Notes on three conjectures involving the digamma and generalized digamma functions

Abstract

In the paper, we solve one conjecture on an inequality involving digamma function, an open problem, and a conjecture on monotonicity of functions involving generalized digamma function. We also prove a new inequality for digamma function.

1 Introduction

In the last years, the $(p,k)$-analogue of the gamma and polygamma functions has been studied intensively by a lot of authors. For historical background of the theory, see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24].

It is well known that:

• a function f is said to be completely monotonic [6, 21] on an interval I if f has derivatives of all orders on I and

  $$ (-1)^{n}f^{(n)}(x)\geq0 $$

  (1)

  for $x\in I$, $n\geq0$, $n\in N$ (due to $0\in N$).

• the Euler gamma function [14,15,16, 20, 22, 23] is defined by

  $$ \varGamma(x)= \int_{0}^{\infty}t^{x-1}e^{-t} \,dt $$

  (2)

  for $x>0$;

• the digamma function [11,12,13, 24] is defined by

  $$ \psi(x)=\frac{\varGamma'(x)}{\varGamma(x)}=-\gamma-\frac {1}{x}+\sum _{n=1}^{\infty}\frac{x}{n(n+x)}, $$

  (3)

  where γ is the Euler–Mascheroni constant [5].

Recently, Díaz and Pariguan [4] defined the generalized gamma function

$$ \varGamma_{k}(x)=\lim_{n\rightarrow\infty}\frac {n!k^{n}(nk)^{\frac{x}{k}-1}}{x(x+k)\cdots (x+(n-1)k)} $$

(4)

for $k>0$ and $x\in C\setminus kZ^{-}$ and the generalized digamma function

$$ \psi_{k}(x)=\frac{\varGamma_{k}'(x)}{\varGamma _{k}(x)}=\frac{\ln(k)-\gamma}{k}-\frac{1}{x}+\sum _{n=1}^{\infty }\frac{x}{nk(nk+x)}. $$

(5)

Very recently, Nantomah, Prempeh, and Twum [8] introduced a new definition of the $(p,k)$-gamma function

$$ \varGamma_{pk}(x)=\frac{(p+1)!k^{p+1}(pk)^{\frac {x}{k}-1}}{x(x+k)\cdots (x+pk)} $$

(6)

for $k>0$ and $x>0$, $p\geq0$, $p\in N$, and the $(p,k)$-digamma function

$$ \psi_{pk}(x)=\frac{\varGamma_{pk}'(x)}{\varGamma _{pk}(x)}=\frac{\ln(pk)}{k}-\sum _{n=0}^{p}\frac{1}{nk+x} $$

(7)

for $k>0$ and $x>0$, $p\geq0$, $p\in N$.

We note that

$$\begin{aligned}& \lim_{k\rightarrow1}\psi_{k}(x) = \psi(x),\qquad \lim _{k\rightarrow 1}\varGamma_{k}(x)=\varGamma(x), \\& \lim_{p\rightarrow\infty}\varGamma_{pk}(x) = \varGamma_{k}(x), \qquad \lim_{p\rightarrow\infty}\psi_{pk}(x)=\psi_{k}(x). \end{aligned}$$

Li Yin, Li-Guo Huang, Zhi-Min Song, and Xiang Kai Dou [19] posed the following conjecture.

Conjecture 1

([19])

For $p>0$ and $k\geq1$, the function

$$ \phi_{pk}(x)=\psi_{pk}(x)+\ln\bigl(e^{\frac{1}{x}-\frac {1}{x+pk+k}}-1\bigr) $$

is strictly decreasing from $(0,\infty)$ onto $(-\infty,\phi_{pk}(k))$.

Li Yin [17] posed the following open problem.

Open Problem 1

([17])

If the function

$$ \delta_{pk\alpha}(x)=x^{\alpha} \biggl[\frac {1}{k}\ln \frac{pkx}{x+k(p+1)}-\psi_{pk}(x) \biggr] $$

is completely monotonic on $(0,\infty)$, then is it true that $\alpha \leq1$?

Yuming Chu, Xiaoming Zhang, and Xiaoming Tang [3] posed the following conjecture.

Conjecture 2

For $b > a > 0$, we have

$$ \bigl(b-L(a, b)\bigr)\psi(b)+\bigl(L(a, b)-a\bigr)\psi(a)>(b-a)\psi (\sqrt{ba}), $$

where $L(a, b)=(b-a)/(\ln(b)-\ln(a))$.

The goal of the paper is to solve Conjecture 1, Conjecture 2, and Open Problem 1.

2 Methods

In this paper, we use methods of mathematical and numerical analysis. We also use the software MATLAB for some computing.

3 Results and discussion

In this section, we disprove Conjecture 1 (see [19]) and Conjecture 2 (see [3]) and prove one new inequality (Theorem 1) and Open Problem 1 (see [17]).

3.1 Disproving Conjecture 1

It is evident that $\phi_{pk}(x)$ is strictly decreasing only if $e^{\phi_{pk}(x)}$ is strictly decreasing. We have

$$ e^{\phi_{pk}(x)}(x)=v_{pk}(x)=e^{\psi _{pk}(x+k)}-e^{\psi_{pk}(x)}. $$

Using Matlab, we obtain Table 1.

Table 1 Values of p, k, $x_{1}$, $x_{2}$, $v_{pk}(x_{1})$, $v_{pk}(x_{2})$

The table shows that $v_{pk}(x_{1})< v_{pk}(x_{2})$ for $0< x_{1}< x_{2}$, $p=100\text{,}000$, $p=100\text{,}010$, $k=1.1$, $k=1.6$, $k=2.1$. So $\phi_{pk}(x)$ is not strictly decreasing on $(0,\infty)$ for $p>0$ and $k>1$.

Remark 1

We note that Conjecture 1 (see [19]) is false since $\lim_{x\rightarrow0^{+}}v'_{pk}(x)>0$ for $p\geq1$ and $k>0$.

Indeed, differentiation of $v_{pk}(x)$ yields

$$ v'_{pk}(x)=e^{\psi_{pk}(x+k)}\psi'_{pk}(x+k)-e^{\psi _{pk}(x)} \psi'_{pk}(x). $$

Because of

$$\begin{aligned} \lim_{x\rightarrow0^{+}}e^{\psi_{pk}(x)}\psi '_{pk}(x) =& \lim_{x\rightarrow0^{+}} e^{\frac{\ln(pk)}{k}-\sum_{n=1}^{p}\frac{1}{nk+x}}e^{-\frac {1}{x}} \Biggl( \frac{1}{x^{2}}+\sum_{n=1}^{p} \frac {1}{(nk+x)^{2}} \Biggr) \\ =&C_{pk}\lim_{t\rightarrow+\infty}e^{-t}t^{2}=0, \end{aligned}$$

where $t=1/x$, and $C_{pk}>0$ is a constant, we obtain

$$ \lim_{x\rightarrow0^{+}}v'_{pk}(x)=e^{\psi _{pk}(k)} \psi'_{pk}(k)>0 $$

for $p\geq1$ and $k>0$. This implies that, for all $k>0$ and $p\geq 1$, there is $x_{pk}>0$ such that $v_{pk}(x)$ is a strictly increasing function on $(0,x_{pk})$. So, $\phi_{pk}(x)$ is a strictly increasing function on $(0,x_{pk})$.

Next, by the mean value theorem we get

$$\begin{aligned} v'_{pk}(x) =&e^{\psi_{pk}(x+k)}\psi '_{pk}(x+k)-e^{\psi_{pk}(x)} \psi'_{pk}(x) \\ =&e^{\psi_{pk}(\xi_{pkx})} \bigl(\psi_{pk}^{'2}( \xi_{pkx})+\psi ''_{pk}( \xi_{pkx}) \bigr)k=e^{\psi_{pk}(\xi_{pkx})}w_{pk}(\xi_{pkx})k, \end{aligned}$$

where $x<\xi_{pkx}<x+k$.

Due to

$$\begin{aligned} w_{pk}(x) =&\frac{1}{x^{4}} \Biggl[ \Biggl(1+\sum _{n=1}^{p}\frac{x^{2}}{(nk+x)^{2}} \Biggr)^{2} -2x \Biggl(1+\sum_{n=1}^{p} \frac{x^{3}}{(nk+x)^{3}} \Biggr) \Biggr] \\ < &\frac{1}{x^{4}} \bigl[(1+p)^{2}-2x \bigr], \end{aligned}$$

we obtain that, for all $k>0$ and $p\geq1$, the function $v'_{pk}(x)$ is a negative function on $((1+p)^{2}/2,+\infty)$. So, $\phi_{pk}(x)$ is a strictly decreasing function on $((1+p)^{2}/2,+\infty)$.

Finally, computer calculations show that, for $p\geq1$ and $k>1$, there is $0< x_{pk}<1$ such that $\phi_{pk}(x)$ is an increasing function on $(0,x_{pk})$ and a decreasing function on $(x_{pk},+\infty)$.

3.2 Proof of Open Problem 1

Let $\delta_{pk\alpha}(x)$ be a completely monotonic function on $(0,\infty)$. Then $(-1)^{n}\delta^{(n)}_{pk\alpha}(x)\geq0$ for $x\in(0,\infty)$ and $\alpha\in R$. So $\delta'_{pk\alpha}(x)\leq 0$ for $x\in(0,\infty)$. A simple computation gives

$$\begin{aligned} \delta'_{pk\alpha}(x) =& \alpha x^{\alpha-1} \biggl[ \frac{1}{k}\ln \frac{pkx}{x+k(p+1)}-\psi_{pk}(x) \biggr] \\ & {}+x^{\alpha} \Biggl[\frac{p+1}{x(x+pk+k)}-\sum _{n=0}^{p}\frac {1}{(nk+x)^{2}} \Biggr]\leq0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} &\alpha \biggl(\frac{1}{k}\ln\frac{pkx}{x+k(p+1)}-\psi _{pk}(x) \biggr) \\ &\quad {}+x \Biggl(\frac{p+1}{x(x+pk+k)}-\sum_{n=0}^{p} \frac {1}{(nk+x)^{2}} \Biggr)\leq0. \end{aligned}$$

Because of (see [17])

$$ \frac{1}{k}\ln\frac{pkx}{x+k(p+1)}-\psi_{pk}(x)>0, $$

we obtain

$$ \alpha\leq d(x)=\frac{x (-\frac{p+1}{x(x+pk+k)}+\sum_{n=0}^{p}\frac{1}{(nk+x)^{2}} )}{\frac{1}{k}\ln\frac {pkx}{x+k(p+1)}-\psi_{pk}(x)} $$

for all $x>0$.

Similarly as in [1], the proof will be done if we show that

$$ \lim_{x\rightarrow0^{+}}d(x)\leq1. $$

Direct computation leads to

$$\begin{aligned} \lim_{x\rightarrow0^{+}}d(x) &=\lim_{x\rightarrow0^{+}} \frac{-\frac{(p+1)x}{x+pk+k}+1+\sum_{n=1}^{p}\frac{x^{2}}{(nk+x)^{2}}}{x [\frac{1}{k}\ln\frac {pkx}{x+k(p+1)}-\frac{\ln(pk)}{k}+\frac{1}{x} +\sum_{n=1}^{p}\frac{1}{nk+x} ]} \\ &=\frac{1}{\lim_{x\rightarrow0^{+}}\frac{x}{k}\ln (\frac {x}{x+pk+k} )+1+\sum_{n=1}^{p}\frac{x}{nk+x}}=1. \end{aligned}$$

Indeed, $\lim_{x\rightarrow0^{+}}d(x)=1$ implies that, for each $\varepsilon>0$, there is $x_{\varepsilon}>0$ such that $d(x_{\varepsilon})<1+\varepsilon$, so $\alpha<1+\varepsilon$, and thus $\alpha\leq1$. This completes the proof.

3.3 Disproving Conjecture 2

We show that Conjecture 2 is false. Let $0< a< b$. Put $y^{2}=a/b$. Then $0< y<1$. Conjecture 2 is equivalent to

$$ \biggl(1+\frac{1-y^{2}}{2\ln(y)} \biggr)\psi(b)- \biggl(\frac {1-y^{2}}{2\ln(y)}+y^{2} \biggr)\psi\bigl(by^{2}\bigr)> \bigl(1-y^{2} \bigr) \psi(by), $$

which can be rewritten as

$$ F(b,y)= \bigl(1-y^{2}+2\ln(y) \bigr) \bigl(\psi(b)-\psi \bigl(by^{2}\bigr) \bigr)-2\ln(y) \bigl(1-y^{2}\bigr) \bigl( \psi(by)-\psi\bigl(by^{2}\bigr) \bigr)< 0. $$

Let b be fixed. We prove that $\lim_{y\rightarrow 0^{+}}F(b,y)=+\infty$. This implies that Conjecture 2 does not valid. Using the well-known formula

$$ \psi(x)=-\gamma-\frac{1}{x}+\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+x}, $$

we obtain

$$ \psi(b)-\psi\bigl(by^{2}\bigr)=\frac{1}{by^{2}}- \frac{1}{b}+\sum_{n=1}^{\infty} \frac{1}{n+by^{2}}-\frac{1}{n+b} $$

(8)

and

$$ \psi(by)-\psi\bigl(by^{2}\bigr)=\frac{1}{by^{2}}- \frac{1}{by}+\sum_{n=1}^{\infty} \frac{1}{n+by^{2}}-\frac{1}{n+by}. $$

(9)

So

$$\begin{aligned} F(b,y) =&\frac{1}{by^{2}} \bigl(1-y^{2}+2\ln(y) \bigr) \bigl(1-y^{2}+by^{2}\varphi_{1}(b,y) \bigr) \\ & {}-\frac{2\ln(y)(1-y^{2})}{by^{2}} \bigl(1-y+by^{2}\varphi _{2}(b,y) \bigr), \end{aligned}$$

where

$$ 0< \varphi_{1}(b,y)< \frac{b(1-y^{2})\pi^{2}}{6} \quad \text{and}\quad 0< \varphi_{2}(b,y)< \frac{by(1-y)\pi^{2}}{6}. $$

The function $F(b,y)$ may be rearranged as

$$\begin{aligned} F(b,y) =&\frac{1}{by^{2}}\bigl[\bigl(1-y^{2}\bigr) \bigl(1-y^{2}+by^{2}\varphi _{1}(b,y)\bigr) \\ &{}+ 2\ln(y) \bigl(by^{2}\varphi _{1}(b,y)+y-y^{3}-b \bigl(1-y^{2}\bigr)y^{2}\varphi_{2}(b,y) \bigr) \bigr]. \end{aligned}$$

This implies that $\lim_{y\rightarrow0^{+}}F(b,y)=+\infty$.

3.4 Proof of Theorem 1

Theorem 1

Let $0< a< b<4/10$. Then

$$ \bigl(b-L(a, b)\bigr)\psi(b)+\bigl(L(a, b)-a\bigr)\psi(a)< (b-a)\psi (\sqrt{ba}). $$

(10)

Proof

It is easily derived that (10) is equivalent to $F(b,y)>0$, where $y^{2}=a/b$, $0< y<1$, and

$$\begin{aligned} F(b,y) =& \bigl(1-y^{2}+2\ln(y) \bigr) \bigl(\psi(b)-\psi \bigl(by^{2}\bigr) \bigr) \\ & {}-2\ln(y) \bigl(1-y^{2}\bigr) \bigl(\psi(by)-\psi \bigl(by^{2}\bigr) \bigr). \end{aligned}$$

Using (8), we obtain

$$\begin{aligned} \psi(b)-\psi\bigl(by^{2}\bigr) =&\frac{1-y^{2}}{by^{2}}+b \bigl(1-y^{2}\bigr)\sum_{n=1}^{\infty} \frac{1}{(n+by^{2})(n+b)} \\ < & \frac{1-y^{2}}{by^{2}}+b\bigl(1-y^{2}\bigr)\sum _{n=1}^{\infty}\frac{1}{(n+by)^{2}} \end{aligned}$$

due to $(n+by^{2})(n+b)>(n+by)^{2}$. So

$$ \psi(b)-\psi\bigl(by^{2}\bigr)< b\bigl(1-y^{2}\bigr) \psi'(by). $$

(11)

Applying (9), we get

$$\begin{aligned} \psi(by)-\psi\bigl(by^{2}\bigr) =&\frac{1-y}{by^{2}}+by(1-y)\sum _{n=1}^{\infty}\frac{1}{(n+by^{2})(n+by)} \\ >& \frac{1-y}{by^{2}}+b(1-y)\sum_{n=1}^{\infty} \frac{1}{(n+by)^{2}} \end{aligned}$$

due to $(n+by^{2})<(n+by)$. So

$$ \psi(by)-\psi\bigl(by^{2}\bigr)>\frac{1-y}{by^{2}} \bigl(1-y+b^{2}y^{3}\psi '(by) \bigr). $$

(12)

It is easy to see that, for $0< y<1$,

$$ s(y)=1-y^{2}+2\ln(y)< 0, $$

which follows from $s(1)=0$ and $s'(y)=2(1-y^{2})/y>0$. This implies that

$$\begin{aligned} F(b,y) >&G(b,y) \\ =&\frac{1}{b^{2}y^{2}}\bigl[\bigl(1-y^{2}+2\ln (y)\bigr)b \bigl(1-y^{2}\bigr)\psi'(by)) \\ & {}-2\ln(y) (1-y) \bigl(1-y^{2}\bigr) \bigl(1-y+b^{2}y^{3} \psi '(by) \bigr)\bigr]. \end{aligned}$$

The inequality $G>0$ is equivalent to

$$ b^{2}y^{2} \bigl(1-y^{2}+2\ln(y) \bigr) \psi'(by))-2\ln(y) (1-y) \bigl(1-y+b^{2}y^{3} \psi'(by) \bigr)>0. $$

(13)

Inequality (13) may be rearranged as

$$ H=\psi'(by)b^{2}y^{2} \bigl(1-y^{2}+2 \ln(y) \bigl(1-y+y^{2}\bigr) \bigr))-2\ln (y) (1-y)^{2}>0. $$

Put $s_{1}(y)=1-y^{2}+2\ln(y)(1-y+y^{2})$. It is easy to see that $s_{1}(y)<0$ for $0< y<1$. Indeed, $s_{1}(y)<0$ is equivalent to

$$ s_{2}(y)=\frac{y^{2}-1}{1-y+y^{2}}-2\ln(y)>0. $$

Due to $s_{2}(1)=0$, it suffices to show that $s'_{2}(y)<0$.

Differentiation leads to

$$ s'_{2}(y)=\frac{-2+3y-2y^{2}+3y^{3}-2y^{4}}{y(1-y+y^{2})^{2}}=\frac {-(1-y)^{2}(2+y+2y^{2})}{y(1-y+y^{2})^{2}}< 0. $$

Using the well-known formula

$$ \psi'(x)=\frac{1}{x^{2}}+\frac{1}{(1+x)^{2}}+\sum _{n=2}^{\infty }\frac{1}{(n+x)^{2}}, $$

we obtain

$$ b^{2}y^{2}\psi'(by)< 1+\frac{b^{2}y^{2}}{(1+by)^{2}}+b^{2}y^{2} \biggl(\frac{\pi^{2}}{6}-1 \biggr). $$

Theorem 1 will be proved if we show

$$\begin{aligned} G(b) =& \biggl(1+\frac{b^{2}y^{2}}{(1+by)^{2}}+b^{2}y^{2} \biggl( \frac {\pi^{2}}{6}-1 \biggr) \biggr) \bigl(1-y^{2}+2\ln(y) \bigl(1-y+y^{2}\bigr) \bigr) \\ &{} -2\ln(y) (1-y)^{2}>0 \end{aligned}$$

for $0< b<4/10$, $0< y<1$. Based on

$$ \frac{dG(b)}{db}= \bigl(1-y^{2}+2\ln(y) \bigl(1-y+y^{2} \bigr) \bigr) \biggl(\frac{2by^{2}}{(1+by)^{3}}+2by^{2} \biggl( \frac{\pi^{2}}{6}-1 \biggr) \biggr)< 0, $$

it suffices to prove that $G(0.4)>0$.

The inequality $G(b)>0$ is equivalent to

$$ 2\ln(y)f(b,y)+g(b,y)>0, $$

(14)

where

$$\begin{aligned}& \begin{aligned} f(b,y)&=\bigl(1-y+y^{2}\bigr) \biggl[(1+by)^{2}+b^{2}y^{2} \biggl(1+(1+by)^{2} \biggl(\frac{\pi^{2}}{6}-1 \biggr) \biggr) \biggr] \\ &\quad {}-(1+by)^{2}(1-y)^{2}, \end{aligned} \\& g(b,y)=\bigl(1-y^{2}\bigr) \biggl[(1+by)^{2}+b^{2}y^{2} \biggl(1+(1+by)^{2} \biggl(\frac{\pi^{2}}{6}-1 \biggr) \biggr) \biggr]. \end{aligned}$$

It is clearly seen that $f(b,y)>0$. So (14) will be done if we prove

$$ h(b,y)=2\ln(y)+\frac{g(b,y)}{f(b,y)}>0 $$

for $b=4/10$ and $0< y<1$. Because of $h(0.4,1)=0$, it suffices to show that $(dh/dy)(0.4,y)<0$ for $0< y<1$. We get

$$ \frac{dh(0.4,y)}{dy}=\frac{2}{y}+\frac{\frac {dg(0.4,y)}{dy}f(0.4,y)-g(0.4,y)\frac{df(0.4,y)}{dy}}{f^{2}(0.4,y)}< 0. $$

(15)

Inequality (15) is equivalent to

$$ u(y)=2f^{2}(0.4,y)+y \biggl(\frac{dg(0.4,y)}{dy}f(0.4,y)-g(0.4,y) \frac {df(0.4,y)}{dy} \biggr)< 0. $$

(16)

Put $a(y)=100u(y)$. Using Taylor’s series and Matlab, we obtain

$$\begin{aligned} a(y) =& (y-1)^{2} \bigl(899.80856587904507327359937090692\,(y-1) \\ &{}- 34.541951843593497556069703611525 \\ &{}+ 2803.0064998600206956003380918157(y-1)^{2} \\ &{} + 3449.9649390326664508411882274417(y-1)^{3} \\ &{} + 2382.8365919732490773391558060343(y-1)^{4} \\ &{} + 1077.4495988779774279297213875464(y-1)^{5} \\ &{} + 341.35184858325869449609722858626(y-1)^{6} \\ &{} + 75.928581022558019892100963581518(y-1)^{7} \\ &{} + 11.561798822762785643694391575539(y-1)^{8} \\ &{}+ 1.1176206646533878730984733813133(y-1)^{9} \\ &{} + 0.05451808120260428649260845762504(y-1)^{10}\bigr) ) \\ =&(y-1)^{2}k(y), \end{aligned}$$

where

$$\begin{aligned} k(y) =& 899.80856587904507327359937090692(y-1) \\ &{} -34.541951843593497556069703611525 \\ &{} +2803.0064998600206956003380918157(y-1)^{2} \\ &{} +3449.9649390326664508411882274417(y-1)^{3} \\ &{} +2382.8365919732490773391558060343(y-1)^{4} \\ &{} +1077.4495988779774279297213875464(y-1)^{5} \\ &{} +341.35184858325869449609722858626(y-1)^{6} \\ &{} +75.928581022558019892100963581518(y-1)^{7} \\ &{} +11.561798822762785643694391575539(y-1)^{8} \\ &{} +1.1176206646533878730984733813133(y-1)^{9} \\ &{} +0.05451808120260428649260845762504(y-1)^{10}. \end{aligned}$$

It is easy to see that $k(y)< kk(y)$, where

$$\begin{aligned} kk(y) =& 899.80856587904507327359937090692(y-1) \\ &{} -34.541951843593497556069703611525 \\ &{} +2803.0064998600206956003380918157(y-1)^{2} \\ &{} +3449.9649390326664508411882274417(y-1)^{3} \\ &{} +2382.8365919732490773391558060343(y-1)^{4} \\ &{} +1077.4495988779774279297213875464(y-1)^{5} \\ &{} +341.35184858325869449609722858626(y-1)^{6} \\ &{} +75.928581022558019892100963581518(y-1)^{7} \\ &{} +11.561798822762785643694391575539(y-1)^{8} \\ &{} +1.1176206646533878730984733813133(y-1)^{9} \\ &{} +0.05451808120260428649260845762504(y-1)^{6}. \end{aligned}$$

To prove that $kk(y)<0$, it suffices to show that $kk(0)\leq0$, $kk(1)\leq0$, $kk'(0)\leq0$, $kk''(1)>0$, $kk''(0)<0$, and $kk''(y)$ is an increasing function on $(0,1)$.

Put $c(y)=kk''(y)$. Direct computation yields

$$\begin{aligned} c(y) =& (22\text{,}759\text{,}659\text{,}395\text{,}315\text{,}613y)/1 \text{,}099\text{,}511\text{,}627\text{,}776 \\ &{} +\bigl(2\text{,}021\text{,}889\text{,}693\text{,}856\text{,}018 \text{,}983(y-1)^{2}\bigr)/70\text{,}368\text{,}744\text{,}177 \text{,}664 \\ &{} +\bigl(23\text{,}693\text{,}367\text{,}246\text{,}178\text{,}465(y-1)^{3} \bigr)/1\text{,}099\text{,}511\text{,}627\text{,}776 \\ &{} +\bigl(738\text{,}027\text{,}641\text{,}132\text{,}151\text{,}741 \text{,}645(y-1)^{4}\bigr)/72\text{,}057\text{,}594\text{,}037 \text{,}927\text{,}936 \\ &{} +\bigl(112\text{,}202\text{,}976\text{,}768\text{,}737 \text{,}799(y-1)^{5}\bigr)/35\text{,}184\text{,}372\text{,}088 \text{,}832 \\ &{} +\bigl(11\text{,}324\text{,}961\text{,}019\text{,}967\text{,}769(y-1)^{7} \bigr)/140\text{,}737\text{,}488\text{,}355\text{,}328 \\ &{} -8\text{,}297\text{,}891\text{,}458\text{,}330\text{,}007/549\text{,}755 \text{,}813\text{,}888. \end{aligned}$$

We now show that $cc(y)=kk'''(y)>0$. We have

$$\begin{aligned} cc(y) =& 57\text{,}465.561379104286260144363041036y \\ &{} +64\text{,}646.975932678647041029762476683(y-1)^{2} \\ &{} +40\text{,}968.763999735354855158409037585(y-1)^{3} \\ &{} +15\text{,}945.002014737184339310260838829(y-1)^{4} \\ &{} +563.28081498530752213582672993653(y-1)^{6} \\ &{} -36\text{,}765.771744908288582109889830463. \end{aligned}$$

Differentiation yields

$$\begin{aligned} cc''(y) =& 16\text{,}898.424449559224740369245409966y^{4} \\ &{} -67\text{,}593.697798236898961476981639862y^{3} \\ &{} +292\text{,}730.57087420154857682064175606y^{2} \\ &{} -204\text{,}461.16215351717255543917417526y \\ &{} +91\text{,}719.816493350586824817582964897. \end{aligned}$$

Using the Cardano formula and Matlab, we get that there are no real roots of $cc''(y)=0$. Due to $cc(0)>0$, we obtain $cc''(y)>0$.

We now show that $v(y)>0$ for $0< y<1$, where $v(y)$ is a tangent line to the function $cc(y)$ at the point $(0.22,cc(0.22))$.

Using Matlab, we have

$$cc(0.22)=1794.965061908937 \quad \mbox{and}\quad cc'(0.22)=149.7626334452943. $$

This implies that

$$ v(y)=1794.965061908937+149.7626334452943(y-0.22). $$

Direct computation yields:

$$\begin{aligned}& v(0)=1762.017282550972 , \qquad v(1)=1911.779915996267 , \\& kk(0)=-2.3294\mathrm{e}{-}14 , \qquad kk(1)=-34.5420 , \qquad kk'(0)=-53.5347 , \\& kk''(0)=-937.2660 , \qquad kk''(1)=5.6060\mathrm{e}{+}03 . \end{aligned}$$

This completes the proof. □

3.5 Open problem

Finally, we give an open problem.

Open Problem 2

Find the best possible real positive constants $b_{0}$, $b_{1}$ such that if $0< a< b\leq b_{0}$, then

$$ \bigl(b-L(a, b)\bigr)\psi(b)+\bigl(L(a, b)-a\bigr)\psi(a)< (b-a)\psi (\sqrt{ba}), $$

and if $0< b_{1}\leq a< b$, then

$$ \bigl(b-L(a, b)\bigr)\psi(b)+\bigl(L(a, b)-a\bigr)\psi(a)>(b-a)\psi (\sqrt{ba}), $$

where $L(a, b)=(b-a)/(\ln(b)-\ln(a))$.

Note 1

Note that our work and [3] show that $4/10\leq b_{0}$ and $2\geq b_{1}$.

4 Conclusion

In this paper, we proved e Open Problem 1 [17] and disproved Conjectures 1 and 2 [3, 19]. We also proved a new inequality (Theorem 1) for the digamma function. Finally, we proposed an Open Problem 2.