{"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 $a(x)$.