Yin et al. Advances in Difference Equations (2018)2018:246 https://doi.org/10.1186/s13662-018-1695-7

R E S E A R C H Open Access Monotonicity, concavity, and inequalities related to the generalized digamma function Li Yin1*, Li-Guo Huang1, Xiu-Li Lin2 and Yong-Li Wang3

*Correspondence: [email protected] Abstract 1College of Science, Binzhou University, Binzhou City, China In this paper, we establish a concave theorem and some inequalities for the Full list of author information is generalized digamma function. Hence, we give complete monotonicity property of a available at the end of the article determinant function involving all kinds of derivatives of the generalized digamma function. MSC: 33B15 Keywords: Generalized digamma function; Complete monotonicity; Inequality; Concavity

1 Introduction It is well known that the Euler gamma function is defined by

 ∞ (x)= tx–1e–t dt, x >0. 0

The logarithmic derivative of (x) is called the psi or digamma function. That is,

d (x) ψ(x)= ln (x)= dx (x) ∞ 1  x =–γ – + , x n(n + x) n=1

where γ = 0.5772 . . . 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 equa- tions, mathematical physics, and theory of infinite series. The reader may see related ref- erences [5, 7, 9, 13, 16, 23–28]. In [6], the k-analogue of the gamma function is defined for k >0andx > 0 as follows:

 ∞ k x–1 – t k(x)= t e k dt 0 x –1 n!kn(nk) k = lim , →∞ n (x)n,k

© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Yin et al. Advances in Difference Equations (2018)2018:246 Page 2 of 9

where limk→1 k(x)=(x). It is natural that the k-analogue of the digamma function is defined for x >0by

 d k(x) ψk(x)= log k(x)= . dx k(x)

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 param- eters.Thisisthefirstobjectinthispaper. 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 >0and x >0,the following identities hold true:   x –1 x (x)=k k , (2.1) k k   ln k 1 x ψ (x)= + ψ . (2.2) k k k k

Proof Using the substitution √t = u and uk = p,weeasilyobtain k k      √  ∞ x–1 t k k x t –( √ ) t √ k k √ k(x)= k k e d k 0 k k  ∞ x x–1 –uk = k k u e du 0  ∞ x –1 x –1 –p = k k p k e dp 0  x –1 x = k k . k

So, we prove formula (2.1). To (2.2), direct computation yields

 x –1 x  (x) [k k ( )] ψ (x)= k = k k x –1 x k(x) [k k ( )]   k ln k 1 x = + ψ .  k k k

Lemma 2.2 ([1, 2]) For x >0,we have

 1 1 1 ψ (x)< + + (2.3) x 2x2 6x3 Yin et al. Advances in Difference Equations (2018)2018:246 Page 3 of 9

and

 1 1 ψ (x)<– – . (2.4) x2 x3

Lemma 2.3 For k, x >0,m ∈ N, we have

∞  1 ψ(m)(x)=(–1)m+1m! (2.5) k (nk + x)m+1 n=0

and  ∞ tm ψ(m) x m+1 e–xt dt k ( )=(–1) –kt . (2.6) 0 1–e

Proof Formula (2.5) may be found in reference [15]. By using formula (9) in reference [15], we have

ψ(m)(x)= lim ψ(m)(x) k p→∞ p,k    ∞ 1–e–k(p+1)t = lim (–1)m+1 tme–xt dt →∞ –kt p 0 1–e  ∞ 1 m+1 tme–xt dt =(–1) –kt . 0 1–e

So, we prove formula (2.6). 

3 Main results 2  ∞ Theorem 3.1 For k >0,the function x ψk(x) is strictly increasing on (0, ).

Proof Using Lemma 2.1,wehave      1  x  1  x ψ (x)= ψ and ψ (x)= ψ . k k2 k k k3 k

∞ ψ(m) m+1 1 Combining with the identity (x)=(–1) m! n=0 (n+x)m+1 ,weget       2 d  2x  x x  x x2ψ (x) = ψ + ψ dx k k2 k k3 k ∞  nk =2x >0. (nk + x)3  n=0

1 ∞ Theorem 3.2 For k >0,the function ψk( x ) is strictly concave on (0, ).

Proof Easy computation results in      d 1 1  1 ψ =– ψ . dx k x x2 k x

Considering Theorem 3.1, we complete the proof.  Yin et al. Advances in Difference Equations (2018)2018:246 Page 4 of 9

Theorem 3.3 For k ≥ √1 = 0.693361 . . . , the function 3 3   1 λ (x)=ψ (x)+ψ k k k x

is strictly concave on (0, ∞).

Proof By differentiation and applying Lemma 2.1,weeasilyobtain     1  1 λ (x)=ψ (x)– ψ , k k x2 k x       2  1 1  1 λ (x)=ψ (x)+ ψ + ψ , k k x3 k x x4 k x

and         x  1  1 k3x4λ (x)=x4ψ +2kxψ + ψ . k k kx kx

Applying Lemma 2.2, k ≥ √1 , and the recurrence relations 3 3      1  1 ψ +1 = ψ – k2x2, kx kx      1  1 ψ +1 = ψ +2k3x3, kx kx

we have         x  1  1 k3x4λ (x)=x4ψ +2kxψ + ψ k k kx kx   k2 k3 kx k2x2 k3x3 < x4 – – +2kx + + x2 x3 1+kx 2(1 + kx)2 6(1 + kx)3 k2x2 k3x3 – – (1 + kx)2 (1 + kx)3 kx   =– 3k2 +9k3x +9k2x2 + k2 3k3 –1 x3 +3k4x4 3(1 + kx)3 <0.

This implies that λk(x) is strictly concave on (0, ∞). 

Theorem 3.4 For x ∈ (0, ∞) and k ≥ √1 , we have 3 3   1 2 ln k +2ψ( 1 ) ψ (x)+ψ ≤ k . (3.1) k k x k

1 ∞ Proof Since the function λk(x)=ψk(x)+ψk( x ) is strictly concave on (0, ), we get

 ≥  ∈ λk(x) λk(1) = 0, x (0, 1], Yin et al. Advances in Difference Equations (2018)2018:246 Page 5 of 9

and

 ≤  ∈ ∞ λk(x) λk(1) = 0, x [1, ).

It follows that λk is increasing on (0, 1] and decreasing on [1, ∞). Hence, λk(x) ≤ λk(1) for x > 0. The proof is complete. 

ln k 1 1 Remark 3.1 Let γk =–ψk(1) = – k – k ψ( k )bethek-analogue of the Euler–Mascheroni constant. It is obvious that limk→1 γk = γ .

Definition 3.1 It is known that the generalized digamma function ψk(x) is strictly increas- + ing on (0, ∞)withψk(0 )ψk(∞) < 0. So, the function has a sole positive root in (0, ∞). We

define this positive root for xk.Thatis,   x ln k + ψ k =0. k

Theorem 3.5 For x ∈ (0, 1) and √1 ≤ k ≤ 1, we have 3 3

ln2 k + γ 2 –2(γ +1)ln k ψ (1 + x)ψ (1 – x) ≤ . (3.2) k k k2

√1 Proof Considering ≤ k ≤ 1 and the definition of xk,wehave 3 3

1 √ x0 ≤ xk ≤ x0, 3 3

where x0 satisfies ψ(x0)=0withx0 = 1.46163 . . . .

Case 1. If x ∈ [xk –1,1),thenwehaveψk(1–x) ≤ 0 ≤ ψk(1+x). This implies that formula (3.2)holds.

Case 2. If x ∈ (0, xk – 1], using the power series expansion

∞ ψ(1 + z)=–γ + (–1)kζ (k)zk–1, |z| < 1, (3.3) k=2

we obtain   ln k 1 x ψ (1 + x) ≥ ψ (k + x)= + ψ 1+ k k k k k  ∞ ln k 1  = + –γ + (–1)kζ (k)xk–1 , k k k=2

∞ where ζ (k)= 1 is the Riemann zeta function. n=1 nk Furthermore, we have

ln k 1 0<–ψ (1 + x) ≤ – + γ – ζ (2)x + ζ (3)x2 . (3.4) k k k Yin et al. Advances in Difference Equations (2018)2018:246 Page 6 of 9

Completely similar to (3.4), we have  ∞ ln k 1  0<–ψ (1 – x) ≤ – + γ + ζ (2)y + ζ (3) xk k k k k=2 ln k 1 ≤ – + γ + ζ (2)x + ζ (3)x2 . (3.5) k k

Combining (3.4)with(3.5), we obtain

ln2 k + γ 2 –2(γ +1)ln k ψ (1 + x)ψ (1 – x) ≤ k k k2

by using ζ (3)x2 <1. 

Theorem 3.6 For x ∈ (0, ∞) and √1 ≤ k ≤ 1, we have 3 3   1 ln2 k + γ 2 –2(γ +1)ln k ψ (x) · ψ ≤ . (3.6) k k x k2

≥ ≥ 1 ≤ ≤ Proof We only need to prove (3.6)forx 1. If x xk,thenwegetψk( x ) 0 ψk(x). It follows that inequality (3.6)holdstrue.

If x ∈ (1, xk]andsettingx =1+z,weget   1 ψ (1 – z) ≤ ψ . k k x

Therefore, we have     1 1 ψ (x) · ψ = ψ (1 + z)ψ k k x k k x

≤ ψk(1 + z)ψk(1 – z) 2 2 ≤ ln k + γ –2(γ +1)ln k k2

by using Theorem 3.5. 

Corollary 3.1 For x ∈ (0, ∞) and √1 ≤ k ≤ 1, we have 3 3

2ψ (x)ψ ( 1 ) ln2 γ 2 γ ln k k x ≥ k + –2( +1) k 1 1 . (3.7) ψk(x)+ψk( x ) k[ln k + ψ( k )]

Proof Applying Theorems 3.4 and 3.6,weobtain

2ψ (x)ψ ( 1 ) ln2 γ 2 γ ln k k x ≥ · k + –2( +1) k 1 1 2 2 1 ψk(x)+ψk( x ) k ψk(x)+ψk( x ) 2 2 ≥ ln k + γ –2(γ +1)ln k k 2 1 . k ln k + ψ( k )

The proof is complete.  Yin et al. Advances in Difference Equations (2018)2018:246 Page 7 of 9

Next, for m, n, j ∈ N,wedefinethefunctionμn by

   (m) (m+j) (m+nj)   ψ (x) ψ (x) ··· ψ (x)   k k k   ψ(m+j) ψ(m+2j) ··· ψ[m+(n+1)j]   k (x) k (x) k (x) μn(x)=  .  . . .   . . .   (m+nj) (m+(n+1)j) ··· (m+2nj)  ψk (x) ψk (x) ψk (x)

Completely similar to the method in [11], the following Theorem 3.7 can be proved.

(n+1)(m+1) Theorem 3.7 For m, n, j ∈ N, then (–1) μn(x) is completely monotonic on (0, ∞).

Proof Using Lemma 2.3,wehave

   m m+j ··· m+nj   u0 u0 u0     m+j m+2j m+(n+1)j 0 0  u u ··· u  n+1 ···  1 1 1  μn(x)=(–1)  . . .  ∞ ∞  . . .  –  –   . . .  n+1 times  m+nj m+(n+1)j m+2nj  un un ··· un

x (u0+u1+···+un) · ek ··· n du0 du1 dun (1 – eui ) i=0    m+j m+nj   um u ··· u   δ(0) δ(0) δ(0)     m+j m+2j m+(n+1)j 0 0  u u ··· u  n+1 ···  δ(1) δ(1) δ(1)  =(–1)  . . .  –∞  –∞  . . .   . . .  n+1 times  m+nj m+(n+1)j ··· m+2nj  uδ(n) uδ(n) uδ(n) x (u0+u1+···+un) · ek ··· n du0 du1 dun, ui i=0(1 – e )

where δ isapermutationon0,1,2,...,n. Let sgn(δ)bethesignofδ, we can obtain

  x ··· n 0 0 k (u0+u1+ +un)  n+1 ··· e m un(x)=(–1) n sgn(δ) ui ∞ ∞ (1 – eui ) –  –  i=0 i=0 n+1 times    j nj  u0 u ··· u   0 0 0   j 2j ··· (n+1)j u1 u1 u1  ·   du0 du1 ··· dun  . . .   . . .   nj (n+1)j 2nj  un un ··· un

  x ··· n+1 0 0 k (u0+u1+ +un) (–1) ··· e ··· m = n (u0u1 un) ui (n +1)!–∞  –∞ i=0(1 – e ) n+1 times    · j j ··· ui – ul du0 du1 dun. 0≤i

Replacing u0, u1,...,un by –u0,–u1,...,–un,weget

 ∞  ∞ x (n+1)(m+1) – (u0+u1+···+un) μn(x)=(–1) ··· e k 0  0  n+1 times   n n · j j ui ··· ui – u du0 du1 dun. l 1–e–ui 0≤i

(n+1)(m+1) This implies that (–1) μn(x) is completely monotonic. 

By taking n = 1, the following Corollary 3.2 can be easily obtained.

Corollary 3.2 For m, j ∈ N, and x >0,we have    (m+j)   ψ(m)(x) ψ (x)   k k   (m+j) (m+2j)  >0. ψk (x) ψk (x)

4Conclusions 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.

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).

Competing interests The authors declare that they have no competing interests.

Authors’ contributions All authors contributed equally to the manuscript and read and approved the final manuscript.

Author details 1College of Science, Binzhou University, Binzhou City, China. 2College of Mathematics Science, Qufu Normal University, Qufu City, China. 3College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao, China.

Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 19 February 2018 Accepted: 2 July 2018

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