A Pick function related to the sequence of volumes of the unit ball in n-space
aa r X i v : . [ m a t h . C A ] D ec A Pick function related to the sequence ofvolumes of the unit ball in n -space ∗ Christian Berg † and Henrik L. Pedersen ‡† Institute of Mathematical Sciences, University of CopenhagenUniversitetsparken 5; DK-2100 København Ø, DenmarkE-mail [email protected] ‡ Department of Basic Sciences and EnvironmentFaculty of Life Sciences, University of CopenhagenThorvaldsensvej 40, DK-1871 Frederiksberg CE-mail [email protected]
November 9, 2018
Abstract
We show that F a ( x ) = ln Γ( x + 1) x ln( ax )is a Pick function for a ≥ f ( x ) = π x/ Γ(1 + x/ ! / ( x ln x ) and show that ln f ( x + 1) is a Stieltjes function and that f ( x + 1) is com-pletely monotonic on ]0 , ∞ [. In particular f ( n ) = Ω / ( n ln n ) n , n ≥ n is the volume of the unit ball inEuclidean n -space. Mathematics Subject Classification : primary 33B15; secondary 30E20, 30E15.Keywords: gamma function, completely monotone function. ∗ Both authors acknowledge support by grant 272-07-0321 from the Danish Research Councilfor Nature and Universe. Introduction and results
Since the appearance of the paper [3], monotonicity properties of the functions F a ( x ) = ln Γ( x + 1) x ln( ax ) , x > , a > n of the unit ball in Euclidean n -space. A recent paperabout inequalities involving Ω n is [2].Let us first consider the case a = 1. In [9] the authors proved that F is a Bernstein function , which means that it is positive and has a completely mono-tonic derivative, i.e., ( − n − F ( n )1 ( x ) ≥ , x > , n ≥ . (2)This extended monotonicity and concavity proved in [4] and [12] respectively.We actually proved a stronger statement than (2), namely that the reciprocalfunction x ln x/ ln Γ( x + 1) is a Stieltjes transform, i.e. belongs to the Stieltjescone S of functions of the form g ( x ) = c + Z ∞ dµ ( t ) x + t , x > , (3)where c ≥ µ is a non-negative measure on [0 , ∞ [ satisfying Z ∞ dµ ( t )1 + t < ∞ . The result was obtained using the holomorphic extension of the function F to the cut plane A = C \ ] − ∞ , µ in (3). Our derivation used the fact that the holomorphic function log Γ( z )only vanishes in A at the points z = 1 and z = 2, a result interesting in itself andincluded as an appendix in [9]. A simpler proof of the non-vanishing of log Γ( z )appeared in [10].In a subsequent paper [10] we proved an almost equivalent result, namely that F is a Pick function, and obtained the following representation formula F ( z ) = 1 − Z ∞ d ( t ) z + t dt, z ∈ A (4)where d ( t ) = ln | Γ(1 − t ) | + ( k −
1) ln tt ((ln t ) + π ) for t ∈ ] k − , k [ , k = 1 , , . . . (5)and d ( t ) tends to infinity when t approaches 1 , , . . . . Since d ( t ) > t > ϕ in the upper half-plane H = { z = x + iy ∈ C | y > } satisfying ℑ ϕ ( z ) ≥ z ∈ H , cf.[11].For a = 2 Anderson and Qiu proved in [4] that F is strictly increasing on[1 , ∞ [, thereby proving a conjecture from [3]. Alzer proved in [2] that F isconcave on [46 , ∞ [. In [14] the concavity was extended to the optimal interval] , ∞ [.We will now describe the main results of the present paper.We also denote by F a the holomorphic extension of (1) to A with an isolatedsingularity at z = 1 /a , which is a simple pole with residue ln Γ(1 + 1 /a ) assuming a = 1, while z = 1 is a removable singularity for F . For details about thisextension see the beginning of section 2. Using the residue theorem we obtain: Theorem 1.1
For a > the function F a has the integral representation F a ( z ) = 1 + ln Γ(1 + 1 /a ) z − /a − Z ∞ d a ( t ) z + t dt, z ∈ A \ { /a } , (6) where d a ( t ) = ln | Γ(1 − t ) | + ( k −
1) ln( at ) t ((ln( at )) + π ) for t ∈ ] k − , k [ , k = 1 , , . . . , (7) and d a (0) = 0 , d a ( k ) = ∞ , k = 1 , , . . . . We have d a ( t ) ≥ for t ≥ , a ≥ / and F a is a Pick function for a ≥ but not for < a < . From this follows the monotonicity property conjectured in [14]:
Corollary 1.2
Assume a ≥ . Then ( − n − F ( n ) a ( x ) > , x > /a, n = 1 , , . . . . (8)In particular, F a is strictly increasing and strictly concave on the interval ]1 /a, ∞ [.The function f ( x ) = (cid:18) π x/ Γ(1 + x/ (cid:19) / ( x ln x ) (9)has been studied because the volume Ω n of the unit ball in R n isΩ n = π n/ Γ(1 + n/ , n = 1 , , . . . . We prove the following integral representation of the extension of ln f ( x + 1)to the cut plane A . This is slightly improved in Remark 2.6 below. heorem 1.3 For z ∈ A we have log f ( z + 1) = −
12 + ln(2 / √ π ) z + ln( √ π )Log( z + 1) + 12 Z ∞ d (( t − / z + t dt. (10) In particular / f ( x + 1) is a Stieltjes function and f ( x + 1) is completelymonotonic. We recall that completely monotonic functions ϕ : ]0 , ∞ [ → R are character-ized by Bernstein’s theorem as ϕ ( x ) = Z ∞ e − xt dµ ( t ) , (11)where µ is a positive measure on [0 , ∞ [ such that the integrals above make sensefor all x > { a n } n ≥ of positive numbers is a Hausdorffmoment sequence if it has the form a n = Z x n dσ ( x ) , n ≥ , (12)where σ is a positive measure on the unit interval. Note that lim n →∞ a n = σ ( { } ).For a discussion of these concepts see [7] or [17]. It is clear that if ϕ is completelymonotonic with the integral representation (11), then a n = ϕ ( n + 1) , n ≥ a n = Z ∞ e − ( n +1) t dµ ( t ) = Z x n dσ ( x ) , where σ is the image measure of e − t dµ ( t ) under e − t . Since lim x →∞ f ( x + 1) = e − / we get Corollary 1.4
The sequence f ( n + 2) = Ω / (( n +2) ln( n +2)) n +2 , n = 0 , , . . . (13) is a Hausdorff moment sequence tending to e − / . A Hausdorff moment sequence is clearly decreasing and convex and by theCauchy-Schwarz inequality is is even logarithmically convex, meaning that a n ≤ a n − a n +1 , n ≥
1. The latter property was obtained in [14] in a different way.4
Properties of the function F a In this section we will study the holomorphic extension of the function F a definedin (1). First a few words about notation. We use ln for the natural logarithm butonly applied to positive numbers. The holomorphic extension of ln from the openhalf-line ]0 , ∞ [ to the cut plane A = C \ ] −∞ ,
0] is denoted Log z = ln | z | + i Arg z ,where − π < Arg z < π is the principal argument. The holomorphic branch of thelogarithm of Γ( z ) for z in the simply connected domain A which equals ln Γ( x )for x > z ). The imaginary part of log Γ( z ) is a continuousbranch of argument of Γ( z ) which we denote arg Γ( z ), i.e.,log Γ( z ) = ln | Γ( z ) | + i arg Γ( z ) , z ∈ A . We shall use the following property of log Γ( z ), cf. [9, Lemma 2.1] Lemma 2.1
We have, for any k ≥ , lim z → t, ℑ z> log Γ( z ) = ln | Γ( t ) | − iπk for t ∈ ] − k, − k + 1[ and lim z → t, ℑ z> | log Γ( z ) | = ∞ for t = 0 , − , − , . . . . The expression F a ( z ) = log Γ( z + 1) z Log( az )clearly defines a holomorphic function in A \ { /a } , and z = 1 /a is a simple poleunless a = 1, where the residue ln Γ(1 + 1 /a ) vanishes. Lemma 2.2
For a > and t ≤ we have lim y → + ℑ F a ( t + iy ) = πd a ( − t ) , (14) where d a is given by (7) .Proof. For − < t < y → + F a ( t + iy ) = ln Γ(1 + t ) t (ln( a | t | ) + iπ ) , hence lim y → + ℑ F a ( t + iy ) = πd a ( − t ). For − k < t < − k + 1 , k = 2 , , . . . we findusing Lemma 2.1 lim y → + F a ( t + iy ) = ln | Γ(1 + t ) | − i ( k − πt (ln( a | t | ) + iπ ) , y → + ℑ F a ( t + iy ) = πd a ( − t ) also in this case.For t = − k, k = 1 , , . . . we have | F a ( − k + iy ) | ≥ | ln | Γ( − k + 1 + iy ) ||| − k + iy || Log( a ( − k + iy )) | → ∞ for y → + because Γ( z ) has poles at z = 0 , − , . . . . Finally, for t = 0 we get (14)from the next Lemma. (cid:3) Lemma 2.3
For a > we have lim z → ,z ∈A | F a ( z ) | = 0 . Proof.
Since log Γ( z + 1) /z has a removable singularity for z = 0 the resultfollows because | Log( az ) | ≥ | ln( a | z | ) | → ∞ for | z | → , z ∈ A . (cid:3) Lemma 2.4
For a > we have the radial behaviour lim r →∞ F a ( re iθ ) = 1 for − π < θ < π, (15) and there exists a constant C a > such that for k = 1 , , . . . and − π < θ < π | F a (( k + ) e iθ ) | ≤ C a . (16) Proof.
We first note that F a ( z ) = F ( z ) Log( z )Log( az ) , (17)and since lim | z |→∞ ,z ∈A Log( z )Log( az ) = 1it is enough to prove the results for a = 1. We do this by using a methodintroduced in [9, Prop. 2.4].Define R k = { z = x + iy ∈ C | − k ≤ x < − k + 1 , < y ≤ } for k ∈ Z and R = ∪ ∞ k =0 R k , S = { z = x + iy ∈ C | x ≤ , | y | ≤ } . By Lemma 2.1 it is clear that M k = sup | θ | <π | F (( k + ) e iθ ) | < ∞ (18)for each k = 1 , , . . . , so it is enough to prove that M k is bounded for k → ∞ .6tieltjes ([16, formula 20]) found the following formula for log Γ( z ) for z inthe cut plane A log Γ( z + 1) = ln √ π + ( z + 1 /
2) Log z − z + µ ( z ) . (19)Here µ ( z ) = ∞ X n =0 h ( z + n ) = Z ∞ P ( t ) z + t dt, where h ( z ) = ( z + 1 /
2) Log(1 + 1 /z ) − P is periodic with period 1 and P ( t ) = 1 / − t for t ∈ [0 , µ ( z ) = 12 Z ∞ Q ( t )( z + t ) dt, (20)where Q is periodic with period 1 and Q ( t ) = t − t for t ∈ [0 , µ is a completely monotonic function. For further properties of Binet’sfunction µ see [13].We claim that | µ ( z ) | ≤ π z ∈ A \ S. In fact, since 0 ≤ Q ( t ) ≤ /
4, we get for z = x + iy ∈ A| µ ( z ) | ≤ Z ∞ dt ( t + x ) + y . For x > Z ∞ dt ( t + x ) + y ≤ Z ∞ dt ( t + 1) = 1 , and for x ≤ , | y | ≥ Z ∞ dt ( t + x ) + y = Z ∞ x dtt + y < Z ∞−∞ dtt + 1 = π. Since F ( z ) = 1 + ln √ π + 1 / z − z + µ ( z ) z Log z , for z ∈ A , we immediately get (15) and | F ( z ) | ≤ z ∈ A \ S for which | z | is sufficiently large. In particular, there exists N ∈ N such that | F (( k + ) e iθ ) | ≤ k ≥ N , ( k + ) e iθ ∈ A \ S. (22)7y continuity the quantity c = sup (cid:8) | log Γ( z ) | | z = x + iy, ≤ x ≤ , ≤ y ≤ (cid:9) (23)is finite.We will now estimate the quantity | F (( k + ) e iθ ) | when ( k + ) e iθ ∈ S , andsince F ( z ) = F ( z ), it is enough to consider the case when ( k + ) e iθ ∈ R k +1 . Todo this we use the relationlog Γ( z + 1) = log Γ( z + k + 1) − k X l =1 Log( z + l ) (24)for z ∈ A and k ∈ N . Equation (24) follows from the fact that the functions onboth sides of the equality sign are holomorphic functions in A , and they agree onthe positive half-line by repeated applications of the functional equation for theGamma function.For z = ( k + ) e iθ ∈ R k +1 we get | log Γ( z + k + 1) | ≤ c by (23), and hence by(24) | log Γ( z + 1) | ≤ c + k X l =1 | Log( z + l ) | ≤ c + kπ + k X l =1 | ln | z + l || . For l = 1 , . . . , k − k − l < | z + l | < k + 2 − l , hence 0 < ln | z + l | < ln( k +2 − l ). Furthermore, 1 / ≤ | z + k | ≤ √
2, hence − ln 2 < ln | z + k | ≤ (ln 2) / | log Γ( z + 1) | ≤ c + kπ + k +1 X j =2 ln j < c + kπ + k ln( k + 1) . From this we get for z = ( k + ) e iθ ∈ R k +1 | F ( z ) | ≤ c + kπ + k ln( k + 1)( k + ) ln( k + ) (25)which tends to 1 for k → ∞ . Combined with (22) we see that there exists N ∈ N such that | F (( k + ) e iθ ) | ≤ k ≥ N , − π < θ < π, which shows that M k from (18) is a bounded sequence. (cid:3) Lemma 2.5
Let a > . For k = 1 , , . . . there exists an integrable function f k,a : ] − k, − k + 1[ → [0 , ∞ ] such that | F a ( x + iy ) | ≤ f k,a ( x ) for − k < x < − k + 1 , < y ≤ . (26)8 roof. For z = x + iy as above we get using (24) | log Γ( z + 1) | ≤ | log Γ( z + k + 1) | + k X l =1 | Log( z + l ) | ≤ L + kπ + k X l =1 | ln | z + l || , where L is the maximum of | log Γ( z ) | for z ∈ R − . We only treat the case k ≥ k = 1 is a simple modification combined with Lemma 2.3.For l = 1 , . . . , k − < | z + l | < k − l , and for l = k − , k ln | x + l | ≤ ln | z + l | ≤ (1 /
2) ln 2, so we find | log Γ( z + 1) | ≤ L + kπ + k X j =2 ln j + | ln | x + k − || + | ln | x + k || , (27)so as f k, we can use the right-hand side of (27) divided by ( k −
1) ln( k − f k,a ( x ) = f k, ( x ) max z ∈ R k | Log z || Log( az ) | . (cid:3) Proof of Theorem 1.1
For fixed w ∈ A \ { /a } we choose ε > , k ∈ N such that ε < | w | , /a < k + and consider the positively oriented contour γ ( k, ε ) in A consisting of the half-circle z = εe iθ , θ ∈ [ − π , π ] and the half-lines z = x ± iε, x ≤ | z | = k + , which closes the contour.By the residue theorem we find12 πi Z γ ( k,ε ) F a ( z ) z − w dz = F a ( w ) + ln Γ(1 + 1 /a )1 /a − w . We now let ε → ε will tend to zero, and by Lemma 2.2 andLemma 2.5 we get12 π Z π − π F a (( k + ) e iθ )( k + ) e iθ − w ( k + ) e iθ dθ + Z − k − d a ( − t ) t − w dt = F a ( w ) + ln Γ(1 + 1 /a )1 /a − w . For k → ∞ the integrand in the first integral converges to 1 for each θ ∈ ] − π, π [ and by Lemma 2.4 Lebesgue’s theorem on dominated convergence can beapplied, so we finally get F a ( w ) = 1 + ln Γ(1 + 1 /a ) w − /a − Z ∞ d a ( t ) t + w dt. The last integral above appears as an improper integral, but we shall seethat the integrand is Lebesgue integrable. We show below that d a ( t ) ≥ ≥ / a the integrability is obvious. The function d a tends to 0 for t → t = 1 so d a is integrableover ]0 , k − < t < k, k ≥ d a ( t ) = (ln( t )) + π (ln( at )) + π d ( t ) + ( k −
1) ln at ((ln( at )) + π ) , (28)and the factor in front of d ( t ) is a bounded continuous function with limit 1 at0 and at infinity. Therefore Z ∞ | d a ( t ) | t dt < ∞ follows from the finiteness of the corresponding integral for a = 1 provided thatwe establish S := ∞ X k =2 ( k − Z kk − dtt ((ln( at )) + π ) < ∞ . Choosing N ∈ N such that aN >
1, we can estimate
S < ∞ X k =1 Z ( k +1) aka dtt (ln ( t ) + π ) < Z Naa dtt (ln ( t ) + π ) + ∞ X k = N Z ( k +1) aka dtt ln ( t )= Z Naa dtt (ln ( t ) + π ) + 1ln( aN ) < ∞ . We next examine positivity of d a .For 0 < t < d a ( t ) = ln | Γ(1 − t ) | t ((ln( at )) + π ) > s ) > < s < k ≥ t ∈ ] k − , k [ the numerator N a in d a can be written N a ( t ) = ln Γ( k − t ) + k − X l =1 ln tat − l , where we have used the functional equation for Γ, hence N a ( t ) ≥ k − X l =1 ln kk − l + ( k −
1) ln a = ( k −
1) ln k − ln Γ( k ) + ( k −
1) ln a, because Γ( k − t ) > t/ ( t − l ) is decreasing for k − < t < k . From (19) weget ln Γ( k ) = ln √ π + ( k − /
2) ln k − k + µ ( k ) (29)10nd in particular for k = 2 µ (2) = 2 −
32 ln 2 − ln √ π. Using (29) we find N a ( t ) ≥ k −
12 ln k − ln √ π − µ ( k )+( k −
1) ln a ≥ k −
12 ln k −
2+ 32 ln 2+( k −
1) ln a, because µ is decreasing on ]0 , ∞ [ as shown by (20).For a ≥ / k − < t < k with k ≥ N a ( t ) ≥ k (1 − ln 2) −
12 ln k + 52 ln 2 − ≥ , because the sequence c k , k ≥ c = 0.We also see that d a ( t ) tends to infinity for t approaching the end points of theinterval ] k − , k [. For z = 1 /a + iy, y > ℑ F a (1 /a + iy ) = − ln Γ(1 + 1 /a ) y + Z ∞ yd a ( t )(1 /a + t ) + y dt. The last term tends to 0 for y → −∞ when0 < a <
1. This shows that F a is not a Pick function for these values of a . (cid:3) Remark 2.6
We proved in Theorem 1.1 that d a ( t ) is non-negative on [0 , ∞ [ for a ≥ /
2. This is not best possible, and we shall explain that the smallest valueof a for which d a ( t ) is non-negative is a = 0 . .. .Replacing k by k + 1 in the numerator N a for d a given by (7), we see that N a ( t ) = ln | Γ(1 − t ) | + k ln( at ) for t ∈ ] k, k + 1[ , k = 1 , , . . . is non-negative if and only ifln(1 /a ) ≤ ln( k + s ) + 1 k ln | Γ(1 − k − s ) | for s ∈ ]0 , , k = 1 , , . . . , and using the reflection formula for Γ this is equivalent to ln(1 /a ) ≤ ρ ( k, s ) forall 0 < s < k = 1 , , . . . , where ρ ( k, s ) = ln( k + s ) − k ln (cid:18) Γ( k + s ) sin( πs ) π (cid:19) . (30)Using Stieltjes’ formula (19), we find that ρ ( k, s ) = 1 + ln( π/ k − (1 /k ) [( s − /
2) ln( s + k ) + ln sin( πs ) − s + µ ( s + k )] (31)11or all s ∈ ]0 ,
1[ and k = 1 , , . . . . For fixed s ∈ ]0 ,
1[ we see that ρ ( k, s ) → k → ∞ , so ln(1 /a ) ≤ d a ( t ). Thiscondition is not sufficient, because for ln(1 /a ) = 1 the inequality 1 ≤ ρ ( k, s ) isequivalent to0 ≥ (1 /
2) ln(2 /π ) + ( s − /
2) ln( s + k ) + ln sin( πs ) − s + µ ( s + k )which does not hold when k is sufficiently large and 1 / < s < k = 1 , , . . . it is easy to verify that the function ρ k ( s ) = ρ ( k, s ) hasa unique minimum m k over ]0 , /a ) = inf { m k , k ≥ } (32)determines the smallest value of a for which d a ( t ) is non-negative. Using Mapleone obtains that m k is decreasing for k = 1 , . . . ,
510 and increasing for k ≥ m = inf m k = 0 . .. corresponding to a =0 . .. . We add that m = 1 . .., m = 1 . .., m =0 . .. . f Proof of Theorem 1.3
The functionln f ( x ) = ( x/
2) ln π − ln Γ(1 + x/ x ln x clearly has a meromorphic extension to A \ z = 1 withresidue ln 2. We denote this meromorphic extension log f ( z ) and havelog f ( z + 1) = ln √ π Log( z + 1) − F (cid:18) z + 12 (cid:19) . Using the representation (6), we immediately get (10). It is well-known that1 / Log( z + 1) is a Stieltjes function, cf. [8, p.130], and the integral representationis 1Log( z + 1) = Z ∞ dt ( z + t )((ln( t − + π ) . (33)It follows that ln( √ ef ( x + 1)) is a Stieltjes function, in particular completelymonotonic, showing that √ ef ( x + 1) belongs to the class L of logarithmicallycompletely monotonic functions studied in [15] and in [6]. Therefore also f ( x + 1)is completely monotonic. (cid:3) Representation of /F a For a > G a ( z ) = 1 /F a ( z ) = z Log( az )log Γ( z + 1) (34)which is holomorphic in A with an isolated singularity at z = 1, which is asimple pole with residue ln a/ Ψ(2) = ln a/ (1 − γ ) if a = 1, while it is a removablesingularity when a = 1. Here Ψ( z ) = Γ ′ ( z ) / Γ( z ) and γ is Euler’s constant. Theorem 4.1
For a > the function G a has the integral representation G a ( z ) = 1 + ln a (1 − γ )( z −
1) + Z ∞ ρ a ( t ) z + t dt, z ∈ A \ { } , (35) where ρ a ( t ) = t ln | Γ(1 − t ) | + ( k −
1) ln( at )(ln | Γ(1 − t ) | ) + (( k − π ) for t ∈ ] k − , k [ , k = 1 , , . . . , (36) and ρ a (0) = 1 /γ, ρ a ( k ) = 0 , k = 1 , , . . . , which makes ρ a continuous on [0 , ∞ [ .We have ρ a ( t ) ≥ for t ≥ , a ≥ a = 0 . .. , cf. Remark 2.6, and G a ( x + 1) is a Stieltjes function for a ≥ but not for < a < .Proof. We notice that for − k < t < − k + 1 , k = 1 , , . . . we get usingLemma 2.1 lim y → + G a ( t + iy ) = t (ln( a | t | ) + iπ )ln | Γ(1 + t ) | − i ( k − π , and for t = − k, k = 1 , , . . . we getlim y → + | G a ( − k + iy ) | = 0because of the poles of Γ, hence lim y → + ℑ G a ( t + iy ) = − πρ a ( − t ) for t < w ∈ A \ { } we choose ε > , k ∈ N such that ε < | w | , < k + and consider the positively oriented contour γ ( k, ε ) in A which was used in theproof of Theorem 1.1.By the residue theorem we find12 πi Z γ ( k,ε ) G a ( z ) z − w dz = G a ( w ) + ln a (1 − γ )(1 − w ) . We now let ε → ε -halfcircle tends to 0 and we get12 π Z π − π G a (( k + ) e iθ )( k + ) e iθ − w ( k + ) e iθ dθ − Z − k − ρ a ( − t ) t − w dt = G a ( w ) + ln a (1 − γ )(1 − w ) . k → ∞ we get (35), leaving the details to the reader. Clearly, ρ a ≥ d a defined in (7) is non-negative. It follows that G a ( x + 1)is a Stieltjes function for a ≥ < a <
1, since in the latter case ℑ G a (1 + iy ) > y > (cid:3) Remark 4.2
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