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The classical Bernoulli polynomials $B_j(t)$ are generated by \begin{equation*} \frac{z\operatorname{e}^{t z}}{\operatorname{e}^z-1}=\sum_{j=0}^{\infty}B_j(t)\frac{z^j}{j!}, \quad |z|<2\pi. \end{equation*} For $\alpha, \beta\in\mathbb{R}$ such that $\alpha\ne\beta$ and $(\alpha,\beta)\not\in\{(0,1),(1,0)\}$, let \begin{equation}\label{Cal-Q(alpha-beta)-dfn}\tag{QAB} \mathcal{Q}_{\alpha,\beta}(x)= \begin{cases} \dfrac{\operatorname{e}^{-\alpha x}-\operatorname{e}^{-\beta x}}{1-\operatorname{e}^{-x}},&x\ne 0;\\ \beta-\alpha,&x=0. \end{cases} \end{equation}

Problem. Prove the integral inequality \begin{equation}\tag{QII} \frac{2\int_{0}^{1/2}\mathcal{Q}_{t,1-t}'(x) |B_{2k+1}(t)|\operatorname{d}\! t}{\int_{0}^{1/2}\mathcal{Q}_{t,1-t}(x) |B_{2k+1}(t)|\operatorname{d}\! t} <\frac{\int_{0}^{1/2}\mathcal{Q}_{t,1-t}''(x) |B_{2k+1}(t)|\operatorname{d}\! t}{\int_{0}^{1/2}\mathcal{Q}_{t,1-t}'(x) |B_{2k+1}(t)|\operatorname{d}\! t} \end{equation} for $k\in\mathbb{N}$ and $x\in(0,\infty)$.

Backgrounds. This problem originates from the paper

G.-Z. Zhang, Z.-H. Yang, and F. Qi, On normalized tails of series expansion of generating function of Bernoulli numbers, Proc. Amer. Math. Soc. (2024), in press; available online at https://doi.org/10.1090/proc/16877.

The monotoncity and convexity of the function $\mathcal{Q}_{\alpha,\beta}(x)$ can be found in the papers

J. Cao, J. L. Lopez-Bonilla, and F. Qi, Three identities and a determinantal formula for differences between Bernoulli polynomials and numbers, Electron. Res. Arch. 32 (2024), no. 1, 224--240; available online at https://doi.org/10.3934/era.2024011

and

B.-N. Guo and F. Qi, Properties and applications of a function involving exponential functions, Commun. Pure Appl. Anal. 8 (2009), no. 4, 1231--1249; available online at https://doi.org/10.3934/cpaa.2009.8.1231.

qifeng618
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  • The derivatives $\mathcal{Q}{t,1-t}'(x)$ and $\mathcal{Q}{t,1-t}''(x)$ in the integral inequality (QII) of the problem are $$\mathcal{Q}{t,1-t}'(x)=\frac{\operatorname{d}!\mathcal{Q}{t,1-t}(x)}{\operatorname{d}!x}\quad\text{and}\quad \mathcal{Q}{t,1-t}''(x)=\frac{\operatorname{d}^2!\mathcal{Q}{t,1-t}(x)}{\operatorname{d}!x^2}.$$ – qifeng618 Jul 16 '24 at 02:33
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    This problem has something to do with the one at https://mathoverflow.net/q/475160/. – qifeng618 Jul 16 '24 at 15:13

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