Monotonicity and inequalities involving the incomplete gamma function

Abstract

In the article, we deal with the monotonicity of the function $x\rightarrow[ (x^{p}+a )^{1/p}-x]/I_{p}(x)$ on the interval $(0, \infty)$ for $p>1$ and $a>0$, and present the necessary and sufficient condition such that the double inequality $[ (x^{p}+a )^{1/p}-x]/a< I_{p}(x)<[ (x^{p}+b )^{1/p}-x]/b$ for all $x>0$ and $p>1$, where $I_{p}(x)=e^{x^{p}}\int_{x}^{\infty}e^{-t^{p}}\,dt$ is the incomplete gamma function.

1 Introduction

Let $a>0$ and $x>0$. Then the classical gamma function $\Gamma(x)$, incomplete gamma function $\Gamma(a, x)$ and psi function $\psi(x)$ are defined by

$$ \Gamma(x)= \int_{0}^{\infty}t^{x-1}e^{-t}\,dt, \qquad \Gamma(a, x)= \int_{x}^{\infty}t^{a-1}e^{-t}\,dt, \qquad \psi(x)=\frac{\Gamma^{\prime}(x)}{\Gamma(x)}, $$

respectively. It is well known that the identities

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

(1.1)

hold for all $x, p>0$.

Recently the bounds for the integral $\int_{x}^{\infty}e^{-t^{p}}\,dt$ or $\int_{0}^{x}e^{-t^{p}}\,dt$ have attracted the attention of many researchers. In particular, many remarkable inequalities for bounding both integrals can be found in the literature [1–12]. Let

$$ I_{p}(x)=e^{x^{p}} \int_{x}^{\infty}e^{-t^{p}}\,dt. $$

(1.2)

Then $I_{2}(x)$ is actually the Mills ratio and it has been investigated by many researchers [13–19], and the functions $I_{3}(x)$ and $I_{4}(x)$ can be used to research the heat transfer problem [20] and electrical discharge in gases [21], respectively.

Komatu [22] and Pollak [23] proved that the double inequality

$$ \frac{1}{\sqrt{x^{2}+2}+x}< I_{2}(x)< \frac{1}{\sqrt{x^{2}+{4}/{\pi}}+x} $$

holds for all $x>0$.

In [24], Gautschi proved that the double inequality

$$ \frac{1}{2} \bigl[ \bigl(x^{p}+2 \bigr)^{1/p}-x \bigr]< I_{p}(x)< \frac {1}{a_{0}} \bigl[ \bigl(x^{p}+a_{0} \bigr)^{1/p}-x \bigr] $$

(1.3)

holds for all $x>0$ and $p>1$, where

$$ a_{0}=\Gamma^{p/(1-p)} \biggl(1+\frac{1}{p} \biggr). $$

(1.4)

An application of inequality (1.3) was given in [25]. Alzer [26] proved that the double inequality

$$ \Gamma \biggl(1+\frac{1}{p} \biggr) \bigl[1- \bigl(1-e^{-\alpha x^{p}} \bigr)^{1/p} \bigr]< I_{p}(x) < \Gamma \biggl(1+\frac{1}{p} \biggr) \bigl[1- \bigl(1-e^{-\beta x^{p}} \bigr)^{1/p} \bigr] $$

holds for all $x>0$ and $p>0$ with $p\neq1$ if and only if $\alpha\geq\max\{1, \Gamma^{-p}(1+1/p)\}$ and $\beta\leq\min\{1, \Gamma^{-p}(1+1/p)\}$.

Motivated by inequality (1.3), in the article we deal with the monotonicity of the function

$$ R(x)=\frac{ (x^{p}+a )^{1/p}-x}{e^{x^{p}}\int_{x}^{\infty }e^{-t^{p}}\,dt}=\frac{ (x^{p}+a )^{1/p}-x}{I_{p}(x)} $$

(1.5)

and prove that the double inequality

$$ \frac{1}{a} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr]< I_{p}(x)< \frac {1}{b} \bigl[ \bigl(x^{p}+b \bigr)^{1/p}-x \bigr] $$

(1.6)

holds for all $x>0$ and $p>1$ if and only if $a\geq2$ and $b\leq a_{0}=\Gamma^{p/(1-p)}(1+1/p)$.

2 Lemmas

In order to prove our main results, we need to introduce an auxiliary function at first.

Let $-\infty\leq a< b\leq\infty$, f and g be differentiable on $(a,b)$, and $g'\neq0$ on $(a,b)$. Then the function $H_{f, g}$ [27, 28] is defined by

$$ H_{f,g}(x)=\frac{f^{\prime}(x)}{g^{\prime}(x)}g(x)-f(x). $$

(2.1)

Lemma 2.1

(See [28], Theorem 9)

Let $\infty\leq a< b\leq\infty$, f and g be differentiable on $(a,b)$ with $f(b^{-})=g(b^{-})=0$ and $g^{\prime}(x)<0$ on $(a,b)$, $H_{f, g}$ be defined by (2.1), and there exists $\lambda\in(a, b)$ such that $f^{\prime}(x)/g^{\prime}(x)$ is strictly increasing on $(a, \lambda)$ and strictly decreasing on $(\lambda, b)$. Then the following statements are true:

• (1) if $H_{f, g}(a^{+})\geq0$, then $f(x)/g(x)$ is strictly decreasing on $(a, b)$;

• (2) if $H_{f, g}(a^{+})<0$, then there exists $x_{0}\in(a, b)$ such that $f(x)/g(x)$ is strictly increasing on $(a, x_{0})$ and strictly decreasing on $(x_{0}, b)$.

Lemma 2.2

(See [29], Theorem 1.25)

Let $-\infty< a< b<\infty$, $f,g:[a,b]\rightarrow{\mathbb{R}}$ be continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x)\neq 0$ on $(a,b)$. If $f^{\prime}(x)/g^{\prime}(x)$ is increasing (decreasing) on $(a,b)$, then so are the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)},\qquad \frac{f(x)-f(b)}{g(x)-g(b)}. $$

If $f^{\prime}(x)/g^{\prime}(x)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.3

The inequality

$$ \Gamma^{1/(1-x)}(1+x)>\frac{1}{2} $$

(2.2)

holds for all $x\in(0, 1)$.

Proof

We clearly see that inequality (2.2) is equivalent to

$$ \log\Gamma(1+x)+(1-x)\log2>0 $$

(2.3)

for $x\in(0, 1)$.

Let

$$ h(x)=\log\Gamma(1+x)+(1-x)\log2. $$

(2.4)

Then simple computations lead to

$$\begin{aligned}& h(1)=0, \end{aligned}$$

(2.5)

$$\begin{aligned}& h^{\prime}(x)=\psi(x+1)-\log2< \psi(2)-\log2=1-\gamma-\log2< 0 \end{aligned}$$

(2.6)

for $x\in(0, 1)$, where $\gamma=0.5772\ldots$ is the Euler-Mascheroni constant.

Therefore, inequality (2.3) follows easily from (2.4)-(2.6). □

Lemma 2.4

The function $\Gamma^{1/x}(1+x)$ is strictly increasing on $(0, \infty)$, and the double inequality

$$ x< \Gamma^{1/x}(1+x)< 1 $$

(2.7)

holds for all $x\in(0, 1)$.

Proof

Let

$$\begin{aligned}& \varphi_{1}(x)=\log\Gamma(1+x), \qquad \varphi_{2}(x)=x,\qquad \varphi(x)=\frac{\varphi_{1}(x)}{\varphi_{2}(x)}= \frac{\log\Gamma(1+x)}{x}, \end{aligned}$$

(2.8)

$$\begin{aligned}& \phi(x)=\log\Gamma(1+x)-x\log x. \end{aligned}$$

(2.9)

Then simple computations lead to

$$\begin{aligned}& \varphi_{1}(0)=\varphi_{2}(0)=0, \end{aligned}$$

(2.10)

$$\begin{aligned}& \phi\bigl(0^{+}\bigr)=\phi(1)=0, \end{aligned}$$

(2.11)

$$\begin{aligned}& \biggl[\frac{\varphi^{\prime}_{1}(x)}{\varphi^{\prime}_{2}(x)} \biggr]^{\prime}=\psi^{\prime}(x+1)>0 \end{aligned}$$

(2.12)

for $x\in(0, \infty)$, and

$$ \phi^{\prime\prime}(x)=\psi^{\prime}(1+x)- \frac{1}{x}< 0 $$

(2.13)

for $x\in(0, 1)$.

It follows from (2.8), (2.10), (2.12), and Lemma 2.2 that $\varphi(x)$ and $e^{\varphi(x)}=\Gamma^{1/x}(1+x)$ is strictly increasing on $(0, \infty)$.

Inequality (2.13) leads to the conclusion that the function $\phi(x)$ is strictly concave on the interval $(0, 1)$ and the inequality

$$ \phi(x)>\phi(0) (1-x)+\phi(1)x $$

(2.14)

holds for all $x\in(0, 1)$.

Therefore, $\phi(x)>0$ and the first inequality of (2.7) holds for all $x\in(0, 1)$ follows from (2.9), (2.11), and (2.14). While the second inequality of (2.7) can be derived from the monotonicity of the function $\Gamma^{1/x}(1+x)$ on the interval $(0, 1)$. □

Lemma 2.5

Let $p>1$ and $x>0$. Then the function $a\rightarrow [ (x^{p}+a )^{1/p}-x ]/a$ is strictly decreasing on $(0, \infty)$.

Proof

Let

$$ \omega_{1}(a)= \bigl(x^{p}+a \bigr)^{1/p}-x,\qquad \omega_{2}(a)=a,\qquad \omega(a)= \frac{\omega_{1}(a)}{\omega_{2}(a)}=\frac{ (x^{p}+a )^{1/p}-x}{a}. $$

(2.15)

Then we clearly see that

$$\begin{aligned}& \omega_{1}(0)=\omega_{2}(0)=0, \end{aligned}$$

(2.16)

$$\begin{aligned}& \biggl[\frac{\omega^{\prime}_{1}(a)}{\omega^{\prime}_{2}(a)} \biggr]^{\prime}=\frac{1-p}{p^{2} (x^{p}+a )^{(2p-1)/p}}< 0 \end{aligned}$$

(2.17)

for all $p>1$, $x>0$ and $a>0$.

Therefore, Lemma 2.5 follows easily from Lemma 2.2 and (2.15)-(2.17). □

Lemma 2.6

Let $p>1$, $a>0$ and $x>0$, $H_{f, g}(x)$ be defined by (2.1), and $f_{1}(x)$ and $g_{1}(x)$ be defined by

$$ f_{1}(x)= \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr]e^{-x^{p}},\qquad g_{1}(x)= \int_{x}^{\infty}e^{-t^{p}}\,dt, $$

(2.18)

respectively. Then $H_{f_{1}, g_{1}}(0^{+})=\Gamma(1+1/p)-a^{1/p}$.

Proof

Let

$$ u=u(x)= \biggl(\frac{x^{p}+a}{x^{p}} \biggr)^{1/p}\in(1, \infty). $$

(2.19)

Then from (2.18) and (2.19) one has

$$\begin{aligned}& f_{1}(0)=a^{1/p},\qquad g_{1}(0)=\frac{1}{p}\Gamma \biggl(\frac{1}{p} \biggr)=\Gamma \biggl(1+\frac {1}{p} \biggr), \end{aligned}$$

(2.20)

$$\begin{aligned}& \frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)}=- \biggl(\frac {x^{p}+a}{x^{p}} \biggr)^{1/p-1}+px^{p} \biggl[ \biggl(\frac {x^{p}+a}{x^{p}} \biggr)^{1/p}-1 \biggr]+1 \\& \hphantom{\frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)}}=1+\frac{(pa-1)u+u^{1-p}-pa}{u^{p}-1}. \end{aligned}$$

(2.21)

It follows from (2.1), (2.20), and (2.21) that

$$\begin{aligned} H_{f_{1}, g_{1}}\bigl(0^{+}\bigr)&=\lim_{x\rightarrow 0^{+}} \frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)}\lim_{x\rightarrow 0^{+}}g_{1}(x)-\lim _{x\rightarrow0^{+}}f_{1}(x) \\ &=\Gamma \biggl(1+\frac{1}{p} \biggr) \biggl[1+\lim_{u\rightarrow \infty} \frac{(pa-1)u+u^{1-p}-pa}{u^{p}-1} \biggr]-a^{1/p} \\ &=\Gamma \biggl(1+\frac{1}{p} \biggr)-a^{1/p}. \end{aligned}$$

 □

3 Main results

Theorem 3.1

Let $p>1$, $a>0$, $x>0$ and $R(x)$ be defined by (1.5). Then the following statements are true:

• (1) if $a\geq2$, then $R(x)$ is strictly increasing on $(0, \infty)$;

• (2) if $a\leq\Gamma^{p}(1+1/p)$, then $R(x)$ is strictly decreasing on $(0, \infty)$;

• (3) if $\Gamma^{p}(1+1/p)< a<2$, then there exists $x_{0}\in(0, \infty)$ such that $R(x)$ is strictly increasing on $(0, x_{0})$ and strictly decreasing on $(x_{0}, \infty)$.

Proof

Let $f_{1}(x)$, $g_{1}(x)$, $u=u(x)\in(1, \infty)$ be defined by (2.18) and (2.19), and $h(u)$ and $h_{1}(u)$ be defined by

$$\begin{aligned}& h(u)=(p-1) (ap-1)u^{2p}-ap^{2}u^{2p-1}+(2p+ap-2)u^{p}+1-p, \end{aligned}$$

(3.1)

$$\begin{aligned}& h_{1}(u)=2(p-1) (ap-1)u^{p}-ap(2p-1)u^{p-1}+2p+ap-2. \end{aligned}$$

(3.2)

Then from (1.2), (1.5), (2.18), (2.21), (3.1), (3.2), and Lemma 2.4 we have

$$\begin{aligned}& R(x)=\frac{f_{1}(x)}{g_{1}(x)}, \end{aligned}$$

(3.3)

$$\begin{aligned}& h(1)=h_{1}(1)=0, \end{aligned}$$

(3.4)

$$\begin{aligned}& \biggl[\frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)} \biggr]^{\prime} =\frac{ \frac{d}{du} [1+\frac{(pa-1)u+u^{1-p}-pa}{u^{p}-1} ]}{ \frac{dx}{du}} = \frac{ (u^{p}-1 )^{1/p-1}}{a^{1/p}u^{2p-1}}h(u), \end{aligned}$$

(3.5)

$$\begin{aligned}& h^{\prime}(u)=pu^{p-1}h_{1}(u), \end{aligned}$$

(3.6)

$$\begin{aligned}& h^{\prime}_{1}(u)=p(p-1)u^{p-2}\bigl[2(ap-1) (u-1)+(a-2)\bigr], \end{aligned}$$

(3.7)

$$\begin{aligned}& \frac{1}{p}< \Gamma^{p} \biggl(1+\frac{1}{p} \biggr)< 2 \end{aligned}$$

(3.8)

for $p>1$.

We divide the proof into four cases.

Case 1: $a\geq2$. Then from (3.4)-(3.7) we clearly see that the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ is strictly increasing on $(0, \infty)$. Therefore, $R(x)$ is strictly increasing on $(0, \infty)$ follows from Lemma 2.2 and (3.3) together with the monotonicity of the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ on the interval $(0, \infty)$ and $f_{1}(\infty)=g_{1}(\infty)=0$.

Case 2: $a\leq1/p$. Then from (3.4)-(3.8) we clearly see that the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ is strictly decreasing on $(0, \infty)$. Therefore, $R(x)$ is strictly decreasing on $(0, \infty)$ follows from Lemma 2.2 and (3.3) together with the monotonicity of the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ on the interval $(0, \infty)$ and $f_{1}(\infty)=g_{1}(\infty)=0$.

Case 3: $1/p< a\leq\Gamma^{p}(1+1/p)$. Then (3.1), (3.2), and Lemma 2.6 lead to

$$\begin{aligned}& \lim_{u\rightarrow\infty}h(u)=\infty,\qquad \lim_{u\rightarrow \infty}h_{1}(u)= \infty, \end{aligned}$$

(3.9)

$$\begin{aligned}& H_{f_{1}, g_{1}}\bigl(0^{+}\bigr)\geq0. \end{aligned}$$

(3.10)

Note that (3.7) can be rewritten as

$$ h^{\prime}_{1}(u)=2p(ap-1) (p-1)u^{p-2}(u-u_{0}) $$

(3.11)

with $u_{0}=1+(2-a)/[2(ap-1)]\in(1, \infty)$.

From (3.11) we clearly see that $h_{1}(u)$ is strictly decreasing on $(1, u_{0})$ and strictly increasing on $(u_{0}, \infty)$. Then from (3.4), (3.6), and (3.9) we know that there exists $\lambda\in(1, \infty)$ such that $h(u)<0$ for $u\in(1, \lambda)$ and $h(u)>0$ for $u\in(\lambda, \infty)$.

From (2.19) we clearly see that the function $x\rightarrow u(x)$ is strictly decreasing from $(0, \infty)$ onto $(1, \infty)$. Then (3.5) and $h(u)<0$ for $u\in(1, \lambda)$ and $h(u)>0$ for $u\in (\lambda, \infty)$ lead to the conclusion that $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ is strictly increasing on $(0, \mu)$ and strictly decreasing on $(\mu, \infty)$, where $\mu=[a/(\lambda^{p}-1)]^{1/p}$.

Therefore, $R(x)$ is strictly decreasing on $(0, \infty)$ follows from (3.3), (3.10), Lemma 2.1(1), and the piecewise monotonicity of the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ on the interval $(0, \infty)$ together with the fact that $g^{\prime}_{1}(x)=-e^{-x^{p}}<0$ and $f_{1}(\infty)=g_{1}(\infty)=0$.

Case 4: $\Gamma^{p}(1+1/p)< a<2$. Then we clearly see that (3.9) and (3.11) again hold. Making use of the same method as in Case 3 we know that there exists $\eta>0$ such that $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ is strictly increasing on $(0, \eta)$ and strictly decreasing on $(\eta, \infty)$.

It follows from Lemma 2.6 that

$$ H_{f_{1}, g_{1}}\bigl(0^{+}\bigr)< 0. $$

(3.12)

Therefore, there exists $x_{0}\in(0, \infty)$ such that $R(x)$ is strictly increasing on $(0, x_{0})$ and strictly decreasing on $(x_{0}, \infty)$ follows from (3.3), (3.12), Lemma 2.1(2), and the piecewise monotonicity of the function $f^{\prime}_{1}(x)/g^{\prime}_{1}(x)$ on the interval $(0, \infty)$ together with the fact that $g^{\prime}_{1}(x)=-e^{-x^{p}}<0$ and $f_{1}(\infty)=g_{1}(\infty)=0$. □

Let $p>1$, $x>0$, $a>0$, $R(x)$, $f_{1}(x)$, $g_{1}(x)$ and $u=u(x)$ be defined by (1.5), (2.18), and (2.19), respectively. Then we clearly see that

$$ f_{1}(\infty)=g_{1}(\infty)=0. $$

(3.13)

It follows from (2.20), (2.21), (3.3), and (3.13) that

$$\begin{aligned}& R\bigl(0^{+}\bigr)=\frac{a^{1/p}}{\Gamma (1+\frac{1}{p} )}, \end{aligned}$$

(3.14)

$$\begin{aligned}& R(\infty)= \lim_{x\rightarrow \infty}\frac{f_{1}(x)}{g_{1}(x)}=\lim _{x\rightarrow \infty}\frac{f^{\prime}_{1}(x)}{g^{\prime}_{1}(x)} \\& \hphantom{R(\infty)}= 1+\lim_{u\rightarrow1^{+}}\frac{(pa-1)u+u^{1-p}-pa}{u^{p}-1}=a. \end{aligned}$$

(3.15)

From (3.14) and (3.15) together with Theorem 3.1 we get Corollary 3.2 immediately.

Corollary 3.2

Let $p>1$, $a, x>0$, $I_{p}(x)$ and $R(x)$ be defined by (1.2) and (1.5), and $x_{0}$ be the unique solution of the equation $R^{\prime}(x)=0$ on the interval $(0, \infty)$ for $\Gamma^{p}(1+1/p)< a<2$. Then the following statements are true:

• (1) if $a\geq2$, then the double inequality

  $$ \frac{1}{a} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr]< I_{p}(x)< a^{-1/p}\Gamma \biggl(1+\frac{1}{p} \biggr) \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr] $$

  holds for all $p>1$ and $x>0$;

• (2) if $0< a\leq\Gamma^{p}(1+1/p)$, then the double inequality

  $$ a^{-1/p}\Gamma \biggl(1+\frac{1}{p} \biggr) \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr]< I_{p}(x)< \frac{1}{a} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr] $$

  holds for all $p>1$ and $x>0$;

• (3) if $\Gamma^{p}(1+1/p)< a<2$, then the two-sided inequality

  $$ \frac{1}{R(x_{0})} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr] \leq I_{p}(x)< \max \biggl\{ \frac{1}{a}, \frac{\Gamma (1+\frac{1}{p} )}{a^{1/p}} \biggr\} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr] $$

  is valid for all $p>1$ and $x>0$.

Theorem 3.3

Let $p>1$, $a, b, x>0$, $I_{p}(x)$ and $a_{0}$ be defined by (1.2) and (1.4), respectively. Then the bilateral inequality

$$ \frac{1}{a} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr]< I_{p}(x)< \frac {1}{b} \bigl[ \bigl(x^{p}+b \bigr)^{1/p}-x \bigr] $$

(3.16)

holds for all $p>1$ and $x>0$ if and only if $a\geq2$ and $b\leq a_{0}$.

Proof

If $a\geq2$ and $b\leq a_{0}$, then inequality (3.16) is valid for all $p>1$ and $x>0$ follows easily from (1.3) and Lemma 2.5.

If the inequality $I_{p}(x)< [ (x^{p}+b )^{1/p}-x ]/b$ takes place for $p>1$ and $x>0$, then (3.14) leads to

$$ \lim_{x\rightarrow 0^{+}}\frac{ (x^{p}+b )^{1/p}-x}{I_{p}(x)}=\frac {b^{1/p}}{\Gamma (1+\frac{1}{p} )}\geq b, $$

which implies $b\leq a_{0}$.

Next, we use the proof by contradiction to prove that $a\geq2$ if the inequality $I_{p}(x)> [ (x^{p}+b )^{1/p}-x ]/a$ holds for all $x>0$ and $p>1$.

From Lemmas 2.3 and 2.4 we clearly see that

$$ \Gamma^{p} \biggl(1+\frac{1}{p} \biggr)< a_{0}< 2. $$

(3.17)

We divide the proof into two cases.

Case 1: $a\leq a_{0}$. Then it follows from the sufficiency of Theorem 3.3 which was proved previously that $I_{p}(x)< [ (x^{p}+b )^{1/p}-x ]/a$ for all $p>1$ and $x>0$.

Case 2: $a_{0}< a<2$. Let $R(x)$ be defined by (1.5), then Theorem 3.1(3), (3.15), and (3.17) lead to the conclusion that there exists $x_{0}\in(0, \infty)$ such that $R(x)$ is strictly decreasing on $(x_{0}, \infty)$ and

$$ \frac{ (x^{p}+a )^{1/p}-x}{I_{p}(x)}=R(x)>R(\infty)=a $$

or

$$ I_{p}(x)< \frac{1}{a} \bigl[ \bigl(x^{p}+a \bigr)^{1/p}-x \bigr] $$

for all $p>1$ and $x\in(x_{0}, \infty)$. □

Let $p>1$, $a>0$, $x>0$, $q=1/p\in(0, 1)$, and $u=x^{p}>0$. Then from (1.1) and (1.2) one has

$$ I_{p}(x)=qe^{u}\Gamma(q, u),\qquad \bigl(x^{p}+a \bigr)^{1/p}-x=(u+a)^{q}-u^{q}, $$

and Corollary 3.2 and Theorem 3.3 can be rewritten as follows.

Corollary 3.4

Let $q\in(0, 1)$, $a>0$, and $u>0$. Then the following statements are true:

• (1) if $a\geq2$, then the double inequality

  $$ \frac{(u+a)^{q}-u^{q}}{qa}< e^{u}\Gamma(q,u)< \frac{\Gamma(1+q) [(u+a)^{q}-u^{q} ]}{qa^{q}} $$

  (3.18)

  holds for all $q\in(0, 1)$ and $u>0$, and inequality (3.18) is reversed if $0< a\leq\Gamma^{1/q}(1+q)$;

• (2) if $\Gamma^{1/q}(1+q)< a<2$, then the two-sided inequality

  $$ \frac{(u+a)^{q}-u^{q}}{q\theta(q, u_{0}, a)}\leq e^{u}\Gamma(q,u)< \max \biggl\{ \frac{1}{a}, \frac{\Gamma(1+q)}{a^{q}} \biggr\} \frac{(u+a)^{q}-u^{q}}{q} $$

  holds for all $q\in(0, 1)$ and $u>0$, where $\theta(q, u_{0}, a)= [(u_{0}+a)^{q}-u^{q}_{0} ]/ [qe^{u_{0}}\Gamma(q, u_{0}) ]$ and $u_{0}$ is the unique solution of the equation

  $$ \frac{d [ \frac{(u+a)^{q}-u^{q}}{qe^{u}\Gamma(q,u)} ]}{du}=0 $$

  on the interval $(0, \infty)$ for $\Gamma^{1/q}(1+q)< a<2$.

Corollary 3.5

Let $a, b, u>0$, $q\in(0, 1)$ and $a_{0}$ be defined by (1.4). Then the double inequality

$$ \frac{(u+a)^{q}-u^{q}}{qa}< e^{u}\Gamma(q, u)< \frac{(u+b)^{q}-u^{q}}{qb} $$

holds for all $q\in(0, 1)$ and $u>0$ if and only if $a\geq2$ and $b\leq a_{0}$.

Let $q\rightarrow0^{+}$ and $Ei(u)=\lim_{q\rightarrow 0^{+}}\Gamma(q, u)$. Then Corollaries 3.4 and 3.5 lead to Remarks 3.6 and 3.7.

Remark 3.6

Let $a>0$ and $u>0$, then the following statements are true:

• (1) if $a\geq2$, then the double inequality

  $$ \frac{\log (1+\frac{a}{u} )}{a}< e^{u}Ei(u)< \log \biggl(1+ \frac {a}{u} \biggr) $$

  (3.19)

  holds for all $u>0$, and inequality (3.19) is reversed if $0< a< e^{-\gamma}$;

• (2) if $e^{-\gamma}< a<2$, then we have the sided inequality

  $$ \frac{e^{u_{0}}Ei(u_{0})}{\log (1+\frac{a}{u_{0}} )}\log \biggl(1+\frac{a}{u} \biggr)\leq e^{u}Ei(u)< \max \biggl\{ \frac{1}{a}, 1 \biggr\} \log \biggl(1+ \frac{a}{u} \biggr) $$

  (3.20)

  for all $u>0$, where $u_{0}$ is the unique solution of the equation

  $$ \frac{d}{du}\frac{\log (1+\frac{a}{u} )}{e^{u}Ei(u)}=0 $$

  (3.21)

  on the interval $(0, \infty)$ for $e^{-\gamma}< a<2$.

Remark 3.7

Let $a, b>0$ and $a_{0}$ be defined by (1.4). Then the double inequality

$$ \frac{\log (1+\frac{a}{u} )}{a}< e^{u}Ei(u)< \frac{\log (1+\frac{b}{u} )}{b} $$

holds for all $u>0$ if and only if $a\geq2$ and $b\leq a_{0}$.

In particular, if $a=1$, then numerical computations show that $u_{0}=0.23855\ldots$ is the unique solution of the equation

$$ \frac{d}{du}\frac{\log (1+\frac{1}{u} )}{e^{u}Ei(u)}=0 $$

and $e^{u_{0}}Ei(u_{0})/\log(1+1/u_{0})=0.83311\ldots>8\text{,}331/10\text{,}000$. Therefore, Remark 3.7 leads to Remark 3.8.

Remark 3.8

The double inequality

$$ \frac{8\text{,}331}{10\text{,}000}\log \biggl(1+\frac{1}{u} \biggr)< e^{u}Ei(u)< \log \biggl(1+\frac{1}{u} \biggr) $$

is valid for all $u>0$.

Remark 3.9

Unfortunately, in the article we cannot deal with the monotonicity for the function $R(x)$ defined by (1.5) and present the bounds for the function $I_{p}(x)$ given by (1.2) in the case of $p\in(0,1)$; we leave it as an open problem to the reader.