Properties of Laplace-Stieltjes-type integrals

Q3 Mathematics Matematychni Studii Pub Date : 2023-12-18 DOI:10.30970/ms.60.2.115-131
M. M. Sheremeta
{"title":"Properties of Laplace-Stieltjes-type integrals","authors":"M. M. Sheremeta","doi":"10.30970/ms.60.2.115-131","DOIUrl":null,"url":null,"abstract":"The properties of Laplace-Stieltjes-type integrals $I(r)=\\int_{0}^{\\infty}a(x)f(xr)dF(x)$ are studied, where $F$ is a non-negative non-decreasing unbounded continuous on the right function on $[0,\\,+\\infty)$,$f(z)=\\sum_{k=0}^{\\infty}f_kz^k$ is an entire transcendental function with $f_k\\ge 0$ for all $k\\ge0$, and a function $a(x)\\ge 0$ on $[0,\\,+\\infty)$ is such that the Lebesgue-Stieltjes integral $\\int_{0}^{K}a(x)f(xr)dF(x)$ exists for every $r\\ge 0$ and$K \\in [0,\\,+\\infty)$.For the maximum of the integrand $\\mu(r)=\\sup\\{a(x)f(xr)\\colon x\\ge 0\\}$ it is proved that if$$\\varliminf\\limits_{x\\to+\\infty}\\frac{f^{-1}\\left(1/a(x)\\right)}{x}=R_{\\mu}$$ then $\\mu(r)<+\\infty$ for $rR_{\\mu}$. The relationship between $R_{\\mu}$ and the radius $R_c$ of convergence of the integral $I(r)$ was found. The concept of the central point $\\nu(r)$ of the maximum of the integrand is introduced and the formula for finding $\\ln \\mu(r)$ over $\\nu(r)$ is proved.Under certain conditions on the function $F$, estimates of $I(r)$ in terms of $\\mu(r)$ are obtained, and in the case when $R_{\\mu}=+\\infty$,in terms of generalized orders, a relation is established between the growth $\\mu(r)$ and $I(r)$ and the decrease of the function $a(x)$.","PeriodicalId":37555,"journal":{"name":"Matematychni Studii","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2023-12-18","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Matematychni Studii","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.30970/ms.60.2.115-131","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"Mathematics","Score":null,"Total":0}
引用次数: 0

Abstract

The properties of Laplace-Stieltjes-type integrals $I(r)=\int_{0}^{\infty}a(x)f(xr)dF(x)$ are studied, where $F$ is a non-negative non-decreasing unbounded continuous on the right function on $[0,\,+\infty)$,$f(z)=\sum_{k=0}^{\infty}f_kz^k$ is an entire transcendental function with $f_k\ge 0$ for all $k\ge0$, and a function $a(x)\ge 0$ on $[0,\,+\infty)$ is such that the Lebesgue-Stieltjes integral $\int_{0}^{K}a(x)f(xr)dF(x)$ exists for every $r\ge 0$ and$K \in [0,\,+\infty)$.For the maximum of the integrand $\mu(r)=\sup\{a(x)f(xr)\colon x\ge 0\}$ it is proved that if$$\varliminf\limits_{x\to+\infty}\frac{f^{-1}\left(1/a(x)\right)}{x}=R_{\mu}$$ then $\mu(r)<+\infty$ for $rR_{\mu}$. The relationship between $R_{\mu}$ and the radius $R_c$ of convergence of the integral $I(r)$ was found. The concept of the central point $\nu(r)$ of the maximum of the integrand is introduced and the formula for finding $\ln \mu(r)$ over $\nu(r)$ is proved.Under certain conditions on the function $F$, estimates of $I(r)$ in terms of $\mu(r)$ are obtained, and in the case when $R_{\mu}=+\infty$,in terms of generalized orders, a relation is established between the growth $\mu(r)$ and $I(r)$ and the decrease of the function $a(x)$.
查看原文
分享 分享
微信好友 朋友圈 QQ好友 复制链接
本刊更多论文
拉普拉斯-斯蒂尔杰斯型积分的性质
研究了拉普拉斯-斯蒂尔杰斯型积分$I(r)=\int_{0}^{\infty}a(x)f(xr)dF(x)$的性质,其中$F$是$[0,\,+\infty)$上的非负非递减无界连续右函数、$f(z)=\sum_{k=0}^{\infty}f_kz^k$是一个对所有$k\ge0$都为$f_k\ge 0$的全超越函数,并且在$[0,\,+\infty)$上有一个函数$a(x)\ge 0$、\,+\infty)$上的函数$a(x)ge 0$是这样的:对于每一个$r\ge 0$和$K\in [0,\,+\infty)$,都存在Lebesgue-Stieltjes积分$\int_{0}^{K}a(x)f(xr)dF(x)$。对于积分的最大值$\mu(r)=\sup\{a(x)f(xr)\colon x\ge 0\}$ 可以证明,如果$$\varliminf\limits_{x\to\+\infty}\frac{f^{-1}left(1/a(x)\right)}{x}=R_{\mu}$,那么对于$rR_{mu}$,$\mu(r)<+\infty$。找到了 $R_\{mu}$ 与积分 $I(r)$ 收敛半径 $R_c$ 之间的关系。引入了积分最大值的中心点 $\nu(r)$ 的概念,并证明了在 $\nu(r)$ 上求 $\ln \mu(r)$ 的公式。在函数 $F$ 的某些条件下,得到了以 $\mu(r)$ 为单位的 $I(r)$ 的估计值,并且在 $R_\{mu}=+\infty$ 的情况下,以广义阶数为单位,建立了函数 $a(x)$ 的增长 $\mu(r)$ 和 $I(r)$ 之间的关系。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
求助全文
约1分钟内获得全文 去求助
来源期刊
Matematychni Studii
Matematychni Studii Mathematics-Mathematics (all)
CiteScore
1.00
自引率
0.00%
发文量
38
期刊介绍: Journal is devoted to research in all fields of mathematics.
期刊最新文献
On the h-measure of an exceptional set in Fenton-type theorem for Taylor-Dirichlet series Almost periodic distributions and crystalline measures Reflectionless Schrodinger operators and Marchenko parametrization Existence of basic solutions of first order linear homogeneous set-valued differential equations Real univariate polynomials with given signs of coefficients and simple real roots
×
引用
GB/T 7714-2015
复制
MLA
复制
APA
复制
导出至
BibTeX EndNote RefMan NoteFirst NoteExpress
×
×
提示
您的信息不完整,为了账户安全,请先补充。
现在去补充
×
提示
您因"违规操作"
具体请查看互助需知
我知道了
×
提示
现在去查看 取消
×
提示
确定
0
微信
客服QQ
Book学术公众号 扫码关注我们
反馈
×
意见反馈
请填写您的意见或建议
请填写您的手机或邮箱
已复制链接
已复制链接
快去分享给好友吧!
我知道了
×
扫码分享
扫码分享
Book学术官方微信
Book学术文献互助
Book学术文献互助群
群 号:481959085
Book学术
文献互助 智能选刊 最新文献 互助须知 联系我们:info@booksci.cn
Book学术提供免费学术资源搜索服务,方便国内外学者检索中英文文献。致力于提供最便捷和优质的服务体验。
Copyright © 2023 Book学术 All rights reserved.
ghs 京公网安备 11010802042870号 京ICP备2023020795号-1