Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 264652, 7 pages http://dx.doi.org/10.1155/2014/264652

Research Article New Inequalities for Gamma and Digamma Functions

M. R. Farhangdoost and M. Kargar Dolatabadi

Department of Mathematics, College of Science, Shiraz University, Shiraz 71475-44776, Iran

Correspondence should be addressed to M. R. Farhangdoost; [email protected]

Received 6 December 2013; Accepted 1 July 2014; Published 12 November 2014

Academic Editor: Vijay Gupta

Copyright © 2014 M. R. Farhangdoost and M. Kargar Dolatabadi. 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.

By using the mean value theorem and logarithmic convexity, we obtain some new inequalities for gamma and digamma functions.

1. Introduction In Section 2, by applying the mean value theorem on Γ(𝑥) 𝜓(𝑥) 𝜓𝑛(𝑥) 𝜁(𝑥) Let , , ,and denote the Euler gamma 󸀠 function, digamma function, polygamma functions, and (log Γ (𝑥)) =𝜓(𝑥) , for 𝑥>0, (5) Riemann zeta function, respectively, which are defined by ∞ −𝑡 𝑥−1 we obtain some new inequalities on gamma and digamma Γ (𝑥) = ∫ 𝑒 𝑡 𝑑𝑡, for 𝑥>0, 0 functions. Γ󸀠 (𝑥) (1) Section 3 is devoted to some new inequalities on 𝜓 (𝑥) = , 𝑥>0, digamma function, by using convex properties of logarithm Γ (𝑥) for of this function. (𝑛) 𝑛 𝜓 (𝑥) Note that in this paper by 𝛾=lim𝑛→∞(∑𝑘=1(1/𝑘) − log(𝑛)) = 0.5772156 ⋅ ⋅ we mean Euler’s constant [5]. ∞ 𝑡𝑛𝑒−𝑥𝑡 (2) = (−1)𝑛+1 ∫ 𝑑𝑡, 𝑥 > 0; 𝑛 = 1, 2, 3, ., −𝑡 for 0 1−𝑒 2. Inequalities for Gamma and Digamma ∞ 1 𝜁 (𝑥) = ∑ , 𝑥>1. Functions by the Mean Value Theorem 𝑥 for (3) 𝑛=1 𝑛 Lemma 1. For 𝑡>0,onehas In the past different papers appeared providing inequalities for the gamma, digamma, and polygamma functions (see [1– −𝜓󸀠󸀠 (𝑡) 18]). <1. 2 (6) By using the mean value theorem to the function log Γ(𝑥) 𝜓󸀠 (𝑡) on [𝑢, 𝑢 +,with 1] 𝑥>0and 𝑢>0,Batir[19]presentedthe following inequalities for the gamma and digamma functions: Proof. By [6,Proposition1],wehave −𝛾 𝜓 (𝑥) ⩽ log (𝑥−1+𝑒 ), for 𝑥>0, 󸀠 󸀠󸀠󸀠 󸀠󸀠 2 1 𝜓 (𝑡) 𝜓 (𝑡) −2[𝜓 (𝑡)] <0, for 𝑡>0. (7) (𝑥) −𝜓(𝑥) < 𝜓󸀠 (𝑥) , 𝑥>1, log 2 for (4) 󸀠󸀠 󸀠 2 𝜋2 Thus the function 𝜓 (𝑡)/𝜓 (𝑡) is strictly decreasing on 𝜓󸀠 (𝑥) ⩾ 𝑒−𝜓(𝑥), 𝑥⩾1. 6𝑒𝛾 for (0, ∞). 2 Journal of Applied Mathematics

By using asymptotic expansions [20, pages 253–256 and We show that the function 𝜃(𝑢) has the following properties: 364], (1) 𝜃(𝑢) is strictly increasing on (0, ∞); 󸀠 󸀠 1 1 1 𝜃 󸀠 𝜓 (𝑡) = + + + ,(0⩽𝜃⩽1), (2) lim𝑢→∞𝜃(𝑢) = ;1/2 2 3 5 (8) 𝑡 2𝑡 6𝑡 30𝑡 󸀠 (3) 𝜃 (𝑢) is strictly decreasing on (0, ∞); 󸀠󸀠 󸀠󸀠 1 1 1 1 𝜃 󸀠󸀠 󸀠 𝜓 (𝑡) =− − − + − ,(0⩽𝜃⩽1). (9) (4) lim𝑢→∞𝜃 (𝑢) =. 0 𝑡2 𝑡3 2𝑡4 6𝑡6 6𝑡8 󸀠 To prove these four properties, since 𝜓 is a decreasing For 𝑡>0,weget 󸀠 function on (0, ∞),weput𝑢=1/𝜓(𝑡),where𝑡>0;by 𝜓󸀠󸀠 (𝑡) formula (13) we have lim =−1. (10) 𝑡→∞𝜓󸀠 (𝑡)2 1 1 𝜓󸀠 ( +𝜃( )) = 𝜓󸀠 (𝑡) . (∗) 𝜓󸀠 (𝑡) 𝜓󸀠 (𝑡) 󸀠󸀠 󸀠 2 Now, the proof follows from the monotonicity of 𝜓 (𝑡)/𝜓 (𝑡) (0, ∞) 󸀠󸀠 󸀠 on and Since by formula (8) we have 𝜓 (𝑡) < 0 and 𝜓 (𝑡) >,for 0 󸀠 󸀠󸀠 all 𝑡>0, then the mapping 𝑡→𝜓(𝑡) from (0, ∞) into 𝜓 (𝑡) 󸀠 󸀠 =−1. (0, ∞) is injective since also 𝜓 (𝑡) →0 and 𝜓 (𝑡) →∞ lim 2 (11) + 𝑡→∞𝜓󸀠 (𝑡) when 𝑡→∞and 𝑡→0, respectively, then the mapping 󸀠 𝑡→𝜓(𝑡) from (0,∞)into(0,∞)isabijectivemap.Clearly, 󸀠 by injectivity of 𝜓 ,wefindthat Theorem 2. Onehasthefollowing: 1 1 𝜃( )=𝑡− , for 𝑡>0. (14) 󸀠 2 𝜓󸀠 (𝑡) 𝜓󸀠 (𝑡) (a) 𝑥 − (1/2) < 1/𝜓 (𝑥) ⩽ 𝑥 + (6/𝜋 )−1for 𝑥⩾1; 2 󸀠 󸀠 2 (b) 1/𝑥 <𝜓(𝑥)𝜓 (𝑥+1)<2/𝑥 for 𝑥>0; Differentiating between both sides of this equation, we get 󸀠 2 󸀠󸀠 4 2 󸀠 (c) [𝜓 (𝑥)] /𝜓 (𝑥) ⩾ −𝜋 /72𝜁(3) for 𝑥⩾1and 𝑥 𝜓 (𝑥 + 2 󸀠 4 󸀠 󸀠󸀠 1)𝜓 (𝑥) < 𝜋 /72𝜁(3) for 𝑥>2; 1 −[(𝜓 (𝑡)) +𝜓 (𝑡)] 𝜃󸀠 ( )= . (15) 󸀠 2 󸀠 󸀠 󸀠 󸀠󸀠 󸀠 󸀠󸀠 (d) ([𝜓 (𝑥+ℎ)] −𝜓(𝑥)𝜓 (𝑥 + ℎ))/ℎ𝜓 (𝑥) > 𝜓 (𝑥 + ℎ) 𝜓 (𝑡) 𝜓 (𝑡) 𝑥>0 ℎ>0 for and ; 󸀠󸀠 󸀠 󸀠 󸀠 2 󸀠 󸀠󸀠 Since by formula (8), 𝜓 (𝑡) <,where 0 𝑡>0,henceformula (𝜓 (𝑥+ℎ)𝜓(𝑥) − [𝜓 (𝑥)] )/ℎ𝜓 (𝑥+ℎ)<𝜓 (𝑥) 󸀠 󸀠 (e) for (15) gives 𝜃 (1/𝜓 (𝑡)) >,forall 0 𝑡>0.Sincethemapping𝑡→ 𝑥>0 ℎ>0 󸀠 󸀠 and ; 1/𝜓 (𝑡) from (0, ∞) to (0, ∞) is also bijective, then 𝜃 (𝑡) > 0 2 󸀠󸀠 󸀠 󸀠 󸀠 󸀠 (f) −𝑥 𝜓 (𝑥) < 𝜓 (𝑥)/𝜓 (𝑥 + 1) and 𝜓 (𝑥 + 1)/𝜓 (𝑥) < for all 𝑡>0, and the proof of (1) is completed. 2 󸀠󸀠 −𝑥 𝜓 (𝑥 + 1) for 𝑥>0; From (8) we have 2 (𝑥+(6/𝜋2)) −𝑥(𝛾+1) (g) ((𝜋 𝑥/6) + 1) 𝑒 ⩽ Γ(𝑥+1)< (2𝑥+ lim 𝜃 (𝑢) (𝑥+(1/2)) −𝑥(1+𝛾) 𝑢→∞ 1) 𝑒 for 𝑥⩾1; (1/𝑥) − 𝜓󸀠(𝑥) < (1/2)𝜓󸀠󸀠(𝑥 + (1/2)) 𝑥>0 1 1 (h) for and = lim 𝜃( )= lim (𝑡 − ) 󸀠 󸀠 −1 󸀠󸀠 𝑡→∞ 𝜓󸀠 (𝑡) 𝑡→∞ 𝜓󸀠 (𝑡) (1/𝑥) − 𝜓 (𝑥) > ((𝜓 ) (1) − 1)𝜓 (𝑥) for 𝑥>1; −1 (16) 𝜓(𝑥+1)> (𝑥 + (1/2)) + 𝜓((𝜓󸀠) (1)) 𝑥⩾1/2 1 (i) log for ; = (𝑡 − ) lim 2 3 5 4 󸀠 −1 󸀠 −1 𝑡→∞ (1/𝑡) + (1/2𝑡 )+(1/6𝑡 )+(1/3𝑡 ) (j) (𝜋 /72𝜁(3)) log(𝑥−(𝜓) (1) + 2) + 𝜓((𝜓 ) (1)) ⩾ 󸀠 −1 1 𝜓(𝑥+1)for 𝑥>(𝜓) (1) − 1. = . 2 Proof. Let 𝑢 be a positive real number and 𝜓(𝑥) defined on the closed interval [𝑢, 𝑢+1].Byusingthemeanvaluetheoremfor Differentiating between both sides of (15),weobtain 󸀠 the function 𝜓(𝑥) on [𝑢, 𝑢 + 1] with 𝑢>0and since 𝜓 is a 1 decreasing function, there is a unique 𝜃 depending on 𝑢 such 𝜃󸀠󸀠 ( ) that 0⩽𝜃=𝜃(𝑢)<1,forall𝑢⩾0;then 𝜓󸀠 (𝑡) (∗∗) 󸀠 󸀠 3 𝜓 (𝑢+1) −𝜓(𝑢) =𝜓 (𝑢+𝜃(𝑢)) , (12) [𝜓 (𝑡)] 2 = [2 (𝜓󸀠󸀠 (𝑡)) −𝜓󸀠 (𝑡) 𝜓󸀠󸀠 (𝑡)]. 󸀠󸀠 󸀠 󸀠 2 𝜓 (𝑡) Since 𝜓(𝑥+1)−𝜓(𝑥)=1/𝑥and 𝜓 (𝑥 + 1) − 𝜓 (𝑥) = −1/𝑥 , 󸀠 󸀠󸀠 we have Since 𝜓 (𝑡) > 0 and 𝜓 (𝑡) <,where 0 𝑡>0,then 󸀠󸀠 󸀠 󸀠 1 𝜃 (1/𝜓 (𝑡)) < 0 for all 𝑡>0. Proceeding as above we conclude 𝜓 (𝑢+𝜃(𝑢)) = , 𝑢>0. 󸀠󸀠 𝑢 for (13) that 𝜃 (𝑡) <,for 0 𝑡>0. This proves (3). Journal of Applied Mathematics 3

For (4), from (8), (9),weconcludethat By using this inequality and the fact that 𝜓(𝑥+1)−𝜓(𝑥) =1/𝑥 2 and [(𝜓󸀠 (𝑡)) +𝜓󸀠󸀠 (𝑡)] 󸀠 󸀠 1 󸀠 󸀠 1 lim 𝜃 (𝑢) = lim 𝜃 ( )= lim − 𝜓 (𝑥+1) −𝜓 (𝑥) =− , (26) 𝑢→∞ 𝑡→∞ 𝜓󸀠 (𝑡) 𝑡→∞ 𝜓󸀠󸀠 (𝑡) 𝑥2

2 [𝜓󸀠 (𝑡)] we obtain =−1− lim =0. 󸀠 󸀠 1 𝑡→∞ 𝜓󸀠󸀠 (𝑡) 𝜓 (𝑡+1) 𝜓 (𝑡) > ,𝑡>0. 𝑡2 (27) (17) Since 𝜃 is strictly increasing on (0, ∞),by(1),itisclearthat 󸀠 Now, we prove the theorem. To prove (a), let 1/𝜓 (1) = 2 1 1 6/𝜋 ⩽𝑡<∞;thenby(1)and(2)wehave 𝜃( )−𝜃( ) 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) 1 (28) 𝜃( )⩽𝜃(𝑡) < lim 𝜃 (𝑡) . (18) + 1 𝜓󸀠 (1) 𝑡→∞ < lim 𝜃 (𝑡) −𝜃(0 )= ,𝑡>0. 𝑡→∞ 2 󸀠󸀠 Equation (13) and 𝜓 (𝑡) < 0 for all 𝑡>0give and then it is clear that (b) holds. 󸀠 For (c), since 𝑡>2, 𝑡 + 𝛿(𝑡) > 1 +𝛿(1),and𝜃 is strictly −1 1 𝜃 (𝑡) =(𝜓󸀠) ( )−𝑡. decreasing on (0, ∞) by (3), then 𝑡 (19) 󸀠 2 𝜃(𝑡) 1 1 [𝜓 (1)] By substituting the value of into (18),weget 𝜃󸀠 ( )<𝜃󸀠 ( )=−1− , 𝜓󸀠 (𝑡+𝛿(𝑡)) 𝜓󸀠 (1) 𝜓󸀠󸀠 (1) (29) 1 󸀠 −1 1 1 1− ⩽(𝜓) ( )−𝑡< lim 𝜃 (𝑡) = . 𝜓󸀠 (1) 𝑡 𝑡→∞ 2 (20) ∀𝑡 > 2.

󸀠 󸀠 󸀠 2 By substituting the value 𝑡=1/𝜓(𝑢) into this inequality, we Since 𝜓(𝑥 + 1) − 𝜓(𝑥) =1/𝑥 and 𝜓 (𝑥 + 1) − 𝜓 (𝑥) = −1/𝑥 , get by using (24),weobtain 4 1 1 6 2 󸀠 󸀠 𝜋 𝑢− < ⩽𝑢+ −1, 𝑡 𝜓 (𝑡+1) 𝜓 (𝑡) < , (30) 2 𝜓󸀠 (𝑢) 𝜋2 (21) 72𝜁 (3)

𝑢⩾1 where 𝑡>2. where . 󸀠 󸀠󸀠 Inordertoprove(b),byusingthemeanvaluetheoremon Since 𝜃 is strictly decreasing on (0, ∞) by (3) and 𝜓 (𝑡) < 󸀠 󸀠 the interval [1/𝜓 (𝑡), 1/𝜓 (𝑡 + 1)], and since 𝜃 is a decreasing 0,forall𝑡>0,wehave function, there exists a unique 𝛿 such that 1 1 𝜃󸀠 ( )⩽𝜃󸀠 ( ), 󸀠 󸀠 (31) 0<𝛿(𝑡) <1, (22) 𝜓 (𝑡) 𝜓 (1) 𝑡⩾1 for 𝑡>0and where . Then it is clear that (c) is true. 1 1 Now we prove (d) and (e) by using the mean value 𝜃( )−𝜃( ) 󸀠 󸀠 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) theorem on [1/𝜓 (𝑡), 1/𝜓 (𝑡 + ℎ)] (𝑡 > 0, ℎ >0),for𝜃,we (23) conclude 1 1 1 =( − )𝜃󸀠 ( ). 1 1 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) 𝜓󸀠 (𝑡+𝛿(𝑡)) 𝜃( )−𝜃( ) 𝜓󸀠 (𝑡+ℎ) 𝜓󸀠 (𝑡) (32) Now, by (14),wehave 1 1 1 =( − )𝜃󸀠 ( ), 1 1 𝜓󸀠 (𝑡+ℎ) 𝜓󸀠 (𝑡) 𝜓󸀠 (𝑡+𝑎) 1− + 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) 0<𝑎<ℎ (24) where . 1 1 1 After brief computation we have =( − )𝜃󸀠 ( ). 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) 𝜓󸀠 (𝑡+𝛿(𝑡)) 1 ℎ𝜓󸀠 (𝑡+ℎ) 𝜓󸀠 (𝑡) 𝜃󸀠 ( )= −1, 𝑡>0. 󸀠 󸀠 󸀠 (33) Since 𝜃 is strictly increasing on (0, ∞),by(1),wehave 𝜓 (𝑡+𝑎) 𝜓 (𝑡) −𝜓 (𝑡+ℎ) 𝜓󸀠 (𝑡+1) −𝜓󸀠 (𝑡) Since 𝑡+𝑎>𝑡for all 𝑎>0, 𝑡>0, and by the monotonicity of 1+ 𝜃󸀠 𝜓󸀠 𝜃󸀠(1/𝜓󸀠(𝑡 + 𝑎))󸀠 <𝜃 (1/𝜓󸀠(𝑡)) 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) and we have ;then (25) 󸀠 󸀠 󸀠 2 1 1 𝜓 (𝑡+ℎ) 𝜓 (𝑡) −[𝜓 (𝑡)] =𝜃( )−𝜃( )>0. <𝜓󸀠󸀠 (𝑡) ,𝑡>0,ℎ>0.(34) 𝜓󸀠 (𝑡+1) 𝜓󸀠 (𝑡) ℎ𝜓󸀠 (𝑡+ℎ) 4 Journal of Applied Mathematics

󸀠 󸀠 By monotonicity of 𝜃 and 𝜓 ,wehave or 1 1 󸀠 −1 𝜃󸀠 ( ) >𝜃󸀠 ( ) . log (𝑥) +𝜓((𝜓) (1))<𝜓(𝑥+𝜃(𝑥)) . (45) 𝜓󸀠 (𝑡+𝑎) 𝜓󸀠 (𝑡+ℎ) (35) 𝜃 𝜓 After some simplification of this inequality (d) is proved. Again using the monotonicity of and , after some simpli- 𝑥⩾1/2 For (f), we put ℎ=1in (e) and (d). fications as for ,wecanrewrite [1, 𝑡] 𝑡>0 For (g), we integrate (a) on for ;thenwehave 1 󸀠 −1 log (𝑥 + )+𝜓((𝜓) (1))<𝜓(𝑥+1) . (46) (𝑡−1) 𝜋2 2 ( +1)−𝛾 log 6 (36) This proves (i). By inequality (c) for 𝑥⩾1,wehave ⩽𝜓(𝑡) < (2𝑡 − 1) −𝛾, 𝑡⩾1; log for 72𝜁 (3) 𝑥+𝜃(𝑥) (𝑥) ⩾ ∫ 𝜓󸀠 (𝑡) 𝑑𝑡 log 4 theproofiscompletedwhenweintegratetheseinequalities 𝜋 (𝜓󸀠)−1(1) on [1, 𝑠],for𝑠>0. (47) 󸀠 72𝜁 (3) −1 By using the mean value theorem for the 𝜓 (𝑡) on [𝑡, 𝑡 + = (𝜓 (𝑥+𝜃(𝑥)) −𝜓((𝜓󸀠) (1))) ; 𝜃(𝑡)],thereisa𝛼(𝑡) depending on 𝑡 such that 0 < 𝛼(𝑡) < 𝜃(𝑡) 𝜋4 𝑡>0 for all ,andso 󸀠 −1 󸀠 −1 since for 𝑥⩾1, 𝜃(𝑥) ⩾ 𝜃(1) =((𝜓 ) (1)−1)) = (𝜓 ) (1)−1, 𝜓󸀠 (𝑡+𝜃(𝑡)) =𝜃(𝑡) 𝜓󸀠󸀠 (𝑡+𝛼(𝑡)) +𝜓󸀠 (𝑡) . (37) from this inequality we find that 󸀠󸀠 𝜓 4 By formula (13) and(2),since is strictly increasing on 𝜋 −1 (0, ∞) (𝑥) +𝜓((𝜓󸀠) (1)) ,wehave 72𝜁 (3) log 𝜓󸀠󸀠 (𝑡+𝛼(𝑡)) 𝜃 (𝑡) (48) −1 ⩾𝜓(𝑥+(𝜓󸀠) (1) −1); 1 󸀠 󸀠󸀠 = −𝜓 (𝑡) < lim 𝜃 (𝑡) 𝜓 (𝑡 + lim 𝜃 (𝑡)), for 𝑡>0, 𝑡 𝑡→∞ 𝑡→∞ 𝑥 𝑥−(𝜓󸀠)−1(1) + 2 𝑥⩾(𝜓󸀠)−1(1) − 1 (38) replacing by ,wegetfor 4 or 𝜋 −1 −1 (𝑥−(𝜓󸀠) (1) +2)+𝜓((𝜓󸀠) (1)) 1 1 1 72𝜁 (3) log −𝜓󸀠 (𝑡) < 𝜓󸀠󸀠 (𝑡 + ), 𝑡>0; (49) 𝑡 2 2 for (39) ⩾𝜓(𝑥+1) , 󸀠󸀠 since 𝜓 is strictly increasing on (0, ∞),by(1),wehave 1 which proves (j). Then the proof is completed. 𝜃 (𝑡) 𝜓󸀠󸀠 (𝑡+𝛼(𝑡)) = −𝜓󸀠 (𝑡) >𝜃(1) 𝜓󸀠󸀠 (𝑡) , 𝑡 (40) Example 3. Consider the matrix 𝑡>1, for 3 1 1 ⋅⋅⋅ 1 [ ] or [1 4 1 ⋅⋅⋅ 1 ] 1 −1 𝐴𝑛 = [ . ] . (50) −𝜓󸀠 (𝑡) > ((𝜓󸀠) (1) −1) 𝜓󸀠󸀠 (𝑡) , 𝑡>1. [ . ] 𝑡 for (41) [1 1 ⋅⋅⋅ 1 𝑛+1] In order to prove (i) and (j), we integrate both sides of (13) over 1⩽𝑢⩽𝑥to obtain By using inequalities (a), we obtain 𝑥 𝑥 1 ∫ 𝜓󸀠 (𝑢+𝜃(𝑢)) 𝑑𝑢 = ∫ 𝑑𝑢. 𝜋2 2 𝑢 (42) ⩽𝜓󸀠 (𝑥) < , 𝑥⩾1. (51) 1 1 𝜋2𝑥+6−𝜋2 2𝑥 − 1 󸀠 Making the change of variable 𝑢=1/𝜓(𝑡) on the left-hand Now, we integrate on [1, 𝑡] (for 𝑡>0) from both sides of (51) side, by (14),wehave to obtain 𝑥+𝜃(𝑥) −𝜓󸀠󸀠 (𝑡) ∫ 𝜓󸀠 (𝑡) 𝑑𝑡 = (𝑥) ; 2 2 log (43) (𝑡−1) 𝜋 (𝜓󸀠)−1(1) 𝜓󸀠 (𝑡) ( +1)−𝛾⩽𝜓(𝑡) < (2𝑡 − 1) −𝛾; log 6 log (52) 𝜓󸀠(𝑡) > 0 𝑡>0 𝜓󸀠(𝑥)𝜓󸀠󸀠(𝑥) − 2[𝜓󸀠󸀠(𝑥)]2 <0 since for all and , 𝑡 𝑛+1 𝑛 we find that, for 𝑥>1, replacing by ( is an integer number) and using the identity 𝜓(𝑛 + 1)𝑛 =𝐻 −𝛾[6] and det 𝐴𝑛 =𝑛!𝐻𝑛 [21], where 𝑥+𝜃(𝑥) 𝑛 󸀠 𝐻𝑛 = ∑ (1/𝑘) is the 𝑛th harmonic number, then we have log (𝑥) < ∫ 𝜓 (𝑡) 𝑑𝑡 𝑘=1 (𝜓󸀠)−1(1) (44) 2 𝑛! 𝑛𝜋 𝑛! 󸀠 −1 ( +1) ⩽𝑛!𝐻 < (2𝑛 + 1) . (53) =𝜓(𝑥+𝜃(𝑥)) −𝜓((𝜓) (1)) log 6 𝑛 log Journal of Applied Mathematics 5

3. New Inequalities for Digamma Function Let V =3and 𝑢=𝑎𝑥+3.Notethat𝜓(3) = (3/2) −𝛾 and by Properties of Strictly Logarithmically (1/𝑎)𝑢 + (1 − (1/𝑎))V =𝑥+3;alsoweobtain Convex Functions [𝜓 (𝑥+3)]𝑎 3 𝑎−1 𝑢−3 3 >( −𝛾) for 𝑥= >− . (58) Definition 4. Apositivefunction𝑓 is said to be logarithmi- 𝜓 (𝑎𝑥) +3 2 𝑎 𝑎 callyconvexonaninterval𝐼 if 𝑓 has derivative of order two on 𝐼 and In order to prove (b), let

󸀠󸀠 (log 𝑓 (𝑥)) ⩾0 (54) 𝑓 (𝑥) = log 𝜓 (𝑎𝑥) +3 − log 𝜓 (3+𝑎) (59) −𝑎 𝜓 (𝑥+3) ; for all 𝑥∈𝐼. log If inequality (54) is strict, for all 𝑥∈𝐼,then𝑓 is said to be 𝜓(4) = (11/6) − 𝛾 𝑓(1) = ((11/6) − 𝛾)−𝑎 strictly logarithmically convex [22]. since ,wehave log . Also Lemma 5. Γ [𝑐, ∞) 𝑐= The function is increasing on ,where 𝜓󸀠 (𝑎𝑥) +3 𝜓󸀠 (𝑥+3) 1/46163 ⋅ ⋅ ⋅ is the only positive zero of 𝜓 [1, 19]. 𝑓󸀠 (𝑥) =𝑎[ − ] . 𝜓 (𝑎𝑥) +3 𝜓 (𝑥+3) (60) Lemma 6. If 𝑥⩾𝑐and 𝑘(𝑥) = 1/𝜓(𝑥),then𝑘 is strictly logarithmically convex on [𝑐, ∞). By Lemma 6,log(1/𝜓(𝑡)) is strictly convex on [𝑐, ∞);then 󸀠󸀠 󸀠 󸀠 (log 𝜓(𝑡)) <0and so (𝜓 (𝑡)/𝜓(𝑡)) <0; this implies that 󸀠 Proof. By differentiation we have (𝜓 (𝑡)/𝜓(𝑡)) is strictly decreasing on [𝑐, ∞).Since𝑎>1and 𝑥 ∈ (0, 1),wehave𝑎𝑥+3>𝑥+3.Then 2 󸀠 󸀠 −𝜓󸀠󸀠 (𝑥) 𝜓 (𝑥) +[𝜓󸀠 (𝑥)] 󸀠󸀠 −𝜓 (𝑥) 󸀠 󸀠 [log 𝑘 (𝑥)] = [ ] = ; 𝜓 (𝑎𝑥+3) 𝜓 (𝑥+3) 𝜓 (𝑥) 2 < . (61) [𝜓 (𝑥)] 𝜓 (𝑎𝑥) +3 𝜓 (𝑥+3) (55) 𝑓󸀠(𝑥) < 0 𝑓(1) = ((11/6) − 𝛾)−𝑎 󸀠 And then ;also log .Then by Lemma 5,weobtain𝜓(𝑥) =Γ (𝑥)/Γ(𝑥),forevery >0 󸀠󸀠 𝑥∈[𝑐,∞)and since 𝜓 (𝑥) < 0 on (0, ∞),thenwehave 11 −𝑎 󸀠󸀠 𝑓 (𝑥) >𝑓(1) = ( −𝛾) (62) (log 𝑘(𝑥)) >0,for𝑥⩾𝑐. log 6 This implies that 1/𝜓(𝑥) is strictly logarithmically convex on [𝑐, ∞). for 𝑎>1and 𝑥 ∈ (0, 1) or 𝑎 𝑎 Theorem 7. Onehasthefollowing: [𝜓 (𝑥+3)] ((11/6) −𝛾) < . (63) 𝑎 𝑎−1 𝜓 (𝑎𝑥+3) 𝜓 (3+𝑎) (a) [𝜓(𝑥 + 3)] /𝜓(𝑎𝑥 + 3) > ((3/2) −𝛾) ,for𝑎>1and 𝑥>−3/𝑎 ; So (b) is proved. 𝑎 𝑎 (b) [𝜓(𝑥 + 3)] /𝜓(𝑎𝑥 + 3) < ((11/6) −𝛾) /𝜓(3 + 𝑎),for By 𝑎>1and 𝑥 ∈ (0, 1); 𝑎 𝑎 𝑎𝑥+3>𝑥+3, for 𝑎 > 1, 𝑥 >1, (c) [𝜓(𝑥 + 3)] /𝜓(𝑎𝑥 + 3) > ((11/6) −𝛾) /𝜓(3 + 𝑎),for 𝑎>1and 𝑥>1; 𝑎𝑥+3<𝑥+3, for 𝑎∈(0, 1) ,𝑥∈(0, 1) , (64) [𝜓(𝑥 +𝑎 3)] /𝜓(𝑎𝑥 + 3) > ((11/6) −𝛾)𝑎/𝜓(3 + 𝑎) (d) ,for 𝑎𝑥+3<𝑥+3, for 𝑎∈(0, 1) , 𝑥>1, 𝑎 ∈ (0, 1) and 𝑥 ∈ (0, 1); 𝑎 𝑎 (e) [𝜓(𝑥 + 3)] /𝜓(𝑎𝑥 + 3) < ((11/6) −𝛾) /𝜓(3 + 𝑎),for (c), (d), and (e) are clear. 𝑎 ∈ (0, 1) and 𝑥>1. Corollary 8. For all 𝑥 ∈ (0, 1) and all integers 𝑛>1,onehas Proof. By Lemma 6 we have, for 𝑎>1, 3 𝑛−1 [𝜓 (𝑥+3)]𝑛 ((11/6) −𝛾)𝑛 ( −𝛾) < < , (65) 𝑢 V 1/𝑝 1/𝑞 2 𝜓 (𝑛𝑥) +3 𝐻 −𝛾 𝜓 [ + ] > [𝜓 (𝑢)] [𝜓 (V)] , 𝑛+2 𝑝 𝑞 (56) 𝑛 where 𝐻𝑛 = ∑𝑘=1(1/𝑘) is the 𝑛th harmonic number. where 𝑝>1, 𝑞>1, (1/𝑝) + (1/𝑞) =1, 𝑢⩾𝑐,andV ⩾𝑐. If 𝑝=𝑎and 𝑞 = 𝑎/(𝑎 −1),then Proof. By [6], for all integers 𝑛⩾1,wehave 1 1 𝜓 (𝑛+1) =𝐻 −𝛾, 𝜓[ 𝑢+(1− ) V]>[𝜓(𝑢)]1/𝑎 [𝜓 (V)]1−(1/𝑎) 𝑛 (66) 𝑎 𝑎 (57) and replacing 𝑎 by 𝑛 in Theorem 7,theproofiscompleted. for 𝑢⩾𝑐and V ⩾𝑐. 6 Journal of Applied Mathematics

Theorem 9. Let 𝑓 be a function defined by Corollary 13. For all 𝑥 ∈ (0, 1) and all 𝑎>0and 𝑏<0,one

𝑎 has [𝜓 (3+𝑏𝑥)] 𝑎 𝑎 𝑓 (𝑥) = ; ∀𝑥>0, [𝜓 (3+𝑏𝑥)] [𝜓 (3 + 𝑏𝑦)] 𝑏 (67) < , [𝜓 (3+𝑎𝑥)] 𝑏 𝑏 (74) [𝜓 (3+𝑎𝑥)] [𝜓 (3 + 𝑎𝑦)] where 3+𝑎𝑥⩾𝑐and 3+𝑏𝑥⩾;thenforall 𝑐 𝑎>𝑏>0or where 3+𝑎𝑥⩾𝑐, 3+𝑏𝑥⩾, 𝑐 3+𝑎𝑦⩾, 𝑐 3+𝑏𝑦⩾,and 𝑐 0>𝑎>𝑏(𝑎>0and 𝑏<0), 𝑓 is strictly increasing (strictly 0<𝑦<𝑥<1. decreasing) on (0, ∞). Remark 14. Taking 𝑎=𝑛and 𝑏=1in Corollary 10,weobtain Proof. Let 𝑔 be a function defined by inequalities of Corollary 8. 𝑔 (𝑥) = log 𝑓 (𝑥) =𝑎log 𝜓 (3+𝑏𝑥) −𝑏log 𝜓 (3+𝑎𝑥) ; (68) Conflict of Interests then The authors declare that there is no conflict of interests 𝜓󸀠 (3+𝑏𝑥) 𝜓󸀠 (3+𝑎𝑥) regarding the publication of this paper. 𝑔󸀠 (𝑥) =𝑎𝑏[ − ]. 𝜓 (3+𝑏𝑥) 𝜓 (3+𝑎𝑥) (69) References By proof of Theorem 7,wehave [1] H. Alzer and S. Ruscheweyh, “A subadditive property of 󸀠󸀠 (log 𝜓 (𝑥)) <0, for 𝑥∈[𝑐, ∞) ; (70) the gamma function,” Journal of Mathematical Analysis and Applications,vol.285,no.2,pp.564–577,2003. 𝑔󸀠(𝑥) > 0 𝑎>𝑏>0 0>𝑎>𝑏(𝑔󸀠(𝑥) < 0 [2] H. Alzer, “Some gamma function inequalities,” Mathematics of this implies that if or Computation,vol.60,no.201,pp.337–346,1993. if 𝑎>0and 𝑏<0);thatis,𝑔 is strictly increasing on (0, ∞) (0, ∞)) 𝑓 [3] A. V. Boyd, “Gurland’s inequality for the gamma function,” (strictly decreasing on .Hence is strictly increasing Scandinavian Actuarial Journal,vol.1960,pp.134–135,1960. on (0, ∞),if𝑎>𝑏>0or 0>𝑎>𝑏(strictly decreasing if 𝑎>0 𝑏<0 [4] J. Bustoz and M. E. H. Ismail, “On gamma function inequalities,” and ). Mathematics of Computation,vol.47,no.176,pp.659–667,1986. Corollary 10. 𝑥 ∈ (0, 1) 𝑎>𝑏>0 0>𝑎>𝑏 [5] C. P. Chen, “Asymptotic expansions of the logarithm of the For all and all or , gamma function in the terms of the polygamma functions,” one has Mathematical Inequalities and Applications,vol.2,pp.521–530, 3 𝑎−𝑏 [𝜓 (3+𝑏𝑥)]𝑎 [𝜓 (3+𝑏)]𝑎 2014. ( −𝛾) < < , 𝑏 𝑏 (71) [6] C.-P.Chen, “Some properties of functions related to the gamma, 2 [𝜓 (3+𝑎𝑥)] [𝜓 (3+𝑎)] psi and tetragamma functions,” Computers & Mathematics with Applications,vol.62,no.9,pp.3389–3395,2011. where 3+𝑏𝑥⩾𝑐, 3+𝑎𝑥⩾𝑐, 3+𝑏⩾𝑐,and3+𝑎⩾𝑐. [7] C. P. Chen, “Unified treatment of several asymptotic formulas for the gamma function,” Numerical Algorithms,vol.64,no.2, Proof. To prove (71), applying Theorem 9 and taking account pp.311–319,2013. of 𝜓(3) = (3/2) − 𝛾,weget𝑓(0) < 𝑓(𝑥) < 𝑓(1) for all 𝑥∈ [8] T. Erber, “The gamma function inequalities of Gurland and (0, 1),andweobtain(71). Gautschi,” vol. 1960, no. 1-2, pp. 27–28, 1960. [9] J. D. Keckic and P. M. Vasic, “Some inequalities for the gamma Corollary 11. For all 𝑥 ∈ (0, 1) and all 𝑎>0and 𝑏<0,one function,” Publications de l’Institut Mathematique´ ,vol.11,pp. has 107–114, 1971.

𝑎 𝑎 𝑎−𝑏 [10] A. Laforgia, “Further inequalities for the gamma function,” [𝜓 (3+𝑏)] [𝜓 (3+𝑏𝑥)] 3 Mathematics of Computation,vol.42,no.166,pp.597–600,1984. < < ( −𝛾) , (72) [𝜓 (3+𝑎)]𝑏 [𝜓 (3+𝑎𝑥)]𝑏 2 [11] D. Lu and X. Wang, “A new asymptotic expansion and some inequalities for the gamma function,” Journal of Number Theory, where 3+𝑎𝑥⩾𝑐, 3+𝑏𝑥⩾𝑐, 3+𝑏⩾𝑐,and3+𝑎⩾𝑐. vol. 140, pp. 314–323, 2014. [12] C. Mortici, “A continued fraction approximation of the gamma Proof. Applying Theorem 9,weget𝑓(1) < 𝑓(𝑥) < 𝑓(0) for function,” Journal of Mathematical Analysis and Applications, all 𝑥 ∈ (0, 1),andweobtain(72). vol. 402, no. 2, pp. 405–410, 2013. [13] C. Mortici, “Best estimates of the generalized Stirling formula,” Corollary 12. For all 𝑥 ∈ (0, 1) and all 𝑎>𝑏>0or 0>𝑎>𝑏, Applied Mathematics and Computation, vol. 215, no. 11, pp. one has 4044–4048, 2010. [14] C. Mortici, V. G. Cristea, and D. Lu, “Completely monotonic 𝑎 𝑎 [𝜓 (3 + 𝑏𝑦)] [𝜓 (3+𝑏𝑥)] functions and inequalities associated to some ratio of gamma < , (73) function,” Applied Mathematics and Computation,vol.240,pp. [𝜓 (3 + 𝑎𝑦)]𝑏 [𝜓 (3+𝑎𝑥)]𝑏 168–174, 2014. 3+𝑎𝑥⩾𝑐3+𝑏𝑥⩾𝑐3+𝑎𝑦⩾ 𝑐 3+𝑏𝑦⩾ 𝑐 [15] C. Mortici, “Estimating gamma function by digamma function,” where , , , ,and Mathematical and Computer Modelling,vol.52,no.5-6,pp.942– 0<𝑦<𝑥<1 . 946, 2010. Journal of Applied Mathematics 7

[16] C. Mortici, “New approximation formulas for evaluating the ratio of gamma functions,” Mathematical and Computer Mod- elling,vol.52,no.1-2,pp.425–433,2010. [17] C. Mortici, “New improvements of the Stirling formula,” Applied Mathematics and Computation,vol.217,no.2,pp.699–704, 2010. [18] C. Mortici, “Ramanujan formula for the generalized Stirling approximation,” Applied Mathematics and Computation,vol. 217, no. 6, pp. 2579–2585, 2010. [19] N. Batir, “Some new inequalities for gamma and polygamma functions,” Journal of Inequalities in Pure and Applied Mathe- matics,vol.6,no.4,article103,2005. [20] B. J. English and G. Rousseau, “Bounds for certain harmonic sums,” Journal of Mathematical Analysis and Applications,vol. 206, no. 2, pp. 428–441, 1997. [21] T. P. Dence and J. B. Dence, “A survey of Euler’s constant,” American Mathematical Monthly,vol.82,pp.225–265,2009. [22] E. Artin, The Gamma Function (Trans M. Butter),Holt,Rinehart and Winston, San Francisco, Calif, USA, 1964. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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