Complete Monotonicity Properties of a Function Involving the Polygamma Function

Eng. Appl. Sci. Letters, Vol. 1(2018), Issue 1, pp. 10-15 Website: https://pisrt.org/psr-press/journals/easl/

COMPLETE MONOTONICITY PROPERTIES OF A FUNCTION INVOLVING THE POLYGAMMA FUNCTION

KWARA NANTOMAH1

Abstract. In this paper, we study completete monotonicity properties of certain functions associated with the polygamma functions. Subsequently, we deduce some inequalities involving diﬀerence of polygamma functions.

Index Terms: Polygamma function; Complete monotonicity; Inequality.

1. Introduction The classical Gamma function, which is an extension of the factorial notation to noninteger values is usually deﬁned as ∫ ∞ Γ(x) = tx−1e−t dt, x > 0, 0 and satisfying the basic property Γ(x + 1) = xΓ(x), x > 0. Its logarithmic derivative, which is called the Psi or digamma function is deﬁned as (see [1] and [2]) ∫ d ∞ e−t − e−xt ψ(x) = ln Γ(x) = −γ + dt, x > 0, (1) − −t dx 0 1 e ∞ 1 ∑ x = −γ − + , x > 0, x k(k + x) k=1 (∑ ) n 1 − where γ = limn→∞ k=1 k ln n = 0.577215664... is the Euler-Mascheroni’s constant. Derivatives of the Psi function, which are called polygamma functions

Received 25-09-2018. Revised 27-10-2018. Accepted 29-10-2018. 1 Corresponding Author ⃝c 2018 Kwara Nantomah. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 10 Complete monotonicity properties of a function involving the polygamma function 11 are given as [1] ∫ ∞ tne−xt ψ(n)(x) = (−1)n+1 dt, x > 0, (2) − −t 0 1 e ∞ ∑ 1 = (−1)n+1n! , x > 0, (k + x)n+1 k=0 satisfying the functional equation [1] (−1)nn! ψ(n)(x + 1) = ψ(n)(x) + , x > 0, (3) xn+1 (0) where n ∈ N0 and ψ (x) ≡ ψ(x). Here, and for the rest of this paper, we use the notations: N = {1, 2, 3, 4,... }, N0 = N ∪ {0} and R = (−∞, ∞). Also, it is well known in the literature that the integral ∫ ∞ n! n −xt n+1 = t e dt, (4) x 0 holds for x > 0 and n ∈ N0. See for instance [1]. In [3], Qiu and Vuorinen established among other things that the function ( ) 1 1 h = ψ x + − ψ (x) − , (5) 1 2 2x is strictly decreasing and convex on (0, ∞). Motivated by this result, Mortici [4] proved a more generalized and deeper result which states that, the function a f = ψ(x + a) − ψ(x) − , a ∈ (0, 1), (6) a x is strictly completely monotonic on (0, ∞). Recall that a function f : (0, ∞) → R is said to be completely monotonic on (0, ∞) if f has derivatives of all order and n (n) (−1) f (x) ≥ 0 for all x ∈ (0, ∞) and n ∈ N0. In this paper, the objective is to extend Mortici’s results to the polygamma functions.

2. Some Lemmas In order to establish our main results, we need the following lemmas.

Lemma 2.1. Let a function qα,β(t) be deﬁned as { e−αt−e−βt ̸ 1−e−t , t = 0, qα,β(t) = (7) β − α, t = 0, where α, β are real numbers such that α ≠ β and (α, β) ∈/ {(0, 1), (1, 0)}. Then qα,β(t) is increasing on (0, ∞) if and only if (β − α)(1 − α − β) ≥ 0 and (β − α)(|α − β| − α − β) ≥ 0. Proof. See [5], [6] or [7]. 12 K. Nantomah

Lemma 2.2. Let a ∈ (0, 1). Then the inequality 1 − e−at a < < 1, (8) 1 − e−t holds for t ∈ (0, ∞).

1−e−at Proof. Note that the function h(t) = 1−e−t which is obtained from Lemma 2.1 by letting α = 0 and β = a ∈ (0, 1) is increasing on (0, ∞). Also, lim h(t) = a and lim h(t) = 1. t→0+ t→∞ Then for t ∈ (0, ∞), we have a = lim h(t) = h(0) < h(t) < h(∞) = lim h(t) = 1, t→0+ t→∞ which gives inequality (8).

3. Main Results We now present our results in this section.

Theorem 3.1. Let fa,k(x) and ha,r(x) be deﬁned for a ∈ (0, 1), k ∈ {2s : s ∈ N0}, r ∈ {2s + 1 : s ∈ N0} and x ∈ (0, ∞) as ak! f (x) = ψ(k)(x + a) − ψ(k)(x) − , (9) a,k xk+1 and ar! h (x) = ψ(r)(x + a) − ψ(r)(x) − . (10) a,r xr+1 Then fa,k(x) and −ha,r(x) are strictly completely monotonic on (0, ∞). Proof. By repeated diﬀerentiations with respect to x, and by using (2) and (4), we obtain − n (n) (k+n) − (k+n) − ( 1) a(k + n)! fa,k (x) = ψ (x + a) ψ (x) k+n+1 ∫ x ∫ ∞ tk+ne−(x+a)t ∞ tk+ne−xt = (−1)k+n+1 dt − (−1)k+n+1 dt − −t − −t 0 1 e 0 1 e ∫ ∞ − (−1)na tk+ne−xt dt. 0 This implies that ∫ ∫ ∫ ∞ tk+ne−xte−at ∞ tk+ne−xt ∞ (−1)nf (n)(x) = − dt + dt − a tk+ne−xt dt a,k 1 − e−t 1 − e−t ∫ 0 [ ] 0 0 ∞ 1 − e−at = − a tk+ne−xt dt − −t 0 1 e > 0, Complete monotonicity properties of a function involving the polygamma function 13 which is as a result of Lemma 2.2. Alternatively, we could proceed as follows. ∫ [ ] ∞ 1 − e−at (−1)nf (n)(x) = − a tk+ne−xt dt a,k 1 − e−t 0∫ [ ] ∞ 1 − e−at 1 − e−t tk+n+1e−xt = a − dt − −t 0 at t 1 e > 0.

1−e−t ∞ ∈ Notice that, since the function t is strictly decreasing on (0, ), then for a 1−e−at 1−e−t (0, 1), we have at > t . Hence fa,k(x) is strictly completely monotonic on (0, ∞). Similarly, we have − n − (n) ( 1) a(r + n)! (r+n) − (r+n) ha,r (x) = r+n+1 + ψ (x) ψ (x + a) x ∫ ∫ ∞ ∞ tr+ne−xt = (−1)na tr+ne−xt dt + (−1)r+n+1 dt 1 − e−t 0 ∫ 0 ∞ tr+ne−(x+a)t − (−1)r+n+1 dt, − −t 0 1 e which implies that ∫ ∫ ∫ ∞ ∞ tr+ne−xt ∞ tr+ne−xte−at (−1)n (−h )(n) (x) = a tr+ne−xt dt + dt − dt a,r 1 − e−t 1 − e−t ∫ 0 [ ] 0 0 ∞ 1 − e−at = a + tr+ne−xt dt − −t 0 1 e > 0.

Hence −ha,r(x) is strictly completely monotonic on (0, ∞). Remark 3.2. Since every completely monotonic function is convex and decreasing, it follows that fa,k(x) is strictly convex and strictly decreasing on (0, ∞). In this way, ha,r(x) is strictly concave and strictly increasing on (0, ∞). Corollary 3.3. The inequality ( ) ak! a 1 < ψ(k)(x+a)−ψ(k)(x) < ψ(k)(a)−ψ(k)(1)+k! + − a , (11) xk+1 xk+1 ak+1 holds for a ∈ (0, 1), k ∈ {2s : s ∈ N0} and x ∈ (1, ∞).

Proof. Since fa,k(x) is decreasing, then for x ∈ (1, ∞) and by applying (3), we obtain (k) (k) 0 = lim fa,k(x) < fa,k(x) < fa,k(1) = ψ (a + 1) − ψ (1) − ak! x→∞ k! = ψ(k)(a) − ψ(k)(1) + − ak!, ak+1 which completes the proof. 14 K. Nantomah

Remark 3.4. In particular, if a = 1 and k = 0 in Corollary 3.3, then we obtain ( ) 2 1 1 1 3 < ψ x + − ψ(x) < + − 2 ln 2, x ∈ (1, ∞). (12) 2x 2 2x 2 Also, if a = 1 and k = 2 in Corollary 3.3, then we obtain 2 ( ) 1 1 1 < ψ′′ x + − ψ′′(x) < + 15 − 12ζ(3), x ∈ (1, ∞), (13) x3 2 x3 where ζ(x) is the Riemann zeta function. Corollary 3.5. The inequality ( ) a 1 ar! ψ(r)(a)−ψ(r)(1)+r! − − a < ψ(r)(x+a)−ψ(r)(x) < , (14) xr+1 ar+1 xr+1 holds for a ∈ (0, 1), r ∈ {2s + 1 : s ∈ N0} and x ∈ (1, ∞).

Proof. Likewise, since ha,r(x) is increasing, then for x ∈ (1, ∞), we obtain

(r) (r) r! ψ (a) − ψ (1) − − ar! = ha,r(1) < ha,r(x) < lim ha,r(x) = 0, ar+1 x→∞ which yields (14). Remark 3.6. If a = 1 and r = 1 in Corollary 3.5, then we obtain 2 ( ) 1 π2 9 1 1 + − < ψ′ x + − ψ′(x) < , x ∈ (1, ∞). (15) 2x2 3 2 2 2x2 Furthermore, if a = 1 and r = 3 in Corollary 3.5, then we obtain 2 ( ) 3 14π4 1 3 + − 99 < ψ′′′ x + − ψ′′′(x) < , x ∈ (1, ∞). (16) x4 15 2 x4 Remark 3.7. If k = 0 in Theorem 3.1, then we obtain the main results of [4] as a special case of the present results. Remark 3.8. This paper is a modiﬁed version of the preprint [8].

Competing Interests The author declares that there are no competing interests regarding the publi- cation of this paper.

References

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Kwara Nantomah Department of Mathematics, Faculty of Mathematical Sciences, University for Development Studies, Navrongo Campus, P. O. Box 24, Navrongo, UE/R, Ghana e-mail: [email protected]