Two double inequalities for k-gamma and k-Riemann zeta functions

Abstract

By using methods in the theory of majorization, a double inequality for the gamma function is extended to the k-gamma function and the k-Riemann zeta function.

MSC:33B15, 26D07, 26B25.

1 Introduction

The Euler gamma function $\mathrm{\Gamma }(x)$ is defined [1] for $x>0$ by

$$\mathrm{\Gamma }(x)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{x-1}dt.$$

(1)

In 2005, by using a geometrical method, Alsina and Tomás [2] proved the following double inequality:

$$\frac{1}{n!}\le \frac{\mathrm{\Gamma }{(1+x)}^n}{\mathrm{\Gamma }(1+nx)}\le 1,x\in [0,1],n\in \mathbb{N}.$$

(2)

In 2009, Nguyen and Ngo [3] obtained the following generalization of the double inequality (2):

$$\frac{{\prod }_{i=1}^n\mathrm{\Gamma }(1+{\alpha }_i)}{\mathrm{\Gamma }(\beta +{\sum }_{i=1}^n{\alpha }_i)}\le \frac{{\prod }_{i=1}^n\mathrm{\Gamma }(1+{\alpha }_{i}x)}{\mathrm{\Gamma }(\beta +({\sum }_{i=1}^n{\alpha }_i)x)}\le \frac{1}{\mathrm{\Gamma }(\beta )},$$

(3)

where $x\in [0,1]$, $\beta \ge 1$, ${\alpha }_i>0$, $n\in \mathbb{N}$.

For $k>0$, the function ${\mathrm{\Gamma }}_k$ is defined [4] by

$${\mathrm{\Gamma }}_k(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{n!k^n{(nk)}^{\frac{x}{k}-1}}{{(x)}_{n,k}},x\in \mathbb{C}\mathrm{\setminus }k{\mathbb{Z}}^{-},$$

(4)

where ${(x)}_{n,k}=x(x+k)(x+2k)\cdots (x+(n-1)k)$.

The above definition is a generalization of the gamma function. For $x\in \mathbb{C}$ with $Re(x)>0$, the function ${\mathrm{\Gamma }}_k(x)$ is given by the integral [4]

$${\mathrm{\Gamma }}_k(x)={\int }_0^{\mathrm{\infty }}{e}^{-\frac{t^k}{k}}{t}^{x-1}dt.$$

(5)

It satisfies the following properties [4–6]:

• (i) ${\mathrm{\Gamma }}_k(k)=1$;

• (ii) ${\mathrm{\Gamma }}_1(x)=\mathrm{\Gamma }(x)$.

For $k>0$, the k-Riemann zeta function is defined [5] by the integral

$${\zeta }_k(x)=\frac{1}{{\mathrm{\Gamma }}_k(x)}{\int }_0^{\mathrm{\infty }}\frac{t^{x-k}}{e^t-1}dt,x>k.$$

(6)

Note that when k tends to 1 we obtain the known Riemann zeta function $\zeta (x)$.

In this note, by using methods on the theory of majorization, we extended the double inequality (3) to the function ${\mathrm{\Gamma }}_k(x)$ and the k-Riemann zeta function, namely, we established the following theorems.

Theorem 1

$$\frac{{\prod }_{i=1}^n{\mathrm{\Gamma }}_k(1+{\alpha }_i)}{{\mathrm{\Gamma }}_k(\beta +{\sum }_{i=1}^n{\alpha }_i)}\le \frac{{\prod }_{i=1}^n{\mathrm{\Gamma }}_k(1+{\alpha }_{i}x)}{{\mathrm{\Gamma }}_k(\beta +({\sum }_{i=1}^n{\alpha }_i)x)}\le \frac{1}{{\mathrm{\Gamma }}_k(\beta )},$$

(7)

where $x\in [0,1]$, $\beta \ge 1$, ${\alpha }_i>0$, $i=1,\dots ,n$, $n\in \mathbb{N}$.

Theorem 2

$$\begin{array}{r}\frac{{\prod }_{i=1}^n{\zeta }_k(k+1+{\alpha }_i){\mathrm{\Gamma }}_k(k+1+{\alpha }_i)}{{\zeta }_k(\beta +k+{\sum }_{i=1}^n{\alpha }_i){\mathrm{\Gamma }}_k(\beta +k+{\sum }_{i=1}^n{\alpha }_i)}\\ \le \frac{{\prod }_{i=1}^n{\zeta }_k(k+1+{\alpha }_{i}x){\mathrm{\Gamma }}_k(k+1+{\alpha }_{i}x)}{{\zeta }_k(\beta +k+({\sum }_{i=1}^n{\alpha }_i)x){\mathrm{\Gamma }}_k(\beta +k+({\sum }_{i=1}^n{\alpha }_i)x)}\\ \le \frac{{({\pi }^2/6)}^n}{{\zeta }_k(\beta +k){\mathrm{\Gamma }}_k(\beta +k)},\end{array}$$

(8)

where $x\in [0,1]$, $\beta \ge 1$, ${\alpha }_i>0$, $i=1,\dots ,n$, $n\in \mathbb{N}$.

Substituting $k=1$ and ${\alpha }_i=1$ ($i=1,\dots ,n$) into (8) and taking into account that $\mathrm{\Gamma }(3)=2$ and $\zeta (2)={\pi }^2/6$, we obtain the following.

Corollary 1

$$\frac{{(2\zeta (3))}^n}{\zeta (1+\beta +n)\mathrm{\Gamma }(1+\beta +n)}\le \frac{{(\zeta (2+x)\mathrm{\Gamma }(2+x))}^n}{\zeta (1+\beta +nx)\mathrm{\Gamma }(1+\beta +nx)}\le \frac{{(\zeta (2))}^n}{\zeta (1+\beta )\mathrm{\Gamma }(1+\beta )},$$

(9)

where $x\in [0,1]$, $\beta \ge 1$, $n\in \mathbb{N}$.

Remark 1 $\zeta (3)$ is Apéry’s constant [7].

2 Definitions and lemmas

We need the following definitions and auxiliary lemmas.

Definition 1 [8, 9]

Let $\mathbf{x}=(x_1,\dots ,x_n)$ and $\mathbf{y}=(y_1,\dots ,y_n)\in {\mathbb{R}}^n$.

• (i) We say y majorizes x (x is said to be majorized by y), denoted by $\mathbf{x}\prec \mathbf{y}$, if ${\sum }_{i=1}^k{x}_{[i]}\le {\sum }_{i=1}^k{y}_{[i]}$ for $k=1,2,\dots ,n-1$ and ${\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{y}_i$, where $x_{[1]}\ge \cdots \ge x_{[n]}$ and $y_{[1]}\ge \cdots \ge y_{[n]}$ are rearrangements of x and y in a descending order.

• (ii) Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$, a function $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a Schur-convex function on Ω if $\mathbf{x}\prec \mathbf{y}$ on Ω implies $\phi (\mathbf{x})\le$ $\phi (\mathbf{y})$. A function φ is said to be a Schur-concave function on Ω if and only if −φ is Schur-convex function on Ω.

Definition 2 [8, 9]

Let $\mathbf{x}=(x_1,\dots ,x_n)$ and $\mathbf{y}=(y_1,\dots ,y_n)\in {\mathbb{R}}^n$, $\alpha \in [0,1]$.

• (i) A set $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ is said to be a convex set if $\mathbf{x},\mathbf{y}\in \mathrm{\Omega }$ implies $\alpha \mathbf{x}+(1-\alpha )\mathbf{y}=(\alpha x_1+(1-\alpha )y_1,\dots ,\alpha x_n+(1-\alpha )y_n)\in \mathrm{\Omega }$.

• (ii) Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$ be a convex set. A function φ: $\mathrm{\Omega }\to \mathbb{R}$ is said to be a convex function on Ω if

  $$\phi (\alpha \mathbf{x}+(1-\alpha )\mathbf{y})\le \alpha \phi (\mathbf{x})+(1-\alpha )\phi (\mathbf{y})$$

for all $\mathbf{x},\mathbf{y}\in \mathrm{\Omega }$. A function φ is said to be a concave function on Ω if and only if −φ is a convex function on Ω.

• (iii) Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$. A function $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a log-convex function on Ω if the function logφ is convex.

Lemma 1 [[8], p.186]

Let $\mathbf{x},\mathbf{y}\in {\mathbb{R}}^n$, $x_1\ge x_2\ge \cdots \ge x_n$, and ${\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{y}_i$. If for some k, $1\le k<n$, $x_i\le y_i$, $i=1,\dots ,k$, $x_i\ge y_i$ for $i=k+1,\dots ,n$, then $\mathbf{x}\prec \mathbf{y}$.

Lemma 2 Let f, g be a continuous nonnegative functions defined on an interval $[a,b]\subset \mathbb{R}$. Then

$$I(x)={\int }_a^{b}g(t){(f(t))}^{x}dt$$

is log-convex on $[0,+\mathrm{\infty })$.

Proof Let $\alpha ,\beta \ge 0$, $0<s<1$ by the Hölder integral inequality [[10], p.140], we have

$$\begin{array}{rl}I(s\alpha +(1-s)\beta )& ={\int }_a^{b}g(t){(f(t))}^{s\alpha +(1-s)\beta }dt={\int }_a^b{(g(t){(f(t))}^{\alpha })}^s{(g(t){(f(t))}^{\beta })}^{1-s}dt\\ \le {({\int }_a^{b}g(t){(f(t))}^{\alpha }dt)}^s{({\int }_a^{b}g(t){(f(t))}^{\beta }dt)}^{1-s}={(I(\alpha ))}^s{(I(\beta ))}^{1-s},\end{array}$$

i.e.

$$logI(s\alpha +(1-s)\beta )\le slogI(\alpha )+(1-s)logI(\beta ),$$

this means that $I(x)$ is log-convex on $[0,+\mathrm{\infty })$. □

Remark 2 When $b=+\mathrm{\infty }$, the results of Lemma 2 presented previously hold true.

Lemma 3 [[8], p.105]

Let g be a continuous nonnegative function defined on an interval $I\subseteq \mathbb{R}$. Then

$$\phi (\mathbf{x})=\prod _{i=1}^{n}g(x_i),\mathbf{x}\in I^n,$$

is Schur-convex on $I^n$ if and only if logg is convex on I.

Lemma 4 Let

$$\mathbf{u}=(u_1,\dots ,u_n,u_{n+1})=(\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1,{\alpha }_1,\dots ,{\alpha }_n)$$

(10)

and

$$\mathbf{v}=(v_1,\dots ,v_n,v_{n+1})=(\beta +\sum _{i=1}^n{\alpha }_i-1,{\alpha }_{1}x,\dots ,{\alpha }_{n}x),$$

(11)

where $x\in [0,1]$, $\beta \ge 1$, ${\alpha }_i>0$, $i=1,\dots ,n$, $n\in \mathbb{N}$. Then $\mathbf{u}\prec \mathbf{v}$.

Proof It is clear that ${\sum }_{i=1}^{n+1}{u}_i={\sum }_{i=1}^{n+1}{v}_i$.

Without loss of generality, we may assume that ${\alpha }_1\ge {\alpha }_2\ge \cdots \ge {\alpha }_n$. So $v_1\ge \cdots \ge v_{n+1}$. The following discussion is divided into two cases:

Case 1. $\beta +({\sum }_{i=1}^n{\alpha }_i)x-1\ge {\alpha }_1$. Notice that $x\in [0,1]$, and ${\alpha }_i>0$, $i=1,\dots ,n$, and we have

$$u_1=\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1\le \beta +\sum _{i=1}^n{\alpha }_i-1=v_1$$

and

$$u_i={\alpha }_{i-1}\ge {\alpha }_{i-1}x=v_i,i=2,\dots ,n+1.$$

Hence from Lemma 1, it follows that $\mathbf{u}\prec \mathbf{v}$.

Case 2. $\beta +({\sum }_{i=1}^n{\alpha }_i)x-1<{\alpha }_1$. Let $u_{[1]}\ge \cdots \ge u_{[n+1]}$ denote the components of u in a decreasing order. There exist $k\in \{2,3,\dots ,n\}$ such that

$${\alpha }_1\ge \cdots \ge {\alpha }_{k-1}\ge \beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1\ge {\alpha }_{k+1}\ge \cdots \ge {\alpha }_n.$$

Notice that $\beta -1\ge 0$, $x\in [0,1]$, and ${\alpha }_i>0$, and if $1\le m\le k-1$, then

$$\sum _{i=1}^m{u}_{[i]}=\sum _{i=1}^m{\alpha }_i\le \beta +\sum _{i=1}^n{\alpha }_i-1\le \sum _{i=1}^m{v}_i.$$

If $n\ge m>k-1$, then

$$\begin{array}{rl}\sum _{i=1}^m{u}_{[i]}& =\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1+\sum _{i=1}^{k-1}{\alpha }_i+\sum _{i=k+1}^m{\alpha }_i(\text{If }m=k,\text{ let }\sum _{i=k+1}^m{\alpha }_i=0)\\ =\beta +(\left(\sum _{i=1}^{m-1}{\alpha }_i\right)x+{\alpha }_{m}x+\left(\sum _{i=m+1}^n{\alpha }_i\right)x)-1+\sum _{i=1}^{k-1}{\alpha }_i+\sum _{i=k+1}^m{\alpha }_i\\ =\beta +\left(\sum _{i=1}^{m-1}{\alpha }_i\right)x-1+(\sum _{i=1}^{k-1}{\alpha }_i+{\alpha }_{m}x+\sum _{i=k+1}^m{\alpha }_i+\left(\sum _{i=m+1}^n{\alpha }_i\right)x)\\ \le \beta +\left(\sum _{i=1}^{m-1}{\alpha }_i\right)x-1+\sum _{i=1}^n{\alpha }_i\\ =\sum _{i=1}^m{v}_i.\end{array}$$

Hence from Definition 1(i), it follows that $\mathbf{u}\prec \mathbf{v}$. □

Lemma 5 Let

$$\mathbf{w}=(w_1,\dots ,w_n,w_{n+1})=(\beta -1,{\alpha }_{1}x,\dots ,{\alpha }_{n}x)$$

(12)

and

$$\mathbf{z}=(z_1,\dots ,z_n,z_{n+1})=(\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1,\underset{n}{\underbrace{0,\dots ,0}}),$$

(13)

where $x\in [0,1]$, $\beta \ge 1$, ${\alpha }_i>0$, $i=1,\dots ,n$, $n\in \mathbb{N}$. Then $\mathbf{w}\prec \mathbf{z}$.

Proof It is clear that ${\sum }_{i=1}^{n+1}{w}_i={\sum }_{i=1}^{n+1}{z}_i$.

The following discussion is divided into two cases:

Case 1. $\beta -1\ge {\alpha }_{1}x$. Notice that $x\in [0,1]$ and ${\alpha }_i>0$, $i=1,\dots ,n$, we have

$$w_1=\beta -1\le \beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1=z_1$$

and

$$w_i={\alpha }_{i-1}x\ge 0=z_i,i=2,\dots ,n+1.$$

Hence from the Lemma 1, it follows that $\mathbf{w}\prec \mathbf{z}$.

Case 2. $\beta -1<{\alpha }_{1}x$. Let $w_{[1]}\ge \cdots \ge w_{[n+1]}$ denote the components of w in a decreasing order. There exist $k\in k=2,\dots ,n$ such that

$${\alpha }_{1}x\ge \cdots \ge {\alpha }_{k-1}x\ge \beta -1\ge {\alpha }_{k+1}x\ge \cdots \ge {\alpha }_{n}x.$$

Now notice that $\beta -1\ge 0$, $x\in [0,1]$ and ${\alpha }_i>0$, we have

$$\begin{array}{c}{w}_{[1]}={\alpha }_{1}x\le \beta +\left(\sum _{i=1}^n{\alpha }_i\right)x-1=z_1,\hfill \\ w_{[i]}={\alpha }_{i}x\ge 0=z_i,i=2,\dots ,k-1,\hfill \\ w_{[k]}=\beta -1\ge 0=z_k\hfill \end{array}$$

and

$$w_{[i]}={\alpha }_{i-1}x\ge 0=z_i,i=k+1,\dots ,n+1.$$

Hence from the Lemma 1, it follows that $\mathbf{w}\prec \mathbf{z}$. □

The Schur-convexity described the ordering of majorization, the order-preserving functions were first comprehensively studied by Issai Schur in 1923. It has important applications in analytic inequalities, combinatorial optimization, special functions, probabilistic, statistical, and so on. See [8, 11–13].

3 Proof of main result

Proof of Theorem 1 Taking $g(t)=e^{-\frac{t^k}{k}}$, $f(t)=t$, $a=0$, $b=+\mathrm{\infty }$, then

$$I(x)={\int }_a^{b}g(t){(f(t))}^{x}dt={\int }_0^{+\mathrm{\infty }}{e}^{-\frac{t^k}{k}}{t}^{x}dt={\mathrm{\Gamma }}_k(x+1).$$

(14)

By Lemma 2, $I(x)$ is log-convex on $[0,+\mathrm{\infty })$, and then from Lemma 3, $\phi (\mathbf{x})={\prod }_{i=1}^{n+1}I(x_i)$ is Schur-convex on ${[0,+\mathrm{\infty })}^{n+1}$. Combining Lemma 4 and Lemma 5, respectively, we have

$$\phi (\mathbf{u})\le \phi (\mathbf{v})$$

and

$$\phi (\mathbf{w})\le \phi (\mathbf{z}),$$

i.e.

$${\mathrm{\Gamma }}_k(\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x)\prod _{i=1}^n{\mathrm{\Gamma }}_k(1+{\alpha }_i)\le {\mathrm{\Gamma }}_k(\beta +\sum _{i=1}^n{\alpha }_i)\prod _{i=1}^n{\mathrm{\Gamma }}_k(1+{\alpha }_{i}x)$$

(15)

and

$${\mathrm{\Gamma }}_k(\beta )\prod _{i=1}^n{\mathrm{\Gamma }}_k(1+{\alpha }_{i}x)\le {\mathrm{\Gamma }}_k(\beta +\left(\sum _{i=1}^n{\alpha }_i\right)x).$$

(16)

Thus, we have proved the double inequality (7).

The proof of Theorem 1 is completed. □

Proof of Theorem 2 Let

$${\xi }_k(x)={\int }_0^{\mathrm{\infty }}\frac{t^{x-k}}{e^t-1}dt,x>k,$$

i.e.

$${\xi }_k(x)={\zeta }_k(x){\mathrm{\Gamma }}_k(x).$$

Taking $g(t)=\frac{t}{e^t-1}$, $f(t)=t$, $a=0$, $b=+\mathrm{\infty }$, then

$$J(x)={\int }_a^{b}g(t){(f(t))}^{x}dt={\int }_0^{+\mathrm{\infty }}\frac{t^{x+1}}{e^t-1}dt={\xi }_k(x+k+1).$$

(17)

By Lemma 2, $J(x)$ is log-convex on $[0,+\mathrm{\infty })$, and then from Lemma 3, $\psi (\mathbf{x})={\prod }_{i=1}^{n+1}J(x_i)$ is Schur-convex on ${[0,+\mathrm{\infty })}^{n+1}$. Combining Lemma 4 and Lemma 5, respectively, we have

$$\psi (\mathbf{u})\le \psi (\mathbf{v})$$

and

$$\psi (\mathbf{w})\le \psi (\mathbf{z}),$$

i.e.

$$\begin{array}{r}{\xi }_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x)\prod _{i=1}^n{\xi }_k(k+1+{\alpha }_i)\\ \le {\xi }_k(\beta +k+\sum _{i=1}^n{\alpha }_i)\prod _{i=1}^n{\xi }_k(k+1+{\alpha }_{i}x)\end{array}$$

and

$$\begin{array}{r}{\xi }_k(\beta +k)\prod _{i=1}^n{\xi }_k(k+1+{\alpha }_{i}x)\\ \le {\xi }_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x){\left(\frac{{\pi }^2}{6}\right)}^n,\end{array}$$

notice that ${\xi }_k(k+1)=\frac{{\pi }^2}{6}$.

Further, we have

$$\begin{array}{r}{\zeta }_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x){\mathrm{\Gamma }}_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x)\prod _{i=1}^n{\zeta }_k(k+1+{\alpha }_i){\mathrm{\Gamma }}_k(k+1+{\alpha }_i)\\ \le {\zeta }_k(\beta +k+\sum _{i=1}^n{\alpha }_i){\mathrm{\Gamma }}_k(\beta +k+\sum _{i=1}^n{\alpha }_i)\prod _{i=1}^n{\zeta }_k(k+1+{\alpha }_{i}x){\mathrm{\Gamma }}_k(k+1+{\alpha }_{i}x)\end{array}$$

(18)

and

$$\begin{array}{r}{\zeta }_k(\beta +k){\mathrm{\Gamma }}_k(\beta +k)\prod _{i=1}^n{\zeta }_k(k+1+{\alpha }_{i}x){\mathrm{\Gamma }}_k(k+1+{\alpha }_{i}x)\\ \le {\zeta }_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x){\mathrm{\Gamma }}_k(\beta +k+\left(\sum _{i=1}^n{\alpha }_i\right)x){\left(\frac{{\pi }^2}{6}\right)}^n.\end{array}$$

(19)

Rearranging (18) and (19) gives the double inequality (8).

The proof of Theorem 2 is completed. □