Journl of Applied Mthemtics Volume 204, Article ID 264652, 7 pges http://dx.doi.org/0.55/204/264652 Reserch Article New Inequlities for Gmm nd Digmm Functions M. R. Frhngdoost nd M. Krgr Doltbdi Deprtment of Mthemtics, College of Science, Shirz University, Shirz 7475-44776, Irn Correspondence should be ddressed to M. R. Frhngdoost; frhng@shirzu.c.ir Received 6 December 203; Accepted July 204; Published 2 November 204 Acdemic Editor: Vijy Gupt Copyright 204 M. R. Frhngdoost nd M. Krgr Doltbdi. This is n open ccess rticle distributed under the Cretive Commons Attribution License, which permits unrestricted use, distribution, nd reproduction in ny medium, provided the originl work is properly cited. By using the men vlue theorem nd logrithmic convexity, we obtin some new inequlities for gmm nd digmm functions.. Introduction Let Γ(x), ψ(x), ψ n (x), ndζ(x) denote the Euler gmm function, digmm function, polygmm functions, nd Riemnn zet function, respectively, which re defined by Γ (x) = e t t x dt, for x>0, 0 ψ (x) = Γ (x) Γ (x), for x>0, ψ (n) (x) = ( ) n+ t n e xt (2) 0 e t dt, for x > 0; n =, 2, 3,..., ζ (x) =, for x>. n=nx (3) In the pst different ppers ppered providing inequlities for the gmm, digmm, nd polygmm functions (see [ 8]). By using the men vlue theorem to the function log Γ(x) on [u, u + ], withx>0nd u>0,btir[9] presentedthe following inequlities for the gmm nd digmm functions: ψ (x) log (x +e γ ), log (x) ψ(x) < 2 ψ (x), ψ (x) π2 6e γ e ψ(x), for x. for x>0, for x>, () (4) In Section 2, by pplying the men vlue theorem on (log Γ (x)) =ψ(x), for x>0, (5) we obtin some new inequlities on gmm nd digmm functions. Section 3 is devoted to some new inequlities on digmm function, by using convex properties of logrithm of this function. Note tht in this pper by γ = lim n ( n k= (/k) log(n)) = 0.577256 we men Euler s constnt [5]. 2. Inequlities for Gmm nd Digmm Functions by the Men Vlue Theorem Lemm. For t>0,onehs Proof. By [6,Proposition],wehve ψ <. (6) ψ 2 ψ ψ 2[ψ ] 2 <0, for t>0. (7) Thus the function ψ /ψ 2 is strictly decresing on (0, ).
2 Journl of Applied Mthemtics By using symptotic expnsions [20, pges 253 256 nd 364], ψ = t + 2t 2 + 6t 3 + θ 30t 5, (0 θ ), (8) ψ = t 2 t 3 2t 4 + 6t 6 θ 6t 8, (0 θ ). (9) For t>0,weget ψ lim =. (0) ψ 2 t Now, the proof follows from the monotonicity of ψ /ψ 2 on (0, ) nd ψ lim =. () ψ 2 t Theorem 2. Onehsthefollowing: () x (/2) < /ψ (x) x + (6/π 2 ) for x ; (b) /x 2 <ψ (x)ψ (x+)<2/x 2 for x>0; (c) [ψ (x)] 2 /ψ (x) π 4 /72ζ(3) for x nd x 2 ψ (x + )ψ (x) < π 4 /72ζ(3) for x>2; (d) ([ψ (x+h)] 2 ψ (x)ψ (x + h))/hψ (x) > ψ (x + h) for x>0nd h>0; (e) (ψ (x+h)ψ (x) [ψ (x)] 2 )/hψ (x+h)<ψ (x) for x>0nd h>0; (f) x 2 ψ (x) < ψ (x)/ψ (x + ) nd ψ (x + )/ψ (x) < x 2 ψ (x + ) for x>0; (g) ((π 2 x/6) + ) (x+(6/π2)) e x(γ+) Γ(x+) < (2x+ ) (x+(/2)) e x(+γ) for x ; (h) (/x) ψ (x) < (/2)ψ (x + (/2)) for x>0nd (/x) ψ (x) > ((ψ ) () )ψ (x) for x>; (i) ψ(x+)>log(x + (/2)) + ψ((ψ ) ()) for x /2; (j) (π 4 /72ζ(3)) log(x (ψ ) () + 2) + ψ((ψ ) ()) ψ(x+)for x>(ψ ) (). Proof. Let u be positive rel number nd ψ(x) defined on the closed intervl [u, u+].by usingthe men vluetheorem for the function ψ(x) on [u, u + ] with u>0nd since ψ is decresing function, there is unique θ depending on u such tht 0 θ=θ(u)<,forllu 0;then ψ (u+) ψ(u) =ψ (u+θ(u)), (2) Since ψ(x+) ψ(x)=/xnd ψ (x + ) ψ (x) = /x 2, we hve ψ (u+θ(u)) =, for u>0. (3) u We showtht the functionθ(u) hs the following properties: () θ(u) is strictly incresing on (0, ); (2) lim u θ(u) = /2; (3) θ (u) is strictly decresing on (0, ); (4) lim u θ (u) = 0. To prove these four properties, since ψ is decresing function on (0, ), weputu = /ψ, wheret > 0;by formul (3) we hve ψ ( ψ +θ( ψ )) = ψ. ( ) Since by formul (8) we hve ψ < 0 nd ψ > 0, for ll t > 0, then the mpping t ψ from (0, ) into (0, ) is injective since lso ψ 0 nd ψ when t nd t 0 +, respectively, then the mpping t ψ from (0, )into(0, )isbijectivemp.clerly, by injectivity of ψ,wefindtht θ( ψ )=t ψ, for t>0. (4) Differentiting between both sides of this eqution, we get θ ( [(ψ ) 2 +ψ ] ψ )= ψ. (5) Since by formul (8), ψ < 0,wheret>0,henceformul (5) gives θ (/ψ ) > 0,forllt>0.Sincethemppingt /ψ from (0, ) to (0, ) is lso bijective, then θ > 0 for ll t>0, nd the proof of () is completed. From (8) we hve lim θ (u) u = lim θ( t ψ )= lim (t t ψ ) = lim (t t (/t) + (/2t 2 )+(/6t 3 )+(/3t 5 ) ) = 2. Differentiting between both sides of (5),weobtin θ ( ψ ) = [ψ ] 3 ψ [2 (ψ ) 2 ψ ψ ]. (6) ( ) Since ψ > 0 nd ψ < 0, wheret > 0,then θ (/ψ ) < 0 for ll t>0. Proceeding s bove we conclude tht θ < 0,fort>0. This proves (3).
Journl of Applied Mthemtics 3 For (4), from (8), (9),weconcludetht lim u θ (u) = lim θ ( [(ψ ) 2 +ψ ] t ψ )= lim t ψ [ψ 2 ] = lim t ψ =0. (7) Now, we prove the theorem. To prove (), let /ψ () = 6/π 2 t< ;thenby()nd(2)wehve θ( ψ ) θ < lim θ. (8) () t Eqution (3) nd ψ < 0 for ll t>0give θ =(ψ ) ( ) t. (9) t By substituting the vlue of θ into (8),weget ψ () (ψ ) ( ) t< lim t θ = t 2. (20) By substituting the vlue t=/ψ (u) into this inequlity, we get u 2 < ψ (u) u+ 6, π2 (2) where u. Inordertoprove(b),byusingthemenvluetheoremon the intervl [/ψ, /ψ (t + )], nd since θ is decresing function, there exists unique δ such tht for t>0nd θ( ψ (t+) ) θ( ψ ) 0<δ <, (22) =( ψ (t+) ψ )θ ( ψ (t+δ) ). Now, by (4),wehve ψ (t+) + ψ =( ψ (t+) ψ )θ ( ψ (t+δ) ). Since θ is strictly incresing on (0, ),by(),wehve + ψ (t+) ψ ψ (t+) ψ =θ( ψ (t+) ) θ( ψ )>0. (23) (24) (25) By using this inequlity nd the fct tht ψ(x+) ψ(x) = /x nd we obtin ψ (x+) ψ (x) = x 2, (26) ψ (t+) ψ >, t > 0. (27) t2 Since θ is strictly incresing on (0, ),by(),itisclertht θ( ψ (t+) ) θ( ψ ) (28) < lim θ θ(0 + )=, t > 0. t 2 nd then it is cler tht (b) holds. For (c), since t>2, t + δ > + δ(), ndθ is strictly decresing on (0, ) by (3), then θ ( ψ (t+δ) )<θ ( ψ () )= [ψ ()] 2 ψ (), t > 2. (29) Since ψ(x + ) ψ(x) = /x nd ψ (x + ) ψ (x) = /x 2, by using (24),we obtin t 2 ψ (t+) ψ π 4 < 72ζ (3), (30) where t>2. Since θ is strictly decresing on (0, ) by (3) nd ψ < 0,forllt>0,wehve θ ( ψ ) θ ( ψ ), (3) () where t. Then it is cler tht (c) is true. Now we prove (d) nd (e) by using the men vlue theorem on [/ψ, /ψ (t + h)] (t > 0, h > 0), forθ, we conclude θ( ψ (t+h) ) θ( ψ ) =( ψ (t+h) ψ )θ ( ψ (t+) ), where 0<<h. After brief computtion we hve (32) θ ( ψ (t+) )= hψ (t+h) ψ ψ ψ, t>0. (33) (t+h) Since t+>tfor ll >0, t>0, nd by the monotonicity of θ nd ψ we hve θ (/ψ (t + )) < θ (/ψ );then ψ (t+h) ψ [ψ ] 2 hψ (t+h) <ψ, t > 0, h > 0. (34)
4 Journl of Applied Mthemtics By monotonicity of θ nd ψ,wehve θ ( ψ (t+) ) >θ ( ψ (t+h) ). (35) After some simplifiction of this inequlity (d) is proved. For (f), we put h=in (e) nd (d). For (g), we integrte () on [, t] for t>0;thenwehve (t ) π2 log ( +) γ 6 (36) ψ < log (2t ) γ, for t ; theproofiscompletedwhenweintegrtetheseinequlities on [, s],fors>0. By using the men vlue theorem for the ψ on [t, t + θ],thereisα depending on t such tht 0 < α < θ for ll t>0,ndso ψ (t+θ) =θ ψ (t+α) +ψ. (37) By formul (3) nd(2),sinceψ is strictly incresing on (0, ),wehve ψ (t+α) θ = t ψ < lim t θ ψ (t + lim t θ ), for t>0, (38) or t ψ < 2 ψ (t + ), for t>0; (39) 2 since ψ is strictly incresing on (0, ),by(),wehve θ ψ (t+α) = t ψ >θ() ψ, (40) for t>, or t ψ > ((ψ ) () ) ψ, for t>. (4) In order to prove (i) nd (j), we integrte both sides of (3) over u xto obtin x ψ x (u+θ(u)) du = du. (42) u Mking the chnge of vrible u=/ψ on the left-hnd side, by (4),wehve x+θ(x) (ψ ) () ψ ψ dt = log (x) ; (43) ψ 2 since ψ > 0 for ll t>0nd ψ (x)ψ (x) 2[ψ (x)] 2 <0, we find tht, for x>, x+θ(x) log (x) < ψ dt (ψ ) () (44) =ψ(x+θ(x)) ψ((ψ ) ()) or log (x) +ψ((ψ ) ()) <ψ(x+θ(x)). (45) Agin using the monotonicity of θ nd ψ, fter some simplifictions s for x /2,wecnrewrite log (x + 2 )+ψ((ψ ) ()) <ψ(x+). (46) This proves (i). By inequlity (c) for x,wehve log (x) = 72ζ (3) π 4 x+θ(x) ψ dt (ψ ) () 72ζ (3) π 4 (ψ (x+θ(x)) ψ((ψ ) ())) ; (47) since for x, θ(x) θ() = ((ψ ) () )) = (ψ ) (), from this inequlity we find tht π 4 72ζ (3) log (x) +ψ((ψ ) ()) ψ(x+(ψ ) () ); (48) replcing x by x (ψ ) () + 2,wegetforx (ψ ) () π 4 72ζ (3) log (x (ψ ) () +2)+ψ((ψ ) ()) ψ(x+), which proves (j). Then the proof is completed. Exmple 3. Consider the mtrix (49) 3 4 A n = [. (50). ] [ n+] By using inequlities (), we obtin π 2 2 π 2 x+6 π 2 ψ (x) < 2x, x. (5) Now, we integrte on [, t] (for t>0) from both sides of (5) to obtin (t ) π2 log ( +) γ ψ < log (2t ) γ; (52) 6 replcing t by n+(n is n integer number) nd using the identity ψ(n + ) = H n γ[6] nd det A n =n!h n [2], where H n = n k= (/k) is the nth hrmonic number, then we hve log ( nπ2 6 +) n! n!h n < log (2n + ) n!. (53)
Journl of Applied Mthemtics 5 3. New Inequlities for Digmm Function by Properties of Strictly Logrithmiclly Convex Functions Definition 4. Apositivefunctionf is sid to be logrithmicllyconvexonnintervli if f hs derivtive of order two on I nd (log f (x)) 0 (54) for ll x I. If inequlity (54) is strict, for ll x I,thenf is sid to be strictly logrithmiclly convex [22]. Lemm 5. The function Γ is incresing on [c, ), wherec= /4663 is the only positive zero of ψ [, 9]. Lemm 6. If x c nd k(x) = /ψ(x), thenk is strictly logrithmiclly convex on [c, ). Proof. By differentition we hve [log k (x)] = [ ψ (x) ψ (x) ] = ψ (x) ψ (x) +[ψ (x)] 2 [ψ (x)] 2 ; (55) by Lemm 5, weobtinψ(x) = Γ (x)/γ(x) > 0, forevery x [c, ) nd since ψ (x) < 0 on (0, ), thenwehve (log k(x)) >0,forx c. This implies tht /ψ(x) is strictly logrithmiclly convex on [c, ). Theorem 7. Onehsthefollowing: () [ψ(x + 3)] /ψ(x + 3) > ((3/2) γ),for>nd x> 3/; (b) [ψ(x + 3)] /ψ(x + 3) < ((/6) γ) /ψ(3 + ), for >nd x (0, ); (c) [ψ(x + 3)] /ψ(x + 3) > ((/6) γ) /ψ(3 + ), for >nd x>; (d) [ψ(x + 3)] /ψ(x + 3) > ((/6) γ) /ψ(3 + ), for (0, ) nd x (0, ); (e) [ψ(x + 3)] /ψ(x + 3) < ((/6) γ) /ψ(3 + ), for (0, ) nd x>. Proof. By Lemm 6 we hve, for >, ψ [ u p + V q ] > [ψ (u)]/p [ψ (V)] /q, (56) where p>, q>, (/p) + (/q) =, u c,ndv c. If p=nd q = /( ),then ψ[ u+( ) V] >[ψ(u)]/ [ψ (V)] (/) (57) for u cnd V c. Let V =3nd u=x+3.notethtψ(3) = (3/2) γ nd (/)u + ( (/))V =x+3;lsoweobtin [ψ (x+3)] >( 3 ψ (x + 3) 2 γ) In order to prove (b), let for x= u 3 f (x) = log ψ (x + 3) log ψ (3+) log ψ (x+3) ; > 3. (58) (59) since ψ(4) = (/6) γ, wehvef() = log((/6) γ). Also f (x) =[ ψ (x + 3) ψ (x + 3) ψ (x+3) ψ (x+3) ]. (60) By Lemm 6, log(/ψ) is strictly convex on [c, ); then (log ψ) < 0 nd so (ψ /ψ) < 0; this implies tht (ψ /ψ) is strictly decresing on [c, ). Since>nd x (0, ),wehvex+3>x+3.then ψ (x+3) ψ (x + 3) < ψ (x+3) ψ (x+3). (6) And then f (x) < 0;lsof() = log((/6) γ).then for >nd x (0, ) or So (b) is proved. By f (x) >f() = log ( 6 γ) [ψ (x+3)] < ψ (x+3) x+3>x+3, for >, x >, (62) ((/6) γ). (63) ψ (3+) x+3<x+3, for (0, ), x (0, ), x+3<x+3, (c), (d), nd (e) re cler. for (0, ), x>, (64) Corollry 8. For ll x (0, ) nd ll integers n>,onehs ( 3 n 2 γ) < [ψ (x+3)]n ψ (nx + 3) < ((/6) γ)n, (65) H n+2 γ where H n = n k= (/k) is the nth hrmonic number. Proof. By [6], for ll integers n,wehve ψ (n+) =H n γ, (66) nd replcing by n in Theorem 7, theproofiscompleted.
6 Journl of Applied Mthemtics Theorem 9. Let f be function defined by [ψ (3+bx)] f (x) = ; x>0, b (67) [ψ (3+x)] where 3+x cnd 3+bx c;thenforll>b>0or 0 > > b( > 0 nd b < 0), f is strictly incresing (strictly decresing) on (0, ). Proof. Let g be function defined by g (x) = log f (x) =log ψ (3+bx) blog ψ (3+x) ; (68) then g (x) =b[ ψ (3+bx) ψ (3+bx) ψ (3+x) ]. (69) ψ (3+x) By proof of Theorem 7,wehve (log ψ (x)) <0, for x [c, ) ; (70) this implies tht g (x) > 0 if >b>0or 0>>b(g (x) < 0 if >0nd b<0);thtis,g is strictly incresing on (0, ) (strictly decresing on (0, )).Hence f is strictly incresing on (0, ), if>b>0or 0>>b(strictly decresing if >0nd b<0). Corollry 0. For ll x (0, ) nd ll >b>0or 0>>b, one hs ( 3 b 2 γ) [ψ (3+bx)] < [ψ (3+x)] < [ψ (3+b)], b b (7) [ψ (3+)] where 3+bx c, 3+x c, 3+b c,nd3+ c. Proof. To prove (7), pplying Theorem 9 nd tking ccount of ψ(3) = (3/2) γ, wegetf(0) < f(x) < f() for ll x (0, ),ndweobtin(7). Corollry. For ll x (0, ) nd ll >0nd b<0,one hs [ψ (3+b)] [ψ (3+bx)] b < b [ψ (3+)] [ψ (3+x)] < (3 b 2 γ), (72) where 3+x c, 3+bx c, 3+b c,nd3+ c. Proof. Applying Theorem 9, wegetf() < f(x) < f(0) for ll x (0, ),ndweobtin(72). Corollry 2. For ll x (0, ) nd ll >b>0or 0>>b, one hs [ψ (3 + by)] [ψ (3 + y)] < [ψ (3+bx)], b b (73) [ψ (3+x)] where 3+x c, 3+bx c, 3+y c, 3+by c,nd 0<y<x<. Corollry 3. For ll x (0, ) nd ll >0nd b<0,one hs [ψ (3+bx)] [ψ (3+x)] < [ψ (3 + by)], b b (74) [ψ (3 + y)] where 3+x c, 3+bx c, 3+y c, 3+by c,nd 0<y<x<. Remrk 4. Tking =nnd b=in Corollry 0,weobtin inequlities of Corollry 8. Conflict of Interests The uthors declre tht there is no conflict of interests regrding the publiction of this pper. References [] H. Alzer nd S. Ruscheweyh, A subdditive property of the gmm function, Journl of Mthemticl Anlysis nd Applictions,vol.285,no.2,pp.564 577,2003. [2] H. Alzer, Some gmm function inequlities, Mthemtics of Computtion,vol.60,no.20,pp.337 346,993. [3] A. V. Boyd, Gurlnd s inequlity for the gmm function, Scndinvin Acturil Journl,vol.960,pp.34 35,960. [4] J. Bustoz nd M. E. H. Ismil, On gmm function inequlities, Mthemtics of Computtion,vol.47,no.76,pp.659 667,986. [5] C. P. Chen, Asymptotic expnsions of the logrithm of the gmm function in the terms of the polygmm functions, Mthemticl Inequlities nd Applictions, vol.2,pp.52 530, 204. [6] C.-P. Chen, Some properties of functions relted to the gmm, psi nd tetrgmm functions, Computers & Mthemtics with Applictions,vol.62,no.9,pp.3389 3395,20. [7] C. P. Chen, Unified tretment of severl symptotic formuls for the gmm function, Numericl Algorithms, vol. 64, no. 2, pp.3 39,203. [8] T. Erber, The gmm function inequlities of Gurlnd nd Gutschi, vol. 960, no. -2, pp. 27 28, 960. [9] J. D. Keckic nd P. M. Vsic, Some inequlities for the gmm function, Publictions de l Institut Mthémtique, vol.,pp. 07 4, 97. [0] A. Lforgi, Further inequlities for the gmm function, Mthemtics of Computtion,vol.42,no.66,pp.597 600,984. [] D. Lu nd X. Wng, A new symptotic expnsion nd some inequlities for the gmm function, Journl of Number Theory, vol. 40, pp. 34 323, 204. [2] C. Mortici, A continued frction pproximtion of the gmm function, Journl of Mthemticl Anlysis nd Applictions, vol. 402, no. 2, pp. 405 40, 203. [3] C. Mortici, Best estimtes of the generlized Stirling formul, Applied Mthemtics nd Computtion, vol. 25, no., pp. 4044 4048, 200. [4] C. Mortici, V. G. Criste, nd D. Lu, Completely monotonic functions nd inequlities ssocited to some rtio of gmm function, Applied Mthemtics nd Computtion, vol.240,pp. 68 74, 204. [5] C. Mortici, Estimting gmm function by digmm function, Mthemticl nd Computer Modelling,vol.52,no.5-6,pp.942 946, 200.
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