# 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 $$b\geq\max\{1, \Gamma^{-p}(1+1/p)\}$$.

## 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 . 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  and electrical discharge in gases , respectively.

Komatu  and Pollak  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  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 .

Alzer  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).

## 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 , 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. (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. (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 , 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)

\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 , 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. (1)

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

2. (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}

□

## 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. (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. (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. (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 $$a\geq2$$.

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## Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125, and 11401191.

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Correspondence to Yu-Ming Chu.

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The authors declare that they have no competing interests.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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