Some New Bounds for Bateman’s G-Function in Terms of the Digamma Function

Abstract

In this paper, we present some new symmetric bounds for Bateman’s G-function and its derivatives, in terms of the digamma and polygamma functions, which are better than some recent results.

1. Introduction

The digamma function is defined by [1]:

$$\psi \left(y\right)={\displaystyle \frac{d}{dy}}ln\mathrm{\Gamma }\left(y\right)={\int }_0^{\infty }\left({\displaystyle \frac{1}{t{e}^t}}-{\displaystyle \frac{e^{(1-y)t}}{e^t-1}}\right)dt,y>0$$

(1)

where $\mathrm{\Gamma }\left(y\right)$ is the ordinary Gamma function and $\psi \left(y\right)$ satisfies the functional equation [1]:

$${\psi }^{\left(h\right)}(1+y)-{\psi }^{\left(h\right)}\left(y\right)={\displaystyle \frac{{(-1)}^{h}h!}{y^{1+h}}},h\in \mathbb{N}\cup \left\{0\right\}$$

(2)

and it has the asymptotic formula:

$$\psi \left(y\right)\sim lny-{\displaystyle \frac{1}{2y}}-\sum _{k=1}^{\infty }{\displaystyle \frac{B_{2k}}{2k{y}^{2k}}},y\to \infty$$

(3)

where ${B_r}^{\prime }s$ are the Bernoulli numbers. Polygamma functions are defined by [1]:

$${\psi }^{\left(h\right)}\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{{(-1)}^{h-1}{t}^h{e}^{(1-y)t}}{e^t-1}}dt,y>0,h\in \mathbb{N}.$$

Bateman’s G-function is given by [2]:

$$G\left(y\right)=\psi \left((1+y)/2\right)-\psi \left(y/2\right),y\ne 0,-1,-2,\cdots$$

and it has the integral formula:

$${\displaystyle \frac{1}{2}}G\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{e^{(1-y)t}}{e^t+1}}dt,y>0$$

(4)

and it satisfies the following relations:

$$G(y+1)={\displaystyle \frac{2}{y}}-G\left(y\right)$$

(5)

The function $G\left(y\right)$ is used in calculating some hypergeometric functions [3]:

$${}_2{F}_1(1,y;y+1;-1)={\displaystyle \frac{y}{2}}G\left(y\right),y\ne 0,-1,-2,\cdots$$

and calculating some alternating series [4]:

$$\sum _{s=0}^{\infty }{\displaystyle \frac{{(-1)}^s}{sb+d}}={\displaystyle \frac{1}{2b}}G\left({\displaystyle \frac{d}{b}}\right),d\ne 0,-b,-2b,\cdots .$$

In 2005, Qiu and Vuorinen [5] presented the following inequality for $G\left(y\right)$:

$${\displaystyle \frac{1}{y}}+{\displaystyle \frac{6+ln(1/256)}{y^2}}<G\left(y\right)<y^{-1}+{\displaystyle \frac{y^{-2}}{2}},2y>1.$$

(6)

In 2016, the following asymptotic formula for $G\left(y\right)$ was introduced by [6]:

$$G\left(y\right)\sim {\displaystyle \frac{1}{y}}+\sum _{k=1}^{\infty }{\displaystyle \frac{(4^k-1)B_{2k}}{k{y}^{2k}}},y\to \infty$$

(7)

and the following symmetric inequality was deduced:

$${\displaystyle \frac{1}{2(y^2+3/4)}}<G\left(y\right)-{\displaystyle \frac{1}{y}}<{\displaystyle \frac{1}{2y^2}},y>0$$

(8)

which refines the lower bound of (6) for $y>\sqrt{{\displaystyle \frac{6ln4-9}{11-8ln4}}}$. In 2017, Mahmoud, Talat, and Moustafa [7] deduced the following symmetric inequality:

$$ln\left({\displaystyle \frac{1}{y+\frac{4}{e^2-4}}}+1\right)+{\displaystyle \frac{2}{y(1+y)}}<G\left(y\right)<ln\left({\displaystyle \frac{1}{1+y}}+1\right)+{\displaystyle \frac{2}{y(1+y)}},y\in (0,\infty ).$$

(9)

In 2019, Hegazi, Mahmoud, Talat, and Moustafa [8] deduced an improvement for inequality (9) by the symmetric inequality:

$$ln\left({\displaystyle \frac{1}{y+\sqrt{\frac{2}{3}}}}+1\right)+{\displaystyle \frac{\left(1+\sqrt{\frac{2}{3}}\right)}{y(y+1)}}<G\left(y\right)<ln\left({\displaystyle \frac{1}{y-\sqrt{\frac{2}{3}}}}+1\right)+{\displaystyle \frac{\left(1-\sqrt{\frac{2}{3}}\right)}{y(y+1)}},$$

(10)

where the upper bound is valid for $y>\sqrt{{\displaystyle \frac{2}{3}}}$ and the lower bound is valid for $y>0$. Nantomah [9] studied Nielsen’s beta function, which is $\beta \left(y\right)={\displaystyle \frac{G\left(y\right)}{2}}$, and deduced the following symmetric inequalities:

$${\displaystyle \frac{1}{2y}}+{\displaystyle \frac{1}{4{(y+1)}^2}}<{\displaystyle \frac{1}{2}}G\left(y\right)<{\displaystyle \frac{1}{2y}}+{\displaystyle \frac{1}{2y^2}},y>0$$

(11)

and

$${\displaystyle \frac{-1}{y^2}}<{\displaystyle \frac{1}{2}}{G}^{\prime }\left(x\right)<{\displaystyle \frac{-1}{2y^2}},y>0.$$

(12)

In 2021, Mahmoud, Talat, Moustafa, and Agarwal [10] deduced an improvement for inequality (8) by:

$${\displaystyle \frac{1}{2y^2+1}}<G\left(y\right)-{\displaystyle \frac{1}{y}}<{\displaystyle \frac{1}{2y^2}},y>0$$

(13)

which also improves the lower and the upper bounds of inequality (11) for $y\in \left(0,\infty \right).$ In 2022, Mahmoud and Almuashi [11] presented the inequality:

$$0<G\left(y\right)+{\displaystyle \frac{1}{3y^3}}-ln\left(1+{\displaystyle \frac{1}{y+1}}\right)-{\displaystyle \frac{2}{y(y+1)}}<{\displaystyle \frac{3}{2y^4}},y>0.$$

(14)

A function S defined on an interval J is said to be completely monotonic (CM) if it has derivatives $S^{\left(i\right)}\left(y\right)$ for every $i\in \mathbb{N}\cup \left\{0\right\}$ such that

$${(-1)}^i{S}^{\left(i\right)}\left(y\right)\ge 0,y\in J;i\in \mathbb{N}\cup \left\{0\right\}.$$

The following condition [12]:

$$S\left(y\right)={\int }_0^{\infty }{e}^{-yu}d\mu \left(u\right),0<y<\infty$$

where $\mu \left(u\right)$ is a non-negative measure on $[0,\infty )$, such that the integral is convergent for $0<y<\infty$, is necessary and sufficient for the function $S\left(y\right)$ to be CM. Let $S\left(y\right)$ be a CM function for $y>0$ and suppose the notation $S(\infty )=\underset{y\to \infty }{lim}S\left(y\right)$. If $y^{\epsilon }[S\left(y\right)-S(\infty )]$ is a CM function for $y>0$ if and only if $\epsilon \in \left[0,\delta \right],$ then the number $\delta \in {\mathbb{R}}^{+}$ is called the CM degree of $S\left(y\right)$ for $y>0$ and denoted by $de{g}_{CM}^y\left[S\left(y\right)\right]=\delta .$ For more information about this topic, see [13,14,15,16,17,18,19,20].

In this paper, we will introduce some CM functions involving $G\left(y\right)$ and $\psi \left(y\right)$. As a consequence, we deduce some new symmetric bounds for the Bateman’s G-function and its derivatives. These relations refine the inequalities (10), and (12)–(14).

The following corollary [19] will be useful in the next part:

Corollary 1.

Suppose that T is a real-valued function defined on $(y_0,\infty )$, $y_0\in \mathbb{R}$ and $\underset{y\to \infty }{lim}T\left(y\right)=0$. Then, for $r\in {\mathbb{R}}^{+}$, $T\left(y\right)<0$, if $T\left(y\right)<T(y+r)$ for all $y>y_0$ and $T\left(y\right)>0$, if $T\left(y\right)>T(y+r)$ for all $y>y_0$.

2. Auxiliary Results

Lemma 1.

Suppose that

$$\alpha =\sqrt{15+2\sqrt{57}},\delta ={\displaystyle \frac{6-31\alpha +{\alpha }^3}{12}}\simeq 0.0884\mathit{and}c={\displaystyle \frac{(\sqrt{57}-7)\alpha }{6}}\simeq 0.503.$$

(15)

Then, we have

$$\psi (y+c)<ln(y-\delta )+{\displaystyle \frac{1}{2\alpha }}\left({\displaystyle \frac{1-\sqrt{\frac{2}{3}}}{y(y+1)}}+ln\left(1+{\displaystyle \frac{1}{y-\sqrt{\frac{2}{3}}}}\right)\right),y>\sqrt{{\displaystyle \frac{2}{3}}}$$

(16)

$$\psi (y-c)<ln(y+\delta -1)-{\displaystyle \frac{1}{2\alpha }}\left(ln\left(1+{\displaystyle \frac{1}{y+\sqrt{\frac{2}{3}}}}\right)+{\displaystyle \frac{1+\sqrt{\frac{2}{3}}}{y(y+1)}}\right),y\ge 9.2$$

(17)

$$\psi (y+c)<ln(y-\delta )+{\displaystyle \frac{1}{2\alpha }}\left({\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{2y^2}}\right),y\ge 0.1$$

(18)

and

$$\psi (y-c)<ln(y+\delta -1)-{\displaystyle \frac{1}{2\alpha }}\left(ln\left(1+{\displaystyle \frac{1}{y+1}}\right)+{\displaystyle \frac{2}{y(y+1)}}-{\displaystyle \frac{1}{3y^3}}\right)y\ge 3.5.$$

(19)

Proof. 

Firstly, we suppose that

$$L\left(y\right)=2\alpha \left(\psi (y+c)-ln(y-\delta )\right)-{\displaystyle \frac{\left(1-\sqrt{\frac{2}{3}}\right)}{y(y+1)}}-ln\left(1+{\displaystyle \frac{1}{y-\sqrt{\frac{2}{3}}}}\right)$$

and we use the functional relation (2) to obtain

$$L^{\prime }(1+y)-L^{\prime }\left(y\right)={\displaystyle \frac{\frac{-l\left(y\right)}{9y^2{(y+1)}^2{(2+y)}^2}}{(6-\sqrt{6}+3y)(3-\sqrt{6}+3y)(-\sqrt{6}+3y)(y-\delta +1)(y-\delta ){(y+c)}^2}},$$

where

$$\begin{array}{ccc}\hfill l\left(y\right)& =& 8\left[119-49\sqrt{6}+21\sqrt{38}-17\sqrt{57}\right]+9\left[120\sqrt{6}+\left(21-5\sqrt{57}\right)\alpha \right]y^7\hfill \\ & & -4[-1218+504\sqrt{6}-216\sqrt{38}+174\sqrt{57}+(-663+85\sqrt{57}+273\sqrt{6}\hfill \\ & & -105\sqrt{38}\left)\alpha \right]y-12[-1076+427\sqrt{6}-129\sqrt{38}+110\sqrt{57}+(-1023+145\sqrt{57}\hfill \\ & & +420\sqrt{6}-180\sqrt{38}\left)\alpha \right]y^2-6[-3912+1484\sqrt{6}-204\sqrt{38}+204\sqrt{57}+(-3414\hfill \\ & & +550\sqrt{57}+1257\sqrt{6}-645\sqrt{38}\left)\alpha \right]y^3-36[-671+205\sqrt{6}-12\sqrt{38}+17\sqrt{57}\hfill \\ & & +\left(-450+85\sqrt{57}+115\sqrt{6}-85\sqrt{38}\right)\alpha ]y^4-18[-621-17\sqrt{6}-3\sqrt{38}+9\sqrt{57}\hfill \\ & & +\left(-387+85\sqrt{57}+24\sqrt{6}-60\sqrt{38}\right)\alpha ]y^5+9[194+360\sqrt{6}-2\sqrt{57}+(189\hfill \\ & & -45\sqrt{57}+19\sqrt{6}+15\sqrt{38}\left)\alpha \right]y^6>0,y>0.\hfill \end{array}$$

Then, $L^{\prime }(1+y)<L^{\prime }\left(y\right)$ for $y>\sqrt{{\displaystyle \frac{2}{3}}}$, and by using the asymptotic Formula (3), we obtain $\underset{y\to \infty }{lim}{L}^{\prime }\left(y\right)=0$. By using Corollary 1, we obtain that $L^{\prime }\left(y\right)>0$ for all $y>\sqrt{{\displaystyle \frac{2}{3}}}$; hence, $L\left(y\right)$ is increasing on $\left(\sqrt{{\displaystyle \frac{2}{3}}},\infty \right)$ with $\underset{y\to \infty }{lim}L\left(y\right)=0.$ Then, $L\left(y\right)<0$ for all $y\in \left(\sqrt{{\displaystyle \frac{2}{3}}},\infty \right)$, and this gives (16). Secondly, we write $M\left(y\right)=-2\alpha \left(\psi (y-c)-ln(y+\delta -1)\right)-ln\left(1+{\displaystyle \frac{1}{y+\sqrt{\frac{2}{3}}}}\right)-{\displaystyle \frac{\left(1+\sqrt{\frac{2}{3}}\right)}{y(y+1)}}$ and then

$$M^{\prime }(1+y)-M^{\prime }\left(y\right)={\displaystyle \frac{\frac{w\left(y-9.2\right)}{9y^2{(y+1)}^2}}{(y+\delta )(y+\delta -1){(y-c)}^2(6+\sqrt{6}+3x)(3+\sqrt{6}+3y)(\sqrt{6}+3y){(2+y)}^2}},$$

where

$$\begin{array}{ccc}\hfill 78125w\left(y\right)& =& -8[-26151619787463\alpha +5(3882587994557-15525870154997\sqrt{6}\hfill \\ & & -15129123915\sqrt{38}-52826589311\sqrt{57}+(1215011554153\sqrt{57}+76652769507\sqrt{6}\hfill \\ & & +347969850045\sqrt{38}\left)\alpha \right)]-20[-6954198232047\alpha +5(915850188594\hfill \\ & & -4571027369148\sqrt{6}-2828108700\sqrt{38}-11845995102\sqrt{57}+(325284476657\sqrt{57}\hfill \\ & & +29096648769\sqrt{6}+80175624015\sqrt{38}\left)\alpha \right)]y\hfill \\ & & -300[-132141971037\alpha +5(14978811620-95903537753\sqrt{6}-35167125\sqrt{38}\hfill \\ & & -184460210\sqrt{57}+\left(6216917347\sqrt{57}+661244295\sqrt{6}+1280195325\sqrt{38}\right)\alpha \left)\right]y^2\hfill \\ & & -3750[-1674854718\alpha +5(156521904-1338475132\sqrt{6}-261780\sqrt{38}\hfill \\ & & -1838292\sqrt{57}+\left(79172758\sqrt{57}+8906559\sqrt{6}+13054365\sqrt{38}\right)\alpha \left)\right]y^3\hfill \\ & & -112500[-5310588\alpha +5(382648-4660362\sqrt{6}-405\sqrt{38}-4294\sqrt{57}\hfill \\ & & +\left(251953\sqrt{57}+26818\sqrt{6}+31130\sqrt{38}\right)\alpha \left)\right]y^4\hfill \\ & & -56250[-606663\alpha +5(29877-582997\sqrt{6}-15\sqrt{38}-321\sqrt{57}+(28853\sqrt{57}\hfill \\ & & +2502\sqrt{6}+2370\sqrt{38}\left)\alpha \right)]y^5-140625[-7707\alpha +5(194-8088\sqrt{6}-2\sqrt{57}\hfill \\ & & +\left(367\sqrt{57}+19\sqrt{6}+15\sqrt{38}\right)\alpha \left)\right]y^6+703125\left[120\sqrt{6}+\left(21-5\sqrt{57}\right)\alpha \right]y^7\hfill \\ & >& 0,y\in [0,\infty ).\hfill \end{array}$$

Then, $M^{\prime }(1+y)>M^{\prime }\left(y\right)$ for $y\ge 9.2$ and $\underset{y\to \infty }{lim}{M}^{\prime }\left(y\right)=0$. Then, in the same way, we have $M^{\prime }\left(y\right)<0$ for all $y\ge 9.2$; hence, $M\left(y\right)$ is strictly decreasing on $\left[9.2,\infty \right)$ with $\underset{x\to \infty }{lim}M\left(y\right)=0$ and then $M\left(y\right)>0$ for all $y\in \left[9.2,\infty \right)$, and this gives (17). Thirdly, we suppose that $N\left(y\right)=2\alpha \left(\psi (y+c)-ln(y-\delta )\right)-{\displaystyle \frac{1}{y}}-{\displaystyle \frac{1}{2y^2}}$ and then

$$N^{\prime }(1+y)-N^{\prime }\left(y\right)={\displaystyle \frac{H\left(y-\frac{1}{10}\right)}{54y^3{(y+1)}^3(y-\delta +1)(y-\delta ){(y+c)}^2}},$$

where

$$\begin{array}{ccc}\hfill 5000H\left(y\right)& =& -28401\alpha +5\left[-21251+\left(2924+731\alpha \right)\sqrt{57}\right]\hfill \\ & & +20\left[-19212\alpha +5\left(-4600+\left(526+497\alpha \right)\sqrt{57}\right)\right]y\hfill \\ & & +200\left[-5685+330\sqrt{57}+\left(-5583+740\sqrt{57}\right)\alpha \right]y^2\hfill \\ & & +2000\left[-880+10\sqrt{57}+\left(-582+85\sqrt{57}\right)\alpha \right]y^3\hfill \\ & & +{10}^4\left[-21\alpha +5\left(-27+\sqrt{57}\alpha \right)\right]y^4<0,y\ge 0.\hfill \end{array}$$

Then, $N^{\prime }(y+1)<N^{\prime }\left(y\right)$ for $y\ge 0.1$ and $\underset{y\to \infty }{lim}{N}^{\prime }\left(y\right)=0$. Then, $N^{\prime }\left(y\right)>0$ for all $y\ge 0.1$; hence, $N\left(y\right)$ is increasing on $\left[0.1,\infty \right)$ with $\underset{y\to \infty }{lim}N\left(y\right)=0.$ Then, $N\left(y\right)<0$ for all $y\in \left[0.1,\infty \right)$ and this gives (18). Finally, we set $U\left(y\right)=-2\alpha \left(\psi (y-c)-ln(y+\delta -1)\right)-ln\left(1+{\displaystyle \frac{1}{y+1}}\right)-{\displaystyle \frac{2}{y(y+1)}}+{\displaystyle \frac{1}{3y^3}}$ and then

$$U^{\prime }(1+y)-U^{\prime }\left(y\right)={\displaystyle \frac{A\left(y-3.5\right)}{54y^4{(y+1)}^4{(y+2)}^2(y+3)(y+\delta )(y+\delta -1){(y-c)}^2}},$$

where

$$\begin{array}{ccc}\hfill 256A\left(y\right)& =& 81602651354+248146684\sqrt{57}+\left(1182278433-2171283485\sqrt{57}\right)\alpha \hfill \\ & & -2\left[-95218106718-220170408\sqrt{57}+\left(-1312734369+2236674425\sqrt{57}\right)\alpha \right)]y\hfill \\ & & -48\left[-4114280651-7019278\sqrt{57}+\left(-53973180+84353100\sqrt{57}\right)\alpha \right)]y^2\hfill \\ & & -96\left[-1243570661-1513678\sqrt{57}+\left(-15935862+22018780\sqrt{57}\right)\alpha \right)]y^3\hfill \\ & & -96\left[-482731332-402816\sqrt{57}+\left(-6323583+7308835\sqrt{57}\right)\alpha \right)]y^4\hfill \\ & & -64\left[-187105772-101692\sqrt{57}+\left(-2639937+2400365\sqrt{57}\right)\alpha \right)]y^5\hfill \\ & & -256\left[-8043111-2646\sqrt{57}+\left(-127950+86710\sqrt{57}\right)\alpha \right)]y^6\hfill \\ & & -1536\left[-147859-26\sqrt{57}+\left(-2736+1330\sqrt{57}\right)\alpha \right)]y^7\hfill \\ & & -256\left[-56942-4\sqrt{57}+\left(-1245+425\sqrt{57}\right)\alpha \right)]y^8\hfill \\ & & -512\left[-810+\left(-21+5\sqrt{57}\right)\alpha \right)]y^9>0,y\in [0,\infty ).\hfill \end{array}$$

Then, $U^{\prime }(1+y)>U^{\prime }\left(y\right)$ for $y\ge 3.5$ and $\underset{y\to \infty }{lim}{U}^{\prime }\left(y\right)=0$. Then, $U^{\prime }\left(y\right)<0$ for all $y\ge 3.5$. Hence, $U\left(y\right)$ is strictly decreasing on $\left[3.5,\infty \right)$ with $\underset{x\to \infty }{lim}U\left(y\right)=0$, and then $U\left(y\right)>0$ for all $y\in \left[3.5,\infty \right)$; this gives (19). □

Lemma 2.

For the values of $\alpha ,\delta$, and c in (15), we have

$${\psi }^{\prime }(y-c)>{\displaystyle \frac{1}{y+\delta -1}}+{\displaystyle \frac{1}{2\alpha y^2}}\forall y\ge 2.5$$

(20)

and

$${\psi }^{\prime }(y+c)>{\displaystyle \frac{1}{y-\delta }}-{\displaystyle \frac{1}{\alpha y^2}}\forall y\ge 0.85.$$

(21)

Proof. 

Letting $V\left(y\right)=-2\alpha \left({\psi }^{\prime }(y-c)-{\displaystyle \frac{1}{y+\delta -1}}\right)+{\displaystyle \frac{1}{y^2}}$ and using (2), we have

$$V(1+y)-V\left(y\right)={\displaystyle \frac{-v(y-2.5)}{54y^2{(y+1)}^2(\delta +y)(y+\delta -1){(y-c)}^2}},$$

where

$$\begin{array}{ccc}\hfill 8v\left(y\right)& =& -47253-96\sqrt{57}+\left(3870+600\sqrt{57}\right)\alpha +432\left(-30+\alpha \right)y^3-1296y^4\hfill \\ & & +4\left(-19654-8\sqrt{57}+\left(1527+110\sqrt{57}\right)\alpha \right)y+8\left(-6021+\left(363+10\sqrt{57}\right)\alpha \right)y^2\hfill \\ & <& 0,y\ge 0.\hfill \end{array}$$

Then, $V(y+1)>V\left(y\right)$ for $y\ge 2.5$ and $\underset{y\to \infty }{lim}V\left(y\right)=0$. By using Corollary 1, we deduce that $V\left(y\right)<0$ for all $y\ge 2.5$, and this gives (20). Next, assuming that $K\left(y\right)=\alpha \left({\psi }^{\prime }(y+c)-{\displaystyle \frac{1}{y-\delta }}\right)+{\displaystyle \frac{1}{y^2}}$ and then

$$K(y+1)-K\left(y\right)={\displaystyle \frac{k(y-0.85)}{54y^2{(y+1)}^2(y-\delta +1)(y-\delta ){(c+y)}^2}},$$

where

$$\begin{array}{ccc}\hfill 16\left({10}^5\right)k\left(y\right)& =& 9\left(-18799911+960000\sqrt{57}+\left(-13305220+2040000\sqrt{57}\right)\alpha \right)\hfill \\ & & +100\left(-3922171+64000\sqrt{57}+\left(-2339232+352000\sqrt{57}\right)\alpha \right)y\hfill \\ & & +4000\left(-115695+\left(-36852+4000\sqrt{57}\right)\alpha \right)y^2-2160000\left(193+32\alpha \right)y^3\hfill \\ & & -7200000\left(39+4\alpha \right)y^4-86400000y^5<0,y\in \left[0,\infty \right).\hfill \end{array}$$

Then, $K(y+1)<K\left(y\right)$ for $y\in \left[0.85,\infty \right)$ and $\underset{y\to \infty }{lim}K\left(y\right)=0$. In the same way, we have $K\left(y\right)>0$ for all $y\ge 0.85$, and this gives (21). □

3. Some CM Functions Involving $\mathit{G}\left(\mathit{y}\right)$ and $\mathsf{\psi }\left(\mathit{y}\right)$

In 2010, Mortici [21] presented a Lemma, which is considered as a powerful tool to construct asymptotic expansions and to measure the rate of convergence. This Lemma has been applied successfully to produce several approximations and inequalities in several papers such as [8]. We used this Lemma in producing the best approximations of Bateman’s G-function in terms of the digamma and logarithmic functions. Now, we will prove the complete monotonicity of some functions involving the function $G\left(y\right)$ depending on this approximation.

Theorem 1.

For the values of $\alpha ,\delta$, and c in (15) and $\beta \in {\mathbb{R}}^{+}$, we have that the function

$$W_{\delta ,\alpha ,\beta }\left(y\right)=2\alpha \left(\psi (y+\beta +\delta )-lny\right)-G(\delta +y)$$

is CM on $(0,\infty )$ if and only if $\beta \ge c$ with $0\le {deg}_{CM}^y\left[W_{\delta ,\alpha ,c}\left(y\right)\right]<1$. Also, the function $-W_{1-\delta ,-\alpha ,-\beta }\left(y\right)$ is CM on $(0,\infty )$ if and only if $\beta \le c$ with $0\le {deg}_{CM}^y\left[-W_{1-\delta ,-\alpha ,-\beta }\left(y\right)\right]<1$

Proof. 

Using (1), (4), and the identity (see [1]) $ln\left({\displaystyle \frac{f}{d}}\right)={\int }_0^{\infty }{\displaystyle \frac{e^{-dt}-e^{-ft}}{t}}dt,f$, $d>0$, we have

$$W_{\delta ,\alpha ,\beta }\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{2e^{-yt}}{t(e^t-1)(e^t+1)}}\phi \left(t\right)dt,$$

where

$$\phi \left(t\right)=\alpha \left(e^{2t}-1\right)-\left[e^{(2-\delta )t}-e^{(1-\delta )t}+\alpha \left(e^{(2-\delta -\beta )t}+e^{(1-\delta -\beta )t}\right)\right]t.$$

Set $\beta \ge c$; then, we obtain

$$\phi \left(t\right)\ge \sum _{h=21}^{\infty }{\displaystyle \frac{f_h{t}^{h+1}}{(h+1)!}}+P\left(t\right),$$

where

$$\begin{array}{ccc}\hfill P\left(t\right)& =& {\displaystyle \frac{\alpha }{27(5!)}}\left[-21+5\sqrt{57}\right]t^5+{\displaystyle \frac{2}{9(6!)}}\left[-8+\left(-21+5\sqrt{57}\right)\alpha \right]t^6\hfill \\ & & +{\displaystyle \frac{1}{27(7!)}}\left[-336+\left(-1367+231\sqrt{57}\right)\alpha \right]t^7\hfill \\ & & -{\displaystyle \frac{8}{243(8!)}}\left[26\left(9+7\sqrt{57}\right)+\left(9657-1449\sqrt{57}\right)\alpha \right]t^8\hfill \\ & & +{\displaystyle \frac{4}{81(9!)}}\left[4644-1092\sqrt{57}+\left(-19245+2852\sqrt{57}\right)\alpha \right]t^9\hfill \\ & & +{\displaystyle \frac{8}{81(10!)}}\left[-4947\alpha +5\left(2519+\left(-473+206\alpha \right)\sqrt{57}\right)\right]t^{10}\hfill \\ & & -{\displaystyle \frac{1}{2187(11!)}}\left[-6745464+1294920\sqrt{57}+\left(-16658275+2016443\sqrt{57}\right)\alpha \right]t^{11}\hfill \\ & & -{\displaystyle \frac{4}{2187(12!)}}\left[-1061082+391738\sqrt{57}+\left(-19431345+2466321\sqrt{57}\right)\alpha \right]t^{12}\hfill \\ & & -{\displaystyle \frac{1}{6561(13!)}}\left[99490248-5977608\sqrt{57}+\left(-546719841+69577313\sqrt{57}\right)\alpha \right]t^{13}\hfill \\ & & -{\displaystyle \frac{2}{19683(14!)}}\left[-835791633\alpha +7\left(96563992+\left(-9232752+14503047\alpha \right)\sqrt{57}\right)\right]t^{14}\hfill \\ & & +{\displaystyle \frac{2}{19683(15!)}}\left[-1589956133\alpha +5\left(-306610032+\left(29439144+46020731\alpha \right)\sqrt{57}\right)\right]t^{15}\hfill \\ & & +{\displaystyle \frac{64}{59049(16!)}}\left[-154541715+8974589\sqrt{57}+\left(-863959089+118275108\sqrt{57}\right)\alpha \right]t^{16}\hfill \\ & & +{\displaystyle \frac{1}{531441(17!)}}[-1102134397287\alpha +85(1771598267\sqrt{57}\alpha -576(-2283069\hfill \\ & & +594071\sqrt{57}\left)\right)]t^{17}+{\displaystyle \frac{2}{59049(18!)}}[40863209452-7099206900\sqrt{57}+(-47394762375\hfill \\ & & +6806308343\sqrt{57}\left)\alpha \right]t^{18}-{\displaystyle \frac{1}{177147(19!)}}[-1035579247861\alpha +57(-9234069080\hfill \\ & & +\left(1596142440+2292009741\alpha \right)\sqrt{57}\left)\right]t^{19}-{\displaystyle \frac{4}{4782969(20!)}}[-1876198637070\hfill \\ & & +547139251390\sqrt{57}+\left(-32385689139321+4199481091305\sqrt{57}\right)\alpha ]t^{20}\hfill \\ & & -{\displaystyle \frac{4}{4782969(21!)}}[14020285258098-1240835687826\sqrt{57}+(-66540894078516\hfill \\ & & +8629702072789\sqrt{57}\left)\alpha \right]t^{21}\hfill \\ & >& 0,t>0\hfill \end{array}$$

and

$$f_h=2^{h+1}\alpha -(h+1)\left[{\left(2-\delta \right)}^h-{\left(1-\delta \right)}^h+\alpha \left({\left(2-\delta -c\right)}^h+{\left(1-\delta -c\right)}^h\right)\right],$$

which leads to

$$f_h>2^{h+1}\alpha -(h+1)\left[{\left(2-\delta \right)}^h+\alpha \left({\left(2-\delta -c\right)}^h+{\left(1-\delta -c\right)}^h\right)\right].$$

Then,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{f_h}{(h+1){\left(2-\delta \right)}^h}}& >& 2\alpha m_h-1-\alpha \left({\left({\displaystyle \frac{2-\delta -c}{2-\delta }}\right)}^h+{\left({\displaystyle \frac{1-\delta -c}{2-\delta }}\right)}^h\right),\hfill \end{array}$$

where $m_h={\displaystyle \frac{{\left(1+\frac{\delta }{2-\delta }\right)}^h}{h+1}}$ is increasing for $h\ge 21$ because

$$m_{h+1}-m_h={\displaystyle \frac{\delta {\left(1+\frac{\delta }{2-\delta }\right)}^h}{(h+1)(h+2)(2-\delta )}}\left(h-{\displaystyle \frac{2(1-\delta )}{\delta }}\right)>0,h\ge 21.$$

Also, $-\alpha \left({\left({\displaystyle \frac{2-\delta -c}{2-\delta }}\right)}^h+{\left({\displaystyle \frac{1-\delta -c}{2-\delta }}\right)}^h\right)$ is increasing for $h\ge 21$ and, consequently, for $h\ge 21$ we obtain

$${\displaystyle \frac{f_h}{(h+1){\left(2-\delta \right)}^h}}>{\displaystyle \frac{\alpha {\left(1+\frac{\delta }{2-\delta }\right)}^{21}}{11}}-1-\alpha \left({\left(1-{\displaystyle \frac{c}{2-\delta }}\right)}^{21}+{\left({\displaystyle \frac{1-\delta -c}{2-\delta }}\right)}^{21}\right)\simeq 0.279>0.$$

We thus obtain that $W_{\delta ,\alpha ,\beta }\left(y\right)$ is CM on $(0,\infty )$ for $\beta \ge c.$ On the contrary, if $W_{\delta ,\alpha ,\beta }\left(y\right)$ is CM on $y>0$, then we have for $y>0:$

$$y{W}_{\delta ,\alpha ,\beta }\left(y\right)=2\alpha y\left(\psi (y+\delta +\beta )-lny\right)-yG(\delta +y)>0.$$

(22)

Using the asymptotic Formula (7), we have $\underset{y\to \infty }{lim}yG(y+\delta )=1$, and by using the asymptotic Formula (3), we have $\underset{y\to \infty }{lim}y\left[\psi (y+\delta +\beta )-lny\right]=\delta +\beta -{\displaystyle \frac{1}{2}}.$ From (22), we conclude that $-1+2\alpha \left(\delta +\beta -{\displaystyle \frac{1}{2}}\right)\ge 0$ and then $\beta \ge c$. Using the asymptotic Formulas (3) and (7), we have $W_{\delta ,\alpha ,c}(\infty )=\underset{y\to \infty }{lim}{W}_{\delta ,\alpha ,c}\left(y\right)=0.$ Now,

$$y{W}_{\delta ,\alpha ,c}\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{2e^{-yt}\mathrm{\Omega }\left(t\right)}{t^2{(e^{2t}-1)}^2}}dt,$$

where

$$\mathrm{\Omega }\left(t\right)=-\alpha {\left(e^{2t}-1\right)}^2+[(2-\delta )e^{(2-\delta )t}-(1-\delta )e^{(1-\delta )t}-(1+\delta )e^{(3-\delta )t}+\delta e^{(4-\delta )t}$$

$$+\alpha \left((1-\delta -c)e^{(1-\delta -c)t}+(2-\delta -c)e^{(2-\delta -c)t}+(1+\delta +c)e^{(3-\delta -c)t}+(\delta +c)e^{(4-\delta -c)t}\right)]t^2$$

with $\mathrm{\Omega }\left(5\right)\simeq 5.56\left({10}^7\right)$ and $\mathrm{\Omega }\left(6\right)\simeq -7.524\left({10}^9\right).$ Then, $y{W}_{\delta ,\alpha ,c}\left(y\right)$ is not CM function; hence, $de{g}_{CM}^y\left[W_{\delta ,\alpha ,c}\left(y\right)\right]<1.$ Now, for $\beta \le c,$ we have $W_{1-\delta ,-\alpha ,-\beta }\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{2e^{-yt}}{t(e^t-1)(e^t+1)}}\varphi \left(t\right)dt,$ where

$$\begin{array}{ccc}\hfill \varphi \left(t\right)& \le & \alpha \left(1-e^{2t}\right)+\left[-e^{(1+\delta )t}+e^{\delta t}+\alpha \left(e^{(\delta +c+1)t}+e^{(\delta +c)t}\right)\right]t\hfill \\ & =& {\displaystyle \frac{\alpha \left(21-5\sqrt{57}\right)t^5}{3240}}+\sum _{h=5}^{\infty }{\displaystyle \frac{F_h{t}^{h+1}}{(h+1)!}}<\sum _{h=5}^{\infty }{\displaystyle \frac{F_h{t}^{h+1}}{(h+1)!}},\hfill \end{array}$$

where

$$F_h=-2^{h+1}\alpha +(h+1)\left[-{(1+\delta )}^h+{\delta }^h+\alpha \left({(\delta +c+1)}^h+{(\delta +c)}^h\right)\right],$$

which leads to

$$F_h<-2^{h+1}\alpha +(h+1){\delta }^h+\alpha (h+1)\left({(\delta +c+1)}^h+{(\delta +c)}^h\right).$$

Then,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{F_h}{(h+1){(\delta +c+1)}^h}}& <& 2\alpha n_h+{\left({\displaystyle \frac{\delta }{\delta +c+1}}\right)}^h+\alpha \left(1+{\left({\displaystyle \frac{\delta +c}{\delta +c+1}}\right)}^h\right),\hfill \end{array}$$

where $n_h={\displaystyle \frac{-{\left(\frac{2}{\delta +c+1}\right)}^h}{h+1}}$ is decreasing for $h\ge 3$ because

$$n_{h+1}-n_h={\displaystyle \frac{-(1-\delta -c){\left(\frac{2}{\delta +c+1}\right)}^h}{(h+1)(h+2)(\delta +c+1)}}\left(h-{\displaystyle \frac{2(\delta +c)}{(1-\delta -c)}}\right)<0,h\ge 3.$$

Also, ${\left({\displaystyle \frac{\delta }{\delta +c+1}}\right)}^h$ and ${\left({\displaystyle \frac{\delta +c}{\delta +c+1}}\right)}^h$ are decreasing for $h\ge 5$ and, consequently, for $h\ge 5,$ we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{F_h}{(h+1){(\delta +c+1)}^h}}& <& {\displaystyle \frac{-\alpha {\left(\frac{2}{\delta +c+1}\right)}^5}{3}}+{\left({\displaystyle \frac{\delta }{\delta +c+1}}\right)}^5+\alpha \left(1+{\left({\displaystyle \frac{\delta +c}{\delta +c+1}}\right)}^5\right)\simeq -0.213<0.\hfill \end{array}$$

Then, $-W_{1-\delta ,-\alpha ,-\beta }\left(y\right)$ is CM on $(0,\infty )$ for $\beta \le c$. Conversely, if $-W_{1-\delta ,-\alpha ,-\beta }\left(y\right)$ is CM, then we obtain for $y>0:$

$$y{W}_{1-\delta ,-\alpha ,-\beta }\left(y\right)=-yG(y-\delta +1)-2\alpha y\left[\psi (y+1-\delta -\beta )-lny\right]<0.$$

(23)

In the same way, we have $\underset{y\to \infty }{lim}yG(y-\delta +1)=1$ and $\underset{y\to \infty }{lim}y\left[\psi (y+1-\delta -\beta )-lny\right]=-(\delta +\beta )+{\displaystyle \frac{1}{2}}.$ From (23), we conclude that $-1+2\alpha \left(\delta +\beta -{\displaystyle \frac{1}{2}}\right)\le 0$; hence, $\beta \le c$. Now,

$$-y{W}_{1-\delta ,-\alpha ,-c}\left(y\right)={\int }_0^{\infty }{\displaystyle \frac{2e^{-yt}\mathrm{\Lambda }\left(t\right)}{t^2{(e^{2t}-1)}^2}}dt,$$

where

$$\begin{array}{ccc}\hfill \mathrm{\Lambda }\left(t\right)& =& -\alpha {\left(e^{2t}-1\right)}^2+[(2-\delta )e^{(2+\delta )t}-(1+\delta )e^{(1+\delta )t}-(1-\delta )e^{(3+\delta )t}+\delta e^{\delta t}\hfill \\ & & +\alpha \left((1+\delta +c)e^{(1+\delta +c)t}+(2-\delta -c)e^{(2+\delta +c)t}+(1-\delta -c)e^{(3+\delta +c)t}+(\delta +c)e^{(\delta +c)t}\right)]t^2\hfill \end{array}$$

with $\mathrm{\Lambda }\left(7.8\right)\simeq 3.55\left({10}^{12}\right)$ and $\mathrm{\Lambda }\left(8\right)\simeq -5.39\left({10}^{12}\right).$ Then, $-y{W}_{1-\delta ,-\alpha ,-c}\left(y\right)$ is not a CM function; hence, $de{g}_{CM}^y[-W_{1-\delta ,-\alpha ,-c}\left(y\right)]<1.$ □

4. Sharp Symmetric Bounds for Bateman’s $\mathit{G}$-Function

Let us mention two important consequences of Theorem 1.

Corollary 2.

For the values of $\alpha ,\delta$, and c in (15) and $\theta \in {\mathbb{R}}^{+}$, we have

$$-2\alpha \left(\psi (y-\theta )-ln(y+\delta -1)\right)<G\left(y\right)<2\alpha \left(\psi (y+\theta )-ln(y-\delta )\right)$$

(24)

with the best constant $\theta =c,$ where the upper bound is valid for $y>\delta$ and the other bound is valid for $y>1-\delta$.

Proof. 

From $W_{1-\delta ,-\alpha ,-c}(y+\delta -1)<0$ for $y>1-\delta$ and $W_{\delta ,\alpha ,c}(y-\delta )>0$ for $y>\delta$ in Theorem 1, we obtain (24) at $\theta =c.$ The two sides of inequality (24) are equivalent to $y{W}_{1-\delta ,-\alpha ,-\theta }(y+\delta -1)<0$ and $y{W}_{\delta ,\alpha ,\theta }(y-\delta )>0$ and, in the same way they were used in proving Theorem 1, we have $\theta \le c$ and $\theta \ge c$ consecutively. Since $\psi \left(y\right)$ is increasing on $(0,\infty )$, then $-\psi \left(y-c\right)\ge -\psi \left(y-\theta \right)$ for $\theta \le c$ and $\psi \left(y+c\right)\le \psi \left(y+\theta \right)$ for $\theta \ge c$. Hence, $\theta =c$ is the best in (24). □

Remark 1.

• Using (16) and (17) yields the upper and the lower bounds of inequality (24), which refines their counterparts of (10) for all $y>\sqrt{{\displaystyle \frac{2}{3}}}$ and $y\ge 9.2$ consecutively.
• Using (18) yields the upper bound of inequality (24), which refines its counterpart of (13) for all $y\ge 0.1.$
• Using (19) yields the lower bound of the inequality (24), which refines its counterpart of (14) for all $y\ge 3.5.$ And the upper bound of inequality (24) refines its counterpart of (14) for all $0.0884<y\le 22.6.$

Corollary 3.

For the values of $\alpha ,\delta$, and c in (15), $\theta \in {\mathbb{R}}^{+}$ and $h\in \mathbb{N}$, we have

$$-2\alpha \left[{(-1)}^h{\psi }^{\left(h\right)}(y-\theta )+{\displaystyle \frac{(h-1)!}{{(y+\delta -1)}^h}}\right]<{(-1)}^h{G}^{\left(h\right)}\left(y\right)<2\alpha \left[{(-1)}^h{\psi }^{\left(h\right)}(y+\theta )+{\displaystyle \frac{(h-1)!}{{(y-\delta )}^h}}\right]$$

(25)

with the best constant $\theta =c,$ where the upper bound is valid for $y>\delta$ and the other bound is valid for $y>1-\delta$.

Proof. 

The left-hand side of inequality (25) is deduced from ${(-1)}^{h+1}{W}_{1-\delta ,-\alpha ,-\theta }^{\left(h\right)}(y+\delta -1)>0$ for $y>1-\delta .$ Then,

$$\begin{gathered} \underset{y\to \infty }{lim}{y}^{h+1}{(-1)}^{h+1}{W}_{1-\delta ,-\alpha ,-\theta }^{\left(h\right)}(y-1+\delta )=2\alpha \underset{y\to \infty }{lim}{y}^{h+1}\left[{(-1)}^h{\psi }^{\left(h\right)}(y-\theta )+{\displaystyle \frac{(h-1)!}{{(y+\delta -1)}^h}}\right] \\ +{(-1)}^h\underset{y\to \infty }{lim}{y}^{h+1}{G}^{\left(h\right)}\left(y\right)\ge 0. \end{gathered}$$

(26)

Using the asymptotic (7), we have $\underset{y\to \infty }{lim}{(-1)}^h{y}^{h+1}{G}^{\left(h\right)}\left(y\right)=h!$, and by using the asymptotic (3), we have

$$\underset{y\to \infty }{lim}{y}^{h+1}\left[{(-1)}^h{\psi }^{\left(h\right)}(y-\theta )+{\displaystyle \frac{(h-1)!}{{(y+\delta -1)}^h}}\right]=h!\left({\displaystyle \frac{1}{2}}-(\delta +\theta )\right).$$

From (4), we conclude that $h!\left(1+2\alpha \left({\displaystyle \frac{1}{2}}-(\delta +\theta )\right)\right)\ge 0$ and then $\theta \le c$. The other side of inequality (25) is deduced from ${(-1)}^{h+1}{W}_{\delta ,\alpha ,\theta }^{\left(h\right)}(y-\delta )>0$ for $y>\delta .$ Then,

$$y^{h+1}{(-1)}^h{W}_{\delta ,\alpha ,\theta }^{\left(h\right)}(y-\delta )={(-1)}^{h+1}{y}^{h+1}{G}^{\left(h\right)}\left(y\right)+2\alpha y^{h+1}\left[{(-1)}^h{\psi }^{\left(h\right)}(y+\theta )+{\displaystyle \frac{(h-1)!}{{(y-\delta )}^h}}\right]>0$$

and in the same way, we obtain $\theta \ge c$. Since ${\psi }^{\prime }\left(y\right)$ is a strictly completely monotonic function on $(0,\infty )$, then ${(-1)}^h{\psi }^{\left(h\right)}\left(y\right)$ are increasing on $(0,\infty )$ for $h=0,1,2,\dots$, and then $\theta =c$ is the best constant in (25). □

Remark 2.

Using Lemma 2, we deduce that the upper and the lower bounds inequality (25) at $h=1$ refines their counterparts of (12) for all $y\ge 2.5$ and $y\ge 0.85$ consecutively.

5. Conclusions

We have deduced some new symmetric bounds for Bateman’s G-function and its derivatives in terms of the digamma and polygamma functions, which are mentioned in (24) and (25). These results improve the results presented by Hegazi, Mahmoud, Talat, and Moustafa; Nantomah, Mahmoud, Talat, Moustafa, and Agarwal; and Mahmoud and Almuashi, which are mentioned in (10) and (12)–(14).