Properties of Laplace-Stieltjes-type integrals

Анотація

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 $r<R_{\mu}$ and $\mu(r)=+\infty$ for $r>R_{\mu}$. 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 $a(x)$.