{"title":"拉普拉斯-斯蒂尔杰斯型积分的性质","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":"{\"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}","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}
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 $a(x)$.