Monotonic functions related to the q-gamma function

Abstract

In this paper, the monotonicity property for two functions involving the logarithmic of the q-gamma function is proven for all $q>0$. As a consequence, sharp inequalities for the q-gamma function are established. Our results are shown to be as a generalization of results which were obtained by Anderson and Qiu (Proc Am Math Soc 125:3355–3362, 1997).

1 Introduction

Euler’s gamma function is defined for positive real numbers x by

$$\begin{aligned} \Gamma (x)=\int _0^\infty t^{x-1}e^{-t}dt,\quad x>0 \end{aligned}$$

which is one of the most important special functions and has many extensive applications in many branches, for example, statistics, physics, engineering, and other mathematical sciences. Anderson and Qiu [1] used the increasing monotonicity of the function

$$\begin{aligned} f(x)={\log \Gamma (x+1)\over x\log x},\quad x>1 \end{aligned}$$

(1.1)

to establish a sharp inequality

$$\begin{aligned} x^{(1-\gamma )x-1}<\Gamma (x)<x^{x-1},\quad x>1 \end{aligned}$$

(1.2)

where $\gamma =0.577215 \ldots$ is the Euler–Mascheroni constant, which has attracted the attention of many researches, because of its simple form, and of its usefulness in practical applications in pure mathematics or other branches of science such as probabilities, engineering, or statistical physics. They conjectured that f is concave on the interval $[1,\infty )$. The concavity of f on $[1,\infty )$ was established by Elbert and Laforgia [2]. A short and simple proof of the increasing of the function f which extended the increasing on $(0,\infty )$, has been presented by Alzer [3]. It is worth mentioning that in 1989, Anderson et al. [4] conjectured that the function

$$\begin{aligned} G(x)={\log \Gamma ({x\over 2}+1)\over x\log x},\quad x\ge 2 \end{aligned}$$

(1.3)

is strictly increasing on $[2,\infty )$. This conjecture was proved by Anderson and Qiu [1].

Many of the classical facts about the ordinary gamma function have been extended to the q-gamma function (see [5–8] and the references given therein). The aim of this paper is to extend the inequality (1.2) to the q-gamma function for all positive real numbers x and q by means of the study of the monotonicity property of the function

$$\begin{aligned} F_q(x)={\log \Gamma _q(x+1)-{x(x-1)\over 2}H(q-1)\log q\over x\log [x]_q-x(x-1)H(q-1)\log q},\quad x>0, \ q>0 \end{aligned}$$

(1.4)

where $[x]_q=(1-q^x)/(1-q)$, $H(\cdot )$ denotes the Heaviside step function and $\Gamma _q(x)$ is the q-gamma function defined as

$$\begin{aligned} \Gamma _q(x)=(1-q)^{1-x}\prod _{n=0}^\infty {1-q^{n+1}\over 1-q^{n+x}},\quad 0<q<1, \end{aligned}$$

(1.5)

and

$$\begin{aligned} \Gamma _q(x)=(q-1)^{1-x}q^{x(x-1)\over 2}\prod _{n=0}^\infty {1-q^{-(n+1)}\over 1-q^{-(n+x)}},\quad q>1. \end{aligned}$$

(1.6)

From the previous definitions, for a positive x and $q\ge 1$, we get

$$\begin{aligned} \Gamma _q(x)=q^{(x-1)(x-2)\over 2}\Gamma _{q^{-1}}(x). \end{aligned}$$

(1.7)

Also, we extend the function G(x) to $F_q(x)$, defined in (1.4), which contains the q-gamma function, for all $q\in (0,\infty )$ and $x\in (0,1)\cup [2,\infty )$. This means that the function G(x) is also increasing on the interval (0, 1). Furthermore, we use these results to establish new inequalities for the q-gamma function.

An important fact for gamma function in applied mathematics as well as in probability is the Stirling’s formula that gives a pretty accurate idea about the size of gamma function. With the Euler–Maclaurin formula, Moak [7] obtained the following q-analogue of Stirling’s formula (see also [9])

$$\begin{aligned} \log \Gamma _q(x)&\sim \left( x-{1\over 2}\right) \log [x]_q+{\text {Li}_2(1-q^x)\over \log q}+{1\over 2}H(q-1)\log q+C_{\hat{q}}\nonumber \\&\quad +\sum _{k=1}^\infty {B_{2k}\over (2k)!}\left( {\log {\hat{q}}\over {\hat{q}}^x-1}\right) ^{2k-1}{\hat{q}}^xP_{2k-3}({\hat{q}}^x),\quad x\rightarrow \infty \end{aligned}$$

(1.8)

where $B_k$ is the Bernoulli numbers,

$$\begin{aligned} {\hat{q}}=\left\{ \begin{array}{l@{\quad }l} q &{} \text {if } \ 0<q\le 1\\ q^{-1}&{} \text {if } \ q\ge 1, \end{array}\right. \end{aligned}$$

$\text {Li}_2(z)$ is the dilogarithm function defined for complex argument z as [10]

$$\begin{aligned} \text {Li}_2(z)=-\int _0^z{\log (1-t)\over t}dt,\quad z\not \in (1,\infty ), \end{aligned}$$

(1.9)

$P_k$ is a polynomial of degree k satisfying

$$\begin{aligned} P_k(z)=(z-z^2)P'_{k-1}(z)+(kz+1)P_{k-1}(z),\quad P_0=P_{-1}=1,\quad k=1,2,\cdots \end{aligned}$$

(1.10)

and

$$\begin{aligned} C_q= & {} {1\over 2}\log (2\pi )+{1\over 2}\log \left( {q-1\over \log q}\right) -{1\over 24}\log q \nonumber \\&+\log \left( \sum _{m=-\infty }^\infty \left( r^{m(6m+1)}-r^{(2m+1)(3m+1)}\right) \right) \end{aligned}$$

(1.11)

where $r=\exp (4\pi ^2/\log q)$. It is easy to see that

$$\begin{aligned} \lim _{q\rightarrow 1}C_q=C_1={1\over 2}\log (2\pi ),\quad \lim _{q\rightarrow 1}{\text {Li}_2(1-q^x)\over \log q}=-x\quad \text {and}\quad P_k(1)=(k+1)! \end{aligned}$$

(1.12)

and so (1.8) when letting $q\rightarrow 1$, tends to the ordinary Stirling’s formula [10]

$$\begin{aligned} \log \Gamma (x)\sim \left( x-{1\over 2}\right) \log x-x+{1\over 2}\log (2\pi )+\sum _{k=1}^\infty {B_{2k}\over 2k(2k-1)}{1\over x^{2k-1}},\quad x\rightarrow \infty . \end{aligned}$$

(1.13)

2 Useful lemmas

In order to prove our main results we need to study the monotonicity properties of some functions which are connected with the q-digamma function $\psi _q(x)$ and its derivative which is defined as the logarithmic derivative of the q-gamma function

$$\begin{aligned} \psi _q(x)={d\over dx}(\log \Gamma _q(x))={\Gamma '_q(x)\over \Gamma _q(x)} \end{aligned}$$

(2.1)

The q-digamma function $\psi _q(x)$ appeared in the work of Krattenthaler and Srivastava [11] when they studied the summations for basic hypergeometric series. Some of its properties are presented and proven in their work. Also, in their work, they proved that $\psi _q(x)$ tends to the digamma function $\psi (x)$ when letting $q\rightarrow 1$. For more details on the q-digamma function (see [12] and the references therein). From (1.5), we get for $0<q<1$ and for all real variable $x>0$

$$\begin{aligned} \psi _q(x)=-\log (1-q)+\log q\sum _{k=1}^\infty {q^{xk}\over 1-q^{k}}, \end{aligned}$$

(2.2)

and from (1.6) we obtain for $q>1$ and $x>0$

$$\begin{aligned} \psi _q(x)=-\log (q-1)+\log q\left[ x-{1\over 2}-\sum _{k=1}^\infty {q^{-xk}\over 1-q^{-k}}\right] . \end{aligned}$$

(2.3)

It is worth mentioning that many papers recently have introduced inequalities related to the q-gamma, q-digamma and q-polygamma functions, see [9, 13–21] and the references therein.

Lemma 2.1

Let x and q be real numbers such that $0<q<1$. Then the function $\log \Gamma _q(x+1)\ge 0$ for all $x\ge 1$ and $\log \Gamma _q(x+1)\le 0$ for all $0\le x\le 1$.

Proof

Replacing x by $x+1$ in (2.2) followed by integrating from 0 to x, the logarithmic of the q-gamma function can be represented as

$$\begin{aligned} \log \Gamma _q(x+1)=-x\log (1-q)+\sum _{k=1}^\infty {q^{(x+1)k}-q^k\over k(1-q^k)} \end{aligned}$$

which can be also rewritten as

$$\begin{aligned} \log \Gamma _q(x+1)=\sum _{k=1}^\infty {q^k\over k(1-q^k)}\alpha (y),\quad y=q^k, \end{aligned}$$

where $\alpha (y)=x(1-y)+y^x-1$ which has the derivative $\alpha '(y)=-x(1-y^{x-1})$. It is clear that $\alpha '(y)\le 0$ if $x\ge 1$ and $\alpha '(y)\ge 0$ if $x\le 1$ which reveals that $\alpha (y)$ is decreasing on (0, 1) if $x\ge 1$ and increasing on (0, 1) if $x\le 1$. Since $\alpha (1)=0$ for all $x\ge 0$, then $\alpha (y)\ge 0$ if $x\ge 1$ and $\alpha (y)\le 0$ if $x\le 1$ which give the desired results.

Lemma 2.2

Let q be a positive real number such that $0<q<1$. Then the function

$$\begin{aligned} f_q(x)=\psi _q''(x+1)-{\log q\over 1-q^x}\psi '_q(x+1) \end{aligned}$$

(2.4)

is strictly positive for all $x\in \mathbb {R}^+$.

Proof

The relation (2.2) and the Cauchy product rule gives

$$\begin{aligned} {\log q\over 1-q^x}\psi '(x+1)=\log ^3q\sum _{k=1}^\infty q^{xk}\sum _{r=1}^k{rq^r\over 1-q^r} \end{aligned}$$

which yields that

$$\begin{aligned} f_q(x)=-\log ^3q\sum _{k=1}^\infty q^{xk}\ell (k) \end{aligned}$$

where

$$\begin{aligned} \ell (k)=\sum _{r=1}^k{rq^r\over 1-q^r}-{k^2q^k\over 1-q^k}. \end{aligned}$$

Forward shift operator gives

$$\begin{aligned} \ell (k+1)-\ell (k)={(k+1)q^{k+1}\over 1-q^{k+1}}-{(k+1)^2q^{k+1}\over 1-q^{k+1}}+{k^2q^{k}\over 1-q^{k}} \end{aligned}$$

which can be simplified as

$$\begin{aligned} \ell (k+1)-\ell (k)={kq^{k}(k(1-q)-q(1-q^k))\over (1-q^k)(1-q^{k+1})}. \end{aligned}$$

Since $1-q^k=(1-q)(1+q+q^2+\cdots +q^{k-1})\le k(1-q)$ for all $k\in \mathbb {N}$, then we get $\ell (k+1)\ge \ell (k)$ for all $k\in \mathbb {N}$ which gives that $\ell (k)\ge \ell (1)=0$ for all $k\in \mathbb {N}$ and so the function $f_q(x)\ge 0$ for all $x>0$.

Lemma 2.3

Let q be a positive real number such that $0<q<1$. Then the function

$$\begin{aligned} h_q(x)=x\psi _q(x+1)-\log \Gamma _q(x+1) \end{aligned}$$

(2.5)

is non-negative and increasing on $[0,\infty )$.

Proof

Differentiation gives

$$\begin{aligned} h'_q(x)=x\psi '_q(x+1)=x\sum _{k=1}^\infty {kq^{(x+1)k}\log ^2q\over 1-q^k}\ge 0 \quad x\ge 0. \end{aligned}$$

Hence, the monotonicity of $h_q$ follows. Obviously, $h_q(0)=0$.

Lemma 2.4

Let q be a positive real number such that $0<q<1$. Then the function

$$\begin{aligned} g_q(x)=x^2\psi '_q(x+1)-2h_q(x)-{x\log q\over 1-q^x}h_q(x) \end{aligned}$$

(2.6)

is strictly positive for all $x\in (0,\infty )$, where $h_q(x)$ is defined as in Lemma 2.3.

Proof

Differentiation gives

$$\begin{aligned} g'_q(x)=x^2\psi ''_q(x+1)-{x^2\log q\over 1-q^x}\psi '_q(x+1)-{(1-q^x+xq^x\log q)\log q\over (1-q^x)^2}h_q(x). \end{aligned}$$

Let $\lambda (y)=y\log y+1-y$ where $y=q^x$. A short calculation shows that

$$\begin{aligned} \lambda (y)=y\sum _{n=2}^\infty {\log (1/y)\over n!}\ge 0,\quad 0<y<1. \end{aligned}$$

(2.7)

Since $h_q(x)\ge 0$ according to Lemma 2.3, then we get $g'_q(x)\ge x^2f_q(x)$ where $f_q(x)$ defined as in Lemma 2.2. This concludes that $g'_q(x)>0$ for all $x>0$ and so that the function $g_q(x)$ is increasing on $(0,\infty )$ for all $0<q<1$. It is clear that from (2.6) and Lemma 2.3 that $\lim _{x\rightarrow 0}g_q(x)=0$ which concludes that $g_q(x)>0$ for all $x>0$ and $0<q<1$.

Lemma 2.5

Let q be a positive real number such that $0<q<1$. Then the function

$$\begin{aligned} H_q(x)=\log [x]_q+{xq^x\log q\over 1-q^x}{\log \Gamma _q(x+1)\over h_q(x)} \end{aligned}$$

(2.8)

is strictly positive on $(0,\infty )$, where $h_q(x)$ is defined as in Lemma 2.3.

Proof

Differentiation gives

$$\begin{aligned} H'_q(x)&=-{q^x\log q\over 1-q^x}+{xq^x\log q\over 1-q^x}{\psi _q(x+1)h_q(x)-h'_q(x)\log \Gamma _q(x+1)\over h^2_q(x)}\\&\quad +{q^x\log q(1-q^x+x\log q)\over (1-q^x)^2}{\log \Gamma _q(x+1)\over h_q(x)}\\&=-{q^x\log q\over (1-q^x)h_q^2(x)}\bigg (h_q^2(x)-x\psi _q(x+1)h_q(x) +x^2\psi '_q(x+1)\log \Gamma _q(x+1)\\&\quad -{h_q(x)\log \Gamma _q(x+1)(1-q^x+x\log q\over 1-q^x}\bigg )\\&=-{q^x\log q\log \Gamma _q(x+1)\over (1-q^x)h_q^2(x)}g(x) \end{aligned}$$

where g(x) defined as in Lemma 2.4. According to the results obtained in Lemmas 2.1 and 2.4, we see that $H'_q(x)\ge 0$ if $x\ge 1$ and $H'_q(x)\le 0$ if $x\le 1$ which yields that $H_q(x)$ is increasing on $[1,\infty )$ and decreasing on (0, 1]. It is obvious from (2.8) that $H_q(1)=0$ which gives that $H_q(x)>0$ for all $x>0$.

Lemma 2.6

Let x and q be positive real numbers. Then the function

$$\begin{aligned} S_q(x)={x\log [x]_q-x(x-1)H(q-1)\log q\over x\log [2x]_q-x(2x-1)H(q-1)\log q} \end{aligned}$$

(2.9)

is strictly increasing on $(0,1/2)\cup (1/2,\infty )$ and $S_q(x)\ge 0$ if $x\in (0,1/2)\cup [1,\infty )$ and $S_q(x)\le 0$ if $x\in (1/2,1]$.

Proof

When $0<q<1$, differentiation gives

$$\begin{aligned} S'_q(x)=-{q^x\log q\over (1+q^{x})\log ^2[x]_q}\beta (x) \end{aligned}$$

(2.10)

where

$$\begin{aligned} \beta (x)=\log [x]_q+{1+q^x\over 1-q^x}\log (1+q^x) \end{aligned}$$

which has the derivative

$$\begin{aligned} \beta '(x)={2q^x\log q\over (1-q^x)^2}\log (1+q^x)<0,\quad x>0. \end{aligned}$$

Since $\lim _{x\rightarrow \infty }\beta (x)=-\log (1-q)>0$ and $\beta '(x)<0$, then $\beta (x)>0$ for all $x>0$ which yields that $S'_q(x)>0$ for all $x\in (0,1)\cup (1,\infty )$ and so the function $S_q(x)$ is increasing on $(0,1/2)\cup (1/2,\infty )$. It is easy to see that $S_q(1)=0$ and $\lim _{x\rightarrow 0}S_q(x)=1$ which give the sign of the function. When $q\ge 1$, we get $S_q(x)=S_{q^{-1}}(x)$. This ends the proof.

3 The main results

In this section, the main results will be provided. At first, we recall that the author in [12] defined the q-analogue of the Euler–Mascheroni constant as

$$\begin{aligned} \gamma _q={1-q\over \log q}\psi _q(1),\quad 0<q<1, \end{aligned}$$

(3.1)

and proved the identity

$$\begin{aligned} \psi _q(x+1)=\psi _q(x)-{q^x\log q\over 1-q^x},\quad x>0. \end{aligned}$$

(3.2)

We are now in a position to prove the following:

Theorem 3.1

Let x and q be positive real numbers. Then the function $F_q(x)$ defined as in (1.4) is strictly increasing on $(0,1)\cup (1,\infty )$ and has the limits:

• 1. $\lim _{x\rightarrow 0}F_q(x)=0$

• 2. $\lim _{x\rightarrow 1}F_q(x)=1-{\hat{q}}^{-1}\gamma _{\hat{q}}$

• 3. $\lim _{x\rightarrow \infty }F_q(x)=1$.

Proof

When $0<q<1$, differentiating (1.4) gives

$$\begin{aligned} (x\log [x]_q)^2F'_q(x)&=x\log [x]_q\psi _q(x+1)-\log [x]_q\log \Gamma _q(x+1) \\&\quad +{xq^x\log q\over 1-q^x}\log \Gamma _q(x+1)\\&=\log [x]_qh_q(x)+{xq^x\log q\over 1-q^x}\log \Gamma _q(x+1)\\&=h_q(x)H_q(x) \end{aligned}$$

where $h_q$ and $H_q$ are defined as in Lemmas 2.3 and 2.5, respectively. Hence, the monotonicity of $F_q$ follows immediately from Lemmas 2.3 and 2.5. When $q\ge 1$, inserting (1.7) into (1.4) yields $F_q(x)=F_{q^{-1}}(x)$ which concludes that $F_q(x)$ is increasing on $(0,1)\cup (1,\infty )$ for all $q>0$.

In order to evaluate the limits, using l’Hôpital’s rule to get

$$\begin{aligned} \lim _{x\rightarrow 0}F_q(x)=\lim _{x\rightarrow 1}\frac{\psi _q(x+1)-{2x-1\over 2}H(q-1)\log q}{\log [x]_q-{xq^x\log q\over 1-q^x}-(2x-1)H(q-1)\log q}=0. \end{aligned}$$

Also, when $0<q<1$, we get

$$\begin{aligned} \lim _{x\rightarrow 1}F_q(x)=\lim _{x\rightarrow 1}\frac{\psi _q(x+1)}{\log [x]_q-{xq^x\log q\over 1-q^x}}=\frac{\psi _q(2)}{-{q\log q\over 1-q}}. \end{aligned}$$

From the relations (3.1) and (3.2), we get

$$\begin{aligned} \lim _{x\rightarrow 1}F_q(x)=1-q^{-1}\gamma _q,\quad 0<q<1. \end{aligned}$$

Since $F_q(x)=F_{q^{-1}}(x)$ when $q\ge 1$, then we get

$$\begin{aligned} \lim _{x\rightarrow 1}F_q(x)=1-q\gamma _{q^{-1}},\quad q\ge 1. \end{aligned}$$

The previous two limits lead to the proof of the second statement. Also, by Moak formula (1.8), we have

$$\begin{aligned} \lim _{x\rightarrow \infty }F_q(x)= & {} \lim _{x\rightarrow \infty }\left[ \frac{(x+{1\over 2})\log [x]_{\hat{q}}+{\text {Li}_2(1-{\hat{q}}^x)\over \log {\hat{q}}}+C_{\hat{q}}}{x\log [x]_{\hat{q}}}+O\left( {{\hat{q}}^x\log {\hat{q}}\over x(1-{\hat{q}}^x)}\right) \right] =1,\\&q>0. \end{aligned}$$

This ends the proof.

Corollary 3.2

Let x and q be positive real numbers. Then the q-gamma function satisfies the inequality

$$\begin{aligned} q^{{x(1-x)\over 2}(2\alpha -1)H(q-1)}[x]_q^{\alpha x-1}\le \Gamma _q(x)\le q^{{x(1-x)\over 2}H(q-1)}[x]_q^{\beta x-1},\quad x\in [1,\infty ) \end{aligned}$$

(3.3)

with the best possible constants $\alpha =1-{\hat{q}}^{-1}\gamma _{\hat{q}}$ and $\beta =1$, where $\gamma _q$ is the q-analogue of the Euler–Mascheroni constant (3.1), and the inequality

$$\begin{aligned} q^{{x(1-x)\over 2}H(q-1)}[x]_q^{\alpha x-1}\le \Gamma _q(x)\le q^{{x(x-1)\over 2}H(q-1)}[x]_q^{\beta x-1},\quad x\in (0,1] \end{aligned}$$

(3.4)

with the best possible constants $\alpha =1$ and $\beta =0$.

Proof

The proof of this corollary comes immediately from Theorem 3.1.

Corollary 3.3

Let $y>x>1$ and q be positive real numbers. Then the q-gamma function satisfies the inequalities

$$\begin{aligned}&{\log \Gamma _q(y+1)-{y(y-1)\over 2}H(q-1)\log q\over \log \Gamma _q(x+1)-{x(x-1)\over 2}H(q-1)\log q}\nonumber \\&\quad >\left( {y\log [y]_q-y(y-1)H(q-1)\log q\over x\log [x]_q-x(x-1)H(q-1)\log q}\right) ^\alpha \end{aligned}$$

(3.5)

for all $\alpha \le 1$ with the best possible constant $\alpha =1$.

Proof

Taking the logarithm of two sides to obtain $\alpha <P(x,y;q)$ where

$$\begin{aligned}&P(x,y;q)\\&\quad ={\log \left( \log \left( q^{{y(1-y)\over 2}H(q-1)}\Gamma _q(y+1)\right) \right) -\log \left( \log \left( q^{{x(1-x)\over 2}H(q-1)}\Gamma _q(x+1)\right) \right) \over \log \left( \log \left( q^{y(1-y)H(q-1)}[y]_q^y\right) \right) -\log \left( \log \left( q^{x(1-x)H(q-1)}[x]_q^x\right) \right) }. \end{aligned}$$

When $0<q<1$, using l’Hośpital rule, one gets

$$\begin{aligned} \lim _{y\rightarrow \infty }P(x,y;q)&=\lim _{y\rightarrow \infty }{\log \left( \log \left( \Gamma _q(y+1)\right) \right) -\log \left( \log \left( \Gamma _q(x+1)\right) \right) \over \log \left( y\log [y]_q\right) -\log \left( x\log [x]_q^x\right) }\\&=\lim _{y\rightarrow \infty }{y\log [y]_q\psi _q(y+1)\over \log \Gamma _q(y+1)(\log [y]_q-{yq^y\log q\over 1-q^y})}\\&=\lim _{y\rightarrow \infty }{y\psi _q(y+1)\over \log \Gamma _q(y+1)}\lim _{y\rightarrow \infty }\left( 1-{yq^y\log q\over (1-q^y)\log [y]_q}\right) ^{-1}\\&=\lim _{y\rightarrow \infty }{\psi _q(y+1)+y\psi '_q(y+1)\over \psi _q(y+1)}\times 1=1. \end{aligned}$$

Here, we use $yq^y\rightarrow 0$ as $y\rightarrow \infty$ and $\lim _{y\rightarrow \infty }y\psi '_q(y+1)=0$ which comes immediately from (2.2). When $q\ge 1$, it is clear that $P(x,y;q)=P(x,y;q^{-1})$.

Theorem 3.4

Let x and q be positive real numbers. Then the function

$$\begin{aligned} G_q(x)=\frac{\log \Gamma _q(1+{x\over 2})-{x(x-2)\over 8}H(q-1)\log q}{x\log [x]_q-x(x-1)H(q-1)\log q} \end{aligned}$$

(3.6)

is strictly increasing on $(0,1)\cup [2,\infty )$ and has the values $G_q(2)=0;~\lim _{x\rightarrow 0}G_q(x)=0$ and $\lim _{x\rightarrow \infty }G_q(x)={1\over 2}$.

Proof

The function $G_q(x)$ after replacing x by 2x can be read as

$$\begin{aligned} G_q(2x)&={1\over 2}\frac{\log \Gamma _q(x+1)-{x(x-1)\over 2}H(q-1)\log q}{x\log [x]_q-x(x-1)H(q-1)\log q}\frac{x\log [x]_q-x(x-1)H(q-1)\log q}{x\log [2x]_q-x(2x-1)H(q-1)\log q}\\&={1\over 2}F_q(x)S_q(x) \end{aligned}$$

where $F_q(x)$ and $S_q(x)$ defined as in (1.4) and (2.9), respectively. Differentiation gives

$$\begin{aligned} 4G'_q(2x)=F'_q(x)S_q(x)+F_q(x)S'_q(x) \end{aligned}$$

It is clear from Theorem 3.1 and Lemma 2.6 that $G'_q(2x)>0$ for all $x\in (0,1/2)\cup [1,\infty )$ which lead to the function $G_q(x)$ is increasing on $(0,1)\cup [2,\infty )$ for all $q>0$. To obtain $\lim _{x\rightarrow \infty }G_q(x)={1\over 2}$, use the l’Hośpital rule and the relations (2.2) and (2.3).

Corollary 3.5

Let x and q be positive real numbers. Then the q-gamma function satisfies the double inequality

$$\begin{aligned} q^{{x(x-2)\over 8}H(q-1)}[x]_q^{-1}(1+q^{x\over 2})\le \Gamma _q\left( {x\over 2}\right) <q^{{x(2-3x)\over 8}H(q-1)}[x]_q^{{x\over 2}-1}(1+q^{x\over 2}) \end{aligned}$$

(3.7)

for all $x\in [2,\infty )$ and satisfies the one-sided inequality

$$\begin{aligned} \Gamma _q\left( {x\over 2}\right) < q^{{x(x-2)\over 8}H(q-1)}[x/2]_q^{-1} \end{aligned}$$

(3.8)

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

Remark 3.6

The function $G_q(x)$ defined as in (3.6) approaches the function G(x) defined as in (1.3) when letting $q\rightarrow 1$ and so the function G(x) is increasing on the interval (0, 1) which is considered an extension of the results obtained for this function by [1].

Conjecture 3.7

The function $G_q(x)$ defined as in (3.6) is strictly increasing on the interval (1, 2] for all $q>0$.

Conjecture 3.8

The function $F_q(x)$ defined as in (1.4) is concave on the interval $(0,1)\cup (1,\infty )$ for all $q>0$.