Open Access

Monotonicity and inequalities involving the incomplete gamma function

Journal of Inequalities and Applications20162016:221

https://doi.org/10.1186/s13660-016-1160-7

Received: 14 June 2016

Accepted: 2 September 2016

Published: 13 September 2016

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.

Keywords

incomplete gamma function gamma function psi function

MSC

33B20 26D07 26D15

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 [112]. 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 [1319], 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. (1)

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

     
  2. (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. (1)

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

     
  2. (2)

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

     
  3. (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. (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. (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. (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. (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. (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. (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. (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.

Declarations

Acknowledgements

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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Computation Sciences, Hunan City University
(2)
Customer Service Center, State Grid Zhejiang Electric Power Research Institute
(3)
Albert Einstein College of Medicine, Yeshiva University

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