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Monotonicity, concavity, and inequalities related to the generalized digamma function
Advances in Difference Equations volume 2018, Article number: 246 (2018)
Abstract
In this paper, we establish a concave theorem and some inequalities for the generalized digamma function. Hence, we give complete monotonicity property of a determinant function involving all kinds of derivatives of the generalized digamma function.
1 Introduction
It is well known that the Euler gamma function is defined by
The logarithmic derivative of \(\Gamma(x)\) is called the psi or digamma function. That is,
where \(\gamma=0.5772\ldots\) is the Euler–Mascheroni constant. The gamma, digamma, and polygamma functions play an important role in the theory of special function, and have many applications in many other branches such as statistics, fractional differential equations, mathematical physics, and theory of infinite series. The reader may see related references [5, 7, 9, 13, 16, 23–28].
In [6], the k-analogue of the gamma function is defined for \(k>0\) and \(x>0\) as follows:
where \(\lim_{k\rightarrow1}\Gamma_{k}(x)=\Gamma(x)\). It is natural that the k-analogue of the digamma function is defined for \(x>0\) by
It is worth noting that Nantomah et al. gave \((p,k)\)-analogue of the gamma and the digamma functions in [15]. Further, they established some inequalities involving these new functions. The reader may see references [14, 15].
Very recently, Alzer and Jameson [2] presented a harmonic mean inequality for the digamma function, and also showed some interesting inequalities. It is natural to ask if one can generalize these results to the generalized digamma function with single parameters. This is the first object in this paper.
The second object of this paper came from the article of Ismail and Laforgia. In [11], they proved complete monotonicity of a determinant function involving the derivatives of the digamma function. Using their idea, we prove that their conclusion is also true for the generalized digamma function. In particular, some of the work about the complete monotonicity of these special functions may be found in [3, 4, 8, 10, 12, 17–22].
2 Lemmas
Lemma 2.1
For \(k>0\) and \(x>0\), the following identities hold true:
Proof
Using the substitution \(\frac{t}{\sqrt[k]{k}}=u\) and \(u^{k}=p\), we easily obtain
So, we prove formula (2.1). To (2.2), direct computation yields
 □
Lemma 2.2
For \(x>0\), we have
and
Lemma 2.3
For \(k,x>0, m\in\mathbb{N}\), we have
and
Proof
Formula (2.5) may be found in reference [15]. By using formula (9) in reference [15], we have
So, we prove formula (2.6). □
3 Main results
Theorem 3.1
For \(k>0\), the function \(x^{2}\psi_{k}'(x)\) is strictly increasing on \((0,\infty)\).
Proof
Using Lemma 2.1, we have
Combining with the identity \(\psi^{(m)}(x)=(-1)^{m+1}m!\sum_{n=0}^{\infty}\frac {1}{(n+x)^{m+1}}\), we get
 □
Theorem 3.2
For \(k>0\), the function \(\psi_{k} (\frac{1}{x} )\) is strictly concave on \((0,\infty)\).
Proof
Easy computation results in
Considering Theorem 3.1, we complete the proof. □
Theorem 3.3
For \(k\geqslant\frac{1}{\sqrt[3]{3}}=0.693361\ldots\) , the function
is strictly concave on \((0,\infty)\).
Proof
By differentiation and applying Lemma 2.1, we easily obtain
and
Applying Lemma 2.2, \(k\geqslant\frac{1}{\sqrt[3]{3}}\), and the recurrence relations
we have
This implies that \(\lambda_{k}(x)\) is strictly concave on \((0,\infty)\). □
Theorem 3.4
For \(x \in(0,\infty)\) and \(k\geqslant\frac{1}{\sqrt[3]{3}}\), we have
Proof
Since the function \(\lambda_{k}(x)=\psi_{k}(x)+\psi_{k} (\frac{1}{x} )\) is strictly concave on \((0,\infty)\), we get
and
It follows that \(\lambda_{k}\) is increasing on \((0,1]\) and decreasing on \([1,\infty)\). Hence, \(\lambda_{k}(x)\leqslant\lambda_{k}(1)\) for \(x>0\). The proof is complete. □
Remark 3.1
Let \(\gamma_{k}=-\psi_{k}(1)=-\frac{\ln k}{k}-\frac{1}{k}\psi (\frac{1}{k} )\) be the k-analogue of the Euler–Mascheroni constant. It is obvious that \(\lim_{k\rightarrow1}\gamma_{k}=\gamma\).
Definition 3.1
It is known that the generalized digamma function \(\psi_{k}(x)\) is strictly increasing on \((0,\infty)\) with \(\psi_{k}(0^{+})\psi_{k}(\infty)<0\). So, the function has a sole positive root in \((0,\infty)\). We define this positive root for \(x_{k}\). That is,
Theorem 3.5
For \(x\in(0,1)\) and \(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\), we have
Proof
Considering \(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\) and the definition of \(x_{k}\), we have
where \(x_{0}\) satisfies \(\psi(x_{0})=0\) with \(x_{0}=1.46163\ldots\) .
Case 1. If \(x\in[x_{k}-1,1)\), then we have \(\psi_{k}(1-x)\leqslant0 \leqslant\psi_{k}(1+x)\). This implies that formula (3.2) holds.
Case 2. If \(x\in(0,x_{k}-1]\), using the power series expansion
we obtain
where \(\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\) is the Riemann zeta function.
Furthermore, we have
Completely similar to (3.4), we have
Combining (3.4) with (3.5), we obtain
by using \(\zeta(3)x^{2}<1\). □
Theorem 3.6
For \(x\in(0,\infty)\) and \(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\), we have
Proof
We only need to prove (3.6) for \(x\geqslant1\). If \(x\geqslant x_{k}\), then we get \(\psi_{k} (\frac{1}{x} )\leqslant 0 \leqslant\psi_{k}(x)\). It follows that inequality (3.6) holds true.
If \(x\in(1,x_{k}]\) and setting \(x=1+z\), we get
Therefore, we have
by using Theorem 3.5. □
Corollary 3.1
For \(x\in(0,\infty)\) and \(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\), we have
Proof
Applying Theorems 3.4 and 3.6, we obtain
The proof is complete. □
Next, for \(m,n,j\in\mathbb{N}\), we define the function \(\mu_{n}\) by
Completely similar to the method in [11], the following Theorem 3.7 can be proved.
Theorem 3.7
For \(m,n,j\in\mathbb{N}\), then \((-1)^{(n+1)(m+1)}\mu_{n}(x)\) is completely monotonic on \((0,\infty)\).
Proof
Using Lemma 2.3, we have
where δ is a permutation on \(0,1,2,\ldots,n\).
Let \(\operatorname{sgn}(\delta)\) be the sign of δ, we can obtain
Replacing \(u_{0},u_{1},\ldots, u_{n}\) by \(-u_{0},-u_{1},\ldots, -u_{n}\), we get
This implies that \((-1)^{(n+1)(m+1)}\mu_{n}(x)\) is completely monotonic. □
By taking \(n=1\), the following Corollary 3.2 can be easily obtained.
Corollary 3.2
For \(m,j\in\mathbb{N}\), and \(x>0\), we have
4 Conclusions
We established a concave theorem and some monotonic properties for the generalized digamma function, and some interesting inequalities were obtained. These conclusions generalize Alzer’s results. On the other hand, we prove a completely monotonic property for the generalized digamma function by using Ismail and Laforgia’s idea.
5 Methods and experiment
Not applicable.
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Acknowledgements
The authors would like to thank the editor and the anonymous referee for their valuable suggestions and comments, which helped us to improve this paper greatly.
Funding
The authors were supported by the National Natural Science Foundation of China (Grant Nos. 11401041, 11705122), the Science and Technology Foundations of Shandong Province (Grant Nos. J16li52 and J14li54) and Science Foundations of Binzhou University (Grant Nos. BZXYL1104 and BZXYL1704).
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Yin, L., Huang, LG., Lin, XL. et al. Monotonicity, concavity, and inequalities related to the generalized digamma function. Adv Differ Equ 2018, 246 (2018). https://doi.org/10.1186/s13662-018-1695-7
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DOI: https://doi.org/10.1186/s13662-018-1695-7
MSC
- 33B15
Keywords
- Generalized digamma function
- Complete monotonicity
- Inequality
- Concavity