Notes on three conjectures involving the digamma and generalized digamma functions.

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.

Introduction

In the last years, the -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, [1-24].

It is well known that: Recently, Díaz and Pariguan [4] defined the generalized gamma function for and and the generalized digamma function

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 for , , (due to ).

the Euler gamma function [14–16, 20, 22, 23] is defined by for ;

the digamma function [11–13, 24] is defined by where γ is the Euler–Mascheroni constant [5].

Very recently, Nantomah, Prempeh, and Twum [8] introduced a new definition of the -gamma function for and , , , and the -digamma function for and , , .

We note that

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

Conjecture 1

([19])

For and , the function is strictly decreasing from onto .

Li Yin [17] posed the following open problem.

Open Problem 1

([17])

If the function is completely monotonic on , then is it true that ?

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

Conjecture 2

For , we have where .

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

Methods

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

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]).

Disproving Conjecture 1

It is evident that is strictly decreasing only if is strictly decreasing. We have Using Matlab, we obtain Table 1.

Table 1

Values of p, k, , , ,

p	k	x_1	v_pk(x_1)	x_2	v_pk(x_2)
100,000	1.1	0.1	0.733034946365922	0.25	0.849008046429035
100,000	1.6	0.1	0.994429244720941	0.25	1.060815738973497
100,000	2.1	0.1	1.121247007413613	0.25	1.157039128396737
100,010	1.1	0.1	0.733034947426004	0.25	0.849008047776431
100,010	1.6	0.1	0.994429244720941	0.25	1.060815740083848
100,010	2.1	0.1	1.121247008239857	0.25	1.157039129298960

The table shows that for , , , , , . So is not strictly decreasing on for and .

Remark 1

We note that Conjecture 1 (see [19]) is false since for and .

Indeed, differentiation of yields Because of where , and is a constant, we obtain for and . This implies that, for all and , there is such that is a strictly increasing function on . So, is a strictly increasing function on .

Next, by the mean value theorem we get where .

Due to we obtain that, for all and , the function is a negative function on . So, is a strictly decreasing function on .

Finally, computer calculations show that, for and , there is such that is an increasing function on and a decreasing function on .

Proof of Open Problem 1

Let be a completely monotonic function on . Then for and . So for . A simple computation gives which is equivalent to Because of (see [17]) we obtain for all .

Similarly as in [1], the proof will be done if we show that Direct computation leads to Indeed, implies that, for each , there is such that , so , and thus . This completes the proof.

Disproving Conjecture 2

We show that Conjecture 2 is false. Let . Put . Then . Conjecture 2 is equivalent to which can be rewritten as Let b be fixed. We prove that . This implies that Conjecture 2 does not valid. Using the well-known formula we obtain and So where The function may be rearranged as This implies that .

Proof of Theorem 1

Theorem 1

Let . Then

Proof

It is easily derived that (10) is equivalent to , where , , and Using (8), we obtain due to . So Applying (9), we get due to . So It is easy to see that, for , which follows from and . This implies that The inequality is equivalent to Inequality (13) may be rearranged as Put . It is easy to see that for . Indeed, is equivalent to Due to , it suffices to show that .

Differentiation leads to Using the well-known formula we obtain Theorem 1 will be proved if we show for , . Based on it suffices to prove that .

The inequality is equivalent to where It is clearly seen that . So (14) will be done if we prove for and . Because of , it suffices to show that for . We get Inequality (15) is equivalent to Put . Using Taylor’s series and Matlab, we obtain where It is easy to see that , where To prove that , it suffices to show that , , , , , and is an increasing function on .

Put . Direct computation yields

We now show that . We have

Differentiation yields

Using the Cardano formula and Matlab, we get that there are no real roots of . Due to , we obtain .

We now show that for , where is a tangent line to the function at the point .

Using Matlab, we have

This implies that

Direct computation yields:

This completes the proof. □

Open problem

Finally, we give an open problem.

Open Problem 2

Find the best possible real positive constants , such that if , then and if , then where .

Note 1

Note that our work and [3] show that and .

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.