Monotonicity of the incomplete gamma function with applications

Abstract

In the article, we discuss the monotonicity properties of the function $x\rightarrow (1-e^{-ax^{p}} )^{1/p}/\int_{0}^{x}e^{-t^{p}}\,dt$ for $a, p>0$ with $p\neq1$ on $(0, \infty)$ and prove that the double inequality $\Gamma(1+1/p) (1-e^{-a x^{p}} )^{1/p} <\int_{0}^{x}e^{-t^{p}}\,dt<\Gamma(1+1/p) (1-e^{-b x^{p}} )^{1/p}$ holds for all $x>0$ if and only if $a\leq\min\{1, \Gamma^{-p}(1+1/p)\}$ and $b\geq\max\{1, \Gamma^{-p}(1+1/p)\}$.

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

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

respectively. It is well known that the identity

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

(1.1)

holds for all $x, p>0$.

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

$$ I_{p}(x)= \int_{0}^{x}e^{-t^{p}}\,dt. $$

(1.2)

Then we clearly see that $I_{1}(x)=1-e^{-t}$ and that $I_{p}(x)$ diverges if $p\leq0$. The functions $I_{3}(x)$ and $I_{4}(x)$ can be used to study the heat transfer problem [13] and electrical discharge in gases [14], respectively.

Komatu [15] and Pollak [16] proved the double inequality

$$ \Gamma \biggl(1+\frac{1}{p} \biggr)-\frac{e^{-x^{2}}}{\sqrt{x^{2}+\frac {4}{\pi}}+x}< I_{2}(x)< \Gamma \biggl(1+\frac{1}{p} \biggr)-\frac {e^{-x^{2}}}{\sqrt{x^{2}+2}+x} $$

for all $x>0$.

Gautschi [17] proved that the double inequality

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

(1.3)

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

An application of inequality (1.3) in radio propagation mode was given in [18].

Alzer [19] proved that $a=\min\{1, \Gamma^{-p}(1+1/p)\}$ and $b=\max\{ 1, \Gamma^{-p}(1+1/p)\}$ are the best possible parameters such that the double inequality

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

(1.4)

holds for all $x>0$ and $p>0$ with $p\neq1$.

Motivated by the Alzer’s inequality (1.4), in this paper, we discuss the monotonicity of the function

$$ x\rightarrow R(a, p; x)=\frac{ (1-e^{-ax^{p}} )^{1/p}}{\int _{0}^{x}e^{-t^{p}}\,dt} $$

(1.5)

and provide an alternative proof of Alzer’s inequality (1.4).

2 Lemmas

In order to prove our main results, we first introduce an auxiliary function. Let $-\infty\leq a< b\leq\infty$, f and g be differentiable on $(a,b)$, and $g'\neq0$ on $(a,b)$. Then the function $H_{f, g}$ [20, 21] is defined by

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

(2.1)

Lemma 2.1

(See [21], Theorem 8)

Let $\infty\leq a< b\leq\infty$, f and g be differentiable on $(a,b)$ with $f(a^{+})=g(a^{+})=0$ and $g^{\prime}(x)>0$ on $(a,b)$, and $H_{f, g}$ be defined by (2.1). Then the following statements are true:

• (1) If $H_{f, g}(b^{-})>0$ and there exists $\lambda\in(a, b)$ such that $f^{\prime}(x)/g^{\prime}(x)$ is strictly decreasing on $(a, \lambda)$ and strictly increasing on $(\lambda, b)$, then there exists $\mu\in(a, b)$ such that $f(x)/g(x)$ is strictly decreasing on $(a, \mu )$ and strictly increasing on $(\mu, b)$;

• (2) If $H_{f, g}(b^{-})<0$ and there exists $\lambda^{\ast}\in(a, b)$ such that $f^{\prime}(x)/g^{\prime}(x)$ is strictly increasing on $(a, \lambda^{\ast})$ and strictly decreasing on $(\lambda^{\ast}, b)$, then there exists $\mu^{\ast}\in(a, b)$ such that $f(x)/g(x)$ is strictly increasing on $(a, \mu^{\ast})$ and strictly decreasing on $(\mu^{\ast}, b)$.

Lemma 2.2

(See [22], 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 $g'(x)\neq0$ on $(a,b)$. If $f^{\prime}(x)/g^{\prime}(x)$ is increasing (decreasing) on $(a,b)$, then so are the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)} \quad \textit{and} \quad\frac{f(x)-f(b)}{g(x)-g(b)}. $$

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

Lemma 2.3

We have

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

for all $x\in(0, 1)$, and the above inequality is reversed for all $x\in (1, \infty)$.

Proof

Let $x>0$, $\gamma=0.577215\cdots$ be the Euler-Mascheroni constant, and

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

(2.2)

Then simple computations lead to

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

(2.3)

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

(2.4)

$$\begin{aligned}& f^{\prime\prime}(x)=\psi^{\prime}(1+x)-\frac{1}{1+x}- \frac{1}{(1+x)^{2}}. \end{aligned}$$

(2.5)

It follows from the identity

$$ \psi^{\prime}(x)=\frac{1}{x}+\frac{1}{2x^{2}}+\frac{1}{6x^{3}}- \frac {\theta}{30x^{5}} \quad(0< \theta< 1), $$

given in [23], and (2.5) that

$$ f^{\prime\prime}(x)< \frac{1}{x+1}+\frac{1}{2(x+1)^{2}}+ \frac {1}{6(x+1)^{3}}-\frac{1}{x+1}-\frac{1}{(x+1)^{2}}=-\frac{3x+2}{6(x+1)^{3}}< 0 $$

(2.6)

for all $x>0$.

Inequality (2.6) implies that $f(x)$ is strictly concave and $f^{\prime }(x)$ is strictly decreasing on the interval $(0, \infty)$.

From the concavity of $f(x)$ and monotonicity of $f^{\prime}(x)$ on the interval $(0, \infty)$, together with (2.3) and (2.4), we clearly see that

$$ f(x)>(1-x)f(0)+xf(1)=0 $$

(2.7)

for all $x\in(0, 1)$ and

$$ f(x)< 0 $$

(2.8)

for all $x\in(1, \infty)$.

Therefore, Lemma 2.3 follows easily from (2.2), (2.7), and (2.8). □

Lemma 2.4

Let $a, p>0$, $I_{p}(x)$ and $H_{f, g}$ be respectively defined by (1.2) and (2.1), and

$$ f(x)= \bigl(1-e^{-ax^{p}} \bigr)^{1/p}. $$

(2.9)

Then the following statements are true:

• (1) $H_{f, I_{p}}(\infty)=\infty$ if $a<1$;

• (2) $H_{f, I_{p}}(\infty)=-1$ if $a>1$.

Proof

From (1.2), (2.1), and (2.9) we get

$$\begin{aligned}& \begin{aligned}[b] H_{f, I_{p}}(x)&=\frac{f^{\prime}(x)}{I^{\prime}_{p}(x)}I_{p}(x)-f(x) \\ &=ax^{p-1}e^{(1-a)x^{p}} \bigl(1-e^{-ax^{p}} \bigr)^{1/p-1} \int _{0}^{x}e^{-t^{p}}\,dt- \bigl(1-e^{-ax^{p}} \bigr)^{1/p}, \end{aligned} \\& H_{f, I_{p}}(\infty)=a\Gamma \biggl(1+\frac{1}{p} \biggr)\lim _{x\rightarrow\infty} \bigl[x^{p-1}e^{(1-a)x^{p}} \bigr]-1 = \textstyle\begin{cases} \infty, & a< 1, \\ -1, & a>1. \end{cases}\displaystyle \end{aligned}$$

 □

3 Main results

Theorem 3.1

Let $a, p>0$ with $p\neq1$, and $R(a, p; x)$ be defined by (1.5). Then the following statements are true:

• (1) if $a\leq\min\{1, 2p/(p+1)\}$, then the function $x\rightarrow R(a, p; x)$ is strictly increasing on $(0, \infty)$;

• (2) if $a\geq\max\{1, 2p/(p+1)\}$, then the function $x\rightarrow R(a, p; x)$ is strictly decreasing on $(0, \infty)$;

• (3) if $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and $p<1$ ($p>1$), then there exists $x_{0}\in(0, \infty)$ such that the function $x\rightarrow R(a, p; x)$ is strictly decreasing (increasing) on $(0, x_{0})$ and strictly increasing (decreasing) on $(x_{0}, \infty)$.

Proof

Let $x>0$, $u=x^{p}>0$, $I_{p}(x)$ and $f(x)$ be respectively defined by (1.2) and (2.9), and

$$ h(u)=a(1-a)pue^{au}+a(p-1)e^{au}+a(a-p)u+a(1-p). $$

(3.1)

Then it follows from (1.2), (1.5), (2.9), and (3.1) that

$$\begin{aligned}& R(a, p; x)=\frac{f(x)}{I_{p}(x)}, \end{aligned}$$

(3.2)

$$\begin{aligned}& f(0)=I_{p}(0)=0, \qquad I^{\prime}_{p}(x)=e^{-x^{p}}>0, \end{aligned}$$

(3.3)

$$\begin{aligned}& \frac{f^{\prime}(x)}{I^{\prime}_{p}(x)}=ax^{p-1}e^{(1-a)x^{p}} \bigl(1-e^{-ax^{p}} \bigr)^{1/p-1} =au^{1-1/p}e^{(1-a)u} \bigl(1-e^{-au} \bigr)^{1/p-1}, \end{aligned}$$

(3.4)

$$\begin{aligned}& \begin{aligned}[b] \biggl[\frac{f^{\prime}(x)}{I^{\prime}_{p}(x)} \biggr]^{\prime}&=a \frac {d}{du} \bigl[u^{1-1/p}e^{(1-a)u} \bigl(1-e^{-au} \bigr)^{1/p-1} \bigr]\frac{du}{dx} \\ &=u^{1-2/p}e^{(1-2a)u} \bigl(1-e^{-au} \bigr)^{1/p-2}h(u), \end{aligned} \end{aligned}$$

(3.5)

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

(3.6)

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

(3.7)

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

(3.8)

$$\begin{aligned}& h^{\prime\prime }(u)=a^{2}\bigl[a(1-a)pu+2p-a(p+1) \bigr]e^{au}=a^{3}(1-a)pe^{au} \biggl[u- \frac {a(p+1)-2p}{a(1-a)p} \biggr]. \end{aligned}$$

(3.9)

We divide the proof into four cases.

Case 1: $a\leq\min\{1, 2p/(p+1)\}$. From $p\neq1$ and (3.9) we know that $h^{\prime}(u)$ is strictly increasing on $(0, \infty)$. Then (3.5), (3.6), and (3.8) lead to the conclusion that $f^{\prime }(x)/I^{\prime}_{p}(x)$ is strictly increasing on $(0, \infty)$. Therefore, $R(a, p; x)$ is strictly increasing on $(0, \infty)$, as follows from Lemma 2.2, (3.2), and (3.3) together with the monotonicity of $f^{\prime}(x)/I^{\prime}_{p}(x)$.

Case 2: $a\geq\max\{1, 2p/(p+1)\}$. From $p\neq1$ and (3.9) we know that $h^{\prime}(u)$ is strictly decreasing on $(0, \infty)$. Then (3.5), (3.6), and (3.8) lead to the conclusion that $f^{\prime }(x)/I^{\prime}_{p}(x)$ is strictly decreasing on $(0, \infty)$. Therefore, $R(a, p; x)$ is strictly decreasing on $(0, \infty)$, as follows from Lemma 2.2, (3.2), and (3.3) together with the monotonicity of $f^{\prime}(x)/I^{\prime}_{p}(x)$.

Case 3: $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and $p<1$. Then we clearly see that $2p/(p+1)< a<1$, and (3.1) and (3.7) lead to

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

(3.10)

$$\begin{aligned}& h^{\prime}(\infty)=\infty. \end{aligned}$$

(3.11)

Let

$$ u_{0}=\frac{a(p+1)-2p}{a(1-a)p}. $$

(3.12)

Then we clearly see that $u_{0}\in(0, \infty)$, and (3.9) leads to the conclusion that $h^{\prime}(u)$ is strictly decreasing on $(0, u_{0})$ and strictly increasing on $(u_{0}, \infty)$.

It follows from (3.8) and (3.11) together with the piecewise monotonicity of $h^{\prime}(u)$ that there exists $u_{1}\in(0, \infty )$ such that $h(u)$ is strictly decreasing on $(0, u_{1})$ and strictly increasing on $(u_{1}, \infty)$. From (3.5), (3.6), and (3.10) together with the piecewise monotonicity of $h(u)$ we know that there exists $\lambda\in(0, \infty)$ such that $f^{\prime}(x)/I^{\prime}_{p}(x)$ is strictly decreasing on $(0, \lambda)$ and strictly increasing on $(\lambda, \infty)$.

Therefore, there exists $x_{0}\in(0, \infty)$ such that the function $x\rightarrow R(a, p; x)$ is strictly decreasing on $(0, x_{0})$ and strictly increasing on $(x_{0}, \infty)$, as follows from Lemma 2.1(1), Lemma 2.4(1), (3.2), (3.3), and the piecewise monotonicity of $f^{\prime }(x)/I^{\prime}_{p}(x)$.

Case 4: $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and $p>1$. Then we clearly see that $1< a<2p/(p+1)$, and (3.1) and (3.7) lead to

$$\begin{aligned}& h(\infty)=-\infty, \end{aligned}$$

(3.13)

$$\begin{aligned}& h^{\prime}(\infty)=-\infty. \end{aligned}$$

(3.14)

Let $u_{0}\in(0, \infty)$ be defined by (3.12). Then from (3.9) we clearly see that $h^{\prime}(u)$ is strictly increasing on $(0, u_{0})$ and strictly decreasing on $(u_{0}, \infty)$. It follows from (3.5), (3.6), (3.8), (3.13), (3.14), and the piecewise monotonicity of $h^{\prime}(u)$ that there exists $\mu\in(0, \infty)$ such that $f^{\prime}(x)/I^{\prime }_{p}(x)$ is strictly increasing on $(0, \mu)$ and strictly decreasing on $(\mu, \infty)$.

Therefore, there exists $x_{0}\in(0, \infty)$ such that the function $x\rightarrow R(a, p; x)$ is strictly increasing on $(0, x_{0})$ and strictly decreasing on $(x_{0}, \infty)$, as follows from Lemma 2.1(2), Lemma 2.4(2), (3.2), (3.3), and the piecewise monotonicity of $f^{\prime }(x)/I^{\prime}_{p}(x)$. □

Remark 3.2

Let $R(a, p; x)$ be defined by (1.5). Then from (1.2), (2.9), and (3.2)-(3.4) we clearly see that

$$\begin{aligned}& R(a, p; \infty)=\frac{1}{\Gamma (1+\frac{1}{p} )}, \\& R\bigl(a, p; 0^{+}\bigr)=\lim_{x\rightarrow0^{+}} \frac{f^{\prime}(x)}{I^{\prime }_{p}(x)}=a\lim_{u\rightarrow0^{+}} \biggl(\frac{1-e^{-au}}{u} \biggr)^{1/p-1}=a^{1/p}. \end{aligned}$$

From Theorem 3.1 and Remark 3.2 we immediately get Corollary 3.3.

Corollary 3.3

Let $a, p>0$ with $p\neq1$, $I_{p}(x)$ and $R(a, p; x)$ be respectively defined by (1.2) and (1.5), and $x_{0}$ be the unique solution of the equation $d[R(a, p; x)]/dx=0$ on the interval $(0, \infty)$ in the case of $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$. Then the following statements are true:

(1) if $a\leq\min\{1, 2p/(p+1)\}$, then we have the double inequality

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

for all $x>0$;

(2) if $a\geq\max\{1, 2p/(p+1)\}$, then we have the double inequality

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

for all $x>0$;

(3) if $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and $p<1$, then we have the double inequality

$$ \min \biggl\{ \Gamma \biggl(1+\frac{1}{p} \biggr), \frac{1}{a^{1/p}} \biggr\} \bigl(1-e^{-ax^{p}} \bigr)^{1/p}< I_{p}(x)\leq \frac{1}{R(a, p; x_{0})} \bigl(1-e^{-ax^{p}} \bigr)^{1/p} $$

for all $x>0$;

(4) if $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and $p>1$, then we have the double inequality

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

for all $x>0$.

Next, we prove Alzer’s inequality (1.4) by using Theorem 3.1, Remark 3.2, and Corollary 3.3.

Theorem 3.4

Let $a, b, p>0$ with $p\neq1$, $a_{0}=\min\{1, \Gamma^{-p}(1+1/p)\}$, $b_{0}=\max\{1, \Gamma^{-p}(1+1/p)\}$, and $I_{p}(x)$ be defined by (1.2). Then the double inequality

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

holds for all $x>0$ if and only if $a\leq a_{0}$ and $b\geq b_{0}$.

Proof

Let $R(a, p; x)$ be defined by (1.5). Then we divide the proof into four steps.

Step 1: $p<1$. We prove that the inequality

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

(3.15)

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

From $p\in(0, 1)$ and Lemma 2.3 we clearly see that

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

(3.16)

If inequality (3.15) holds for all $x>0$, then (1.5) and Remark 3.2, together with (3.16), lead to the conclusion that

$$\begin{aligned}& R(a, p; x)< \Gamma^{-1} \biggl(1+\frac{1}{p} \biggr),\qquad a^{1/p}=R\bigl(a, p; 0^{+}\bigr)\leq\Gamma^{-1} \biggl(1+\frac{1}{p} \biggr), \\& a\leq\Gamma^{-p} \biggl(1+\frac{1}{p} \biggr)=a_{0}. \end{aligned}$$

Next, we prove inequality (3.15) for all $x>0$ if $a\leq a_{0}$. We divide the proof into two cases.

Case 1.1: $a\leq2p/(1+p)$. Then from (3.16) and Corollary 3.3(1) we clearly see that $a\leq\min\{1, 2p/(1+p)\}$ and inequality (3.15) holds for all $x>0$.

Case 1.2: $2p/(1+p)< a\leq a_{0}=\Gamma^{-p}(1+1/p)$. Then (3.16) and Corollary 3.3(3) lead to the conclusion that $\min\{1, 2p/(1+p)\}< a<\max \{1, 2p/(1+p)\}$ and

$$ I_{p}(x)>\min \biggl\{ \Gamma \biggl(1+\frac{1}{p} \biggr), \frac {1}{a^{1/p}} \biggr\} \bigl(1-e^{-ax^{p}} \bigr)^{1/p} =\Gamma \biggl(1+\frac{1}{p} \biggr) \bigl(1-e^{-ax^{p}} \bigr)^{1/p} $$

for all $x>0$.

Step 2: $p>1$. We prove that inequality (3.15) holds for all $x>0$ if and only if $a\leq a_{0}$.

From $p\in(0, 1)$ and Lemma 2.3 we clearly see that

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

(3.17)

If $a\leq a_{0}$, then inequality (3.17) and Corollary 3.3(1) lead to the conclusion that $a\leq\min\{1, 2p/(1+p)\}$ and inequality (3.15) holds for all $x>0$.

Next, we prove by contradiction that $a\leq a_{0}$ if inequality (3.15) holds for all $x>0$. We divide the proof into two cases.

Case 2.1: $a\geq2p/(p+1)$. Then (3.17) and Corollary 3.3(2) lead to the conclusion that $a\geq\max\{1, 2p/(p+1)\}$ and the opposite direction inequality of (3.15) holds for all $x>0$.

Case 2.2: $1=a_{0}< a<2p/(p+1)$. Then inequality (3.17) and Theorem 3.1(3), together with Remark 3.2, lead to the conclusion that $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and there exists $x_{0}\in(0, \infty )$ such that

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

for $x\in(x_{0}, \infty)$, which contradicts with (3.15).

Step 3: $p<1$. We prove that the inequality

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

(3.18)

holds for all $x>0$ if and only if $a\geq b_{0}$.

From $p<1$ and Lemma 2.3 we clearly see that

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

(3.19)

If $a\geq b_{0}$, then (3.19) and Corollary 3.3(2) lead to the conclusion that $a\geq\max\{1, 2p/ (p+1)\}$ and inequality (3.18) holds for all $x>0$.

Next, we prove by contradiction that $a\geq b_{0}$ if inequality (3.18) holds for all $x>0$. We divide the proof into two cases.

Case 3.1: $a\leq2p/(1+p)$. Then (3.19) and Corollary 3.3(1) lead to the conclusion that $a\leq\min\{1, 2p/(p+1)\}$ and the opposite direction inequality of (3.18) holds for all $x>0$.

Case 3.2: $2p/(p+1)< a< b_{0}=1$. Then (3.19) and Theorem 3.1(3), together with Remark 3.2, lead to the conclusion that $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and there exists $x_{0}$ such that the opposite direction inequality of (3.18) holds for $x\in(x_{0}, \infty)$.

Step 4: $p>1$. We prove that inequality (3.18) holds for all $x>0$ if and only if $a\geq b_{0}$.

From $p>1$ and Lemma 2.3 we clearly see that

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

(3.20)

If inequality (3.18) holds for all $x>0$, then (1.2), (1.5), Remark 3.2, and (3.20) lead to

$$\begin{aligned}& \Gamma \biggl(1+\frac{1}{p} \biggr)R(a, p; x)>1,\qquad \Gamma \biggl(1+ \frac {1}{p} \biggr)R\bigl(a, p; 0^{+}\bigr)=a^{1/p}\Gamma \biggl(1+\frac{1}{p} \biggr)\geq1, \\& a\geq\Gamma^{-p} \biggl(1+\frac{1}{p} \biggr)=b_{0}. \end{aligned}$$

Next, we prove that inequality (3.18) holds for all $x>0$ if $a\geq b_{0}$. We divide the proof into two cases.

Case 4.1: $a\geq2p/(p+1)$. Then (3.20) and Corollary 3.3(2) lead to the conclusion that $a\geq\max\{1, 2p/(p+1)\}$ and inequality (3.18) holds for all $x>0$.

Case 4.2: $b_{0}\leq a<2p/(p+1)$. Then (3.20) and Corollary 3.3(4) lead to the conclusion that $\min\{1, 2p/(p+1)\}< a<\max\{1, 2p/(p+1)\}$ and

$$ I_{p}(x)< \max \biggl\{ \frac{1}{a^{1/p}}, \Gamma \biggl(1+ \frac{1}{p} \biggr) \biggr\} \bigl(1-e^{-ax^{p}} \bigr)^{1/p}= \Gamma \biggl(1+\frac {1}{p} \biggr) \bigl(1-e^{-ax^{p}} \bigr)^{1/p} $$

for all $x>0$. □

Let $q=1/p$, and $u=x^{p}$. Then (1.1) and (1.2), together with Corollary 3.3, lead to Corollary 3.5.

Corollary 3.5

Let $a>0$, $q>0$ with $q\neq1$, and $u_{0}$ be the unique solution of the equation

$$ \frac{d}{du} \biggl[\frac{ (1-e^{-au} )^{q}}{\Gamma(q)-\Gamma(q, u)} \biggr]=0 $$

on the interval $(0, \infty)$ in the case of $\min\{1, 2/(q+1)\}< a<\max \{1, 2/(q+1)\}$. Then the following statements are true:

(1) if $a\leq\min\{1, 2/(q+1)\}$, then we have the double inequality

$$ 1-\frac{ (1-e^{-au} )^{q}}{a^{q}\Gamma(1+q)}< \frac{\Gamma(q, u)}{\Gamma(q)}< 1- \bigl(1-e^{-au} \bigr)^{q} $$

for all $u>0$;

(2) if $a\geq\max\{1, 2/(q+1)\}$, then we have the double inequality

$$ 1- \bigl(1-e^{-au} \bigr)^{q}< \frac{\Gamma(q, u)}{\Gamma(q)}< 1- \frac { (1-e^{-au} )^{q}}{a^{q}\Gamma(1+q)} $$

for all $u>0$;

(3) if $\min\{1, 2/(q+1)\}< a<\max\{1, 2/(q+1)\}$ and $q>1$, then we have the double inequality

$$ 1-\frac{\Gamma(q)-\Gamma(q, u_{0})}{\Gamma(q) (1-e^{-au_{0}} )^{q}} \bigl(1-e^{-au} \bigr)^{q}\leq \frac{\Gamma(q, u)}{\Gamma(q)} < 1-\min \biggl\{ \frac{1}{a^{q}\Gamma(1+q)}, 1 \biggr\} \bigl(1-e^{-au} \bigr)^{q} $$

for all $u>0$;

(4) if $\min\{1, 2/(q+1)\}< a<\max\{1, 2/(q+1)\}$ and $q<1$, then we have the double inequality

$$ 1-\max \biggl\{ \frac{1}{a^{q}\Gamma(1+q)}, 1 \biggr\} \bigl(1-e^{-au} \bigr)^{q}< \frac{\Gamma(q, u)}{\Gamma(q)} \leq1-\frac{\Gamma(q)-\Gamma(q, u_{0})}{\Gamma(q) (1-e^{-au_{0}} )^{q}} \bigl(1-e^{-au} \bigr)^{q} $$

for all $u>0$.

Note that

$$\begin{aligned}& \lim_{q\rightarrow0^{+}} \bigl[\Gamma(q) \bigl(1- \bigl(1-e^{-au} \bigr)^{q} \bigr) \bigr]=-\log \bigl(1-e^{-au} \bigr), \end{aligned}$$

(3.21)

$$\begin{aligned}& \lim_{q\rightarrow0^{+}} \biggl[\Gamma(q) \biggl(1-\frac{ (1-e^{-au} )^{q}}{a^{q}\Gamma(1+q)} \biggr) \biggr]=\log a-\gamma -\log \bigl(1-e^{-au} \bigr). \end{aligned}$$

(3.22)

Let $Ei(u)=\Gamma(0, u)$ be the exponential integral. Then Corollary 3.5(1) and (2), together with (3.21) and (3.22), immediately lead to Corollary 3.6.

Corollary 3.6

We have the double inequality

$$ \log a-\gamma-\log \bigl(1-e^{-au} \bigr)< Ei(u)< -\log \bigl(1-e^{-au} \bigr) $$

(3.23)

for all $u>0$ and $0< a\leq1$, and inequality (3.23) is reversed for all $u>0$ if $a\geq2$.