Binet's convergent factorial series in the theory of the Gamma function
aa r X i v : . [ m a t h . F A ] F e b Convergent Binet series for the Gamma function
Piet Van Mieghem ∗ Delft University of Technologyv1: 2 February 2021v2: 9 February 2021
Abstract
We investigate Binet’s convergent series for the logarithm of the Gamma function and contributeto the classical theory of the Gamma function by correcting an unfortunate error in Binet’s originalcomputation. We compare the corrected Binet series with Stirling’s asymptotic expansion anddiscuss the advantage of both series.
In his truly comprehensive “memoir” [2, Section 3, p. 223] in 1839, Jacques Binet defines his function µ ( z ) in relation to the Gamma function Γ ( z ) aslog Γ ( z ) = (cid:18) z − (cid:19) log z − z + 12 log (2 π ) + µ ( z ) (1)Binet has derived two integral representations of his function [5, p. 248-251], for Re ( z ) > µ ( z ) = 2 Z ∞ arctan (cid:0) tz (cid:1) e πt − dt (2)and µ ( z ) = 12 Z ∞ e − zt t (cid:18) e − t − e − t − t (cid:19) dt (3)There exist more representations of µ ( z ), but here we merely concentrate on an elegant, convergingfactorial series due to Binet [2, p. 234], in his notation,2 µ ( z ) = I (1) z + 1 + I (2)2 ( z + 1) ( z + 2) + I (3)3 ( z + 1) ( z + 2) ( z + 3) + · · · (4) ∗ Faculty of Electrical Engineering, Mathematics and Computer Science, P.O Box 5031, 2600 GA Delft, The Nether-lands; email : [email protected]. The reference in Whittaker and Watson [5, footnote, p. 248] with page numbers from 123 to 143 suggests an ordinaryarticle, but the correct page numbers pp. 123-343 point to a book-size treatise. Apart from adding his own beautifulcontributions, Binet has reviewed the knowledge about the Beta-function before 1839. The focal point at that time werethe many properties of Euler’s Beta integral from which properties of the Gamma function were derived. Today, onthe other hand and perhaps after Weierstrass’s product and Hankel’s contour integral, the theory concentrates on theGamma function, with applications to the Beta function. I ( m ) = Z x ( x + 1) ( x + 2) · · · ( x + m −
1) (2 x − dx Binet [2, p. 234] lists the first few coefficients, I (1) = , I (2) = , I (3) = and I (4) = .Unfortunately, Binet, whose derivation is rather involved due to his use of multiple integrals, made acomputational error and the expansion (4) is not entirely correct. Many (e.g. Wikipedia, Whittakerand Watson [5, p. 253], [4, p. 30]) have copied Binet’s original formula (4). Our main result is thederivation of the Binet’s corrected expansion in Theorem 1 in Section 3.We will first discuss the main properties of Binet’s function µ ( z ) in Section 2. After this prepara-tion, we present Binet’s corrected expansion in Section 3, give two proofs and two different forms forits coefficients. Finally, Section 4 discusses and compares Stirling’s asymptotic expansion and Binet’sconvergent expansion. µ ( z ) The definition (1) of Binet’s function µ ( z ) directly shows, for a positive integer z = n , that µ ( n ) = n + log ( n − − (cid:18) n − (cid:19) log ( n ) −
12 log (2 π )The sequence µ (1) ≃ . µ (2) ≃ . µ (3) ≃ . µ (4) ≃ . µ (5) ≃ . µ (10) ≃ . µ ( n ) ≈ n . The precise decay is given in(22) below. Maximum at real values of z . Binet’s integral (3) can be rewritten as µ ( z ) = Z ∞ e − zt t (cid:18) e t − − t + 12 (cid:19) dt where 0 ≤ e t − − t + ≤ for real, non-negative t . Since the integrand is positive, we observe that µ ( x ) > x . In addition, for a complex number z = σ + iT , the above integral showsthat | µ ( z ) | ≤ Z ∞ e − σt t (cid:18) e t − − t + 12 (cid:19) dt = µ ( σ )In other words, the maximum absolute value of Binet’s function µ ( z ) for Re ( z ) > Logarithmic behavior around z = 0. Combining the definition (1) of Binet’s function µ ( z ) and thefunction equation Γ ( z + 1) = z Γ ( z ) of the Gamma function, µ ( z ) = log Γ ( z + 1) − (cid:18) z + 12 (cid:19) log z + z −
12 log (2 π )shows that the regime for z → µ ( z ) ∼ −
12 log z −
12 log (2 π ) (5)Thus, µ ( z ) has a logarithmic singularity at z = 0 and lim z → µ ( z ) = ∞ , due to the pole of Γ ( z ) at z = 0. 2 ifference µ ( z + 1) − µ ( z ). The definition (1) of Binet’s function µ ( z ) becomes, with the functionalequation of the Gamma function Γ ( z + 1) = z Γ ( z ), µ ( z + 1) = log Γ ( z ) + log z − (cid:18) z + 12 (cid:19) log ( z + 1) + z + 1 −
12 log (2 π )For any complex z , the forwards difference ∆ µ ( z ) = µ ( z + 1) − µ ( z ) equals µ ( z + 1) − µ ( z ) = (cid:18) z + 12 (cid:19) log zz + 1 + 1 (6) Reflection formula of Binet’s function µ ( z ). The reflection formula of the Gamma functionΓ ( z ) Γ (1 − z ) = π sin πz in the definition (1) of Binet’s function µ ( z ) leads to µ ( − z ) = − (cid:26) log Γ ( z ) − (cid:18) z − (cid:19) log ( z ) + z −
12 log (2 π ) (cid:27) − log sin πz + (cid:18) z + 32 (cid:19) iπ − log (2)After recognizing the definition (1) of µ ( z ) between brackets { . } , we find the corresponding “reflection”formula for Binet’s function µ ( − z ) = − µ ( z ) − log sin πz (cid:18) z + 32 (cid:19) iπ (7)which extends the Re ( z ) > z ) < µ ( z ) for any complex number z . In addition, the Binet reflection formula (7) and its definition(1) illustrates that Binet’s function µ ( z ) has only logarithmic singularities at negative integers z = − k (with k ∈ N ) including z = 0. In passing by, this observation suggests us to consider the exponentiation M ( z ) = e µ ( z ) , which equals, for Re ( z ) > M ( z ) = e log Γ( z ) − ( z − ) log z + z − log(2 π ) = 1 √ π Γ ( z ) z − ( z − ) e z and for Re ( z ) < M ( − z ) = e − µ ( z ) − log sin πz + ( z + ) iπ = 1 M ( z ) 2sin πz e ( z + ) iπ The function M ( z ) is analytic everywhere, except for the simple poles at the negative integers z = − k and an essential singularity at z = 0. Thus, apart from z = 0, M ( z ) has precisely the same poles asthe Gamma function Γ ( z ). Complex integral for Binet’s function µ ( z ). In Appendix B, we deduce the complex integral (intwo ways) µ ( z ) = − πi Z c + i ∞ c − i ∞ π sin πs ζ ( s ) s z s ds with − < c < s ) < ζ ( s ) = 2(2 π ) s − sin πs Γ(1 − s ) ζ (1 − s ) indicates that ζ ( − s ) = O (Γ( s )).However, neglecting this restriction and using ζ ( − k ) = ( − k k +1 B k +1 and the odd Bernoulli numbers B k +1 = 0, for k >
0, leads to Stirling’s asymptotic approximation [1, 6.1.41] in the Poincar´e sense(see e.g. [4]) µ ( z ) ≃ K X k =1 ( − k ζ ( − k ) k z − k = K X k =1 B k +1 k ( k + 1) z k = K X m =1 B m (2 m −
1) (2 m ) z m − (9)3hich is useful and surprisingly accurate up to some finite K ≤ K ∗ ( z ) that depends upon z and isroughly equal to the minimum k -value of | B k +1 | k ( k +1) | z | k , but divergent if K → ∞ . On the other hand,provided | z | <
1, the contour can be closed over the Re ( s ) > s = 0and s = 1 are encountered whose residues are computed in Appendix B, resulting in µ ( z ) = ∞ X k =2 ( − k ζ ( k ) k z k − z (log z − γ ) −
12 log z −
12 log(2 π ) (10)where the Taylor series of log Γ( z + 1) around z = 0,log Γ( z + 1) = − γz + ∞ X k =2 ( − k ζ ( k ) k z k follows directly from Weierstrass’ product1Γ( z + 1) = e γ z ∞ Y n =1 (cid:16) zn (cid:17) e − z/n (11)In contrast to the Stirling’s series P Kk =1 ( − k ζ ( − k ) k z − k for some finite K ≤ K ∗ ( z ) in (9), violation ofthe restriction | z | < Theorem 1
A convergent factorial series of Binet’s function µ ( z ) is µ ( z ) = 1 z ∞ X m =1 c m Q m − k =1 ( z + k ) for Re ( z ) > where the rational coefficients are c m = ( − m − m m X k =1 kS ( k ) m ( k + 2) ( k + 1) (13) and S ( k ) m is the Stirling Number of the First Kind. The first few coefficients c m in (13) are c = , c = 0, c = − , c = − , c = − , c = − , c = − , which are smaller in absolute value than 1, but c m < − m >
8. Moreimportantly, since the coefficients c m < m >
2, any truncation at m = K > µ ( z ).We present two proofs. The first one essentially follows Binet’s original proof in [2, p. 339] andthe second is more a combinatorial proof. Proof (Binet):
Binet [2, p. 339] substitutes e − t = 1 − u or t = − log (1 − u ) in the latterintegral, 2 µ ( z ) = − Z (1 − u ) z − u (cid:18) − u log (1 − u ) + 2 u log (1 − u ) (cid:19) du (14) Likely, this – today less common – substitution was suggested from Beta function theory and Newton’s binomialexpansion. g ( u ) = 2 − u log (1 − u ) + 2 u log (1 − u ) = 2 (cid:26) − u ) + u log (1 − u ) (cid:27) − u log (1 − u )in a Taylor series around u = 0. Instead of following Binet, who has used integrals rather thanStirling numbers S ( k ) m , we invoke the Taylor expansion (25) for n = 2, derived in the Appendix A andconvergent for | u | <
1, 1log (1 − u ) + u log (1 − u ) = − − ∞ X m =1 m X k =1 k ! S ( k ) m ( k + 2)! ! ( − u ) m m !and the Taylor series (23) − u log (1 − u ) = 1 + ∞ X m =1 m X k =1 S ( k ) m k + 1 ! ( − u ) m m !to obtain g ( u ) = ∞ X m =1 m X k =1 kS ( k ) m ( k + 2) ( k + 1) ! ( − u ) m m !Introduced in Binet’s function (14) and reversing summation and integral, justified because a Taylorseries can be term-wise integrated within its radius of convergence,2 µ ( z ) = ∞ X m =1 m X k =1 kS ( k ) m ( k + 2) ( k + 1) ! ( − m − m ! Z (1 − u ) z − u m − du using the Beta integral R u p − (1 − u ) q − du = Γ( p )Γ( q )Γ( p + q ) , valid for Re ( p ) > q ) >
0, yields aconverging series, for Re ( z ) > µ ( z ) = 12 ∞ X m =1 m X k =1 kS ( k ) m ( k + 2) ( k + 1) ! ( − m − m Γ ( z )Γ ( z + m )With Γ( z + m )Γ( z ) = Q m − k =0 ( z + k ), we arrive at (12). (cid:3) Second Proof:
Assuming the existence of a factorial series (12) and introducing that series (12)in the difference formula (6) provides a factorial expansion for the function1 − (cid:18) z + 12 (cid:19) log (cid:18) z (cid:19) = − ∞ X m =1 mc m Q mk =0 ( z + k ) (15)We expand now both sides of (15) into powers of z . Introducing the Taylor series in z of log (cid:0) z (cid:1) = P ∞ k =1 ( − k − k z k in the left-hand side of (15), we have for | z | > − (cid:18) z + 12 (cid:19) log (cid:18) z (cid:19) = 12 ∞ X k =1 ( − k z k +1 (cid:18) k ( k + 1) ( k + 2) (cid:19) Nielsen [3, band I, p. 68] derives 1 Q mk =0 ( z + k ) = ∞ X k = m ( − m − k S ( m ) k z k +1 (16)5here S ( n ) k denotes the Stirling Number of the Second Kind [1, Sec. 24.1.3 and 24.1.4], which we usein the right-hand side of (15) ∞ X m =1 mc m Q mk =0 ( z + k ) = ∞ X m =1 mc m ∞ X k = m ( − m − k S ( m ) k z k +1 = ∞ X k =1 k X m =1 mc m ( − m S ( m ) k ( − k z k +1 Equating corresponding powers in z of both sides in (15) yields k X m =1 mc m ( − m − S ( m ) k = 12 k ( k + 1) ( k + 2)After multiplying both sides by S ( k ) j , summing over k ∈ [1 , j ], we have12 j X k =1 kS ( k ) j ( k + 1) ( k + 2) = j X k =1 k X m =1 mc m ( − m − S ( k ) j S ( m ) k We reverse the k - and m -summation12 j X k =1 kS ( k ) j ( k + 1) ( k + 2) = j X m =1 mc m ( − m − j X k = m S ( k ) j S ( m ) k ! invoke the second orthogonality formula j X k = m S ( k ) j S ( m ) k = δ mj (17)for the Stirling numbers [1, sec. 24.1.4], and we find again (13). (cid:3) For z →
0, Binet’s original expansion (4) is finite. Unfortunately, the logarithmic behavior (5)of µ ( z ) around z = 0 indicates that Binet’s original expansion (4) is erroneous, but the behavior µ ( z ) → ∞ for z → µ ( z ) around z = 0 shows that lim z → zµ ( z ) = 0. Binet’sconvergent series (12), written as zµ ( z ) = ∞ X m =1 c m Q m − k =1 ( z + k ) = 112 − ∞ X m =2 | c m +1 | Q mk =1 ( z + k )illustrates, with lim z → zµ ( z ) = 0 that0 = ∞ X m =1 c m ( m − − ∞ X m =3 | c m | ( m − P Km =1 c m ( m − converges very slowly with increasing K . For positive real z , it holds that P ∞ m =3 | c m | Q m − k =1 ( z + k ) ≤ P ∞ m =3 | c m | ( m − = . Hence, Binet’s convergent series (12) perfectly agrees withthe bound [5, p. 249] log Γ ( z ) ≤ (cid:18) z − (cid:19) log z − z + 12 log (2 π ) + 112 z (18)6 Stirling’s asymptotic and Binet’s convergent series
In this section, we demonstrate the intimate connection between Stirling’s asymptotic (and divergent)series (9) and Binet’s convergent series (12) by applying the same combinatorial method as in thesecond proof of Theorem 1. After substituting Nielsen’s expansion (16) into Binet’s convergent series(12) µ ( z ) = ∞ X m =1 c m Q m − k =0 ( z + k ) = ∞ X m =0 c m +1 ∞ X k = m ( − m − k S ( m ) k z k +1 and after reversal of the summations (and ignoring divergence), we arrive at µ ( z ) = P ∞ k =0 h k z k +1 , withcoefficient h k = k X m =0 ( − m − k S ( m ) k c m +1 Equating corresponding powers of z in the above and in the Stirling’s asymptotic (and divergent)series µ ( z ) = P ∞ k =1 B k +1 k ( k +1) z k in (9) indicates that h k = ( − k k X m =0 ( − m S ( m ) k c m +1 = B k +2 ( k + 1) ( k + 2) (19)Thus, the formula (19) expresses the Bernoulli numbers B n in terms of the Binet coefficients c m . Weinvert this formula (19) to find the coefficients c m . After multiplying both sides by S ( k ) j , summingover k ∈ [0 , j ], we have j X k =0 ( − k S ( k ) j B k +2 ( k + 1) ( k + 2) = j X k =0 k X m =0 ( − m S ( k ) j S ( m ) k c m +1 Reversing the k - and m -sum and applying the second orthogonality formula (17) yields j X k =0 ( − k S ( k ) j B k +2 ( k + 1) ( k + 2) = j X m =0 ( − m j X k = m S ( k ) j S ( m ) k ! c m +1 = j X m =0 ( − m ( δ jm ) c m +1 = ( − j c j +1 from which Theorem 2 follows: Theorem 2
Besides the coefficient c m = ( − m − m P mk =1 kS ( k ) m ( k +2)( k +1) in (13) of the convergent expansion(12) of Binet’s function µ ( z ) = z P ∞ m =1 c m Q m − k =1 ( z + k ) for Re ( z ) > , these coefficients can be expressedin terms of Bernoulli numbers as c m = ( − m − m − X k =0 ( − k S ( k ) m − B k +2 ( k + 1) ( k + 2) = ( − m − m − X k =0 S ( k ) m − B k +2 ( k + 1) ( k + 2) (20)In summary, Stirling’s asymptotic (and divergent) expansion (9) can be compared to Binet’s con-vergent series µ ( z ) = 112 z − z + 11260 z + ∞ X m =4 B m (2 m −
1) (2 m ) z m − (21) µ ( z ) = 112 z − z ( z + 1) ( z + 2) − z ( z + 1) ( z + 2) ( z + 3) + ∞ X m =5 c m Q m − k =0 ( z + k ) (22)7he first two coefficients in Stirling’s asymptotic (9) and Binet’s convergent (12) expansion are thesame. While Stirling’s expansion (9) is an asymptotic and approximate series, Binet’s series (12) isexact and of a same computational complexity. The Bernoulli numbers B m are alternating, whileall c m but the first coefficient c are negative. The ratio (cid:12)(cid:12)(cid:12) c m B m (cid:12)(cid:12)(cid:12) for m > m = 8and from m >
8, that ratio (cid:12)(cid:12)(cid:12) c m B m (cid:12)(cid:12)(cid:12) decreases and tends to zero. Comparison of the error e Binet ( K ) = µ ( z ) − P Km =1 c m Q m − k =0 ( z + k ) and e Stirling ( K ) = (cid:12)(cid:12)(cid:12) µ ( z ) − P Km =1 B m (2 m − m ) z m − (cid:12)(cid:12)(cid:12) , for integer z = n ∈ [0 , e Stirling ( K ) < e Binet ( K ) for K < m n and e Stirling ( K ) > e Binet ( K ) for k > m n , where the integer m n = [6 . n − . x ] is the integerpart of x . Thus, with a same number of terms evaluated, we may conclude that Stirlings’ asymptoticseries (9) is superior for larger, real z , perhaps, because the highest order in z , i.e. O (cid:0) z − K (cid:1) is abouttwice as high as O (cid:0) z − K (cid:1) in (12). Only for relatively small real z or high accuracy evaluations, Binet’sconvergent series (12) is more accurate.The major advantage of Binet’s series (12) over Stirling’s asymptotic (9) lies in its convergence forall Re ( z ) towards µ ( z ), which allows incorporation into integrals and series and may lead to othersharp bounds and approximations. Acknowledgement
I am very grateful to Professor R. B. Paris for pointing to a few misprints inan early version.
References [1] M. Abramowitz and I. A. Stegun.
Handbook of Mathematical Functions . Dover Publications, Inc., New York, 1968.[2] J. P. M. Binet. M´emoire sur les int´egrales d´efinites Eul´eriennes et sur leur application `a la th´eorie des suites ainsiqu‘ `a l`´evaluation des functions des grands nombres.
Journal de l‘ ´Ecole Polytechnique , XVI:123–343, July 1839.[3] N. Nielsen.
Die Gammafunktion: Band I. Handbuch der Theorie der Gammafunktion und Band II. Theorie desIntegrallogarithmus und verwandter Transzendenten . B. G. Teubner, Leipzig 1906; republished by Chelsea, NewYork, 1956.[4] R. B. Paris and D. Kaminski.
Asymptotics and Mellin-Barnes Integrals , volume 85 of
Encyclopedia of Mathematicsand its Applications . Cambridge University Press, U. K., 2001.[5] E. T. Whittaker and G. N. Watson.
A Course of Modern Analysis . Cambridge University Press, Cambridge, UK,cambridge mathematical library edition, 1996.
A Taylor series of x log n (1+ x ) for integer n Integrating the double generating function(1 + x ) u = e u log(1+ x ) = ∞ X m =0 m X k =0 S ( k ) m m ! u k x m of the Stirling numbers S ( k ) m of the First Kind [1, Sec. 24.1.3 and 24.1.4] with respect to u results, for | x | <
1, in e b log(1+ x ) − e a log(1+ x ) log (1 + x ) = ∞ X m =0 m X k =0 S ( k ) m b k +1 − a k +1 k + 1 x m m !8n particular, for b = 1 and a = 0, we obtain the Taylor series, valid for | x | < x log (1 + x ) = ∞ X m =0 m X k =0 S ( k ) m k + 1 ! x m m ! = 1 + ∞ X m =1 m X k =1 S ( k ) m k + 1 ! x m m ! (23)We generalize the above. The n -fold integral of e uλ equals Z ba du Z u a du . . . Z u n − a du n e u n λ = 1( n − Z ba ( b − u ) n − e uλ du Let t = b − u , followed by y = λt , then Z ba ( b − u ) n − e uλ du = e λb λ n Z λ ( b − a )0 y n − e − y dy and the integral can be executed leading to1( n − Z ba ( b − u ) n − e uλ du = e λb λ n − e − λ ( b − a ) m X k =0 ( λ ( b − a )) k k ! ! On the other hand, e uλ = P ∞ k =0 λ k u k k ! and the n -fold integration of u k is Z ba du Z u a du . . . Z u n − a du n u kn = 1( n − Z ba ( b − u ) n − u k du (24)and Z ba ( b − u ) n − u k du = b n Z ba (cid:16) − ub (cid:17) n − u k du = b n + k Z ab (1 − w ) n − w k dw which simplifies considerably if a = 0, due to the Beta integral R (1 − w ) n − w k dw = Γ( n )Γ( k +1)Γ( n + k +1) .Thus, choosing a = 0 leads to 1 λ n e λb − m X k =0 ( λb ) k k ! ! = b n ∞ X k =0 λ k b k ( n + k )!Further, with λ = log (1 + x ) and introducing the Taylor series λ k = log k (1 + x ) = k ! P ∞ m = k S ( k ) m x m m ! ,valid for x | <
1, yields, after reversal of the k - and m -sum,1log n (1 + x ) e log(1+ x ) b − n − X k =0 (log (1 + x ) b ) k k ! ! = b n ∞ X m =0 m X k =0 k ! S ( k ) m b k ( k + n )! ! x m m !valid for | x | < b = 1 to, x log n (1 + x ) = 1 n ! + n − X k =1 k ! log n − k (1 + x ) + ∞ X m =1 m X k =1 k ! S ( k ) m ( k + n )! ! x m m ! (25)For n = 1 in (25), we find again (23).Applying n -fold integration to the generating function of the Stirling numbers S ( k ) m of the FirstKind, Q m − k =0 ( x − k ) = P mk =0 S ( k ) m x k , yields Z b du Z u du . . . Z u n − du n m − Y k =0 ( u n − k ) = m X k =0 S ( k ) m k !( n + k )! b k + n n − Z b ( b − u ) n − m − Y k =0 ( u − k ) du = m X k =0 S ( k ) m k !( n + k )! b k + n which links Stirling numbers to the general integral form used by Binet [2, p. 339] in the seriesexpansion of his Binet function µ ( z ). B Complex integral for Binet’s function µ ( z ) From Weierstrass’s product (11) of the Gamma function, Wittacker and Watson [5, p. 277] deducethe formula, valid for all a ,log Γ ( a )Γ ( z + a ) = − z Γ ′ ( a )Γ ( a ) + 12 πi Z q + i ∞ q − i ∞ π sin πs ζ ( s, a ) s z s ds with 1 ≤ q ≤ ζ ( s, a ) = P ∞ n =0 1( n + a ) s is the Hurwitz Zeta-function, which reduces for a = 1 to the RiemannZeta-function ζ ( s ). Thus, for a = 1 and Γ ′ (1)Γ(1) = − γ , where γ is the Euler constant, we havelog Γ ( z + 1) = − γz − πi Z q + i ∞ q − i ∞ π sin πs ζ ( s ) s z s ds with 1 ≤ q ≤ < c = Re ( s ) <
1, we encounter a double pole at s = 1,because ζ ( s ) = s − + γ + O ( s −
1) around s = 1 and a zero of sin πs . The residue at s = 1 followsfrom Cauchy’s integral theorem k ! d k f ( z ) dz k (cid:12)(cid:12)(cid:12) z = z = πi R C ( z ) f ( ω ) dω ( ω − z ) k +1 , where f ( z ) is analytic withinthe contour C ( z ) that encloses the point z and we obtain12 πi Z q + i ∞ q − i ∞ π sin πs ζ ( s ) s z s ds = 12 πi Z c + i ∞ c − i ∞ π sin πs ζ ( s ) s z s ds + lim s → dds z s s πζ ( s ) ( s − sin πs ! because the function between brackets is analytic at s = 1. Executing the derivative, dds z s s πζ ( s ) ( s − sin πs ! = z s π ( s − s sin πs (cid:18)(cid:18) log z − s (cid:19) ζ ( s ) ( s −
1) + ζ ′ ( s ) ( s −
1) + 2 ζ ( s ) − πζ ( s ) ( s −
1) cos πs sin πs (cid:19) and using the Taylor expansions of ( s − ζ ( s ) around s = 1 gives uslim s → dds z s s πζ ( s ) ( s − sin πs ! = − z (log z − γ )and we obtain, for 0 < c < z + 1) = z log z − z − πi Z c + i ∞ c − i ∞ π sin πs ζ ( s ) s z s ds Moving the line of integration over the double pole at s = 0 to the left yields, for − < c ′ < z + 1) = z log z − z − πi Z c ′ + i ∞ c ′ − i ∞ π sin πs ζ ( s ) s z s ds − lim s → dds (cid:18) πsζ ( s )sin πs z s (cid:19) dds (cid:18) sζ ( s )sin πs z s (cid:19) = z s ζ ( s ) s sin πs (cid:26) s − π cos πs sin πs + log z + ζ ′ ( s ) ζ ( s ) (cid:27) Since π cot ( πx ) = x − P ∞ n =1 ζ (2 n ) x n − , we find that lim s → s − π cot ( πs ) = 0 andlim s → dsds (cid:18) πsζ ( s )sin πs z s (cid:19) = ζ (0) (cid:26) log z + ζ ′ (0) ζ (0) (cid:27) With ζ (0) = − and ζ ′ (0) = − log(2 π ), we arrive at lim s → dsds (cid:16) πsζ ( s )sin πs z s (cid:17) = − { log z + log(2 π ) } andlog Γ ( z + 1) = (cid:18) z + 12 (cid:19) log z − z + 12 log(2 π ) − πi Z c ′ + i ∞ c ′ − i ∞ π sin πs ζ ( s ) s z s ds with − ≤ c ′ ≤ z ) = (cid:0) z − (cid:1) log z − z + log (2 π ) + µ ( z ), we find (8).We present a second, shorter derivation of (8) by employing the inverse Mellin transform1 e πt − πi Z c + i ∞ c − i ∞ Γ ( s ) ζ ( s ) (2 πt ) − s ds for c > µ ( z ) = 2 Z ∞ arctan (cid:0) tz (cid:1) e πt − dt = 2 12 πi Z c + i ∞ c − i ∞ Γ ( s ) ζ ( s ) (2 π ) − s (cid:18)Z ∞ arctan (cid:18) tz (cid:19) t − s dt (cid:19) ds Partial integration, followed by a substitution u = (cid:0) tz (cid:1) and the use of the Beta integral and theGamma reflection formula results in − Z ∞ arctan (cid:18) tz (cid:19) t − s dt = z − s π − s ) sin πs and µ ( z ) = − z i Z c + i ∞ c − i ∞ ζ ( s ) Γ ( s ) (2 πz ) − s (1 − s ) sin πs ds Using the functional equation ζ ( s ) = 2(2 π ) s − sin πs Γ(1 − s ) ζ (1 − s ) yields µ ( z ) = − πi Z c + i ∞ c − i ∞ π sin πs ζ (1 − s )(1 − s ) z − s ds with 1 < c < w = 1 − ss