文摘
In the article, we prove that the double inequalities $$ \frac{\sqrt{\pi}e^{-x}}{\sqrt{2(x+a)}}< K_{0}(x)< \frac{\sqrt{\pi }e^{-x}}{\sqrt{2(x+b)}},\qquad 1+ \frac{1}{2(x+a)}< \frac {K_{1}(x)}{K_{0}(x)}< 1+\frac{1}{2(x+b)} $$ hold for all \(x>0\) if and only if \(a\geq1/4\) and \(b=0\) if \(a, b\in[0, \infty)\), where \(K_{\nu}(x)\) is the modified Bessel function of the second kind. As applications, we provide bounds for \(K_{n+1}(x)/K_{n}(x)\) with \(n\in\mathbb{N}\) and present the necessary and sufficient condition such that the function \(x\mapsto\sqrt {x+p}e^{x}K_{0}(x)\) is strictly increasing (decreasing) on \((0, \infty)\).Keywordsmodified Bessel functiongamma functionmonotonicityMSC33B1026A481 IntroductionThe modified Bessel function of the first kind \(I_{\nu}(x)\) is a particular solution of the second-order differential equation $$ x^{2}y^{\prime\prime}(x)+xy^{\prime}(x)- \bigl(x^{2}+ \nu^{2} \bigr)y(x)=0, $$ and it can be expressed by the infinite series $$ I_{\nu}(x)=\sum_{n=0}^{\infty} \frac{1}{n!\Gamma(\nu+n+1)} \biggl(\frac {x}{2} \biggr)^{2n+\nu}. $$While the modified Bessel function of the second kind \(K_{\nu}(x)\) is defined by $$ K_{\nu}(x)=\frac{\pi (I_{-\nu}(x)-I_{\nu}(x) )}{2\sin(\pi\nu)}, $$ (1.1) where the right-hand side of the identity of (1.1) is the limiting value in case ν is an integer.The following integral representation formula and asymptotic formulas for the modified Bessel function of the second kind \(K_{\nu}(x)\) can be found in the literature [1], 9.6.24, 9.6.8, 9.6.9, 9.7.2: $$\begin{aligned}& K_{\nu}(x)= \int_{0}^{\infty}e^{-x\cosh(t)}\cosh(\nu t)\,dt\quad (x>0), \end{aligned}$$ (1.2)$$\begin{aligned}& K_{0}(x)\sim-\log x\quad (x\rightarrow0), \end{aligned}$$ (1.3)$$\begin{aligned}& K_{\nu}(x)\sim\frac{1}{2}\Gamma(\nu) \biggl(\frac{x}{2} \biggr)^{-\nu}\quad (\nu >0 ,x\rightarrow0), \end{aligned}$$ (1.4)$$\begin{aligned}& K_{\nu}(x)\sim\sqrt{\frac{\pi}{2x}}e^{-x} \biggl[1+ \frac{4\nu ^{2}-1}{8x}+\frac{ (4\nu^{2}-1 ) (4\nu^{2}-9 )}{2!(8x)^{2}}+\cdots \biggr]\quad (x\rightarrow \infty). \end{aligned}$$ (1.5)From (1.2) we clearly see that $$\begin{aligned}& K_{0}(x)= \int_{0}^{\infty}e^{-x\cosh(t)}\,dt= \int_{1}^{\infty}\frac {e^{-xt}}{\sqrt{t^{2}-1}}\,dt, \end{aligned}$$ (1.6)$$\begin{aligned}& K^{\prime}_{0}(x)=- \int_{1}^{\infty}\frac{te^{-xt}}{\sqrt{t^{2}-1}}\,dt=-K_{1}(x). \end{aligned}$$ (1.7)Recently, the bounds for the modified Bessel function of the second kind \(K_{\nu}(x)\) have attracted the attention of many researchers. Luke [2] proved that the double inequality $$ \frac{8\sqrt{x}}{8x+1}< \sqrt{\frac{2}{\pi}}e^{x}K_{0}(x)< \frac {16x+7}{(16x+9)\sqrt{x}} $$ (1.8) holds for all \(x>0\).Gaunt [3] proved that the double inequality $$ \frac{1}{\sqrt{x+\frac{1}{2}}}< \frac{\Gamma (x+\frac{1}{2} )}{\Gamma(x+1)}< \sqrt{\frac{2}{\pi}}e^{x}K_{0}(x)< \frac{1}{\sqrt{x}} $$ (1.9) takes place for all \(x>0\), where \(\Gamma(x)=\int_{0}^{\infty }t^{x-1}e^{-t}\, dx\) is the classical gamma function.In [4], Segura proved that the double inequality $$ \frac{\nu+\sqrt{x^{2}+\nu^{2}}}{x}< \frac{K_{\nu+1}(x)}{K_{\nu}(x)}< \frac {\nu+\frac{1}{2}+\sqrt{x^{2}+ (\nu+\frac{1}{2} )^{2}}}{x} $$ (1.10) holds for all \(x>0\) and \(\nu\geq0\).Bordelon and Ross [5] and Paris [6] provided the inequality $$ \frac{K_{\nu}(x)}{K_{\nu}(y)}>e^{y-x} \biggl(\frac{x}{y} \biggr)^{\nu} $$ (1.11) for all \(\nu>-1/2\) and \(y>x>0\).Laforgia [7] established the inequality $$ \frac{K_{\nu}(x)}{K_{\nu}(y)}>e^{y-x} \biggl(\frac{x}{y} \biggr)^{-\nu} $$ (1.12) for all \(y>x>0\) if \(\nu\in(0, 1/2)\), and inequality (1.12) is reversed if \(\nu\in(1/2, \infty)\).Baricz [8] presented the inequality $$ \frac{K_{\nu}(x)}{K_{\nu}(y)}>e^{y-x} \biggl(\frac{x}{y} \biggr)^{-1/2} $$ for all \(y>x>0\) and \(\nu\in(-\infty, -1/2)\cup(1/2, \infty)\).Motivated by inequality (1.9), in the article, we prove that the double inequality $$ \frac{\sqrt{\pi}e^{-x}}{\sqrt{2(x+a)}}< K_{0}(x)< \frac{\sqrt{\pi }e^{-x}}{\sqrt{2(x+b)}} $$ holds for all \(x>0\) if and only if \(a\geq1/4\) and \(b=0\) if \(a, b\in [0, \infty)\). As applications, we provide bounds for \(K_{n+1}(x)/K_{n}(x)\) with \(n\in\mathbb{N}\) and present the necessary and sufficient condition such that the function \(x\mapsto\sqrt {x+p}e^{x}K_{0}(x)\) is strictly increasing (decreasing) on \((0, \infty)\).2 LemmasIn order to prove our main results, we need two lemmas which we present in this section.
