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. (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 [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. (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

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

 □

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. (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 \(a\geq2\).