Complete Monotonicity Properties of a Function Involving the Polygamma Function

1. Introduction

The classical Gamma function, which is an extension of the factorial notation to noninteger values is usually defined as \begin{align*} \Gamma(x)&=\int_{0}^{\infty} t^{x-1}e^{-t}\,dt, \quad x>0, \end{align*} and satisfying the basic property \begin{align*} \Gamma(x+1)&=x\Gamma(x), \quad x>0. \end{align*} Its logarithmic derivative, which is called the Psi or digamma function is defined as (see [1] and [2]) where \(\gamma=\lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n}\frac{1}{k}- \ln n \right)=0.577215664...\) is the Euler-Mascheroni's constant. Derivatives of the Psi function, which are called polygamma functions are given as [1] satisfying the functional equation [1] where \(n\in \mathbb{N}_0\) and \(\psi^{(0)}(x)\equiv \psi(x)\). Here, and for the rest of this paper, we use the notations: \(\mathbb{N}=\{1,2,3,4,\dots \}\), \(\mathbb{N}_0=\mathbb{N}\cup \{0\}\) and \(\mathbb{R}=(-\infty, \infty)\). Also, it is well known in the literature that the integral holds for \(x>0\) and \(n\in \mathbb{N}_0\). See for instance [1]. In [3], Qiu and Vuorinen established among other things that the function is strictly decreasing and convex on \((0,\infty)\). Motivated by this result, Mortici [4] proved a more generalized and deeper result which states that, the function is strictly completely monotonic on \((0,\infty)\). Recall that a function \(f:(0,\infty)\rightarrow \mathbb{R}\) is said to be completely monotonic on \((0,\infty)\) if \(f\) has derivatives of all order and \((-1)^{n}f^{(n)}(x)\geq0\) for all \(x\in(0,\infty)\) and \(n\in \mathbb{N}_0\). In this paper, the objective is to extend Mortici's results to the polygamma functions.

2. Some Lemmas

In order to establish our main results, we need the following lemmas.

Lemma 2.1. Let a function \(q_{\alpha, \beta}(t)\) be defined as

where \(\alpha\), \(\beta\) are real numbers such that \(\alpha \neq \beta\) and \((\alpha, \beta)\notin \{ (0,1), (1,0) \}\). Then \(q_{\alpha, \beta}(t)\) is increasing on \((0,\infty)\) if and only if \((\beta-\alpha)(1-\alpha-\beta)\geq0\) and \((\beta-\alpha)(|\alpha-\beta|-\alpha-\beta)\geq0\).

Proof. See [5], [6] or [7].

Lemma 2.2. Let \(a\in(0,1)\). Then the inequality

holds for \(t\in(0,\infty)\).

Proof. Note that the function \(h(t)=\frac{1-e^{-at}}{1-e^{-t}}\) which is obtained from Lemma 2.1 by letting \(\alpha=0\) and \(\beta=a\in(0,1)\) is increasing on \((0,\infty)\). Also, \begin{equation*} \lim_{t\rightarrow 0^+}h(t)=a \quad \text{and} \quad \lim_{t\rightarrow \infty}h(t)=1 . \end{equation*} Then for \(t\in(0,\infty)\), we have \begin{equation*} a=\lim_{t\rightarrow 0^+}h(t)=h(0)< h(t)< h(\infty)=\lim_{t\rightarrow \infty}h(t)=1 , \end{equation*} which gives inequality (8).

3. Main Results

We now present our results in this section.

Theorem 3.1. Let \(f_{a,k}(x)\) and \(h_{a,r}(x)\) be defined for \(a\in(0,1)\), \(k\in \{ 2s: s\in \mathbb{N}_0 \}\), \(r\in \{ 2s+1: s\in \mathbb{N}_0 \}\) and \(x\in(0,\infty)\) as

and Then \(f_{a,k}(x)\) and \(-h_{a,r}(x)\) are strictly completely monotonic on \((0,\infty)\).

Proof. By repeated differentiations with respect to \(x\), and by using (2) and (4), we obtain \begin{align*} f_{a,k}^{(n)}(x)&=\psi^{(k+n)}(x+a) - \psi^{(k+n)}(x) - \frac{(-1)^{n}a(k+n)!}{x^{k+n+1}} \\ &= (-1)^{k+n+1}\int_{0}^{\infty}\frac{t^{k+n}e^{-(x+a)t}}{1-e^{-t}}\,dt - (-1)^{k+n+1}\int_{0}^{\infty}\frac{t^{k+n}e^{-xt}}{1-e^{-t}}\,dt \\ & \quad - (-1)^na\int_{0}^{\infty}t^{k+n}e^{-xt}\,dt. \end{align*} This implies that \begin{align*} (-1)^nf_{a,k}^{(n)}(x)&= -\int_{0}^{\infty}\frac{t^{k+n}e^{-xt}e^{-at}}{1-e^{-t}}\,dt + \int_{0}^{\infty}\frac{t^{k+n}e^{-xt}}{1-e^{-t}}\,dt - a\int_{0}^{\infty}t^{k+n}e^{-xt}\,dt \\ % &= \int_{0}^{\infty} \left[\frac{1-e^{-at}}{1-e^{-t}} - a \right]t^{k+n}e^{-xt}\,dt \\ &>0, \end{align*} which is as a result of Lemma 2.2. Alternatively, we could proceed as follows. \begin{align*} (-1)^nf_{a,k}^{(n)}(x)&= \int_{0}^{\infty} \left[\frac{1-e^{-at}}{1-e^{-t}} - a \right]t^{k+n}e^{-xt}\,dt \\ &= a\int_{0}^{\infty} \left[\frac{1-e^{-at}}{at} - \frac{1-e^{-t}}{t} \right]\frac{t^{k+n+1}e^{-xt}}{1-e^{-t}}\,dt \\ &>0. \end{align*} Notice that, since the function \(\frac{1-e^{-t}}{t}\) is strictly decreasing on \((0,\infty)\), then for \(a\in(0,1)\), we have \(\frac{1-e^{-at}}{at} > \frac{1-e^{-t}}{t}\). Hence \(f_{a,k}(x)\) is strictly completely monotonic on \((0,\infty)\). Similarly, we have \begin{align*} - h_{a,r}^{(n)}(x) &=\frac{(-1)^{n}a(r+n)!}{x^{r+n+1}} + \psi^{(r+n)}(x) - \psi^{(r+n)}(x+a) \\ &= (-1)^na\int_{0}^{\infty}t^{r+n}e^{-xt}\,dt + (-1)^{r+n+1}\int_{0}^{\infty}\frac{t^{r+n}e^{-xt}}{1-e^{-t}}\,dt \\ & \quad - (-1)^{r+n+1}\int_{0}^{\infty}\frac{t^{r+n}e^{-(x+a)t}}{1-e^{-t}}\,dt , \end{align*} which implies that \begin{align*} (-1)^n \left(- h_{a,r}\right)^{(n)}(x) &= a\int_{0}^{\infty}t^{r+n}e^{-xt}\,dt + \int_{0}^{\infty}\frac{t^{r+n}e^{-xt}}{1-e^{-t}}\,dt - \int_{0}^{\infty}\frac{t^{r+n}e^{-xt}e^{-at}}{1-e^{-t}}\,dt \\ &= \int_{0}^{\infty} \left[a + \frac{1-e^{-at}}{1-e^{-t}} \right]t^{r+n}e^{-xt}\,dt \\ &>0 . \end{align*} Hence \(-h_{a,r}(x)\) is strictly completely monotonic on \((0,\infty)\).

Remark 3.2. Since every completely monotonic function is convex and decreasing, it follows that \(f_{a,k}(x)\) is strictly convex and strictly decreasing on \((0,\infty)\). In this way, \(h_{a,r}(x)\) is strictly concave and strictly increasing on \((0,\infty)\).

Corollary 3.3. The inequality

holds for \(a\in(0,1)\), \(k\in \{ 2s: s\in \mathbb{N}_0 \}\) and \(x\in(1,\infty)\).

Proof. Since \(f_{a,k}(x)\) is decreasing, then for \(x\in(1,\infty)\) and by applying (3), we obtain \begin{align*} 0=\lim_{x\rightarrow \infty}f_{a,k}(x) < f_{a,k}(x) < f_{a,k}(1)&=\psi^{(k)}(a+1) - \psi^{(k)}(1) - ak! \\ &=\psi^{(k)}(a) - \psi^{(k)}(1) + \frac{k!}{a^{k+1}} - ak! , \end{align*} which completes the proof.

Remark 3.4. In particular, if \(a=\frac{1}{2}\) and \(k=0\) in Corollary 3.3, then we obtain

Also, if \(a=\frac{1}{2}\) and \(k=2\) in Corollary 3.3, then we obtain where \(\zeta(x)\) is the Riemann zeta function.

Corollary 3.5. The inequality

holds for \(a\in(0,1)\), \(r\in \{ 2s+1: s\in \mathbb{N}_0 \}\) and \(x\in(1,\infty)\).

Proof. Likewise, since \(h_{a,r}(x)\) is increasing, then for \(x\in(1,\infty)\), we obtain \begin{equation*} \psi^{(r)}(a) - \psi^{(r)}(1) - \frac{r!}{a^{r+1}} - ar! = h_{a,r}(1) < h_{a,r}(x) < \lim_{x\rightarrow \infty}h_{a,r}(x)=0, \end{equation*} which yields (14).

Remark 3.6. If \(a=\frac{1}{2}\) and \(r=1\) in Corollary 3.5, then we obtain

Furthermore, if \(a=\frac{1}{2}\) and \(r=3\) in Corollary 3.5, then we obtain

Remark 3.7. If \(k=0\) in Theorem 3.1, then we obtain the main results of [4] as a special case of the present results.

Remark 3.8. This paper is a modified version of the preprint [8].

Competing Interests

The author declares that there are no competing interests regarding the publication of this paper.