slug: xu-yang-2026-mills-remainder-cm
bibkey: xuyang2026millsratio
doi: 10.1186/s13660-026-03505-9
triage: theorem
motivation_gids:
  - D5/S3/Analytic/XuYangMillsRemainder/Result.result
url: https://doi.org/10.1186/s13660-026-03505-9

Complete monotonicity of the Xu–Yang Mills remainders

Problem

H.-Z. Xu and Z.-H. Yang, Higher order monotonicity involving the Mills ratio with applications, Journal of Inequalities and Applications 2026:98, DOI 10.1186/s13660-026-03505-9.

Mills-ratio definition, printed p. 2:

The Mills ratio [3] is defined by

$$R\equiv R(x)=\frac{1-\Phi (x)}{\varphi (x)}>=e^{x^2/2}{\int }_x^{\infty }{e}^{-t^2/2}dt,x\ge 0,$$

which satisfies the differential equation of first order:

$$R^{\prime }=xR-1.(1.1)$$

Here $\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }$ and $\Phi (x)={\int }_{-\infty }^x\varphi (t)dt$ are the standard Gaussian density and cumulative distribution function used in the preceding source sentences.

The bound family in (1.5), printed p. 2, is

$${\phi }_{p,q,r}(x)=\frac{p}{\sqrt{x^2+q}+rx},p,q>0,r\ge 0.$$

Equation (1.8), printed p. 3, with its complete surrounding statement:

Note that all bounds of the form ${\phi }_{p,q,r}(x)$ can be derived from the inequality

$$Y_{a,b}(x):=(x^2+a)R^2(x)+bxR(x)<(>)c(1.8)$$

for all $x>0$, $a,b\in \mathbb{R}$, where

$$c=\underset{x\to 0}{\lim }{Y}_{a,b}(x)=\frac{a\pi }{2}>\text{or}c=\underset{x\to \infty }{\lim }{Y}_{a,b}(x)=b+1.$$

Equation (3.7), printed p. 6, gives the coefficients following the asymptotic expansion in Theorem 3.1:

where

$$u_n\equiv u_n(a,b)=(b+1)\frac{2^n\Gamma (n+1/2)}{\Gamma (1/2)}>+(n-a)(n-1)!\sum _{k=0}^{n-1}>\frac{2^k\Gamma (k+1/2)}{\Gamma (1/2)k!}.(3.7)$$

In particular, $u_1=b+2-a$, $u_2=3b+7-2a$, $u_3=15b+36-7a$, and $u_4=105b-36a+249$.

Conjecture 7.1, printed p. 27, complete statement:

Moreover, inspired by Theorem 4.2, we propose the following conjecture.

Conjecture 7.1 Let

$$D_n(x,a)=Y_{a,a-2}(x)-\left(a-1+\sum _{k=1}^n>\frac{(-1{)}^k{u}_k}{x^{2k}}\right),$$

where $u_k$ is defined by (3.7). Then $x\mapsto (-1{)}^n{D}_n(x,-1)$ is CM on $(0,\infty )$.

Formal reading

The namespace is D5.S3.Analytic.XuYangMillsRemainder.Result. The definitions millsRatio, Y, u and D retain the displayed Gaussian integral, polynomial expression, Gamma coefficients and finite remainder. Here $\Gamma$ is Euler’s Gamma function, $k!$ is the factorial of the nonnegative integer $k$, and all integrals use Lebesgue measure. CompletelyMonotoneOn f means $f$ is $C^{\infty }$ on $(0,\infty )$ and every signed derivative $(-1{)}^m{f}^{(m)}(x)$ is nonnegative there. claim quantifies over every integer $n\ge 2$; result states claim. The source does not display a lower bound for $n$ in Conjecture 7.1. The restriction $n\ge 2$ is necessary: $n=1$ fails, as explained in Triage.

Motivation

The declaration D5/S3/Analytic/XuYangMillsRemainder/Result.result settles the all-index assertion for $n\ge 2$. The source note is D5/L/Asymptotics/xuyang2026millsratio. Theorem 4.2 already treats $n=2,3,4$; the additional range is $n\ge 5$.

Gap

An asymptotic expansion alone does not control a remainder for every positive $x$, nor its derivatives of every order. A positive Laplace representation supplies both the global sign and all derivative signs. The Gamma coefficients must agree exactly with the coefficients removed from that representation.

Route

Write $h(t)=\exp (-t^2/2)$ and let $Q(t)={\int }_0^{t}h(s)h(t-s)ds$, the convolution on the positive half-line. For $x>0$,

$$R(x)={\int }_0^{\infty }{e}^{-xt}h(t)dt,R(x{)}^2={\int }_0^{\infty }{e}^{-xt}Q(t)dt,$$

$$Q(t)=t{\int }_0^1{e}^{-(1+v^2)t^2/4}dv.$$

Two integrations by parts, and the fundamental theorem applied to $v\exp (-(1+v^2)t^2/4)$ and $v(1-v^2)\exp (-(1+v^2)t^2/4)$, give

$$H(x):=(x^2-1)R(x{)}^2-3xR(x)+2={\int }_0^{\infty }{e}^{-xt}K(t)dt,$$

$$K(t)=\frac{t^5}{8}{\int }_0^1{v}^2(1-v^2)e^{-(1+v^2)t^2/4}dv.$$

For nonnegative integer $j$, put

$$I_j={\int }_0^1(1+v^2{)}^{j}dv,J_j={\int }_0^1{v}^2(1-v^2)(1+v^2{)}^{j}dv,A_j=\frac{(2j+5)!}{84^{j}j!}{J}_j.$$

The recurrence $(2j+3)I_{j+1}-2(j+1)I_j=2^{j+1}$, obtained by integrating the derivative of $v(1+v^2{)}^{j+1}$, identifies the original Gamma coefficient as $u_{j+3}(-1,-3)=-A_j$. Also $u_1=u_2=0$. Thus $H=D_2(\cdot ,-1)$.

For nonnegative integer $r$ and $v\ge 0$, define the finite exponential remainder

$$E_r(v)=(-1{)}^r\left(e^{-v}-\sum _{j=0}^{r-1}\frac{(-v{)}^j}{j!}\right).$$

The sum is empty when $r=0$, so $E_0(v)=\exp (-v)$. Taylor’s theorem gives $0\le E_r(v)\le v^r/r!$ and the exact recursion $E_{r+1}(v)=v^r/r!-E_r(v)$. Set

$$K_r(t)=\frac{t^5}{8}{\int }_0^1{v}^2(1-v^2)E_r((1+v^2)t^2/4)dv.$$

For $t\ge 0$, $0\le K_r(t)\le t^{2r+5}$. Integrating the finite recursion, using ${\int }_0^{\infty }{t}^m\exp (-xt)dt=m!/x^{m+1}$, identifies

$$(-1{)}^r{D}_{r+2}(x,-1)={\int }_0^{\infty }{e}^{-xt}{K}_r(t)dt.$$

Every polynomially weighted moment is integrable for $x>0$. On a neighborhood of $x$ it is dominated using $\exp (-xt/2)$. Differentiation under the integral yields

$$(-1{)}^m\frac{d^m}{d{x}^m}{\int }_0^{\infty }{e}^{-xt}{K}_r(t)dt={\int }_0^{\infty }{t}^m{e}^{-xt}{K}_r(t)dt\ge 0.$$

All orders of differentiability follow from the same domination. Since $(-1{)}^{r+2}=(-1{)}^r$, this is the requested complete monotonicity.

Falsifier

A discrepancy in the Gaussian normalization, the coefficient identity, the finite remainder recursion or differentiation under the integral would invalidate the connection to the original conjecture. A finite numerical sign check alone would not establish the all-index assertion.

Evidence

PROVED. D5.S3.Analytic.XuYangMillsRemainder.Result.result states claim without additional assumptions: complete monotonicity for every integer $n\ge 2$ and every positive argument, with all derivative orders.

The experiment is published in the pinned experiment directory.

• COMPUTED. Script xu-yang-c71.py, python3 xu-yang-c71.py, exit code 0; SHA-256 bdb10be43e6f8eb65b15cce53567ab6b21b82f720208d22b8f6b9392f93a6c3b. The tested scope is $n\le 6$, $m\le 4$ and $x\in \{0.3,0.7,1,2,4,8,15\}$, at 60 decimal digits. The values $u_1,\dots ,u_6$ at $(-1,-3)$ are $0,0,-2,-30,-396,-5460$. The tested signs are positive for $n=0$ and $n=2,\dots ,6$; $n=1$ fails. The minimum observed margins for $n=2,\dots ,6$ are approximately $8.865\times 10^{-9}$, $1.623\times 10^{-9}$, $2.084\times 10^{-10}$, $2.440\times 10^{-11}$ and $2.395\times 10^{-12}$.

The supplied literature check covered the Semantic Scholar citation list, web searches of the title and DOI, and comparison with Salazar, arXiv:2606.31647, and Yang, arXiv:2507.10954. No settlement was found in that scope. No new network search is asserted.

Triage

What the settlement shows

The mechanism is positivity of a whole family of exact Laplace kernels. The finite Taylor recursion preserves positivity after every truncation; it supplies uniform statements in $n$ and derivative order, rather than finite numerical evidence. The delivered theorem asserts the paper’s CM condition (signed derivatives $\ge 0$); the strict positivity mentioned in the preregistration is not asserted by the Lean theorem.

• PROVED. D5.S3.Analytic.XuYangMillsRemainder.Result.result covers every $n\ge 2$; the range beyond Theorem 4.2 is $n\ge 5$.
• PROVED in prose. The lower index $n=1$ fails. The original formula (3.7) gives $u_1(-1,-3)=u_2(-1,-3)=0$, hence $D_1=D_2$. Theorem 4.2(i), printed p.14, states that $-D_2(\cdot ,a)$ is CM exactly when $a\ge -3/4$. At $a=-1$ this condition fails, so $-D_1(\cdot ,-1)$ is not CM.
• PROVED in the source. The $n=2,3,4$ cases overlap Theorem 4.2(i)–(iii), printed p.14. At $a=-1$ its respective conditions $a\le -1$, $a\le -3/4$ and $a\le -13/23$ all hold.
• COMPUTED. The finite margins and their complete tested scope are given in Evidence; they assert no continuous-domain bound.
• OPEN. The analogous classification for general $a$ is not settled. No assertion outside $n\ge 2$, $a=-1$ follows from the stated conclusion.

ASSUMED-UNVERIFIED

Exhaustive literature coverage, Google Scholar, citation-index completeness and priority are unverified. Finite precision computations do not establish uniform signs or complete monotonicity.