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MATHEMATICS OF COMPUTATION Volume 74, Number 250, Pages 723–742 S 0025-5718(04)01675-8 Article electronically published on May 18, 2004

SOME PROPERTIES OF THE AND PSI FUNCTIONS, WITH APPLICATIONS

S.-L. QIU AND M. VUORINEN

Abstract. In this paper, some monotoneity and concavity properties of the gamma, and psi functions are obtained, from which several asymptotically sharp inequalities follow. Applying these properties, the authors improve some n n well-known results for the volume Ωn of the unit ball B ⊂ R ,thesurface n−1 area ωn−1 of the unit sphere S , and some related constants.

1. Introduction For real and positive values and y, the Euler gamma function, the beta and psi (or polygamma) functions are defined as  ∞ Γ(x)Γ(y) Γ(x) (1.1) Γ(x)= tx−1e−tdt, B(x, y)= ,ψ(x)= , 0 Γ(x + y) Γ(x) respectively. For the extensions to complex variables and for the basic properties of these functions, see [AS], [AAR], [Mi], [T], and [WW]. Over the past half century, many authors have obtained various properties and inequalities for these very important functions (see [A1], [A2], [A5], [G], [Ke], [K2], [L], [MSC] and bibliographies therein). Formulas for the volumes of geometric bodies sometimes involve the gamma function. This topic and related inequalities have been studied recently in [AQ], [A1]–[A4], [BP], [EL]. Let Bn and Sn−1 be the unit ball and unit sphere in Rn, n n−1 respectively, Ωn be the volume of B ,andωn−1 denote the surface area of S . Set Ω0 =1.ItiswellknownthatΩn is increasing for 2 ≤ n ≤ 5 and decreasing for n ≥ 5, while ωn−1 is increasing for 2 ≤ n ≤ 7 and decreasing for n ≥ 7(see[BH, pp. 263–264] and [AVV1, p. 38]): 2π πn/2 πn/2 (1.2) Ωn = Ωn−2 = ,ωn−1 = nΩn =2 . n Γ(1 + n/2) Γ(n/2) Throughout this paper, we let γ =0.57721 ··· = −ψ(1) denote the Euler- Mascheroni constant, (2n−1)!! = (2n−1)(2n−3) ···1forn ∈ N = {k; k is a natural number},  π/2  π/2 n−2 (2−n)/(n−1) (1.3) In = sin tdt, Jn = sin tdt, 0 0

Received by the editor April 2, 2002 and, in revised form, September 27, 2003. 2000 Mathematics Subject Classification. Primary 33B15; Secondary 26B15, 26D15, 51M25. Key words and phrases. Gamma function, beta function, psi function, monotoneity, concavity, inequalities.

c 2004 American Mathematical Society 723

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(1.4)  1/(n−1)  1/(n−1) ωn−1 n−1 1 ωn−1 An = n ,Bn = bn ,Dn = , 2 cn n − 1 cn

1−n (1.5) bn = Jn/(n − 1) and cn =(2Jn) ωn−2, for n ∈ N with n ≥ 2. These constants have applications in some fields of math- ematics such as geometry of Grassmannian subspaces of Rn [KR], optimization theory [B], geometric function theory [AVV1, pp. 234–246], [V], [Vu], as well as geometry of spaces of constant curvature [BH]. These are some examples of the fields where the gamma function is frequently used. The accumulated literature on the gamma function is so vast that it is difficult for someone working chiefly in these specialized areas of research to find the property of the gamma function which, in this particular application, would lead to the desired conclusion. This is exactly how we were led to explore the gamma function, our main research area being geometric function theory; see [V], [Vu], [AVV1], [AQ]. Some of the latest developments on special functions that have been studied in this context are reviewed in [AVV2]. The statement of some of our main results, which we think have independent interest as such, now follows. Theorem A. For n =1, 2, 3, ...   π Ωn π 2 ≤ < 2 . 2n + π − 2 Ωn−1 2n +1 √ √ Theorem B. For x ≥ 1 , let P (x)=max{ πx, 4x − 1}. Then for x ≥ 1 x(1−γ)x−1P (x) ≤ Γ(x)P (x) ≤ 2Γ(x +1/2). Theorem C. For x>1 1 1 1 2γ − 1 log x − − <ψ(x) < log x − − . 2x 12x2 2x 2x2 Theorems A, B, and C are special cases of the more technical Theorems 1.17, 1.12, and 2.1, respectively. We next proceed to review some earlier results and to discuss our preliminary and main results. Various properties of Ωn and ωn−1 have been proved in these contexts. For example, it was shown in [B] that for n ≥ 3,   2π Ωn 2π (1.6) ≤ ≤ n +1 Ωn−1 n and   2π ωn−1 2π (1.7) ≤ ≤ , n − 1 ωn−2 n − 2 and a new property of Ωn was obtained in [AQ]. Several inequalities for An,Bn,cn, Dn,In and Jn were shown or stated in [AVV1]. For example, it was proved that for n ≥ 3,   π π (1.8)

(1.10) lim An/n =1, lim Bn = lim Dn =2, n→∞ n→∞ n→∞

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and

(1.11) 1 0, define functions f5,f6 and f7 on (0, 1) by   1 1 Γ(x + s) β(x) − f (s) s[f (s) − α(x)] f (s)= log x − log ,f(s)= 5 ,f(s)= 5 , 5 1 − s s Γ(x) 6 s 7 1 − s

respectively, where α(x)=ψ(x +1)− log x and β(x)=logx − ψ(x).Thenf5 and f6 are both strictly decreasing on (0, 1) with ranges (α(x),β(x)) and (ξ(x),η(x)), respectively, while f7 is strictly increasing from (0, 1) onto (0,θ(x)),where 1 1 ξ(x)=β(x) − α(x),η(x)= ψ(x) − β(x) and θ(x)=α(x) − ψ(x +1). 2 2 In particular, for x>0 and s ∈ (0, 1), Γ(x + s) (1.14) exp{s(s − 1)β(x)} < exp{s(1 − s)ρ(s, x)} < xsΓ(x) < exp{s(s − 1)ω(s, x)}≤exp{s(s − 1)α(x)}, where ρ(s, x)=max{sξ(x) − β(x), −α(x)+(1− s)θ(x)/s} and ω(s, x)=min{sη(x) − β(x), −α(x)}. The above theorem yields some inequalities of Γ(x). Some similar or related results have been proved recently in [AQ], [BP], [EL], [A5], [K2], [Me1]–[Me3]. 1.15. Remarks. (1) For n ∈ N,letG(n, k) be the so-called Grassmannian,andlet τn denote the invariant measure on G(n, 1), that is, on the set of all straight lines through the origin (cf. [KR, p. 206]). Set g(n)=τn(G(n, 1)). Then the functions f1(x), f4(x)andf5(s) in Theorem 1.12 are related to g(n). Therefore, Theorem 1.12 is related to some results in [KR]. For example, it was proved in [KR,√ p. 214] that g(n) is increasing for n ∈ N, while Theorem 1.12(1) says that g(n)/ n is also strictly increasing for n ∈ N since g(n)= nπ/2f1(n/2) by the expression of g(n) (see [KR, p. 214]).

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(2) The double inequality (1.14) and its Corollary 3.2 are related to a result in [LL], which gives the estimates  1/(s−1) (1 − s) log(x + s/2)

for s ∈ (0, 1) and x>0, where Ds(x)=logΓ(x +1)− log Γ(x + s). For some other related results, see [A3, p. 365], [Me1], [MSC], etc. 1/x 1.16. Theorem. (1) The function g1(x) ≡ [xB(x, 1/2)] is strictly decreasing and convex from (0, ∞) onto (1, 4). (2) There exists a unique x1 ∈ (1.5, 2) such that the function   1/(2x)   1 1 1 1 1 g2(x) ≡ B x, B , , 2x +1 2 2 4x 2

for x ∈ (0, ∞), is strictly decreasing on (0,x1] and increasing on [x1, ∞),with 2 g2(1/2) = π /4 and lim g2(x)=2. x→∞ (3) There exists a unique x2 ∈ (33, 34) such that the function   1/(2x) 1 g3(x) ≡ 2 1+ g2(x), 2x

for x ∈ (0, ∞), is strictly decreasing on (0,x2] and increasing on [x2, ∞),with 2 g3(1/2) = π and lim g3(x)=2. x→∞ As an example of the applications of the above two theorems, we shall prove the following improvements upon some results for Ωn, ωn−1, An, Bn, bn, In and Jn. −2 1.17. Theorem. (1) For n ∈ N \{1, 2}, the function F1(n) ≡ 2n − πIn is strictly increasing with F1(3) = 6 − π and lim F1(n)=3.Inparticular,forn ≥ 3, n→∞   π π (1.18) ≤ In < , 2n − (6 − π) 2n − 3 with equality iff n =3. 2 (2) The function F2(n) ≡ n − 2π[ωn−2/ωn−1] is strictly increasing for n ∈ N \{1, 2},withF2(3) = 3 − (π/2) and lim F2(n)=3/2.Inparticular,forn ≥ 3, n→∞   π ωn−1 π (1.19) 2 ≤ < 2 , 2n − 6+π ωn−2 2n − 3 with equality iff n =3. 2 (3) The function F3(n) ≡ n−2π[Ωn−1/Ωn] is strictly increasing for n ∈ N,with F3(1) = 1 − π/2 and lim F3(n)=−1/2.Inparticular,forn ∈ N, n→∞   π Ωn π (1.20) 2 ≤ < 2 , 2n + π − 2 Ωn−1 2n +1 with equality iff n =1. (4) Let n ∈ N.Thenbn is strictly decreasing for n ≥ 2,andBn is strictly increasing for n ≥ 2.Inparticular,forn ≥ 2, (1.21)  1/(n−1) π π 1/(n−1) π ≤ Bn < 2, (n − 1) ≤ Jn ≤ (n − 1) min 2 , , 2 2 2 with equality in each instance iff n =2.

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(5) Let a =1−log 2, b =3−0.5B(1/4, 1/2) = 0.37794 ···, c =[π2 −3(log 4)2]/24 =0.171007 ···,andn ∈ N. Then the function F4(n) ≡ (n−1)(n−Jn −a) is strictly increasing for n ≥ 3,withF4(3) = d ≡ 2(b − a)=0.14217 ··· and lim F4(n)=c. n→∞ In particular, for n ≥ 3,

(1.22) d ≤ (n − 1)(n − Jn − a)

with equality iff n =3. (6) Let n ∈ N. Then the function F5(n) ≡ An/n is strictly decreasing for 2 ≤ n ≤ 5 and increasing for n ≥ 5.Inparticular,

(1.23) 2 1 1 ≤ A2/2=π /8, for 2 ≤ n ≤ 5, lim An =1,C≤ An n→∞ n n < 1, for n ≥ 5, where √  1/4 1 π 2 Γ(1/8) C = A5 = =0.8411396629 ···. 5 10 3 Γ(5/8)

(7) Let n ∈ N.ThenDn is strictly decreasing for 2 ≤ n ≤ 69 and increasing for n ≥ 69.Inparticular,

2 ≤ D2 = π , for 2 ≤ n ≤ 69, (1.24) lim Dn =2,D≤ Dn n→∞ < 2, for n ≥ 69, where √  1/68 2π 33! Γ(1/136) D = D69 = =1.9853487779 ···. 68 67!! Γ(69/136)

2. Some properties of ψ(x) In this section, we obtain some properties of ψ(x), which are also needed in the proofs of the theorems stated in Section 1.

2.1. Theorem. (1) The function h1(x) ≡ ψ(x +1/2) − ψ(x) − [1/(2x)] is strictly decreasing and convex from (0, ∞) onto (0, ∞). Moreover, h2(x) ≡ h1(1/x) is convex on (0, ∞). (2) The function h3(x) ≡ xh1(x) is strictly decreasing from (0, ∞) onto (0, 1/2). 2 (3) The function h4(x) ≡ x h1(x) is strictly increasing from (0, ∞) onto (0, 1/8). In particular, for x>1,   1 1.5 − log 4 1 1 1 (2.2) ψ(x)+ + <ψ x + <ψ(x)+ + . 2x x2 2 2x 8x2

2 (4) The function h5(x) ≡ x [ψ(x)−log x]+(x/2) is strictly decreasing and convex from (0, ∞) onto (−1/12, 0).Inparticular,forx>1, 1 1 1 2γ − 1 (2.3) log x − − <ψ(x) < log x − − . 2x 12x2 2x 2x2 (5) Let η and θ be functions defined in Theorem 1.12(3). Then η and θ are both strictly decreasing from (0, ∞) onto (0, ∞).

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Proof. (1) Differentiation gives    ∞ −xt  −h6(x) ≡  1 −  1 1 − te h1(x)= 2 ψ x + ψ (x)+ 2 = 2 −t/2 dt, x 2 2x 2x 0 1+e where  ∞ −u ue du − 1 h6(x)= −u/(2x) , 0 1+e 2 since [AS, 6.4.1]  ∞ n −xt (n) − n+1 t e (2.4) ψ (x)=( 1) −t dt. 0 1 − e  Clearly, h6 is strictly decreasing from (0, ∞)onto(0, 1/2). Hence h1 is strictly increasing from (0, ∞)onto(−∞, 0), and the monotoneity and convexity of h1 follow. + By [AS, 6.3.5], h1(x)=ψ(x +1/2) − ψ(x +1)+[1/(2x)] so that h1(0 )=∞. The limiting value lim h1(x) = 0 follows from [AS, 6.3.18]. x→∞ The convexity of h2 follows from the monotoneity of h6,since   2 h2(x)=−h1(1/x)/x = h6(1/x). (2) Put y =1/x.Then  h2(y) h2(y)  h3(x)= and d = h2(y), y dy (y)

and hence, the monotoneity of h3 follows from the convexity of h2 and the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25]. By [AS, 6.3.5], h3(x) can be written as   x 1 h3(x)= 2ψ x + − ψ(x +1)− ψ(x) , 2 2 + and hence, h3(0 )=1/2. It follows from [AS, 6.3.18] that lim h3(x)=0. x→∞ 2 (3) Write h4(x)=h2(y)/y ,wherey =1/x,andlet    ∞ −u 1 ue du − 1 h7(y)=h6 = −uy/2 ,h8(y)=y. y 0 1+e 2

Then h7(0) = h8(0) = 0, and by (2.4),  h2(y) h7(y) (2.5) d 2 = , dy (y ) 2h8(y)   ∞ 2 −u(1+y/2) h7(y) ≡ 1 u e (2.6)  = h9(y) −uy/2 2 du. h8(y) 2 0 (1 + e )

Clearly, h9 is strictly decreasing in y on (0, ∞). Hence the monotoneity of h4 follows from the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25]. + Clearly, h4(0 ) = lim xh3(x) = 0. By l’Hˆopital’s Rule, (2.5) and (2.6), x→0  ∞ h2(y) 1 2 −u 1 lim h4(x) = lim 2 = h9(0) = u e du = . x→∞ y→0 y 16 0 8 Inequality (2.2) is clear.

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(4) Using [AS, 6.3.21], we obtain by differentiation

 2  1 1 h (x)=h10(x) ≡ 2x[ψ(x) − log x]+x ψ (x) − + 5 x 2  ∞ − 1 2tdt 1 = 2x + 2 2 2πt + 2x 0 (t + x )(e − 1) 2  ∞ 2 1 4xtdt + x 2 + 2 2 2 2πt 2x 0 (t + x ) (e − 1)  ∞  ∞ 3 tdt − tdt =4x 2 2 2 2πt 4x 2 2 2πt 0 (t + x ) (e − 1) 0 (t + x )(e − 1)  ∞ 3  ∞ 3 − t dt − u du = 4x 2 2 2 2πt = 4x 2 2 2πux , 0 (t + x ) (e − 1) 0 (1 + u ) (e − 1) 2πux where u = t/x.Sincex −→ x/(e − 1) is strictly decreasing on (0, ∞), h10 is strictly increasing on (0, ∞). Hence the monotoneity and convexity of h5 follow. + Clearly, h5(0 )=0,h5(1) = (1/2) − γ. By [AS, 6.3.18], h5(∞)=−1/12. (5) By differentiation and (2.4), we get    ∞ − −xt  1  − 1  xt(2 t)e − 2xη (x)=2x ψ (x) + ψ (x) = −t dt 2 2 x 0 1 − e  ∞ − −u 1 u(2 u/x)e − = −u/x du 2, x 0 1 − e which is strictly increasing from (0, ∞)onto(−∞, 0) since the function v → v(2 − v)/(1 − e−v) is strictly decreasing from (0, ∞)onto(−∞, 2). This yields the monotoneity of η. By [AS, 6.3.5], we can rewrite η(x)as 1 1 1 η(x)= ψ(x +1)+ − log x + ψ(x +1)− , 2 2x2 x from which we see that η(0+)=∞. The limiting value lim η(x) = 0 follows from x→∞ [AS, 6.3.18 and 6.4.12]. Next, similarly, we have  ∞ −(x+1)t  ∞ −u  xt(2 + t)e − 1 u(2 + u/x)e − 2xθ (x)= −t dt 2= u/x du 2, 0 1 − e x 0 e − 1 which is strictly increasing from (0, ∞)onto(−2, 0) since v → v(2 + v)/(ev − 1) is strictly decreasing from (0, ∞)onto(0, 2). Hence the monotoneity of θ follows. Clearly, θ(0+)=∞. By [AS, 6.3.18 and 6.4.12], we obtain lim θ(x)=0.  x→∞

2.7. Remarks. (1) The double inequality (2.3) improves the inequality obtained in [ABRVV, Theorem 3.1]. (2) It is well known that [AS, 6.1.3] n  1 (2.8) γ = lim dn,dn = − log n, n→∞ k=1 k and by [AS, 6.3.2], dn − γ = ψ(n +1)− log n.

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S. R. Tims and J. A. Tyrrell [TT] obtained the bounds 1 1 (2.9)

for dn − γ, which was improved by R. M. Young [Y] as 1 1 (2.10)

where α = {1/[2(1 − γ)]}−1=0.1826 ··· and β =1/6. Let h5 be as in Theorem 2.1(4). Then

2 1 n 2 n h5(n)=n ψ(n +1)− log n − + = n (dn − γ) − , n 2 2

and hence our Theorem 2.1(4) implies the following estimates for dn − γ: 2.13. Corollary. For n ∈ N, 1 α 1 β (2.14) −

3. Proofs of the main theorems In this section we prove the theorems stated in Section 1. 3.1. Proof of Theorem 1.12. (1) Logarithmic differentiation gives  f1(x)/f1(x)=h1(x),

where h1 is as in Theorem 2.1(1), and hence the monotoneity and log-concavity of f1 follow from Theorem 2.1(1). By [AS, 6.1.15 and 6.1.37], we have

+ √ Γ(x +1/2) f1(0 ) = lim x =0 x→0 Γ(x +1) and e−(x+1/2)(x +1/2)x lim f1(x) = lim √ =1. x→∞ x→∞ xe−xxx−1/2  Since f2(x)/f2(x)=−h4(1/x), where h4 is as in Theorem 2.1(3), the log- convexity of f2 follows from Theorem 2.1(3). Next, by differentiation,   2 f3(x)=−f1(x)/f1(x) = −h1(x)/f1(x), which is strictly increasing on (0, ∞) by Theorem 2.1(1), and hence the convexity of f3 follows.

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(2) Set y =Γ(x +1/2)/Γ(x). Then   1 f  (x)=1− 2y2 ψ x + − ψ(x) 4 2 and   2 1  1 f4 (x)=− ψ x + − ψ(x) f8(x), 2y2 2 where    ψ (x +1/2) − ψ (x) f8(x)=f (x)=2+ , 9 [ψ(x +1/2) − ψ(x)]2   −1 1 h4(x) f9(x)=2x − ψ x + − ψ(x) =2 2 h3(x)+1/2

and h3 and h4 are as in Theorem 2.1. By Theorem 2.1(2)-(3), f9 is strictly increasing  from (0, ∞)onto(0, 1/2), so that f8(x) > 0forx ∈ (0, ∞), and hence f4 (x) < 0for x ∈ (0, ∞). This yields the concavity of f4. Furthermore, by part (1) and Theorem 2.1(2),   f4(x) > lim f4(x) x→∞   2 1 =1− 2 lim f1(x) · x ψ x + − ψ(x) x→∞ 2   2 1 =1− 2 lim f1(x) lim h3(x)+ =0 x→∞ x→∞ 2

for x ∈ (0, ∞), and hence the monotoneity of f4 follows. + 2 By part (1), f4(0 ) = lim [x − xf1(x) ] = 0. Applying l’Hˆopital’s Rule, we x→0 obtain 2 2 1 − f1(x) 1 − f2(t) lim f4(x) = lim = lim x→∞ x→∞ 1/x t→0 t  2 = −2 lim f2(t)f2(t) = 2 lim f2(t) h4(1/t) t→0 t→0 2 = 2 lim f1(x) h4(x)=1/4. x→∞

The second and third inequalities in (1.13) follow from the monotoneity of f4 on (0, ∞) and the log-convexity of f2 on (0, 1), while the first and fourth inequalities in (1.13) hold by [AQ, Theorem 1.5]. The equality case is clear. The asymptotic sharpness of the third and fourth inequalities in (1.13) follows from the monotoneity of f1 and [AQ, Theorem 1.5]. (3) For x>0, let 1 Γ(x + s) f10(s)=logx − log ,f11(s)=1− s, s Γ(x) 2 f12(s)=sψ(x + s) − log Γ(x + s)+logΓ(x)andf13(s)=s .

Then f5(s)=f10(s)/f11(s),f10(1) = f11(1) = f12(0) = f13(0) = 0, and   f10(s) f12(s) f12(s) 1   = ,  = ψ (x + s). f11(s) f13(s) f13(s) 2  Since ψ (t) is strictly decreasing on (0, ∞), the monotoneity of f5 follows from the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25].

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By l’Hˆopital’s Rule,

− f12(s) f5(1 ) = lim = ψ(x +1)− log x = α(x), s→1 f13(s) and

+ + log Γ(x + s) − log Γ(x) f5(0 )=f10(0 )=logx − lim s→0 s =logx − ψ(x)=β(x). 2 Let f14(s)=sβ(x) − sf5(s), f15(s)=s ,  2 f16(s)=ψ (x + s)(1 − s) − 2{[log x − ψ(x + s)](1 − s) +s log x − log Γ(x + s)+logΓ(x)}, 3  and f17(s)=(1− s) .Thenf6(s)=f14(s)/f15(s), f14(0) = f15(0) = f14(0) =  f15(0) = f16(1) = f17(1) = 0 and   f14(s) f16(s) f16(s) 1   = ,  = − ψ (x + s). f15(s) 2f17(s) f17(s) 3 It is well known that ψ is strictly increasing on (0, ∞). Hence the monotoneity of f6 follows from the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25]. By l’Hˆopital’s Rule, we get   + f14(s) f14(s) f16(0) 1  f6(0 ) = lim  = lim  = = ψ (x) − β(x)=η(x). s→0 f15(s) s→0 f15(s) 2f17(0) 2 − The limiting value f6(1 ) follows from the result for f5. Next, let

Γ(x + s) 2 f18(s)=s log x − log − s(1 − s)α(x)andf19(s)=(1− s) . Γ(x)   Then f7(s)=f18(s)/f19(s),f18(1) = f19(1) = f18(1) = f19(1) = 0 and  f18(s) 1   = α(x) − ψ (x + s), f19(s) 2 which is strictly increasing in s on (0, 1) by the well-known monotoneity of ψ. Hence the monotoneity of f7 follows from the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25]. + − Clearly, f7(0 ) = 0. By l’Hˆopital’s Rule, f7(1 )=θ(x). The inqualities in (1.14) are clear.

3.2. Corollary. (1) For x>1 and s ∈ (0, 1),     s(s − 1) 1 Γ(x + s) s(s − 1) 1 exp 1+ < (1 − s)/2 and s ∈ (0, 1),   Γ(x + s) 1 − s (3.3) >xs exp − s(1 − s)β >xs exp{−sλ(s)}. Γ(x) 2

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Proof. By (2.3) and [AS, 6.3.5], we have     1 1 1 1 β(x) < 1+ and α(x) > 1 − , 2x 6x 2x 6x and hence part (1) follows from (1.14) Next, by (2.4),  ∞ −xt  ∞ −u  −  − xte − 1 ue du xβ (x)=1 xψ (x)=1 −t dt =1 −u/x , 0 1 − e x 0 1 − e which is strictly increasing from (0, ∞)onto(−∞, 0) since v → v/(1−e−v) is strictly increasing from (0, ∞)onto(1, ∞). Hence β is strictly decreasing and convex from (0, ∞) onto itself. This, together with the first inequality in (2.3), yields (3.3). 

3.4. Proof of Theorem 1.16. (1) By (1.1), we can rewrite g1(x)as  √ 1/x √ 1/x πxΓ(x) πΓ(x +1) g1(x)= = . Γ(x +1/2) Γ(x +1/2) Logarithmic differentiation gives  (3.5) g1(x)/g1(x)=g4(x) ≡ g5(x)/g6(x), 2 where g6(x)=x and   1 Γ(x +1/2) g5(x)=x ψ(x +1)− ψ x + +log√ , 2 πΓ(x +1)

with g5(0) = g6(0) = 0, lim g5(x)=−∞,and x→∞    g5(x) 1   1 (3.6)  = g7(x) ≡ ψ (x +1)− ψ x + . g6(x) 2 2   Since ψ (x) is strictly increasing, g7(x) > 0. Hence g4 is strictly increasing on (0, ∞) by the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25]. By l’Hˆopital’s Rule, (3.5), (3.6), and [AS, 6.4.2, 6.4.4, 6.4.10 and 23.2.24],   2 + + 1   1 π g4(0 )=g7(0 )= ψ (1) − ψ = −ζ(2) = − 2 2 6 and lim g4(x) = lim g7(x)=0. x→∞ x→∞ Here ζ(x) is the Riemann function. Hence g4 is strictly increasing from (0, ∞) 2  onto (−π /6, 0), so that the monotoneity of g1 follows from (3.5). Since g1(x)=  −g1(x) · [−g4(x)], g1 is strictly increasing on (0, ∞) and the convexity of g1 follows. By l’Hˆopital’s Rule and [AS, 6.3.2, 6.3.3 and 6.3.18],   + log Γ(x +1)− log Γ(x +1/2) + (log π)/2 g1(0 )=explim x→0 x =exp(ψ(1) − ψ(1/2)) = 4 and   1 lim g1(x)=exp lim ψ(x +1)− ψ x + =1. x→∞ x→∞ 2

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(2) By (1.1) and [AS, 6.1.15], g2(x) can be rewritten as √ √ 1/(2x) πΓ(1 + 1/(4x)) π Γ(x) g2(x)= . Γ((3/2) + 1/(4x)) 2 Γ(x +1/2) Logarithmic differentiation gives   2  4x g2(x) 1 (3.7) = g8(x) ≡ 2x ψ(x) − ψ x + g2(x) 2 √     πΓ(x) 1 3 1 − 2log + ψ + − ψ +1 . 2Γ(x +1/2) 4x 2 4x Since       1 1 1 1 3 g (x)=2x ψ(x) − ψ x + + ψ +1 − ψ + , 8 2 4x2 4x 4x 2  which is positive for x ∈ (0, ∞) by the monotoneity of ψ on (0, ∞), g8 is strictly increasing on (0, ∞). Let h3 be as in Theorem 2.1. Then g8 can be rewritten as (3.8) √     πΓ(x) 1 3 1 g8(x)=−2h3(x) − 1 − 2log + ψ + − ψ +1 . 2Γ(x +1/2) 4x 2 4x Therefore, by [AS, 6.3.18 and 6.1.37],    + 1 3 1 g8(0 )=−2 + log 4 + lim 2logx +log + +1 x→0 4x 2 4x 1 1 − + = −∞ [1/(2x)] + 3 [1/(2x)] + 2

and   √ 3 π Γ(x) lim g8(x)=−1+ψ − ψ(1) − 2log − 2 lim log x→∞ 2 2 x→∞ Γ(x +1/2)   −x e 1 =1− log π − 2 lim log 1+ = ∞. x→∞ x 2x

Hence g8 has a unique zero x1 ∈ (0, ∞) such that g8(x) < 0forx ∈ (0,x1)and g8(x) > 0forx>x1, so that the piecewise monotoneity of g2 follows from (3.7). By [AS, 6.3.7 and 6.3.8], we have         2 1 1 1 ψ − ψ = ψ 1 − − ψ 3 6 3   6   π 1 1 = π cot + ψ − ψ 3 3 6       π 1 1 1 1 1 1 = π cot + ψ + ψ + +log2 − ψ 3 2 6 2  6 2 6 π 1 2 1 = π cot + ψ − ψ +log2, 3 2 3 6 so that     2 1 π ψ − ψ =2π cot +log4. 3 6 3

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Consequently, by [AS, 6.3.2, 6.3.4 and 6.3.5],     √     3 3 πΓ(3/2) 10 7 g8 =3ψ − ψ(2) − 2log + ψ − ψ 2 2  2Γ(2)  6  6 π 2 1 =3(1− log 4) − 2log + ψ +1 − ψ +1 4 3 6     3 2 1 = − − log 4 − 2logπ + ψ − ψ 2 3 6 3 π = − − 2log(2π)+2π cot +log4=−0.16186 ···< 0. 2 3 On the other hand, by [AS, 6.3.2, 6.3.4, 6.1.12, 6.3.5 and 6.3.8],       5 2 5 8 1 g8(2) = 4 log 4 − − 2log + ψ + − 8+ψ 3 3  8 5  8 196 5 1 = 3 log 4 + 2 log 3 − + ψ − ψ 15 8 8     196 1 1 1 = 3 log 4 + 2 log 3 − + ψ + − ψ 15  8  2 8 196 1 1 =2log12− +2 ψ − ψ . 15 4 8 It follows from [GR, 8.366.4 and 8.363.6, pp. 944–945] that   1 π ψ = −γ − − 3log2 4 2 and   7     1 π π  kπ kπ ψ = −γ − 3log2− cot + cos log 2sin 8 2 8 k=1 4 8 π π √ sin(π/8) = −γ − 4log2− cot + 2log 2 8 sin(3π/8) π π √ π = −γ − 4log2− cot + 2logtan , 2 8 8 so that     1 1 π π π √ π ψ − ψ = − +log2+ cot + 2logcot , 4 8 2 2 8 8 and hence, 196 π √ π g8(2)=3log4+2log3− − π + π cot +2 2logcot =0.225224 ···> 0. 15 8 8

Consequently, x1 ∈ (3/2, 2). Clearly,   √ 1 π π Γ(1/2) π2 g2 = · · = . 2 2 2 Γ(1) 4

It follows from [AS, 6.1.37] that lim g2(x)=2. x→∞

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(3) Logarithmic differentiation gives (3.9) 2  4x g3(x) 2 = g9(x) ≡ g8(x)+ − 2 − log 4 g3(x )   2x +1     1 Γ(x +1/2) 1 1 1 =2x ψ(x) − ψ x + +2log √ + ψ + − ψ − 4x. 2 πΓ(x) 4x 2 4x By [AS, 6.4.1], for each x>0,  ∞ −xt  ∞ −[1+1/(4x)]t  te dt 1 te dt g8(x)=2x −t/2 + 2 −t/2 0 1+e 4x 0 1+e  ∞  ∞ te−xtdt 2 (xt)e−xtd(xt) > 2x −t/2 = −t/2 0 1+e x 0 1+e  ∞ −u  ∞ 2 ue du 1 −u 1 = −u/(2x) > ue du = , x 0 1+e x 0 x and hence, 4 1 4 4x2 +1 g (x)=g (x) − > − = > 0, 9 8 (2x +1)2 x (2x +1)2 x(2x +1)2 + so that g9 is strictly increasing on (0, ∞). Since g9(0 )=−∞ and lim g9(x)=∞, x→∞ g9 has a unique zero x2 such that g9(x) < 0forx ∈ (0,x2)andg9(x) > 0for x ∈ (x2, ∞). This, together with (3.9), yields the piecewise monotoneity of g3. Applying the formulas [AS, 6.3.2, 6.3.4, 6.3.5 and 6.1.12] n−1  n − 1 1 − − 1 (3.10) ψ(n)= γ + ,ψ n + = γ log 4 + 2 − , k=1 k 2 k=1 2k 1 (3.11)   1 1 · 3 · 5 ···(2n − 1)√ Γ n + = π, 2 2n and [GR, 8.370 and 8.372]     ∞ x +1 x  (−1)k (3.12) ψ − ψ =2 , 2 2 k=0 x + k we get 1 1 (2n − 1)!! g9(n)=g10(n) ≡ log 2 − 2+ log 2n n (n − 1)! ∞ − k n ( 1) − 1 − 1 +2 − k=0 2nk +1 k=1 k(2k 1) n for n ∈ N, and hence, by the properties of the Leibniz series and by computation, ∞ − k 33 − 1 65!! ( 1) − 1 − 1 g10(33) = log 2 2+ log +2 − 33 32! k=0 66k +1 k=1 k(2k 1) 33 30000 − k 33 − 1 65!! ( 1) − 1 − 1 < log 2 2+ log +2 − 33 32! k=0 66k +1 k=1 k(2k 1) 33 = −0.000381 ···< 0

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and ∞ − k 34 − 1 67!! ( 1) − 1 − 1 g10(34) = log 2 2+ log +2 − 34 33! k=0 68k +1 k=1 k(2k 1) 34 30001 − k 34 − 1 67!! ( 1) − 1 − 1 > log 2 2+ log +2 − 34 33! k=0 68k +1 k=1 k(2k 1) 34 =0.000065 ···> 0.

Hence x2 ∈ (33, 34). 2 Finally, it is clear that g3(1/2) = 4g2(1/2) = π and lim g3(x) = lim g2(x)=2. x→∞ x→∞ 

3.13. ProofofTheorem1.17. (1) By the expression   1 n − 1 1 In = B , , 2 2 2

F1(n) can be rewritten as    n 2  Γ 2 n − 1 F1(n)=2n − 4  =4f4 +2 n−1 2 Γ 2

by (1.1), where f4 is as in Theorem 1.12. Hence the monotoneity of F1 and the limiting value lim F1(n) = 3 follow from Theorem 1.12(2). Clearly, F1(1) = 6 − π. n→∞ Thus, for n ≥ 3, 6 − π ≤ F1(n) < 3, so that (1.18) holds. (2) By (1.2),    n−1 1 2    n − 1 Γ 2 + 2 1 n − 1 F2(n)=2 −  + =2f4 +1, 2 n−1 2 2 Γ 2

which is strictly increasing for n ≥ 3withF2(3) = 3 − π/2 and lim F2(n)=3/2 n→∞ by Theorem 1.12(2). Hence the result follows. (3) It follows from (1.2) that  n Ωn−1 1 Γ 2 +1 ωn = √  = , Ωn π n+1 ωn+1 Γ 2 so that  2 ωn F3(n)=n − 2π = F2(n +2)− 2. ωn+1 Hence part (3) follows from part (2). (4) It follows from (1.2), the expression [AVV1, (2.32), p. 41]   1 1 1 (3.14) Jn = B , 2 2(n − 1) 2 and [AS, 6.1.15] that  √ 1   π Γ 2(n−1) √ 1 (3.15) bn =  = πF6 , 2(n − 1) 1 1 2(n − 1) Γ 2(n−1) + 2

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where F6(x)=Γ(x +1)/Γ(x +1/2). Since

 F6(x)=F6(x)[ψ(x +1)− ψ(x +1/2)] > 0

by the monotoneity of ψ(x), F6 is strictly increasing on (0, ∞). Hence the mono- toneity of bn follows from (3.15), and √ (3.16) 1= πF6(0)

Next, let g1 be as in Theorem 1.16(1). Then by (1.4) and (3.14),

  n−1    1 1 1 1 Bn = B , = g1 , 2(n − 1) 2(n − 1) 2 2(n − 1)

and hence the monotoneity of Bn follows from Theorem 1.16(1). Moreover, for n ≥ 2,  π + (3.17) = B2 ≤ Bn < lim Bn = g1(0 )=2. 2 n→∞

The second double inequality in (1.21) follows from (3.16) and (3.17). (5) We know that n − Jn − a has the expression

 π/2  1/(n−1) t n − Jn − a = 1 − (sin t) tan dt. 0 2

(Cf. [AVV1, p. 43].) Hence    π/2 1 t (3.18) F4(n)= F7 ,t tan dt, 0 n − 1 2

x where F7(x, t)=[1− (sin t) ]/x. Since the function

d [1 − (sin t)x] ≡ dx x 1 F8(x, t) d =(sint) log dx (x) sin t

is strictly decreasing in x on (0, ∞) for each fixed t ∈ (0,π/2), it follows from the Monotone l’Hˆopital’s Rule [AVV1, Theorem 1.25] that F7 is strictly decreasing in x on (0, ∞). Hence F4 is strictly increasing for n ≥ 3. Clearly, F4(3) = 2(3 − J3 − a)=2(b − a). Set x =1/[2(n − 1)]. Then, by (3.14),   1 1 1 1 F4(n)= +1− B x, − a 2x 2x 2 2  1 √ Γ(x +1) = x log 4 − π  +1 . 4x2 1 Γ x + 2

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Hence, by l’Hˆopital’s Rule,

F4(∞) ≡ lim F4(n) n→∞ √ 1 x log 4 + 1 − [ πΓ(x +1)/Γ(x +1/2)] = lim 4 x→0 x2     1 1 √ Γ(x +1) 1 = lim log 4 − π  ψ(x +1)− ψ x + 8 x→0 x 1 2 Γ x + 2 √ π Γ(x +1) = − lim  8 x→0 1 Γ x + 2     2   1 1 × ψ(x +1)− ψ x + + ψ(x +1)− ψ x + 2 2   1 π2 1 1 π2 = − (log 4)2 + − ψ = − (log 4)2 = c. 8 6 2 8 3 The inequality (1.22) is clear. (6) By (1.2), (1.4), (1.5) and [AS, 6.1.15], we have    n/2 n−1 1/(n−1) 1 π Γ 2 Γ 2(n−1) An =   . n n 1 1 2 Γ 2 Γ 2(n−1) + 2

Putting x =(n − 1)/2 and applying [AS, 6.1.15], we get

(3.19)  √ √ 1/(2x) 1 π 1 π Γ(x) Γ 4x 1 F5(n)=   = g2(x), 2 2x +1 2 1 1 1 2 Γ x + 2 Γ 4x + 2

where g2 is as in Theorem 1.16(2). Next, computation gives (3.20) √ √ 1/3 π π Γ(3/2) Γ(1 + 1/16) F5(4) = =0.8890941475 ··· 2 2 Γ(2) Γ(1 + 9/16) and (3.21) √ √ 1/4 π π Γ(2) Γ(1 + 1/8) F5(5) = =0.8411396629 ···. 2 2 Γ(2 + 1/2) Γ(1 + 5/8)

The result for F5 now follows from (3.19), (3.20), (3.21) and Theorem 1.16(2). The limiting value lim (1/n)An = 1 follows from Theorem 1.16(2), and the n→∞ inequality in (1.23) is clear. (7) Put x =(n − 1)/2. Then, by (3.19),

n/(n−1) 1 n/(n−1) n An (3.22) Dn =2 An =2 · = g3(x), n − 1 n − 1 n

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where g3 is as in Theorem 1.16(3). Hence, by Theorem 1.16(3), Dn is strictly decreasing for 2 ≤ n ≤ 67 and strictly increasing for n ≥ 69. By computation, we obtain  1/66 67/66 2 1 67/2 Γ(33) Γ(1/132) D67 = A67 = π 66 66 Γ(33 + 1/2) Γ(67/132) √  1/66 2π 32! Γ(1/132) = =1.98535339227 ···, 66 65!! Γ(67/132)  1/67 68/67 2 1 68/2 Γ(33 + 1/2) Γ(1/134) D68 = A68 = π 67 67 Γ(34) Γ(68/134)  1/67 1 π69/134 65!! Γ(1/134) = =1.98534941481 ···, 67 233/67 33! Γ(68/134)  1/68 69/68 2 1 69/2 Γ(34) Γ(1/136) D69 = A69 = π 68 68 Γ(34 + 1/2) Γ(69/136) √  1/68 2π 33! Γ(1/136) = =1.98534877792 ···, 68 67!! Γ(69/136)  1/69 70/69 2 1 70/2 Γ(34 + 1/2) Γ(1/138) D70 = A70 = π 69 69 Γ(35) Γ(70/138)  1/69 1 π71/138 67!! Γ(1/138) = =1.98535124396 ···. 69 234/69 34! Γ(70/138)

These values, together with the conclusion just proved above, yield the piecewise monotoneity of Dn. 2 Clearly, D2 = π . By (3.22) and (1.23), lim Dn = 2. Hence (1.24) holds.  n→∞ Acknowledgments The authors are indebted to the referee, whose remarks led to several improve- ments in the final text. References

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President’s Office, Hangzhou Institute of Electronics Engineering (HIEE), Hang- zhou 310037, Peoples Republic of China E-mail address: sl [email protected] Department of Mathematics, University of Turku, Vesilinnankatu 5, FIN-20014, Turku, Finland E-mail address: [email protected]

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