Monotonicity properties of q -digamma and q -trigamma functions

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Monotonicity properties of q-digamma and q-trigamma functions Necdet Batir Department of Mathematics, Faculty of Sciences and Arts, Nevs¸ehir Haci Bektas¸ Veli University, Nevs¸ehir, Turkey Received 21 May 2014; received in revised form 23 December 2014; accepted 29 December 2014 Available online 14 January 2015 Communicated by Paul Nevai

Abstract Some complete monotonicity results for q-polygamma functions are proved. Our results extend positivity of some functions containing q-polygamma functions to complete monotonicity property. Also, we give two new inequalities for q-trigamma function. c 2015 Elsevier Inc. All rights reserved. ⃝

MSC: primary 33B15; 26D15 Keywords: q-digamma function; q-psi function; q-trigamma function; q-gamma function; q-extensions; Inequalities

1. Introduction As it is known, the gamma function 0(x) is defined by the improper integral  ∞ 0(x) = t x−1 e−t dt, x > 0. 0

The most important function related to the gamma function is the digamma or psi function ψ, ′ (x) . The derivatives which is defined as the logarithmic derivative of Γ , namely ψ(x) = Γ0(x) ′ ′′ ψ (x), ψ (x), . . . are known to be the polygamma functions in the literature. Particularly ψ ′ and E-mail addresses: [email protected], [email protected]. http://dx.doi.org/10.1016/j.jat.2014.12.013 c 2015 Elsevier Inc. All rights reserved. 0021-9045/⃝

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ψ ′′ are called the trigamma and tetragamma functions, respectively. The q-analogue of 0(x), denoted by Γq (x), was introduced by Jackson [16] as Γq (x) = (1 − q)1−x

∞  1 − q n+1 , 1 − q n+x n=0

0
(1.1)

and Γq (x) = (q − 1)1−x q x(x−1)/2

∞  1 − q −(n+1) , 1 − q −(n+x) n=0

q>1

(1.2)

for x > 0. It was proved in [19] that lim Γq (x) = lim Γq (x) = 0(x).

q→1−

q→1+

Γ ′ (x)

Similarly the q-digamma or q-psi function ψq (x) is defined by ψq (x) = Γqq (x) . The derivatives ψq′ , ψq′′ , . . . are called the q-polygamma functions. In particular the functions ψq′ and ψq′′ are called q-trigamma and q-tetragamma functions, respectively. In [17] it was shown that limq→1− ψq (x) = limq→1+ ψq (x) = ψ(x). From the definitions (1.1) and (1.2) one can easily deduce that ψq (x) = − log(1 − q) + log q

∞  q nx , 1 − qn n=1

0
(1.3)

and 

∞ 1  q −nx ψq (x) = − log(q − 1) + log q x − − 2 n=1 1 − q −n

 ,

q > 1.

Differentiation of (1.3) and (1.4) gives  ∞  nq nx  2   log q ; 0 < q < 1,   1 − qn n=1 ′ ψq (x) = ∞   nq −nx  2  log q + log q ; q > 1.   1 − q −n n=1

(1.4)

(1.5)

Alzer and Grinshpan [3] proved that ψq′ is completely monotonic on (0, ∞), that is, (−1)n (ψq′ (x))(n) > 0,

x > 0, q > 0, n = 0, 1, 2, . . . .

(1.6)

We recall that a function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and (−1)n f (n) (x) ≥ 0 for all x ∈ I and all integers n ≥ 0. These functions have important applications in probability and numerical analysis. In particular, completely monotonic functions involving the gamma and q-gamma functions are very important because they enable us to estimate the polygamma and q-polygamma functions. For more information for these functions, we refer to Chapter IV of [29]. A short calculation gives for x > 0 and q > 0 Γq (x + 1) =

1 − qx Γq (x) 1−q

(1.7)

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and Γq (x) = q

(x−1)(x−2) 2

Γq −1 (x).

(1.8)

If we take logarithm of both sides of (1.7) and (1.8) and then differentiate, we find ψq (x + 1) − ψq (x) =

q x · log q , qx − 1

ψq′ (x + 1) − ψq′ (x) = − ψq′′ (x + 1) − ψq′′ (x) =

(1.9)

q −x · log2 q , (1 − q −x )2

q −x · (1 + q −x ) log3 q , (1 − q −x )3

(1.10) (1.11)

  3 ψq (x) = x − log q + ψ1/q (x), 2

(1.12)

′ ψq′ (x) = log q + ψ1/q (x),

(1.13)

and

′′ ψq′′ (x) = ψ1/q (x).

We refer [3,1,4,12,14,15,18] for basic properties of the q-gamma and q-digamma functions. Recently, many monotonicity and complete monotonicity properties and inequalities for the gamma and digamma functions have been extended to the q-gamma and q-digamma functions; see, for example, [3,1,9–11,20,23,24,26,27,25,22] and references therein. In [2] H. Alzer proved  2 ψ ′′ (x) + ψ ′ (x) > 0

(1.14)

for x > 0. The author rediscovered it in [5] and used it to prove interesting inequalities for the digamma function, see [7,6,5]. Alzer and Grinshpan [3] obtained a q-analogue of (1.14) and proved that  2 ψq′′ (x) + ψq′ (x) > 0

(1.15)

for q > 1 and x > 0. In [20], F. Qi showed that the function given in (1.15) is completely monotonic for q > 1 on (0, ∞). The author [8, Lemma 1] provided another q-extension of (1.14) and proved that 2  ψq′′ (x) + ψq′ (x) − (log q) · ψq′ (x) > 0

(1.16)

for all q > 0 and x > 0. Note that putting q = 1 in (1.16) one obtains (1.14). The inequality given in (1.16) played a central role in the proofs of inequalities in [8]. Our first aim in this work is to show the function in (1.16) is completely monotonic for all q > 0 on (0, ∞). The author [8, Theorem 2.1] proved that the function Fq defined by 1

1 − q x+ 2 Fq (x) = ψq (x + 1) − log 1−q

is positive for q > 1. Our second aim is to prove that Fq (x) is completely monotonic for q > 0 on (0, ∞).

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In [5] the author proved that 1

1 − e− x +

1 1 < ψ ′ (x) < e x − 1. 2 x

(1.17)

In [13] the authors employed the second inequality in (1.17) to improve some inequalities for polygamma functions. In [21] the authors proved that the function f (x) = e1/x − ψ ′ (x) is completely monotonic on (0, ∞) and applied this to bound the modified Bessel function I p (x). Our final aim in this work is to obtain a q-analogue of (1.17). 2. Main results The following theorems are our main results. Theorem 2.1. Let q > 0 and  2 G q (x) = ψq′′ (x) + ψq′ (x) − log q · ψq′ (x). Then G q is completely monotonic on (0, ∞). Proof. We shall use an idea from [3]. Let q > 1. Then by [8] G q (x) > 0 for q > 0 and x > 0. So, to prove Theorem 2.1, we only need to show (−1)k G q(k) (x) > 0

(2.1)

for k = 1, 2, 3, . . . , and x, q > 0. Applying (1.5) we obtain G q (x) log2 q

=

∞  i−1 

c j (q, x)ci− j (q, x) −

i=2 j=1

=

 ∞  i−1  i=2

∞ 

(i − 1)ci (q, x)

i=2

 c j (q, x)ci− j (q, x) − (i − 1)ci (q, x) ,

j=1

where ci (q, x) =

i · q −i x · log q . 1 − q −i

Hence, in order to prove (2.1) it is enough to see (−1)k

d k G q (x) > 0, d x k log2 q

or equivalently (−1)k

∞  i−1 k  ∂ [c j (q, x)ci− j (q, x)] i=2 j=1

∂xk

≥ (−1)k

∞  ∂ k [ci (q, x)] (i − 1) . ∂xk i=2

If we take the partial derivatives here, we see that this is implied by i−1  j=1

(q j

j (i − j) i(i − 1) ≤ , i− j − 1)(1 − q ) (log q)(1 − q i )

i ≥ 2,

(2.2)

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which is proved in [3]. This proves that G q is completely monotonic for q > 1 on (0, ∞). But if we use (1.13) we see that G q (x) = G 1/q (x), and therefore G q is completely monotonic for q > 0 on (0, ∞).  I would like to add another proof of Theorem 2.1 which was provided by an anonymous referee. Second proof of Theorem 2.1. Let q > 1. Applying (1.10) and (1.11) we get G q (x + 1) − G q (x) =

q −x log2 q · g(x), (1 − q −x )2

where g(x) = −2ψq′ (x) + log q +

q −x · log2 q + (1 − q −2x ) · log q . (1 − q −x )2

By using (1.10), easy calculations give g(x + 1) − g(x) =

log q (1 − q −x )2 (1 − q −x−1 )2

· h(x)

(2.3)

where q 2 · h(x) = (2q − 2q 2 + q(q + 1) log q) · q −x + (2q 2 − 2 − 4q · log q) · q −2x + (2 − 2q + (q + 1) · log q) · q −3x . h is completely monotonic on (0, ∞) as linear combination with positive coefficients of expressions exp(−kx log q) with k = 1, 2, 3. Then (2.3) shows that g(x + 1) − g(x) is completely monotonic because 1 − q −x = 1 − exp(−x log q) is a Bernstein function and then the reciprocal (and its positive powers) are completely monotonic, see [28] or [29]. Also, g(x + n + 1) − g(x + n) is completely monotonic for n = 0, 1, 2, . . . and the identity g(x + n) − g(x) =

n−1  [g(x + k + 1) − g(x + k)] k=0

gives that g(x + n) − g(x) is completely monotonic. Finally for n → ∞ we conclude that −g(x) is completely monotonic, i.e., G q (x) − G q (x + 1) is completely monotonic on (0, ∞) for q > 1. The same procedure as for g proves that G q (x) is completely monotonic. From the identity G q (x) = G 1/q (x) it follows that G q (x) is completely monotonic on (0.∞) for all q > 0, which was the assertion of Theorem 2.1. We recall that a function f : I ⊆ (−∞, ∞) → [0, ∞) is called a Bernstein function on I if f (t) has derivatives of all orders and f ′ (t) is completely monotonic on I ; see [28]. Theorem 2.2. Let q > 0 and Fq be defined by   1 1 − q x+ 2 . Fq (x) = ψq (x + 1) − log 1−q Then Fq is completely monotonic on (0, ∞).

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Proof. In Theorem 2.1 of [8] it was proved that Fq (x) > 0 for q > 0 and x > 0. Now let q > 1. Differentiation of Fq (x) gives Fq′ (x) = ψq′ (x + 1) −

log q 1

1 − q −x− 2

.

Since ψq′ (x + 1) = log q + log2 q

∞  n · q −n(x+1)

1 − q −n

n=1

,

and 1 1

1 − q −x− 2

=1+

∞ 

q −n(x+1/2)

n=1

we get Fq′ (x) = − log q

∞  q −3n/2 an (q) n=1

1 − q −n

· e−nx log q ,

(2.4)

where an (q) = q n − 1 − n · q n/2 · log q. It is very easy to show that an (q) is positive for all n ∈ N. From Eq. (2.4) together with this fact we conclude that −Fq′ (x) is completely monotonic, which necessarily implies that Fq (x) is completely monotonic for q > 1 since Fq (x) is positive for x > 0 [8]. Employing the first identity in (1.13) we arrive at Fq (x) = F1/q (x) for all x > 0 and q > 0, that is, Fq is completely monotonic on (0, ∞) for q > 0.  Corollary 2.3. Let x and q be positive real numbers. Then the following inequalities hold: 1

1

1 − q x+ 2 1 − q x+ 2 α∗ + log < ψq (x + 1) < α ∗ + log , 1−q 1−q √ where α∗ = 0 and α ∗ = ψq (1) + log(1 + q) are the best possible constants, and β∗ +

log q 1−q

−x− 12

< ψq′ (x + 1) < β ∗ +

log q 1

1 − q −x− 2

,

(2.5)

(2.6)

where q log q β∗ = ψq′ (1) + √ q −q

and

β∗ = 0

are the best possible constants. Proof. By Theorem 2.2 Fq is strictly decreasing and Fq′ is strictly increasing. From (1.3) and (1.5) we find lim ψq (x) = − log(1 − q),

x→∞

0 < q < 1,

(2.7)

and lim ψq′ (x) = log q,

x→∞

q > 1.

(2.8)

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Let 0 < q < 1. Then, by using (2.7) we obtain   lim Fq (x) = lim

x→∞

x→∞

1

1 − q x+ 2 ψq (x + 1) − log 1−q



= − log(1 − q) + log(1 − q) = 0. If q > 1, then by using (1.12) and (2.7), we obtain      1 1 1 − q x+ 2 log q − log lim Fq (x) = lim ψ1/q (x + 1) + x − x→∞ x→∞ 2 1−q   1−q = − log(1 − 1/q) + lim log = 0. x→∞ q −x+1/2 − q

(2.9)

(2.10)

Combining monotonic decrease of Fq with (2.9), (2.10) and (2.8) we obtain 0 = α∗ = lim Fq (x) < Fq (x) < Fq (0) = α ∗ , x→∞

which is equivalent to (2.5). Now let us prove (2.6). If q > 1 applying (2.8) we get   log q ′ ′ = 0. lim F (x) = lim ψq (x + 1) + 1 x→∞ q x→∞ q −x− 2 − 1 If 0 < q < 1, then 1/q > 1, and hence applying (1.13) and (2.8) yields   log q ′ ′ lim F (x) = lim log q + ψ1/q (x + 1) + = 0. 1 x→∞ q x→∞ q −x− 2 − 1

(2.11)

(2.12)

Taking into account monotonic increase of Fq′ , and the limit values in (2.11) and (2.12), we get for q > 0 and x ≥ 0 β∗ = Fq′ (0) < Fq′ (x) < lim Fq′ (x) = β ∗ = 0, x→∞

which is equivalent to (2.6).



Theorem 2.4. Let q > 0 and x > 0. Then log q log q < ψq′ (x + 1) < , α−x 1−q 1 − q β−x where the constants   log q 1 · log 1 − ′ α= log q ψq (1) and    1 q log q   − · log ; if 0 < q < 1,  log q q −1   β= 1 log q    · log ; if q > 1 log q q −1 are the best possible.

(2.13)

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Proof. Let us define   1 log q Hq (x) = x + · log 1 − ′ . log q ψq (x + 1)

(2.14)

Hq is a well-defined function because the expression under log is positive. This is clear for 0 < q < 1, and for q > 1 because ψq′ (x) > log q by (2.8) and the fact that ψq′ is decreasing. First we shall show that Hq is strictly increasing on (0, ∞) for q > 0. Differentiation of Hq yields  2 ψq′′ (x + 1) + ψq′ (x + 1) − log q · ψq′ (x + 1) . Hq′ (x) =  2 ψq′ (x + 1) − log q · ψq′ (x + 1) Utilizing (1.13) we see that  2 ′ (x + 1) > 0. ψq′ (x + 1) − log q · ψq′ (x + 1) = ψq′ (x + 1)ψ1/q Hence, by Theorem 2.1 we get Hq′ (x) > 0 for x > 0 and q > 0. Secondly, we need to show the limit values    1 q log q   · log ; if 0 < q < 1, − log q q −1   lim Hq (x) = (2.15) x→∞ 1 log q    · log ; if q > 1. log q q −1 Let 0 < q < 1. Then we get 1 · log lim Hq (x) = x→∞ log q



 lim

x→∞

log q q − −x−1 q ·q ψq′ (x + 1) x

From (1.5) we can easily deduce that  2   log q ; if 0 < q < 1, lim q −x ψq′ (x) = 1 − q x→∞  0; if q > 1, consequently we get for 0 < q < 1   q log q 1 · log . lim Hq (x) = − x→∞ log q q −1 Now let q > 1. Then upon using (1.13) we see that   ′ (x + 1) (1/q)−x−1 ψ1/q 1 lim Hq (x) = . · log lim x→∞ x→∞ q log q + qψ ′ (x + 1) log q 1/q ′ (x + 1) = 0, and Since limx→∞ ψ1/q ′ lim (1/q)−x−1 ψ1/q (x + 1) =

x→∞

log2 (1/q) 1 − 1/q

 .

(2.16)

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by (2.16), we get   1 log q . lim Hq (x) = · log x→∞ log q q −1 This proves (2.15). Combining monotonic increase of Hq with (2.15) we get for all q > 0 and x >0   1 log q α= · log 1 − ′ = Hq (0) < Hq (x) < lim Hq (x) = β (2.17) x→∞ log q ψq (1) or 1 1 − q α−x 1 − q β−x < ′ < , log q ψq (x + 1) log q which is equivalent to (2.13).



Our last theorem provides a q-analogue of (1.17). Theorem 2.5. Let x > 0. Then the following inequalities hold: If q > 1     log q q −x · log2 q log q ′ 1 − exp − + < ψq (x) < exp −1 1 − q −x 1 − q −x (1 − q −x )2

(2.18)

and if 0 < q < 1  q x · log2 q log q + 1 + log q − exp < ψq′ (x) 1 − qx (1 − q x )2   log q < −1 + log q + exp − . 1 − qx 

(2.19)

Proof. Let q > 1. Applying the mean value theorem to eψq (t) on the interval [x, x + 1], we get eψq (x+1) − eψq (x) = ψq′ (x + δ)eψq (x+δ) ,

0 < δ < 1.

(2.20)

We define u(x) = ψq′ (x)eψq (x) , x > 0, q > 1. Differentiation gives u ′ (x) = {ψq′′ (x) + [ψq′ (x)]2 }eψq (x) . By (1.15) u is strictly increasing. Since     log q exp{ψq (x + 1)} − exp{ψq (x)} = exp{ψq (x)} exp −1 1 − q −x by (1.9), we conclude from (2.20) that     log q ′ ψq (x) ψq (x) ψq (x)e
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Remark 2.7. Letting q → 1 in Theorem 2.2 leads to the observation that F1 (x) = ψ(x + 1) − log(x + 1/2) is completely monotonic on (0, ∞). For 0 ≤ α one finds easily the integral representation  ∞ f α (x) = ψ(x + 1) − log(x + α) = e−xt δα (t)dt, 0

with δα (t) =

exp((1 − α)t) − exp(−αt) − t t (exp(t) − 1)

and one observes that δα (t) > 0 for t > 0 if and only if 0 ≤ α ≤ 1/2. In particular one has the integral representation of the completely monotonic function F1 . Remark 2.8. If we let q → 1 in (2.17), we obtain −

1 ψ ′ (1)

= H1 (0) < H1 (x) = x −

1 1 < lim H1 (x) = − x→∞ + 1) 2

ψ ′ (x

or 1 1 < ψ ′ (x + 1) < , x + 1/2 x + 6/π 2 since ψ ′ (1) = π 2 /6, which is a new inequality as far as we know. Acknowledgments I would like to gratefully and sincerely thank the anonymous referees for their very valuable comments and useful corrections on this work, which improved the quality of the paper significantly. Furthermore, I thank very much an anonymous referee for providing the second proof of Theorem 2.1 and the remarks given in Remark 2.7. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

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