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New Inequalities for Gamma and Digamma Functions 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() () () () Let , , ,and denote the Euler gamma function, digamma function, polygamma functions, and (log Γ ()) =() , for >0, (5) Riemann zeta function, respectively, which are defined by ∞ − −1 we obtain some new inequalities on gamma and digamma Γ () = ∫ , for >0, 0 functions. Γ () (1) Section 3 is devoted to some new inequalities on () = , >0, digamma function, by using convex properties of logarithm Γ () for of this function. () () Note that in this paper by =lim→∞(∑=1(1/) − log()) = 0.5772156 ⋅ ⋅ we mean Euler’s constant [5]. ∞ − (2) = (−1)+1 ∫ , > 0; = 1, 2, 3, ., − for 0 1− 2. Inequalities for Gamma and Digamma ∞ 1 () = ∑ , >1. Functions by the Mean Value Theorem for (3) =1 Lemma 1. For >0,onehas In the past different papers appeared providing inequalities for the gamma, digamma, and polygamma functions (see [1– − () 18]). <1. 2 (6) By using the mean value theorem to the function log Γ() () on [, +,with 1] >0and >0,Batir[19]presentedthe following inequalities for the gamma and digamma functions: Proof. By [6,Proposition1],wehave − () ⩽ log (−1+ ), for >0, 2 1 () () −2[ ()] <0, for >0. (7) () −() < () , >1, log 2 for (4) 2 2 Thus the function ()/ () is strictly decreasing on () ⩾ −(), ⩾1. 6 for (0, ∞). 2 Journal of Applied Mathematics By using asymptotic expansions [20, pages 253–256 and We show that the function () has the following properties: 364], (1) () is strictly increasing on (0, ∞); 1 1 1 () = + + + ,(0⩽⩽1), (2) lim→∞() = ;1/2 2 3 5 (8) 2 6 30 (3) () is strictly decreasing on (0, ∞); 1 1 1 1 () =− − − + − ,(0⩽⩽1). (9) (4) lim→∞ () =. 0 2 3 24 66 68 To prove these four properties, since is a decreasing For >0,weget function on (0, ∞),weput=1/(),where>0;by () formula (13) we have lim =−1. (10) →∞ ()2 1 1 ( +( )) = () . (∗) () () 2 Now, the proof follows from the monotonicity of ()/ () (0, ∞) on and Since by formula (8) we have () < 0 and () >,for 0 all >0, then the mapping →() from (0, ∞) into () =−1. (0, ∞) is injective since also () →0 and () →∞ lim 2 (11) + →∞ () when →∞and →0, respectively, then the mapping →() from (0,∞)into(0,∞)isabijectivemap.Clearly, by injectivity of ,wefindthat Theorem 2. Onehasthefollowing: 1 1 ( )=− , for >0. (14) 2 () () (a) − (1/2) < 1/ () ⩽ + (6/ )−1for ⩾1; 2 2 (b) 1/ <() (+1)<2/ for >0; Differentiating between both sides of this equation, we get 2 4 2 (c) [ ()] / () ⩾ − /72(3) for ⩾1and ( + 2 4 1) () < /72(3) for >2; 1 −[( ()) + ()] ( )= . (15) 2 (d) ([ (+ℎ)] −() ( + ℎ))/ℎ () > ( + ℎ) () () >0 ℎ>0 for and ; 2 Since by formula (8), () <,where 0 >0,henceformula ( (+ℎ)() − [ ()] )/ℎ (+ℎ)< () (e) for (15) gives (1/ ()) >,forall 0 >0.Sincethemapping→ >0 ℎ>0 and ; 1/ () from (0, ∞) to (0, ∞) is also bijective, then () > 0 2 (f) − () < ()/ ( + 1) and ( + 1)/ () < for all >0, and the proof of (1) is completed. 2 − ( + 1) for >0; From (8) we have 2 (+(6/2)) −(+1) (g) (( /6) + 1) ⩽ Γ(+1)< (2+ lim () (+(1/2)) −(1+) →∞ 1) for ⩾1; (1/) − () < (1/2)( + (1/2)) >0 1 1 (h) for and = lim ( )= lim ( − ) −1 →∞ () →∞ () (1/) − () > (( ) (1) − 1) () for >1; −1 (16) (+1)> ( + (1/2)) + (() (1)) ⩾1/2 1 (i) log for ; = ( − ) lim 2 3 5 4 −1 −1 →∞ (1/) + (1/2 )+(1/6 )+(1/3 ) (j) ( /72(3)) log(−() (1) + 2) + (( ) (1)) ⩾ −1 1 (+1)for >() (1) − 1. = . 2 Proof. Let be a positive real number and () defined on the closed interval [, +1].Byusingthemeanvaluetheoremfor Differentiating between both sides of (15),weobtain the function () on [, + 1] with >0and since is a 1 decreasing function, there is a unique depending on such ( ) that 0⩽=()<1,forall⩾0;then () (∗∗) 3 (+1) −() = (+()) , (12) [ ()] 2 = [2 ( ()) − () ()]. 2 () Since (+1)−()=1/and ( + 1) − () = −1/ , we have Since () > 0 and () <,where 0 >0,then 1 (1/ ()) < 0 for all >0. Proceeding as above we conclude (+()) = , >0. for (13) that () <,for 0 >0. This proves (3). Journal of Applied Mathematics 3 For (4), from (8), (9),weconcludethat By using this inequality and the fact that (+1)−() =1/ 2 and [( ()) + ()] 1 1 lim () = lim ( )= lim − (+1) − () =− , (26) →∞ →∞ () →∞ () 2 2 [ ()] we obtain =−1− lim =0. 1 →∞ () (+1) () > ,>0. 2 (27) (17) Since is strictly increasing on (0, ∞),by(1),itisclearthat Now, we prove the theorem. To prove (a), let 1/ (1) = 2 1 1 6/ ⩽<∞;thenby(1)and(2)wehave ( )−( ) (+1) () 1 (28) ( )⩽() < lim () . (18) + 1 (1) →∞ < lim () −(0 )= ,>0. →∞ 2 Equation (13) and () < 0 for all >0give and then it is clear that (b) holds. For (c), since >2, + () > 1,and +(1) is strictly −1 1 () =() ( )−. decreasing on (0, ∞) by (3), then (19) 2 () 1 1 [ (1)] By substituting the value of into (18),weget ( )< ( )=−1− , (+()) (1) (1) (29) 1 −1 1 1 1− ⩽() ( )−< lim () = . (1) →∞ 2 (20) ∀ > 2. 2 By substituting the value =1/() into this inequality, we Since ( + 1) − () =1/ and ( + 1) − () = −1/ , get by using (24),weobtain 4 1 1 6 2 − < ⩽+ −1, (+1) () < , (30) 2 () 2 (21) 72 (3) ⩾1 where >2. where . Inordertoprove(b),byusingthemeanvaluetheoremon Since is strictly decreasing on (0, ∞) by (3) and () < the interval [1/ (), 1/ ( + 1)], and since is a decreasing 0,forall>0,wehave function, there exists a unique such that 1 1 ( )⩽ ( ), (31) 0<() <1, (22) () (1) ⩾1 for >0and where . Then it is clear that (c) is true. 1 1 Now we prove (d) and (e) by using the mean value ( )−( ) (+1) () theorem on [1/ (), 1/ ( + ℎ)] ( > 0, ℎ,for >0) ,we (23) conclude 1 1 1 =( − ) ( ). 1 1 (+1) () (+()) ( )−( ) (+ℎ) () (32) Now, by (14),wehave 1 1 1 =( − ) ( ), 1 1 (+ℎ) () (+) 1− + (+1) () 0<<ℎ (24) where . 1 1 1 After brief computation we have =( − ) ( ). (+1) () (+()) 1 ℎ (+ℎ) () ( )= −1, >0. (33) Since is strictly increasing on (0, ∞),by(1),wehave (+) () − (+ℎ) (+1) − () Since +>for all >0, >0, and by the monotonicity of 1+ (1/( + )) <(1/()) (+1) () and we have ;then (25) 2 1 1 (+ℎ) () −[ ()] =( )−( )>0. < () ,>0,ℎ>0.(34) (+1) () ℎ (+ℎ) 4 Journal of Applied Mathematics By monotonicity of and ,wehave or 1 1 −1 ( ) > ( ) . log () +(() (1))<(+()) . (45) (+) (+ℎ) (35) After some simplification of this inequality (d) is proved. Again using the monotonicity of and , after some simpli- ⩾1/2 For (f), we put ℎ=1in (e) and (d). fications as for ,wecanrewrite [1, ] >0 For (g), we integrate (a) on for ;thenwehave 1 −1 log ( + )+(() (1))<(+1) . (46) (−1) 2 2 ( +1)− log 6 (36) This proves (i). By inequality (c) for ⩾1,wehave ⩽() < (2 −) 1 −, ⩾1; log for 72 (3) +() () ⩾ ∫ () log 4 theproofiscompletedwhenweintegratetheseinequalities ()−1(1) on [1, ],for>0. (47) 72 (3) −1 By using the mean value theorem for the () on [, + = ( (+()) −(() (1))) ; ()],thereisa() depending on such that 0 < () < () 4 >0 for all ,andso −1 −1 since for ⩾1, () ⩾ (1) =(( ) (1)−1)) = ( ) (1)−1, (+()) =() (+()) + () . (37) from this inequality we find that 4 By formula (13) and(2),since is strictly increasing on −1 (0, ∞) () +(() (1)) ,wehave 72 (3) log (+()) () (48) −1 ⩾(+() (1) −1); 1 = − () < lim () ( + lim ()), for >0, →∞ →∞ −()−1(1) + 2 ⩾()−1(1) − 1 (38) replacing by ,wegetfor 4 or −1 −1 (−() (1) +2)+(() (1)) 1 1 1 72 (3) log − () < ( + ), >0; (49) 2 2 for (39) ⩾(+1) , since is strictly increasing on (0, ∞),by(1),wehave 1 which proves (j). Then the proof is completed. () (+()) = − () >(1) () , (40) Example 3. Consider the matrix >1, for 3 1 1 ⋅⋅⋅ 1 [ ] or [1 4 1 ⋅⋅⋅ 1 ] 1 −1 = [ . ] . (50) − () > (() (1) −1) () , >1. [ . ] for (41) [1 1 ⋅⋅⋅ 1 +1] In order to prove (i) and (j), we integrate both sides of (13) over 1⩽⩽to obtain By using inequalities (a), we obtain 1 ∫ (+()) = ∫ . 2 2 (42) ⩽ () < , ⩾1. (51) 1 1 2+6−2 2 − 1 Making the change of variable =1/() on the left-hand Now, we integrate on [1, ] (for >0) from both sides of (51) side, by (14),wehave to obtain +() − () ∫ () = () ; 2 2 log (43) (−1) ()−1(1) () ( +1)−⩽() < (2 −) 1 −; log 6 log (52) () > 0 >0 ()() − 2[()]2 <0 since for all and , +1 we find that, for >1, replacing by ( is an integer number) and using the identity ( + 1) = −[6] and det =! [21], where +() = ∑ (1/) is the th harmonic number, then we have log () < ∫ () =1 ()−1(1) (44) 2 ! ! −1 ( +1) ⩽! < (2 +) 1 . (53) =(+()) −(() (1)) log 6 log Journal of Applied Mathematics 5 3. New Inequalities for Digamma Function Let V =3and =+3.Notethat(3) = (3/2) − and by Properties of Strictly Logarithmically (1/) + (1 − (1/))V =+3;alsoweobtain Convex Functions [ (+3)] 3 −1 −3 3 >( −) for = >− . (58) Definition 4. Apositivefunction is said to be logarithmi- () +3 2 callyconvexonaninterval if has derivative of order two on and In order to prove (b), let (log ()) ⩾0 (54) () = log () +3 − log (3+) (59) − (+3) ; for all ∈. log If inequality (54) is strict, for all ∈,then is said to be (4) = (11/6) − (1) = ((11/6) − )− strictly logarithmically convex [22].
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