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
$$\psi_{k}'(x)=\frac{1}{k^{2}}\psi' \biggl(\frac{x}{k}\biggr) \quad \mbox{and}\quad \psi _{k}''(x)= \frac{1}{k^{3}}\psi''\biggl(\frac{x}{k} \biggr). $$
Combining with the identity \(\psi^{(m)}(x)=(-1)^{m+1}m!\sum_{n=0}^{\infty}\frac {1}{(n+x)^{m+1}}\), we get
$$\begin{aligned} \frac{d}{dx} \bigl(x^{2}\psi_{k}'(x) \bigr) =&\frac{2x}{k^{2}} \psi' \biggl(\frac{x}{k} \biggr)+ \frac{x^{2}}{k^{3}} \psi '' \biggl(\frac{x}{k} \biggr) \\ =&2x\sum_{n=0}^{\infty}\frac{nk}{(nk+x)^{3}}>0. \end{aligned}$$
□
Theorem 3.2
For
\(k>0\), the function
\(\psi_{k} (\frac{1}{x} )\)
is strictly concave on
\((0,\infty)\).
Proof
Easy computation results in
$$\frac{d}{dx} \biggl(\psi_{k} \biggl(\frac{1}{x} \biggr) \biggr)=-\frac {1}{x^{2}}\psi_{k}' \biggl( \frac{1}{x} \biggr). $$
Considering Theorem 3.1, we complete the proof. □
Theorem 3.3
For
\(k\geqslant\frac{1}{\sqrt[3]{3}}=0.693361\ldots\) , the function
$$\lambda_{k}(x)=\psi_{k}(x)+\psi_{k} \biggl( \frac{1}{x} \biggr) $$
is strictly concave on
\((0,\infty)\).
Proof
By differentiation and applying Lemma 2.1, we easily obtain
$$\begin{aligned}& \lambda_{k}'(x)=\psi_{k}'(x)- \frac{1}{x^{2}}\psi_{k}' \biggl(\frac{1}{x} \biggr), \\& \lambda_{k}''(x)=\psi_{k}''(x)+ \frac{2}{x^{3}}\psi_{k}' \biggl(\frac {1}{x} \biggr)+\frac{1}{x^{4}}\psi_{k}'' \biggl( \frac{1}{x} \biggr), \end{aligned}$$
and
$$k^{3}x^{4}\lambda_{k}''(x)=x^{4} \psi'' \biggl(\frac{x}{k} \biggr)+2kx \psi' \biggl(\frac{1}{kx} \biggr)+\psi'' \biggl(\frac{1}{kx} \biggr). $$
Applying Lemma 2.2, \(k\geqslant\frac{1}{\sqrt[3]{3}}\), and the recurrence relations
$$\begin{aligned}& \psi' \biggl(\frac{1}{kx}+1 \biggr)=\psi' \biggl(\frac{1}{kx} \biggr)-k^{2}x^{2}, \\& \psi'' \biggl(\frac{1}{kx}+1 \biggr)= \psi'' \biggl(\frac{1}{kx} \biggr)+2k^{3}x^{3}, \end{aligned}$$
we have
$$\begin{aligned} k^{3}x^{4}\lambda_{k}''(x) =&x^{4} \psi'' \biggl(\frac{x}{k} \biggr)+2kx\psi ' \biggl(\frac{1}{kx} \biggr)+\psi'' \biggl(\frac{1}{kx} \biggr) \\ < & x^{4} \biggl(-\frac{k^{2}}{x^{2}}-\frac{k^{3}}{x^{3}} \biggr)+2kx \biggl[\frac {kx}{1+kx}+\frac{k^{2}x^{2}}{2(1+kx)^{2}} +\frac{k^{3}x^{3}}{6(1+kx)^{3}} \biggr] \\ &{}-\frac{k^{2}x^{2}}{(1+kx)^{2}} -\frac{k^{3}x^{3}}{(1+kx)^{3}} \\ =&-\frac{kx}{3(1+kx)^{3}} \bigl[3k^{2}+9k^{3}x+9k^{2}x^{2}+k^{2} \bigl(3k^{3}-1\bigr)x^{3}+3k^{4}x^{4} \bigr] \\ < &0. \end{aligned}$$
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
$$ \psi_{k}(x)+\psi_{k} \biggl(\frac{1}{x} \biggr)\leqslant\frac{2\ln k+2\psi (\frac{1}{k} )}{k}. $$
(3.1)
Proof
Since the function \(\lambda_{k}(x)=\psi_{k}(x)+\psi_{k} (\frac{1}{x} )\) is strictly concave on \((0,\infty)\), we get
$$\lambda_{k}'(x)\geqslant\lambda_{k}'(1)=0, \quad x\in(0,1], $$
and
$$\lambda_{k}'(x)\leqslant\lambda_{k}'(1)=0, \quad x\in[1,\infty). $$
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,
$$\ln k +\psi \biggl(\frac{x_{k}}{k} \biggr)=0. $$
Theorem 3.5
For
\(x\in(0,1)\)
and
\(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\), we have
$$ \psi_{k}(1+x)\psi_{k}(1-x)\leqslant \frac{\ln^{2} k+\gamma^{2}-2(\gamma +1)\ln k}{k^{2}}. $$
(3.2)
Proof
Considering \(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\) and the definition of \(x_{k}\), we have
$$\frac{1}{\sqrt[3]{3}}x_{0}\leqslant x_{k} \leqslant x_{0}, $$
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
$$ \psi(1+z)=-\gamma+\sum_{k=2}^{\infty}(-1)^{k} \zeta(k)z^{k-1},\quad |z|< 1, $$
(3.3)
we obtain
$$\begin{aligned} \psi_{k}(1+x) \geqslant&\psi_{k}(k+x)=\frac{\ln k}{k}+ \frac{1}{k}\psi \biggl(1+\frac{x}{k} \biggr) \\ =& \frac{\ln k}{k}+\frac{1}{k} \Biggl[-\gamma+\sum _{k=2}^{\infty }(-1)^{k}\zeta(k)x^{k-1} \Biggr], \end{aligned}$$
where \(\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\) is the Riemann zeta function.
Furthermore, we have
$$ 0< -\psi_{k}(1+x)\leqslant-\frac{\ln k}{k}+ \frac{1}{k}\bigl[\gamma-\zeta (2)x+\zeta(3)x^{2}\bigr]. $$
(3.4)
Completely similar to (3.4), we have
$$\begin{aligned} 0 < &-\psi_{k}(1-x)\leqslant -\frac{\ln k}{k}+\frac{1}{k} \Biggl[\gamma +\zeta(2)y+\zeta(3)\sum_{k=2}^{\infty}x^{k} \Biggr] \\ \leqslant& -\frac{\ln k}{k}+\frac{1}{k}\bigl[\gamma+\zeta(2)x+ \zeta(3)x^{2}\bigr]. \end{aligned}$$
(3.5)
Combining (3.4) with (3.5), we obtain
$$ \psi_{k}(1+x)\psi_{k}(1-x)\leqslant\frac{\ln^{2} k+\gamma^{2}-2(\gamma +1)\ln k}{k^{2}} $$
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
$$ \psi_{k}(x)\cdot\psi_{k} \biggl( \frac{1}{x} \biggr)\leqslant\frac{\ln^{2} k+\gamma^{2}-2(\gamma+1)\ln k}{k^{2}}. $$
(3.6)
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
$$\psi_{k}(1-z)\leqslant\psi_{k} \biggl(\frac{1}{x} \biggr). $$
Therefore, we have
$$\begin{aligned} \psi_{k}(x)\cdot\psi_{k} \biggl(\frac{1}{x} \biggr) =& \psi_{k}(1+z)\psi _{k} \biggl(\frac{1}{x} \biggr) \\ \leqslant& \psi_{k}(1+z)\psi_{k}(1-z) \\ \leqslant& \frac{\ln^{2} k+\gamma^{2}-2(\gamma+1)\ln k}{k^{2}} \end{aligned}$$
by using Theorem 3.5. □
Corollary 3.1
For
\(x\in(0,\infty)\)
and
\(\frac{1}{\sqrt[3]{3}}\leqslant k \leqslant1\), we have
$$ \frac{2\psi_{k}(x)\psi_{k} (\frac{1}{x} )}{\psi_{k}(x)+\psi _{k} (\frac{1}{x} )}\geqslant\frac{\ln^{2} k+\gamma^{2}-2(\gamma +1)\ln k}{k[\ln k +\psi (\frac{1}{k} )]}. $$
(3.7)
Proof
Applying Theorems 3.4 and 3.6, we obtain
$$\begin{aligned} \frac{2\psi_{k}(x)\psi_{k} (\frac{1}{x} )}{\psi_{k}(x)+\psi _{k} (\frac{1}{x} )} \geqslant& 2\cdot\frac{\ln^{2} k+\gamma^{2}-2(\gamma+1)\ln k}{k^{2}}\frac{1}{\psi _{k}(x)+\psi_{k} (\frac{1}{x} )} \\ \geqslant& \frac{\ln^{2} k+\gamma^{2}-2(\gamma+1)\ln k}{k^{2}}\frac {k}{\ln k+\psi (\frac{1}{k} )}. \end{aligned}$$
The proof is complete. □
Next, for \(m,n,j\in\mathbb{N}\), we define the function \(\mu_{n}\) by
$$ \mu_{n}(x)= \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} \psi_{k}^{(m)}(x) &\psi_{k}^{(m+j)}(x)&\cdots&\psi_{k}^{(m+nj)}(x)\\ \psi_{k}^{(m+j)}(x)&\psi_{k}^{(m+2j)}(x)&\cdots&\psi_{k}^{[m+(n+1)j]}(x) \\ \vdots&\vdots& & \vdots\\ \psi_{k}^{(m+nj)}(x)&\psi_{k}^{(m+(n+1)j)}(x)& \cdots&\psi_{k}^{(m+2nj)}(x) \end{array}\displaystyle \right \vert . $$
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
$$\begin{aligned} \mu_{n}(x) =&(-1)^{n+1}\underbrace{ \int_{-\infty}^{0}\cdots \int _{-\infty}^{0}}_{n+1\text{ times}} \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} u_{0}^{m} &u_{0}^{m+j}&\cdots&u_{0}^{m+nj}\\ u_{1}^{m+j}&u_{1}^{m+2j}&\cdots&u_{1}^{m+(n+1)j} \\ \vdots&\vdots& & \vdots\\ u_{n}^{m+nj}&u_{n}^{m+(n+1)j}& \cdots&u_{n}^{m+2nj} \end{array}\displaystyle \right \vert \\ &{}\cdot\frac{e^{\frac{x}{k}(u_{0}+u_{1}+\cdots+u_{n})}}{\prod_{i=0}^{n}(1-e^{u_{i}})}\,du_{0}\,du_{1}\cdots \,du_{n} \\ =&(-1)^{n+1}\underbrace{ \int_{-\infty}^{0}\cdots \int_{-\infty }^{0}}_{n+1 \text{ times}} \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} u_{\delta(0)}^{m} &u_{\delta(0)}^{m+j}&\cdots&u_{\delta(0)}^{m+nj}\\ u_{\delta(1)}^{m+j}&u_{\delta(1)}^{m+2j}&\cdots&u_{\delta (1)}^{m+(n+1)j} \\ \vdots&\vdots& & \vdots\\ u_{\delta(n)}^{m+nj}&u_{\delta(n)}^{m+(n+1)j}& \cdots&u_{\delta(n)}^{m+2nj} \end{array}\displaystyle \right \vert \\ &{}\cdot\frac{e^{\frac{x}{k}(u_{0}+u_{1}+\cdots+u_{n})}}{\prod_{i=0}^{n}(1-e^{u_{i}})}\,du_{0}\,du_{1}\cdots \,du_{n}, \end{aligned}$$
where δ is a permutation on \(0,1,2,\ldots,n\).
Let \(\operatorname{sgn}(\delta)\) be the sign of δ, we can obtain
$$\begin{aligned} u_{n}(x) =&(-1)^{n+1}\underbrace{ \int_{-\infty}^{0}\cdots \int_{-\infty }^{0}}_{n+1 \text{ times}} \frac{e^{\frac{x}{k}(u_{0}+u_{1}+\cdots+u_{n})}}{\prod_{i=0}^{n}(1-e^{u_{i}})} \operatorname{sgn}(\delta)\prod_{i=0}^{n}u_{i}^{m} \\ &{}\cdot \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} u_{0}^{0} &u_{0}^{j}&\cdots&u_{0}^{nj}\\ u_{1}^{j}&u_{1}^{2j}&\cdots&u_{1}^{(n+1)j} \\ \vdots&\vdots& & \vdots\\ u_{n}^{nj}&u_{n}^{(n+1)j}& \cdots&u_{n}^{2nj} \end{array}\displaystyle \right \vert \,du_{0} \,du_{1}\cdots \,du_{n} \\ =&\frac{(-1)^{n+1}}{(n+1)!}\underbrace{ \int_{-\infty}^{0}\cdots \int _{-\infty}^{0}}_{n+1 \text{ times}} \frac{e^{\frac{x}{k}(u_{0}+u_{1}+\cdots+u_{n})}}{\prod_{i=0}^{n}(1-e^{u_{i}})}(u_{0}u_{1} \cdots u_{n})^{m} \\ &{}\cdot\prod_{0\leqslant i< l\leqslant n}\bigl(u_{i}^{j}-u_{l}^{j} \bigr)\,du_{0}\,du_{1}\cdots \,du_{n}. \end{aligned}$$
Replacing \(u_{0},u_{1},\ldots, u_{n}\) by \(-u_{0},-u_{1},\ldots, -u_{n}\), we get
$$\begin{aligned} \mu_{n}(x) =&{(-1)^{(n+1)(m+1)}}\underbrace{ \int_{0}^{\infty}\cdots \int_{0}^{\infty}}_{n+1 \text{ times}} e^{-\frac{x}{k}(u_{0}+u_{1}+\cdots+u_{n})} \\ &{}\cdot\prod_{0\leqslant i< l\leqslant n}\bigl(u_{i}^{j}-u_{l}^{j} \bigr){\prod_{i=0}^{n}\frac {u_{i}^{n}}{1-e^{-u_{i}}}} \,du_{0}\,du_{1}\cdots \,du_{n}. \end{aligned}$$
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
$$ \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{}} \psi_{k}^{(m)}(x) &\psi_{k}^{(m+j)}(x)\\ \psi_{k}^{(m+j)}(x)&\psi_{k}^{(m+2j)}(x) \end{array}\displaystyle \right \vert >0. $$