Monotonicity and absolute monotonicity for the two-parameter hyperbolic and trigonometric functions with applications

Abstract

In this paper, we present the monotonicity and absolute monotonicity properties for the two-parameter hyperbolic and trigonometric functions. As applications, we find several complete monotonicity properties for the functions involving the gamma function and provide the bounds for the error function.

1 Introduction

Let $p, q\in\mathbb{R}$ and $a, b>0$ with $a\neq b$. Then the Stolarsky mean $S_{p, q}(a,b)$ [1] is given by

$$ S_{p, q}(a,b)= \textstyle\begin{cases} [\frac{q (a^{p}-b^{p} )}{p (a^{q}-b^{q} )} ]^{1/(p-q)}, &pq(p-q)\neq0, \\ {}[\frac{a^{p}-b^{p}}{p(\log a-\log b)} ]^{1/p}, &p\neq0, q=0, \\ {}[\frac{a^{q}-b^{q}}{q(\log a-\log b)} ]^{1/q}, &p=0, q\neq 0, \\ \exp [\frac{a^{p}\log a-b^{p}\log b}{a^{p}-b^{p}}-\frac{1}{p} ], &p=q\neq0, \\ \sqrt{ab}, &p=q=0. \end{cases} $$

It is well known that $S_{p, q}(a,b)$ is continuous and symmetric on the domain $\{(p, q, a, b): p, q\in\mathbb{R}, a>0, b>0\}$ and strictly increasing with respect to its parameters $p, q\in \mathbb{R}$ for fixed $a, b>0$ with $a\neq b$. Many bivariate means are particular cases of the Stolarksy mean, and many remarkable inequalities and properties for this mean can be found in the literature [2–13]. We clearly see that the value $S_{p,q}(a,b)$ in the case of $pq(p-q)=0$ is the limit of the case of $pq(p-q)\neq0$.

Let $b>a>0$ and $t=\log\sqrt{b/a}\in(0, \infty)$. Then the Stolarsky mean $S_{p, q}(a,b)$ can be expressed by a hyperbolic function as follows:

$$ S_{p, q}(a,b)=\sqrt{ab}H_{p, q}(t), $$

(1.1)

where

$$ H_{p, q}(t)= \textstyle\begin{cases} (\frac{q\sinh(pt)}{p\sinh(qt)} )^{1/(p-q)}, &pq(p-q)\neq0, \\ (\frac{\sinh(pt)}{pt} )^{1/p}, &p\neq0, q=0, \\ (\frac{\sinh(qt)}{qt} )^{1/q}, &p=0, q\neq0, \\ \exp (t\coth(pt)-\frac{1}{p} ), &p=q\neq0, \\ 1, &p=q=0, \end{cases} $$

(1.2)

is the two-parameter hyperbolic sine function [14].

Let $p, q\in[-2, 2]$ and $t\in(0, \pi/2)$. Then the two-parameter trigonometric sine function $T_{p, q}(t)$ [14] is given by

$$ T_{p, q}(t)= \textstyle\begin{cases} (\frac{q\sin(pt)}{p\sin(qt)} )^{1/(p-q)}, &pq(p-q)\neq0, \\ (\frac{\sin(pt)}{pt} )^{1/p}, &p\neq0, q=0, \\ (\frac{\sin(qt)}{qt} )^{1/q}, &p=0, q\neq0, \\ \exp (t\cot(pt)-\frac{1}{p} ), &p=q\neq0, \\ 1, &p=q=0. \end{cases} $$

(1.3)

The main purpose of this paper is to deal with the monotonicity of the functions $t\mapsto[\log H_{p, q}(t)]/t$ and $t\mapsto[\log H_{p, q}(t)]/t^{2}$ on the interval $(0, \infty)$ and with the absolute monotonicity of the functions $t\mapsto\log T_{p, q}(t)$, $t\mapsto [\log T_{p, q}(t)]/t$ and $t\mapsto[\log T_{p, q}(t)]/t^{2}$ on the interval $(0, \pi/2)$. As applications, we shall present several complete monotonicity properties for the functions involving the gamma function and provide bounds for the error function.

2 Main results

Theorem 2.1

Let $p, q\in\mathbb{R}$, $t>0$, and $H_{p, q}(t)$ be defined by (1.2). Then the function $t\mapsto[\log H_{p, q}(t)]/t$ is strictly increasing (decreasing) and strictly concave (convex) from $(0, \infty)$ onto $(0, (p+q)/(|p|+|q|))$ ($((p+q)/(|p|+|q|), 0)$) if $p+q>0$ (<0).

Proof

We only prove the desired result in the case of $pq(p+q)\neq0$; the other cases can be derived easily from the continuity and limit values. Let

$$\begin{aligned}& f_{1}(t)=t \biggl[\frac{p\cosh(pt)}{\sinh(pt)}-\frac{q\cosh(qt)}{\sinh (qt)} \biggr]-\log \sinh\bigl(\vert p\vert t\bigr) \\& \hphantom{f_{1}(t)={}}{}+\log\sinh\bigl(\vert q\vert t\bigr)+\log \vert p\vert -\log{ \vert q\vert }, \\& f_{2}(t)=tf^{\prime}_{1}(t)-2f_{1}(t), \\& F_{1}(u)=\frac{u}{\sinh(u)},\qquad F_{2}(u)= \frac{u^{3}\cosh(u)}{\sinh^{3}(u)}. \end{aligned}$$

Then elaborated computations lead to

$$\begin{aligned}& f_{1}\bigl(0^{+}\bigr)=f_{2}\bigl(0^{+} \bigr)=\lim_{t\rightarrow0^{+}}\frac{\log H_{p, q}(t)}{t}=0, \end{aligned}$$

(2.1)

$$\begin{aligned}& \log H_{p,q}(t)= \frac{1}{p-q}\log \biggl(\frac{q\sinh(pt)}{p\sinh(qt)} \biggr) =\frac{1}{p-q}\log \biggl(\frac{|q|\sinh(|p|t)}{|p|\sinh(|q|t)} \biggr) \\& \hphantom{\log H_{p,q}(t)}= \frac{|p|-|q|}{p-q}t+\frac{1}{p-q}\log \biggl[ \frac{|q| (1-e^{-2|p|t} )}{|p| (1-e^{-2|q|t} )} \biggr], \\& \lim_{t\rightarrow\infty}\frac{\log H_{p, q}(t)}{t}=\frac{|p|-|q|}{p-q}= \frac{p+q}{|p|+|q|}, \end{aligned}$$

(2.2)

$$\begin{aligned}& \biggl[\frac{\log H_{p, q}(t)}{t} \biggr]^{\prime}=\frac{f_{1}(t)}{(p-q)t^{2}}= \frac {p+q}{(|p|+|q|)t^{2}}\times\frac{f_{1}(t)}{|p|-|q|}, \end{aligned}$$

(2.3)

$$\begin{aligned}& f^{\prime}_{1}(t)=\frac{1}{t} \biggl[\frac{(qt)^{2}}{\sinh^{2}(qt)}- \frac {(pt)^{2}}{\sinh^{2}(pt)} \biggr], \\& \frac{f^{\prime}_{1}(t)}{|p|-|q|}= \frac {F^{2}_{1}(qt)-F^{2}_{1}(pt)}{(|p|-|q|)t}=\frac {F^{2}_{1}(|qt|)-F^{2}_{1}(|pt|)}{|pt|-|qt|} \\& \hphantom{\frac{f^{\prime}_{1}(t)}{|p|-|q|}} = -\bigl[F_{1}\bigl(|qt|\bigr)+F_{1}\bigl(|pt|\bigr)\bigr] \frac{F_{1}(|qt|)-F_{1}(|pt|)}{|qt|-|pt|}, \end{aligned}$$

(2.4)

$$\begin{aligned}& \biggl[\frac{\log H_{p, q}(t)}{t} \biggr]^{\prime\prime}=\frac{f_{2}(t)}{(p-q)t^{3}}= \frac {p+q}{(|p|+|q|)t^{3}}\times\frac{f_{2}(t)}{|p|-|q|}, \end{aligned}$$

(2.5)

$$\begin{aligned}& f^{\prime}_{2}(t)=\frac{2}{t} \biggl[\frac{(pt)^{3}\cosh(pt)}{\sinh ^{3}(pt)}- \frac{(qt)^{3}\cosh(qt)}{\sinh^{3}(qt)} \biggr], \\& \frac{f^{\prime}_{2}(t)}{|p|-|q|}=\frac{2[F_{2}(|pt|)-F_{2}(|qt|)]}{|pt|-|qt|}, \end{aligned}$$

(2.6)

$$\begin{aligned}& F^{\prime}_{1}(u)=-\frac{\cosh(u)}{\sinh^{2}(u)}\bigl[u-\tanh(u) \bigr]< 0, \end{aligned}$$

(2.7)

$$\begin{aligned}& F^{\prime}_{2}(u)=-\frac{3u^{3}}{\sinh^{4}(u)} \biggl[\frac{\sinh (2u)}{2u}- \frac{2+\cosh(2u)}{3} \biggr]< 0 \end{aligned}$$

(2.8)

for $u>0$, where the inequality in (2.8) is the Cusa-type inequality given in [15].

It follows from (2.1), (2.4), and (2.6)-(2.8) that

$$ \frac{f_{1}(t)}{|p|-|q|}>0 $$

(2.9)

and

$$ \frac{f_{2}(t)}{|p|-|q|}< 0 $$

(2.10)

for $t\in(0, \infty)$.

Therefore, Theorem 2.1 follows easily from (2.1)-(2.3), (2.5), (2.9), and (2.10). □

Theorem 2.2

Let $p, q\in\mathbb{R}$ and $t>0$, and let $H_{p, q}(t)$ be defined by (1.2). Then the function $t\mapsto[\log H_{p, q}(t)]/t^{2}$ is strictly decreasing (increasing) from $(0, \infty)$ onto $(0, (p+q)/6)$ ($((p+q)/6, 0)$) if $p+q>0$ (<0).

Proof

Let $g_{1}(t)=[\log H_{p, q}(t)]/t$ and $g_{2}(t)=t$. Then we clearly see that

$$ \frac{g^{\prime}_{1}(t)}{g^{\prime}_{2}(t)}=g^{\prime}_{1}(t)= \biggl[ \frac{\log H_{p, q}(t)}{t} \biggr]^{\prime}, $$

(2.11)

and (2.1) leads

$$ \frac{\log H_{p, q}(t)}{t^{2}}=\frac{g_{1}(t)}{g_{2}(t)}=\frac {g_{1}(t)-g_{1}(0^{+})}{g_{2}(t)-g_{2}(0^{+})}. $$

(2.12)

From Theorem 2.1, (2.11), (2.12), and the well-known monotone form of l’Hôpital’s rule [16] we know that the function $t\mapsto[\log H_{p, q}(t)]/t^{2}$ is strictly decreasing (increasing) on $(0, \infty)$ if $p+q>0$ (<0).

It follows from l’Hôpital’s rule and (2.2) that

$$ \lim_{t\rightarrow0^{+}}\frac{\log H_{p, q}(t)}{t^{2}}=\frac{p+q}{6} \quad \mbox{and}\quad \lim_{t\rightarrow \infty}\frac{\log H_{p, q}(t)}{t^{2}}=0. $$

 □

From (1.1) and Theorem 2.2 we get the following corollary.

Corollary 2.1

For $a, b>0$ with $a\neq b$, we have the double inequality

$$ \sqrt{ab}< (>)\, S_{p, q}(a,b)< (>)\, \sqrt{ab} e^{\frac{p+q}{24}(\log b-\log a)^{2}} $$

if $p+q>0$ (<0).

Letting $b>a>0$, $t=\log\sqrt{b/a}>0$, and $(p,q)=(1,0),(1,1),(3/2,1/2)$ in Corollary 2.1, we get the following corollary.

Corollary 2.2

We have the inequalities

$$ \frac{\sinh(t)}{t}< e^{t^{2}/6},\qquad e^{t\cosh(t)-1}< e^{t^{2}/3}, \qquad \frac{2\cosh(t)+1}{3}< e^{t^{2}/3} $$

for all $t>0$.

Next, we recall the definition of absolutely monotonic function [17]. A real-valued function f is said to be absolutely monotonic on the interval I if f has derivatives of all orders on I and

$$ f^{(n)}(x)>0 $$

for all $x\in I$ and $n\geq0$.

Theorem 2.3

Let $p, q\in[-2, 2]$ and $t\in(0, \pi/2)$, and let $T_{p, q}(t)$ be defined by (1.3). Then the functions $t\rightarrow\log T_{p, q}(t)$, $t\rightarrow[\log T_{p, q}(t)]/t$, and $t\rightarrow[\log T_{p, q}(t)]/t^{2}$ are absolutely monotonic on $(0, \pi/2)$ if $p+q<0$. Moreover, the functions $t\rightarrow-\log T_{p, q}(t)$, $t\rightarrow-[\log T_{p,q}(t)]/t$, and $t\rightarrow-[\log T_{p, q}(t)]/t^{2}$ are absolutely monotonic on $(0,\pi/2)$ if $p+q>0$.

Proof

We only prove the desired result in the case of $pq(p+q)\neq0$; the other cases can be derived easily from the continuity and limit values.

Let $i=0,1,2$. Then from (1.3) and the power series formula

$$ \log\frac{\sin(t)}{t}=-\sum_{n=1}^{\infty} \frac{2^{2n-1}|B_{2n}|}{n(2n)!}t^{2n},\quad |t|< \pi, $$

listed in [18], 4.3.71, we get

$$\begin{aligned}& \log T_{p, q}(t)=\frac{1}{p-q}\log \biggl( \frac{q\sin(pt)}{p\sin(qt)} \biggr)=\frac {1}{p-q}\log \biggl(\frac{|qt|\sin(|pt|)}{|pt|\sin(|qt|)} \biggr) \\& \hphantom{\log T_{p, q}(t)}=-(p+q)t^{2}\sum_{n=1}^{\infty} \frac{2^{2n}|B_{2n}| (p^{2n}-q^{2n} )}{2n(2n)! (p^{2}-q^{2} )}t^{2n-2}, \\& \frac{\log T_{p, q}(t)}{t^{i}}=-(p+q)t^{2-i}\sum_{n=1}^{\infty} \frac{2^{2n}|B_{2n}| (p^{2n}-q^{2n} )}{2n(2n)! (p^{2}-q^{2} )}t^{2n-2}, \end{aligned}$$

(2.13)

where $B_{n}$ are the Bernoulli numbers.

Therefore, Theorem 2.3 follows easily from (2.13). □

Let $(p,q)=(1,0), (1,1), (3/2,1/2)$ in Theorem 2.3. Then we immediately get the following corollary.

Corollary 2.3

We have the inequalities

$$\begin{aligned}& \biggl(\frac{2}{\pi} \biggr)^{4t^{2}/\pi^{2}}< \frac{\sin (t)}{t}< e^{-t^{2}/6}, \end{aligned}$$

(2.14)

$$\begin{aligned}& 1-\frac{4t^{2}}{\pi^{2}}< \frac{t}{\tan(t)}< 1-\frac{t^{2}}{3}, \\& 3^{-4t^{2}/\pi^{2}}< \frac{2\cos(t)+1}{3}< e^{-t^{2}/3} \end{aligned}$$

(2.15)

for all $t\in(0,\pi/2)$.

Remark 2.1

The second inequality in (2.14) was first proved by Yang [19], and the double inequality (2.15) can be found in [20], which is better than the Redheffer-type inequality in Theorem 3 of [21].

Remark 2.2

Bhayo and Sándor [22], equation (3.3), presented the double inequality

$$ 1-\frac{4t^{2}}{\pi^{2}}< \frac{t}{\tan(t)}< \frac{\pi^{2}}{8}- \frac{t^{2}}{2} $$

(2.16)

for all $t\in(0, \pi/2)$. The second inequality in (2.16) is better than the second inequality in (2.15) for $t\in(\sqrt{3\pi^{2}/4-6},\pi/2)$.

3 Applications

Recall that a real-valued function f is said to be completely monotonic [23] on the interval I if f has derivatives of all order on I and

$$ (-1)^{n}f^{(n)}(x)\geq0 $$

for all $n\geq0$ and $x\in I$. The set of all completely monotonic functions on I is denoted by $\operatorname{CM}[I]$. A positive function f is said to be logarithmically completely monotonic on the interval I if its logarithm logf is completely monotonic on I. The class of all logarithmically completely monotonic functions on I is denoted by $\operatorname{LCM}[I]$. The famous Bernstein theorem [17] implies that the function

$$ f(x)= \int_{0}^{\infty}e^{-xt}g(t)\, dt $$

is completely monotonic on $(0, \infty)$ if and only if $g(t)\geq0$ for all $t\in(0,\infty)$ if $g(t)$ is continuous on $(0, \infty)$.

Theorem 3.1

Let $s, t, r\in\mathbb{R}$, $\rho=\min\{s, t, r\}$, $x\in(-\rho, \infty)$, let $\Gamma(u)=\int_{0}^{\infty}e^{-t}t^{u-1}\,dt$ ($u>0$) be the gamma function, $\psi(u)=\Gamma^{\prime}(u)/\Gamma(u)$ be the psi function, and the function $x\rightarrow v(s, t, r; x)$ be defined by

$$ v(s, t, r; x)= \textstyle\begin{cases} e^{-\psi(x+r)} [\frac{\Gamma(x+t)}{\Gamma(x+s)} ]^{1/(t-s)}, &t\neq s, \\ e^{-\psi(x+r)}\lim_{t\rightarrow s} [\frac{\Gamma(x+t)}{\Gamma (x+s)} ]^{1/(t-s)}=e^{\psi(x+s)-\psi(x+r)}, &t=s. \end{cases} $$

(3.1)

Then $v(s, t, r; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $r\leq \min\{s, t\}$, and $1/v(s, t, r; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $r\geq(s+t)/2$.

Proof

We only prove the desired result in the case of $t\neq s$ because the case of $t=s$ can be derived easily from the continuity and limit values.

Let $L(a,b)=(b-a)/(\log b-\log a)$ be the logarithmic mean of two distinct positive real numbers a and b, $u>0$, $y=|(t-s)u/2|$, and $p(s, t, r; u)$ and $q(s,t,r;u)$ be respectively defined by

$$\begin{aligned}& p(s,t,r;u)=\frac{\log e^{(\rho-r)u}-\log\frac{e^{(\rho-s)u}-e^{(\rho-t)u}}{(t-s)u}}{u}, \\& q(s, t, r; u)=\frac{e^{(\rho-r)u}-\frac{e^{(\rho-s)u}-e^{(\rho-t)u}}{(t-s)u}}{1-e^{-u}}. \end{aligned}$$

Then we clearly see that

$$\begin{aligned}& p(s,t,r;u)=-r-\frac{1}{u}\log\frac{e^{-su}-e^{-tu}}{(t-s)u} =-r+\frac{t+s}{2}- \frac{|t-s|}{2} \biggl(\frac{1}{y}\log\frac{\sinh (y)}{y} \biggr), \end{aligned}$$

(3.2)

$$\begin{aligned}& q(s, t, r; u)=\frac{u}{1-e^{-u}}L \biggl(e^{(\rho-r)u}, \frac{e^{(\rho-s)u}-e^{(\rho-t)u}}{(t-s)u} \biggr)p(s,t,r; u). \end{aligned}$$

(3.3)

It follows from (1.2) and Theorem 2.1 that the function $y\rightarrow[\log(\sinh(y)/y)]/y$ is strictly increasing from $(0, \infty)$ onto $(0, 1)$. Then (3.2) leads to the conclusion that

$$ \min\{s, t\}-r=-r+\frac{t+s}{2}-\frac{|t-s|}{2}< p(s, t, r; u)< -r+ \frac{t+s}{2}. $$

Therefore,

$$ p(s, t, r; u)\geq0 $$

(3.4)

for all $u>0$ if and only if $r\leq\min\{s, t\}$, and

$$ p(s, t, r; u)\leq0 $$

(3.5)

for all $u>0$ if and only if $r\geq(s+t)/2$.

From (3.1) and the integral formulas

$$\begin{aligned}& \log\Gamma(x)= \int_{0}^{\infty}\frac{1}{u} \biggl((x-1)e^{-u}-\frac {e^{-u}-e^{-xu}}{1-e^{-u}} \biggr)\,du, \\& \psi(x)= \int_{0}^{\infty} \biggl(\frac{e^{-u}}{u}- \frac {e^{-xu}}{1-e^{-u}} \biggr)\,du, \end{aligned}$$

given in [18], 6.1.50, 6.3.21, we get

$$\begin{aligned} \log v(s,t,r;x)&=\frac{\log\Gamma(x+t)-\log\Gamma(x+x)}{t-s}-\psi (x+r) \\ &= \int_{0}^{\infty}\frac{e^{-xu}}{1-e^{-u}} \biggl[ \frac {e^{-tu}-e^{-su}}{(t-s)u}+e^{-ru} \biggr]\,du \\ &= \int_{0}^{\infty}e^{-(x+\rho)u}q(s, t, r; u)\,du. \end{aligned}$$

(3.6)

Therefore, Theorem 3.1 follows easily from (3.3)-(3.6) and the Bernstein theorem. □

Remark 3.1

Qi and Guo [24] gave a sufficient condition for $v(s, t, r; x)\in \operatorname{LCM}[(-\rho, \infty)]$ and a necessary and sufficient condition for $1/v(s, t, r; x)\in \operatorname{LCM}[(-\rho, \infty)]$ by using different methods.

Theorem 3.2

Let $a, b, c\in\mathbb{R}$, $\rho=\min\{a, b, c\}$, $x\in(-\rho, \infty)$, and let the function $x\rightarrow U(a, b, c; x)$ be defined by

$$ U(a, b, c; x)= \textstyle\begin{cases} \frac{1}{x+c} (\frac{\Gamma(x+a)}{\Gamma(x+b)} )^{1/(a-b)},& b\neq a, \\ \lim_{b\rightarrow a}\frac{1}{x+c} (\frac{\Gamma(x+a)}{\Gamma (x+b)} )^{1/(a-b)}=\frac{1}{x+c}e^{\psi(x+a)},& b=a. \end{cases} $$

(3.7)

Then $U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $c\leq(a+b-\max\{|a-b|, 1\})/2$, and $1/U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $c\geq(a+b-\min\{|a-b|, 1\})/2$.

Proof

We only prove the desired result in the case of $b\neq a$ because the case of $b=a$ can be derived easily from the continuity and limit values.

We clearly see that $U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $-[\log U(a, b, c; x)]^{\prime}\in \operatorname{CM}[(-\rho, \infty)]$ and that $1/U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$ if and only if $[\log U(a, b, c; x)]^{\prime}\in \operatorname{CM}[(-\rho, \infty)]$.

Let $t>0$, $H_{p,q}(t)$ be defined by (1.2), and $p(a, b, c; t)$ and $q(a, b, c; t)$ be respectively defined by

$$ p(a, b, c; t)=\frac{\log e^{(\rho-c)t}-\log\frac{e^{(\rho-a)t}-e^{(\rho-b)t}}{(b-a) (1-e^{-t} )}}{t} $$

and

$$ q(a, b, c; t)=e^{(\rho-c)t}-\frac{e^{(\rho-a)t}-e^{(\rho-b)t}}{(b-a) (1-e^{-t} )}. $$

Then we clearly see that

$$\begin{aligned} p(a,b,c;t)=&-c-\frac{1}{t}\log\frac{e^{-at}-e^{-bt}}{(b-a) (1-e^{-t} )} \\ =&\frac{a+b-1}{2}-c-\frac{1}{t}\log \biggl[\frac{\sinh (\vert \frac {(b-a)t}{2}\vert )}{|b-a|\sinh (\frac{t}{2} )} \biggr] \\ =&\frac{a+b-2c-1}{2}-\frac{|b-a|-1}{2}\frac{\log H_{|b-a|,1}(t/2)}{t/2} \end{aligned}$$

(3.8)

and

$$ q(a, b, c; t)=tL \biggl(e^{(\rho-c)t}, \frac{e^{(\rho-a)t}-e^{(\rho-b)t}}{(b-a) (1-e^{-t} )} \biggr)p(a, b, c; t). $$

(3.9)

It follows from Theorem 2.1 and (3.8) that the function $t\rightarrow p(a, b, c; t)$ is strictly monotonic on $(0, \infty)$ and

$$ p\bigl(a, b, c; 0^{+}\bigr)=\frac{a+b-2c}{2},\qquad p(a, b, c; \infty)=\frac{a+b-2c}{2}-\frac{|b-a|-1}{2}. $$

(3.10)

The monotonicity of the function $t\rightarrow p(a, b, c; t)$ on the interval $(0, \infty)$ and (3.10) lead to the conclusion that

$$ p(a, b, c;t)\geq(\leq)\, 0 $$

(3.11)

for all $t\in(0, \infty)$ if and only if $\min(\max)\{p(a, b, c; 0^{+}), p(a, b, c; \infty)\}\geq(\leq)\, 0$, that is, $c\leq (\geq)\,(a+b-\max(\min)\{|a-b|, 1\})/2$.

From (3.7) and the formulas

$$ \psi(x)= \int_{0}^{\infty} \biggl(\frac{e^{-t}}{t}- \frac {e^{-xt}}{1-e^{-t}} \biggr)\,dt,\qquad \frac{1}{x}= \int_{0}^{\infty}e^{-xt}\,dt $$

we have

$$\begin{aligned} - \bigl(\log U(a, b, c; x) \bigr)^{\prime}&=\frac{1}{x+c}- \frac{\psi(x+b)-\psi (x+a)}{b-a} \\ &= \int_{0}^{\infty}e^{-(x+c)t}\,dt- \int_{0}^{\infty}\frac {e^{-(x+a)t}-e^{-(x+b)t}}{(b-a) (1-e^{-t} )}\,dt \\ &= \int_{0}^{\infty}e^{-(x+\rho)t}q(a, b, c; t)\,dt. \end{aligned}$$

(3.12)

Therefore, Theorem 3.2 follows from (3.9), (3.11), (3.12), and the Bernstein theorem. □

Remark 3.2

Qi [25] presented a sufficient condition for $U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$ or $1/U(a, b, c; x)\in \operatorname{LCM}[(-\rho, \infty)]$.

Theorem 3.3

Let $\operatorname {erf}(x)=2\int_{0}^{x}e^{-t^{2}}\,dt/\sqrt{\pi}$ be the error function. Then we have the double inequality

$$ \frac{4}{\sqrt{\pi}}\arctan\frac{2e^{\sqrt{3}x}+1}{\sqrt{3}}+1-2\sqrt {\pi}< \operatorname {erf}(x) < \frac{4}{\sqrt{\pi}}\arctan\frac{2e^{\sqrt{3}x}+1}{\sqrt{3}}-\frac {4\sqrt{\pi}}{3} $$

for all $x>0$.

Proof

It follows from the third inequality in Corollary 2.2 that

$$ e^{-u^{2}}-\frac{3}{2\cosh(\sqrt{3}u)+1}< 0 $$

(3.13)

for $u>0$.

Let

$$\begin{aligned} F(x)&=\frac{2}{\sqrt{\pi}} \int_{0}^{x} \biggl(e^{-u^{2}}- \frac{3}{2\cosh (\sqrt{3}u)+1} \biggr)\,du \\ &=\operatorname {erf}(x)-\frac{6}{\sqrt{\pi}} \int_{0}^{x}\frac{1}{2\cosh(\sqrt{3}u)+1}\,du. \end{aligned}$$

(3.14)

Then

$$ F(0)=0, \qquad F(\infty)=1-\frac{2\sqrt{\pi}}{3}. $$

(3.15)

It follows from (3.13)-(3.15) that

$$ \frac{6}{\sqrt{\pi}} \int_{0}^{x}\frac{1}{2\cosh(\sqrt{3}u)+1}\,du+1- \frac {2\sqrt{\pi}}{3}< \operatorname {erf}(x)< \frac{6}{\sqrt{\pi}} \int_{0}^{x}\frac{1}{2\cosh(\sqrt{3}u)+1}\,du $$

(3.16)

for $x>0$.

Therefore, Theorem 3.3 follows easily from (3.16). □