$, the kernel\nexponent is $n\\sigma/(\\sigma^2+n)$, whose per-step increment is\n$n(n-\\sigma^2)/(\\sigma^2+n)^2 \\approx 1$ for $\\sigma \\ll \\sqrt n$, exceeding\n$\\log 2 \\approx 0.693$ until $\\sigma \\approx 0.36\\sqrt n$. Hence the product\nrises from $O(1)$ at $\\sigma = 3$ to a peak of order $e^{0.066\\sqrt n}$ at\n$\\sigma \\approx 0.36\\sqrt n$, and only then falls (to\n$e^{\\sqrt n/2}\\cdot 2^{-(2\\sqrt n - 2)} \\approx 0$ at $\\sigma_1 = 2\\sqrt n$).\nThe subsidiary claim ``after $O(\\log n)$ steps it becomes $O(1)$'' fails\ndirectly: at $\\sigma = 15$ ($\\approx 3 + O(\\log n)$) the product is\n$\\approx 375$ and still growing.\n\n\\section{Numerical evidence (mpmath, dps 25--40)}\n\\subsection{The peak tracks $\\sigma \\approx 0.36\\sqrt n$ and height $\\sim e^{0.066\\sqrt n}$}\nThe product $|k_n(w-1)|\\cdot|\\sum_{k<300}\\Lambda(k)/k^w|$ at $w = \\sigma +\ni\\sqrt n$, maximized over $\\sigma \\in [3,\\,2\\sqrt n]$ (step 2):\n\\begin{center}\n\\begin{tabular}{lcccr}\n\\toprule\n$n$ & peak at $\\sigma$ & $\\sigma/\\sqrt n$ & peak product & $\\log(\\text{peak})/\\sqrt n$ \\\\\n\\midrule\n$10^3$ & 11 & 0.348 & 4.40 & 0.0468 \\\\\n$10^4$ & 35 & 0.350 & 494 & 0.0620 \\\\\n$10^5$ & 111 & 0.351 & $1.56\\times 10^{9}$ & 0.0669 \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n\n\\subsection{Envelope integral of the right tail}\n$\\int_{3}^{2\\sqrt n} |k_n(w-1)\\cdot L'/L(w)|\\,d\\sigma$ (step 2):\n\\begin{center}\n\\begin{tabular}{lcc}\n\\toprule\n$n$ & envelope integral & peak product at $\\sigma$ \\\\\n\\midrule\n$10^4$ & $1.05\\times 10^{4}$ & 494 at $\\sigma = 35$ \\\\\n$10^5$ & $\\mathbf{5.87\\times 10^{10}}$ & $1.56\\times 10^{9}$ at $\\sigma = 111$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\nThe claimed bound is $|I_{\\mathrm{II}}| = O(\\sqrt n) \\approx 316$ at $n = 10^5$.\n\n\\subsection{Direct factor evaluation at the peak ($n = 10^5$, $T = \\sqrt n = 316.2$)}\n\\begin{center}\n\\begin{tabular}{lc}\n\\toprule\nquantity & value \\\\\n\\midrule\n$|k_n(100+317i)| = |(1+1/(100+317i))^{100000}|$ & $4.39\\times 10^{39}$ \\\\\n$|\\zeta'/\\zeta(101+317i)|$ & $\\approx 1.8\\times 10^{-31}$ \\\\\nproduct & $\\approx 8\\times 10^{8}$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\n(The zeta value is from mpmath's finite-difference $\\zeta'/\\zeta$,\ncross-checked against the Dirichlet series $\\sum \\Lambda(k)/k^s$,\n$2.7\\times 10^{-31}$.)\n\n\\section{The mechanism, exactly}\nAt $T = \\sqrt n$, (6.12) bounds the kernel ratio by $e > 2$, so the\nfactor-2 gain of (6.10) cannot dominate. The argument works exactly when\nthe kernel ratio drops below 2, i.e. when $n/T^2 < \\log 2$, i.e.\n\\[\nT > \\sqrt n/\\sqrt{\\log 2} \\approx 1.2012\\,\\sqrt n .\n\\]\nWith $T = c\\sqrt n$, the kernel ratio is $\\le e^{1/c^2} < 2$ and the product\n\\emph{does} decrease per step:\n\n\\medskip\n\\noindent product $\\log P$ at a $\\sigma$-grid, $n = 10^5$:\n\\begin{center}\n\\begin{tabular}{lrrrrrr}\n\\toprule\n$T$ & $\\sigma{=}3$ & $\\sigma{=}10$ & $\\sigma{=}30$ & $\\sigma{=}60$ & $\\sigma{=}100$ & $\\sigma{=}200$ \\\\\n\\midrule\n$\\sqrt n$ & $-0.03$ & $+2.17$ & $+8.08$ & $+15.51$ & $+20.86$ & $+3.71$ \\\\\n$1.5\\sqrt n$ & $-1.91$ & $-3.10$ & $-8.10$ & $-15.92$ & $-27.32$ & $-63.66$ \\\\\n$2\\sqrt n$ & $-1.59$ & $-4.91$ & $-13.80$ & $-27.21$ & $-45.41$ & $-93.63$ \\\\\n\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $T = \\sqrt n$ the product grows to its peak at $\\sigma \\approx 0.36\\sqrt n$\n(the stepping is broken); at $T = 1.5\\sqrt n$ and $2\\sqrt n$ it decreases\nmonotonically from $\\sigma = 3$ (the stepping works, and the right tail is\ngenuinely small).\n\n\\section{Assessment}\n\\begin{itemize}\n\\item \\textbf{A proof gap, not (apparently) a theorem falsification.} The\nclosed contour identity $I_2 = S_f - \\lambda_n(T) + \\mathrm{triv}$ is exact,\nand measured $|S_f|$ values are $\\sim 13$--$20$ (Ma\\'slanka's computations;\nindependent verification to $n \\le 5000$ in this project). The top and bottom\nhorizontals (II), (IV) are conjugate sides: their real parts cancel outright,\nand the integrand's phase winds several hundred radians over the peak region,\nso the oscillatory integral can be tiny even though the envelope is\n$\\sim e^{0.066\\sqrt n}$. The conclusion $S_f(n) = \\lambda_n(\\sqrt n) +\nO(\\sqrt n \\log n)$ is consistent with all measurements and likely true.\n\\item \\textbf{Side-wise envelope bounds genuinely are $\\sim e^{0.066\\sqrt n}$}\nat $T = \\sqrt n$. Any assembly that charges the horizontal sides by kernel\n$\\times |L'/L| \\times$ length (as an explicit $B(n)$ must) inherits the\nbroken stepping and undercharges by $\\sim 10^{7}\\times$ at $n = 10^5$.\n\\item \\textbf{The repair is cheap and verified}: $T = c\\sqrt n$ with\n$c > 1.2012$ restores the argument with the same proof structure, at the\nprice of the range $N_1 \\to (T_0/c)^2$ in finite-height applications.\n\\item \\textbf{Literature status}: the arXiv v1--v4 revision notes mention only\na correction to Theorem 1.1's remainder term; a web search found no erratum\nor discussion of a gap in Theorem 6.1 or (6.12). The issue may be\nundocumented.\n\\end{itemize}\n\n\\section{Questions for the author}\n\\begin{enumerate}\n\\item Is there a repair of the stepping at $T = \\sqrt n$ short of raising $T$\n(something missed here)?\n\\item Was the intended contour height different in drafts (e.g. $T = 2\\sqrt n$,\nwhich makes the argument work verbatim)?\n\\item Does the conclusion of Theorem 6.1 stand in your view (e.g. via the\n(II)/(IV) conjugate cancellation plus oscillation), and is there a written\nargument anywhere?\n\\item For explicit finite-height applications needing traced constants, is the\n$T = c\\sqrt n$ repair the right course, or is there a subtler bound for the\nright tail that keeps $T = \\sqrt n$?\n\\end{enumerate}\n\n\\section{Reproduction}\n\\label{sec:repro}\n\\texttt{E-031-right-tail-stepping.py} (Python 3 + mpmath, dps 25) computes:\n\\begin{itemize}\n\\item[V1:] the product profile and the peak (position and height) across\n$n \\in \\{10^3, 10^4, 10^5\\}$;\n\\item[V2:] the right-tail envelope integral $\\int_3^{2\\sqrt n}$ (step 2) at\n$n \\in \\{10^4, 10^5\\}$;\n\\item[V3:] the product log-profiles at $T = c\\sqrt n$, $c \\in \\{1, 1.5, 2\\}$,\n$n = 10^5$.\n\\end{itemize}\nRun: \\texttt{python E-031-right-tail-stepping.py --force} (output:\n\\texttt{E-031-right-tail-stepping.csv} + a JSON sidecar with the check\nverdicts; the tables above are reproduced verbatim, \\texttt{evidence.csv}\nholds the same numbers in tidy form).\n\nThe zeta factor is the Dirichlet series $\\sum_{2 \\le k < 300} \\Lambda(k)/k^w$\n(absolutely convergent for $\\operatorname{Re} w \\ge 4$; the $k \\ge 300$ tail\ncontributes $< 4\\times 10^{-5}$ at $\\sigma = 3$ and less elsewhere),\ncross-checked at $\\sigma = 100$ against mpmath's direct finite-difference\n$\\zeta'/\\zeta$ ($1.8\\times 10^{-31}$ vs series $2.7\\times 10^{-31}$).\n\n\\section*{Honesty notes}\nThe envelope integral is an \\emph{upper} bound on the true side integral\n$|I_{\\mathrm{II}}\\text{'s right part}| \\le \\int |f|$; the true complex\nintegral may be smaller via oscillation (see the Assessment). The envelope is\nthe quantity every side-wise assembly charges, so it is the right measure for\nthe provability of explicit $B(n)$-style bounds. All computations are\ndeterministic (mpmath, fixed dps) and reproducible from the script alone.\n\n\\end{document}\n",
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