On the proximate order and lower proximate order for meromorphic functions

For a function $\Gamma(r)=\exp\left\{\displaystyle\int\nolimits_1^r\dfrac{\gamma(t)}{t}dt\right\},$ $\gamma(r)$ is a proximate order, we deduce the expression $\Gamma(r)=r^{\gamma(r)}L(r),$ where $L(r)$ is a slowly varying function on $[1,+\infty),$ i.e., $rL'(r)/L(r)\to 0$ as $r\to+\infty....

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Збережено в:
Бібліографічні деталі
Дата:2026
Автори: Zabolotskyy, M., Zabolotskyy, T., Mostova, M., Заболоцький, Микола, Заболоцький, Тарас, Мостова, Мар'яна
Формат: Стаття
Мова:Українська
Опубліковано: Institute of Mathematics, NAS of Ukraine 2026
Онлайн доступ:https://umj.imath.kiev.ua/index.php/umj/article/view/9342
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Назва журналу:Ukrains’kyi Matematychnyi Zhurnal

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Ukrains’kyi Matematychnyi Zhurnal
Опис
Резюме:For a function $\Gamma(r)=\exp\left\{\displaystyle\int\nolimits_1^r\dfrac{\gamma(t)}{t}dt\right\},$ $\gamma(r)$ is a proximate order, we deduce the expression $\Gamma(r)=r^{\gamma(r)}L(r),$ where $L(r)$ is a slowly varying function on $[1,+\infty),$ i.e., $rL'(r)/L(r)\to 0$ as $r\to+\infty.$ We define the notions of proximate order $\rho(r)$ and proximate lower order $\lambda(r)$ of a function $f$ meromorphic in $\mathbb{C}$ such that either $\underline{\Delta}(D)=\liminf_{r\to+\infty}T(r,f)/D(r)>0$ and $\overline{\Delta}(H)=\limsup_{r\to+\infty}T(r,f)/H(r)=+\infty$ or $\underline{\Delta}(D)=0$ and $\overline{\Delta}(H)<+\infty,$ where $D(r)=\exp\left\{\displaystyle\int\nolimits_1^r\dfrac{\rho(t)}{t}dt\right\}$ and $H(r)=\exp\left\{\displaystyle\int\nolimits_1^r\dfrac{\lambda(t)}{t}dt\right\}.$ The obtained results reveal the incorrectness of Lemma 1 and Theorem 1 from the paper by S. H. Dwivedi [S. H. Dwivedi,  Compos. Math., 22, No. 1, 39–48 (1970)]. We also generalize the statement of Lemma 2 in the cited paper as follows: If a function $\phi(r)$ is such that $r\phi'(r)/\phi(r)\to\phi_0$ as $r \to +\infty,$ then $\displaystyle\int\nolimits_1^r\dfrac{\phi(t)}{t^{1+\alpha}}dt \sim \dfrac{\phi(r)}{(\phi_0-\alpha)r^\alpha}$ for $0 \le \alpha < \phi_0$ and $\displaystyle\int\nolimits_r^{+\infty}\dfrac{\phi(t)}{t^{1+\alpha}}dt \sim \dfrac{\phi(r)}{(\alpha-\phi_0)r^\alpha}$ for $\alpha > \phi_0$ as $r \to +\infty.$
DOI:10.3842/umzh.v78i7-8.9342