In [36], Elezović et al. derived that
$$ \psi _{1}(x)< e^{-\psi (x)} $$
(4.1)
by the fact that the function \(e^{\psi (x+t)}-x\) is decreasing on \((0,\infty )\) for all \(t>0\). In addition, [26, Lemma 1.2] provides a different proof of (4.1). Some extensions of (4.1) for higher-order derivatives of \(\psi (x)\) can be found in [25, 27]. For example, it is given in [27, Theorem 2.1] that the inequality (4.1) was generalized to
$$ e^{-n\psi (x+\beta )}< \frac{\psi _{n}(x)}{(n-1)!}< e^{-n\psi (x+ \alpha )} $$
(4.2)
for \(x>0\) and \(n\in \mathbb{N}\), where \(\beta =1/2\) and \(\alpha =0\). In particular, inequality (4.2) was proved again by using monotonicity of functions involving the polygamma functions (see [37, Corollary 1]).
We introduce the divided differences of psi and polygamma functions (see [38]). For \(n\in \mathbb{N}\), \(s,t\in \mathbb{R} \), \(r=\min \{s,t\}\) and \(x\in (-r,\infty )\), we define
$$ \phi _{n}(x)= \textstyle\begin{cases} \frac{\psi _{n-1}(x+s)-\psi _{n-1}(x+t)}{t-s}, & t\neq s; \\ \psi _{n}(x+t), & t=s. \end{cases} $$
(4.3)
For the sake of consistency, we set \(\psi _{0}(x)=-\psi (x)\).
Using Theorem 3.1 and inequality (4.2), we establish the following result.
Corollary 3
For \(\beta \in \mathbb{R}\) and \(n\in \mathbb{N}\), let the function \(f_{3}(x)=(n-1)!e^{-n\psi (x+\beta )}-\psi _{n}(x)\) be defined on \((\max (-\beta ,0),\infty )\). Then the function \(f_{3}(x)\) is decreasing on \((-\beta ,\infty )\) if \(\beta \leq 0\); and is increasing on \((0,\infty )\) if \(\beta \geq \frac{1}{2}\).
Proof
A simple computation gives
$$ f'_{3}(x)=\psi _{n+1}(x)-n(n-1)!\psi _{1}(x+\beta )e^{-n\psi (x+ \beta )}. $$
For \(\beta \leq 0\), from the right-hand side of (4.2), we get
$$ f'_{3}(x)< \psi _{n+1}(x)-n\psi _{1}(x+\beta )\psi _{n}(x+\beta ), $$
and therefore, in the view of Theorem 3.1, we have \(f'_{3}(x)<0\). By the same spirit, the left-hand side of (4.2) and Theorem 3.1 imply the case \(\beta \geq \frac{1}{2}\).
This completes the proof. □
Remark 4
For \(\lambda \neq 0\), \(s,t\in \mathbb{R} \) and \(r=\min \{s,t\}\), define the function Ψ for \(x\in (-r,\infty )\)
$$ \Psi (x;\lambda ,s,t)= \textstyle\begin{cases} [\frac{\Gamma (x+t)}{\Gamma (x+s)} ]^{ \frac{1}{\lambda (t-s)}}, & t\neq s; \\ e^{\frac{1}{\lambda }\psi (x+s)}, & t=s. \end{cases} $$
(4.4)
It was shown in [36] that the function \(\Psi (x;1,s,t)\) is convex on \((-r,\infty )\) for \(|t-s|<1\) and concave on the same interval for \(|t-s|>1\). Since
$$ \Psi ''(x;\lambda ,s,t)= \frac{1}{\lambda ^{2}}\Psi (x; \lambda ,s,t) \bigl( \phi _{1}^{2}(x)-\lambda \phi _{2}(x)\bigr), $$
we deduce from [39, Theorem 3.1]:
(1) For \(0<|t-s|<1\), the function \(\Psi (x;\lambda ,s,t)\) is convex on \((-r,\infty )\) if and only if \(\lambda \neq 0 \leq 1\) and concave on the same interval if and only if \(\lambda \geq \frac{1}{|t-s|}\);
(2) For \(|t-s|>1\), the function \(\Psi (x;\lambda ,s,t)\) is convex on \((-r,\infty )\) if and only if \(\lambda \neq 0 \leq \frac{1}{|t-s|}\) and concave on the same interval if and only if \(\lambda \geq 1\);
(3) For \(|t-s|=1\), the function \(\Psi (x;\lambda ,s,t)\) is convex on \((-r,\infty )\) if and only if \(\lambda \neq 0 \leq 1\) and concave on the same interval if and only if \(\lambda \geq 1\);
(4) For \(s=t\), the function \(\Psi (x;\lambda ,s,t)\) is convex on \((-r,\infty )\) if and only if \(\lambda \neq 0 \leq 1\).
In addition, it was proved in [36] that
$$ \Psi (x;1,s,t) \phi _{1}(x)< 1 $$
(4.5)
holds for \(x>-r\) if \(|t-s|<1\) and its reversed inequality is valid on \((-r,\infty )\) if \(|t-s|>1\). Obviously, (4.5) is a generalization of (4.1).
In the following, we will prove the monotonicity of the function \(z(x;\lambda ,s,t)=\Psi (x;\lambda ,s,t) \phi _{n}(x)\) and therefore extend (4.5) or the right-hand side of (4.2).
Theorem 4.1
For \(\lambda \neq 0\), \(s, t \in \mathbb{R}\), \(r=\min (s,t)\) and \(n \in \mathbb{N}\), the function \(z(x;\lambda ,s,t)\) has the following monotonic properties:
(1) For \(0<|t-s|<1\), the function \(z(x;\lambda ,s,t)\) is increasing on \((-r,\infty )\) if and only if \(1/\lambda \geq n\) and decreasing on the same interval if and only if \(1/\lambda \leq n|t-s|\);
(2) For \(|t-s|>1\), the function \(z(x;\lambda ,s,t)\) is increasing on \((-r,\infty )\) if and only if \(1/\lambda \geq n|t-s|\) and decreasing on the same interval if and only if \(1/\lambda \leq n\);
(3) For \(|t-s|=1\), the function \(z(x;\lambda ,s,t)\) is increasing on \((-r,\infty )\) if and only if \(1/\lambda \geq n\) and decreasing on the same interval if and only if \(1/\lambda \leq n\);
(4) For \(s=t\), the function \(z(x;\lambda ,s,t)\) is increasing on \((-r,\infty )\) if and only if \(1/\lambda \geq n\) and decreasing on the same interval if and only if \(1/\lambda \leq 0\).
Proof
Differentiating \(z(x;\lambda ,s,t)\) yields
$$ z'(x;\lambda ,s,t)=\Psi (x;\lambda ,s,t) \biggl(\frac{1}{\lambda } \phi _{1}(x) \phi _{n}(x)-\phi _{n+1}(x) \biggr). $$
This in combination with Theorem [39, Theorem 3.1] easily establishes the Theorem. □
Using Theorem 4.1, we have the following:
Corollary 4
For \(s,t\in \mathbb{R} \), \(r=\min \{s,t\}\) and \(n \in \mathbb{N}\), we have the inequality
$$ \Psi \biggl(x;\frac{1}{n},s,t \biggr) \phi _{n}(x)< (n-1)! $$
(4.6)
for \(x>-r\) if \(|t-s|<1\) and its reversed inequality is valid on \((-r,\infty )\) if \(|t-s|>1\).
Proof
Obviously, we only assume \(s\neq t\). In view of Theorem 4.1, we only need to check
$$ \lim_{x \rightarrow \infty } \Psi \biggl(x;\frac{1}{n},s,t \biggr) \phi _{n}(x)=(n-1)!. $$
(4.7)
Applying the asymptotic formula (1.5), we obtain
$$ \lim_{x \rightarrow \infty }\frac{e^{n\psi (x)}}{x^{n}}=1. $$
Therefore, this together with [24, Lemma 4] establishes
$$ \lim_{x \rightarrow \infty }e^{n\psi (x+c)} \phi _{n}(x)=(n-1)!, \quad \text{for all } c\in \mathbb{R}. $$
(4.8)
According to [13, Corollary 1.4], the inequality
$$ e^{\psi (x+r)}< \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{1/(t-s)}< e^{ \psi (x+\frac{s+t}{2} )} $$
(4.9)
holds for \(x>-r\), so that this combined with (4.8) yields (4.7).
Hence we complete the proof of this Theorem. □
Theorem 4.2
For \(s,t\in \mathbb{R} \), \(r=\min \{s,t\}\) and \(c\in (-r,\infty )\), we have the double inequality
$$ \frac{e^{G_{s,t}(X_{s,t})}}{e^{H_{s,t}(X_{s,t})}}< \frac{e^{G_{s,t}(x)}}{e^{H_{s,t}(x)}}< \frac{\sqrt{2\pi e} e^{A_{c,s,t}-(s+t)/2}}{\Gamma (c+\frac{s+t}{2} )} $$
for \(x>X_{s,t}\) if \(|t-s|<1\) and its reversed inequality is valid on \((X_{s,t}, \infty )\) if \(|t-s|>1\), where \(X_{s,t}\) is the only zero of \(1+\ln \Psi (x;1,s,t)\) on \((-r,\infty )\),
$$\begin{aligned}& G_{s,t}(x)= \textstyle\begin{cases} \frac{1}{t-s}\int _{c}^{x} \ln [ \frac{\Gamma (u+t)}{\Gamma (u+s)} ]\,du, & t\neq s, \\ \int _{c}^{x} \psi (u+s)\,du, & t=s; \end{cases}\displaystyle \\& A_{c,s,t}= \textstyle\begin{cases} \int _{c}^{\infty } \frac{1}{t-s}\ln [ \frac{\Gamma (u+t)}{\Gamma (u+s)} ]-\psi (u+\frac{s+t}{2} )\,du, & t\neq s, \\ 0, & t=s; \end{cases}\displaystyle \end{aligned}$$
(4.10)
and
$$ H_{s,t}(x)=\Psi (x;1,s,t)\ln \Psi (x;1,s,t)-x. $$
Proof
Let \(g_{s,t}(x)=e^{G_{s,t}(x)}\), \(h_{s,t}(x)=\ln \Psi (x;1,s,t) \) and \(f_{s,t}(x)=g_{s,t}(x)e^{x-h_{s,t}(x) e^{h_{s,t}(x)}}\). Since \(g_{s,t}'(x)=g_{s,t}(x) h_{s,t}(x)\) and \(h_{s,t}'(x)=\phi _{1}(x)\), we obtain
$$ f_{s,t}'(x)=g_{s,t}(x)e^{x-h_{s,t}(x) e^{h_{s,t}(x)}} \bigl(1-\phi _{1}(x)e^{h_{s,t}(x)}\bigr) \bigl(1+h_{s,t}(x) \bigr). $$
(4.11)
Using the asymptotic formula (see [4])
$$ \frac{\Gamma (x+t)}{\Gamma (x+s)}=x^{t-s} \biggl(1- \frac{(s-t)(s+t-1)}{2x}+O \biggl(\frac{1}{x^{2}} \biggr) \biggr), \quad x \rightarrow \infty , $$
(4.12)
we get \(\lim_{x\rightarrow \infty }h_{s,t}(x)=\infty \), and therefore by \(h_{s,t}'(x)>0\) and \(\lim_{x\rightarrow -r}h_{s,t}(x)=-\infty \), we conclude that \(1+h_{s,t}(x)\) has a unique zero on \((-r,\infty )\).
Hence thanks to \(h_{s,t}'(x)>0\), Corollary 4 and (4.11), we have the following statements:
(i) For \(|t-s|<1\), \(f_{s,t}(x)\) is increasing on \((X_{s,t},\infty )\) and decreasing on \((-r,X_{s,t})\),
(ii) For \(|t-s|>1\), \(f_{s,t}(x)\) is decreasing on \((X_{s,t},\infty )\) and increasing on \((-r,X_{s,t})\).
On the one hand, we check that
$$\begin{aligned}& \lim_{x\rightarrow \infty } \int _{c}^{x} \psi \biggl(u+\frac{s+t}{2} \biggr)\,du +x- h_{s,t}(x)e^{h_{s,t}(x)} \\& \quad =\frac{1}{2}+ \frac{1}{2} \ln (2\pi )- \frac{s+t}{2}- \ln \Gamma \biggl(c+ \frac{s+t}{2} \biggr). \end{aligned}$$
(4.13)
Case 1. \(s\neq t\). Now we derive the asymptotic formula of \(h_{s,t}(x)e^{h_{s,t}(x)}\). Taking the logarithm in (4.12), we get
$$ \begin{aligned} h_{s,t}(x)&= \frac{ [\ln \Gamma (x+t)-\ln \Gamma (x+s) ]}{t-s} \\ & =\ln x+\frac{1}{t-s}\ln \biggl(1-\frac{(s-t)(s+t-1)}{2x}+O \biggl( \frac{1}{x^{2}} \biggr) \biggr),\quad x\rightarrow \infty . \end{aligned} $$
(4.14)
Together with
$$ \ln (1+x)=x-\frac{x^{2}}{2}+O\bigl(x^{3}\bigr),\quad x\rightarrow 0, $$
we can rewrite (4.14) as
$$\begin{aligned} h_{s,t}(x)&= \frac{ [\ln \Gamma (x+t)-\ln \Gamma (x+s) ]}{t-s} \\ &=\ln x+ \frac{t+s-1}{2x}+O \biggl(\frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , \end{aligned}$$
(4.15)
which implies that
$$ h_{s,t}(x)e^{h_{s,t}(x)}=x \biggl(\ln x+ \frac{t+s-1}{2x}+O \biggl( \frac{1}{x^{2}} \biggr) \biggr)e^{\frac{t+s-1}{2x}+O ( \frac{1}{x^{2}} )}, \quad x \rightarrow \infty . $$
(4.16)
Therefore, by the aid of
$$ e^{x}=1+x+O\bigl(x^{2}\bigr),\quad x\rightarrow 0, $$
we obtain
$$ \begin{aligned} h_{s,t}(x)e^{h_{s,t}(x)}={}&x \ln x+\frac{t+s-1}{2}\ln x+ \frac{t+s-1}{2}+\frac{(t+s-1)^{2}}{4x} \\ &{} +O \biggl(\frac{\ln x}{x} \biggr)+O \biggl(\frac{1}{x^{2}} \biggr),\quad x \rightarrow \infty . \end{aligned} $$
(4.17)
Combining (1.4) with (4.17), we deduce that
$$\begin{aligned} &\ln \Gamma \biggl(x+\frac{s+t}{2} \biggr)+x- h_{s,t}(x)e^{h_{s,t}(x)} \\ \begin{aligned}&\quad =x \ln \biggl(1+\frac{s+t}{2x} \biggr) + \frac{t+s-1}{2} \ln \biggl(1+ \frac{s+t}{2x} \biggr) \\ &\qquad {}+\frac{1}{2}-(t+s) +\frac{1}{2} \ln (2\pi )+\frac{1}{12} \frac{1}{x+\frac{s+t}{2}} \end{aligned} \\ &\qquad {}-\frac{(t+s-1)^{2}}{4x}+O \biggl(\frac{\ln x}{x} \biggr)+O \biggl( \frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , \end{aligned}$$
(4.18)
which implies (4.13).
Case 2. \(s=t\). Using (1.4) and the asymptotic formula (see [4])
$$ \psi \biggl(x+\frac{s+t}{2} \biggr)=\ln x+ \frac{s+t-1}{2x}+O \biggl( \frac{1}{x^{2}} \biggr),\quad x\rightarrow \infty , $$
(4.19)
we can easily prove (4.13).
On the other hand, we show that
$$ \lim_{x\rightarrow \infty } \biggl( G_{s,t}(x)- \int _{c}^{x} \psi \biggl(u+ \frac{s+t}{2} \biggr)\,du \biggr)= \textstyle\begin{cases} A_{c,s,t}, & t\neq s, \\ 0, & t=s; \end{cases} $$
(4.20)
where
$$ \int _{c}^{\infty } \frac{1}{t-s}\ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]-\psi \biggl(u+\frac{s+t}{2}\biggr)\,du=A_{c,s,t}. $$
Note that the case \(t=s\) is obvious. Then using (4.15) and (4.19), we get
$$ \frac{1}{t-s}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]- \psi \biggl(x+ \frac{s+t}{2} \biggr)=O \biggl(\frac{1}{x^{2}} \biggr),\quad x \rightarrow \infty , $$
which implies the exitance of constants C and \(X>0\) such that
$$ \biggl\vert \frac{1}{t-s}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]-\psi \biggl(x+\frac{s+t}{2} \biggr) \biggr\vert \leq C \biggl\vert \frac{1}{x^{2}} \biggr\vert $$
for all \(x>X\). It follows that
$$ \lim_{x\rightarrow \infty } \int _{x}^{\infty }\frac{1}{t-s}\ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]-\psi \biggl(u+\frac{s+t}{2} \biggr)\,du=0, $$
so that \(A_{c,s,t}\) is well defined. Hence, (4.20) is proved.
Finally, taking into consideration (4.13) and (4.20), we have
$$\begin{aligned}& \lim_{x\rightarrow \infty } \bigl( G_{s,t}(x)+x- h_{s,t}(x)e^{h_{s,t}(x)} \bigr) \\& \quad = \textstyle\begin{cases} \frac{1}{2}+\frac{1}{2} \ln (2\pi )-\frac{s+t}{2}- \ln \Gamma (c+ \frac{s+t}{2} )+A_{c,s,t}, \quad t\neq s; \\ \frac{1}{2}+\frac{1}{2} \ln (2\pi )-s- \ln \Gamma (c+s ),\quad t=s. \end{cases}\displaystyle \end{aligned}$$
(4.21)
Applying the monotonicity of \(f_{s,t}(x)\) and (4.21), we complete the proof of this Theorem. □
Remark 5
Let \(0.785003\leq s < t\). Using the inequality (see [13, Corollary 1.4])
$$ \psi (x+s)< \frac{1}{t-s} \bigl[\ln \Gamma (x+t)-\ln \Gamma (x+s) \bigr]< \psi \biggl(x+\frac{s+t}{2} \biggr),\quad x>-s, $$
we have \(1+h_{s,t}(0)>1+\psi (s)\geq 0\), so that by \(h_{s,t}'(x)>0\), Corollary 4 and (4.11) again, we conclude that \(f_{s,t}(x)\) is increasing on \((0,\infty )\) if \(|t-s|<1\) and decreasing on the same interval if \(|t-s|>1\). Similarly, we have the inequality
$$ \begin{aligned} \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{ \frac{1}{t-s}}\ln \biggl[\frac{\Gamma (x+t)}{\Gamma (x+s)} \biggr]^{ \frac{1}{t-s}}< {}& \biggl[\frac{\Gamma (t)}{\Gamma (s)} \biggr]^{ \frac{1}{t-s}}\ln \biggl[\frac{\Gamma (t)}{\Gamma (s)} \biggr]^{ \frac{1}{t-s}}+x \\ &{} + \frac{1}{t-s} \int _{0}^{x} \ln \biggl[ \frac{\Gamma (u+t)}{\Gamma (u+s)} \biggr]\,du \end{aligned} $$
for \(x>0\) if \(|t-s|<1\) and its reversed inequality is valid on \((0,\infty )\) if \(|t-s|>1\).