Inequalities for the Polygamma Function 1 Introduction

Mathematica Aeterna, Vol. 3, 2013, no. 4, 249 - 252 Inequalities for the polygamma function Banyat Sroysang Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, Pathumthani 12121 Thailand [email protected] Abstract In this paper, we prersent some inequalities involving the polygamma function. Mathematics Subject Classification: 26D15 Keywords: Polygamma function, inequality 1 Introduction The Euler gamma function Γ is defined by ∞ Γ(x)= tx−1e−tdt, Z0 where x> 0. The polygamma function ψ of order n ∈ N is defined by dn+1 Ψ (x)= ln Γ(x), n dxn+1 where x> 0. The polygamma function may be represented as ∞ tne−xt Ψ (x)=(−1)n+1 dt, n 1 − e−t Z0 where n ∈ N and x> 0. In 2006, Laforgia and Natalini [1] proved that 2 Ψm(x)Ψn(x) ≥ Ψ m+n (x) 2 250 Banyat Sroysang m+n N where m, n, 2 ∈ and x> 0. In 2011, Sulaiman [2] gave the inequalities as follows. 1/p 1/q Ψ (x)Ψ (x) ≥ Ψ m + n (x) (1) m n p q m n ∈ N 1 1 where m, n, p + q , x> 0, p> 1 and p + q = 1. xp yq Ψ1/p(px)Ψ1/q(qx) ≥ Ψ + (2) n n n p q 1 1 ≤ 1 1 where n is a positive odd integer, x,y > 1, x + y 1, p> 1 and p + q = 1. xp yq −Ψ1/p(px)Ψ1/q(qx) ≤ Ψ + (3) n n n p q 1 1 ≤ 1 1 where n is a positive even integer, x,y > 1, x + y 1, p> 1 and p + q = 1. In this paper, we present the generalizations for the inequalities (1), (2) and (3). 2 Results Theorem 2.1. Let n1, n2, ..., nk ∈ N, x > 0 and p1,p2, ...pk > 1 be such that k ni ∈ N and k 1 =1. Then i=1 pi i=1 pi P P k 1/pi ≥ k ni Ψni (x) Ψ (x). i=1 pi i=1 Y P Proof. By the generalized Hölder inequality, k ni ∞ i=1 −xt k ni +1 t pi e i=1 p P k ni x − i dt Ψ ( )=( 1)P −t i=1 pi 0 1 − e P Z n ∞ k i − x t k ni+1 t pi e pi − i=1 pi dt =( 1)P −t 1/p 0 (1 − e ) i i=1 Z Y k ∞ n −xt 1/pi k ni+1 t i e ≤ − i=1 pi dt ( 1)P −t 0 1 − e i=1 Z k Y 1/pi = Ψni (x). i=1 Y We note on Theorem 2.1 that if k = 2 then we obtain the inequality (1). Inequalities for the polygamma function 251 Theorem 2.2. Let n be a positive odd integer, and let x1, x2, ..., xk > 1 and k k k 1 p1,p2, ...p > 1 be such that x ≤ x and =1. Then k i=1 i i=1 i i=1 pi k P Q k P xpi Ψ1/pi (p x ) ≥ Ψ i . n i i n p i=1 i=1 i ! Y X Proof. We note that k k xpi x ≤ i . (4) i p i=1 i=1 i Y X By the generalized Hölder inequality, p x i − k i k ∞ t i=1 xpi tne pi i P dt Ψn = −t p 0 1 − e i=1 i ! Z X ∞ − k tne t i=1 xi ≤ Q dt −t 0 1 − e Z ∞ − k tne t i=1 xi ≤ P dt −t 0 1 − e Z n ∞ k −x t t pi e i dt = −t 1/p 0 (1 − e ) i i=1 Z Y k ∞ − 1/p tne pixit i ≤ dt −t 0 1 − e i=1 Z Yk 1/pi = Ψn (pixi). i=1 Y We note on Theorem 2.2 that if k = 2 then we obtain the inequality (2). Theorem 2.3. Let n be a positive even integer, and let x1, x2, ..., xk > 1 k k k 1 and p1,p2, ...p > 1 be such that x ≤ x and =1. Then k i=1 i i=1 i i=1 pi k P Q k P xpi (−1)k+1 Ψ1/pi (p x ) ≤ Ψ i . n i i n p i=1 i=1 i ! Y X 252 Banyat Sroysang Proof. By the inequality (4) and by the generalized Hölder inequality, p x i − k i k ∞ t i=1 xpi tne pi i − P dt Ψn = −t p 0 1 − e i=1 i ! Z X ∞ − k tne t i=1 xi ≥ − Q dt −t 0 1 − e Z ∞ − k tne t i=1 xi ≥ − P dt −t 0 1 − e Z n ∞ k −x t t pi e i − dt = −t 1/p 0 (1 − e ) i i=1 Z Y k ∞ − 1/p tne pixit i ≥ − dt −t 0 1 − e i=1 Z Y k k+1 1/pi =(−1) Ψn (pixi). i=1 Y We note on Theorem 2.3 that if k = 2 then we obtain the inequality (3). References [1] A. Laforgia and P. Natalini, Turan-type inequalites for some special func- tions, J. Ineq. Pure Appl. Math., 2006, 7(1), Art. 32. [2] W. T. Sulaiman, Turan inequalites for the digamma and polygamma func- tions, South Asian J. Math., 2011, 1(2), 49–55. Received: February, 2013.

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