A new recurrence relation for the Li/Keiper constants in terms of - arXiv

A recurrence relation for the Li/Keiper constants in terms of the Stieltjes constants Donal F. Connon [email protected] 10 February 2009 Abstract A recurrence relation for the Li/Keiper constants in terms of the Stieltjes constants is derived in this paper. In addition, we also report a formula for the Stieltjes constants in terms of the higher derivatives, ς ( n ) (0) , of the Riemann zeta function evaluated at s = 0 . A formula for the Stieltjes constants in terms of the (exponential) complete Bell polynomials containing the eta constants ηn as the arguments is also derived. CONTENTS

Page

1. Introduction

1

2. A brief survey of the Stieltjes constants

3

3. A recurrence relation for the Li/Keiper constants

6

4. The eta constants

27

5. A formula for the Stieltjes constants in terms of the higher derivatives of the Riemann zeta function ς ( n ) (0)

34

6. A formula for the Stieltjes constants in terms of the (exponential) complete Bell polynomials containing the eta constants η n as the arguments

39

Appendix A: A brief survey of the (exponential) complete Bell polynomials

42

1. Introduction The Riemann xi function ξ ( s) is defined as (1.1)

1 2

ξ ( s) = s ( s − 1) π

−s

2

Γ( s / 2)ς ( s )

and we see from the functional equation for the Riemann zeta function ς ( s ) (1.2)

ς (1 − s) = 2(2π ) − s Γ( s) cos(π s / 2)ς ( s)

that ξ ( s ) satisfies the functional equation (1.3)

ξ ( s ) = ξ (1 − s)

In 1996, Li [24] defined the sequence of numbers (λ n ) by (1.4)

1 d n n −1 [ s log ξ ( s )] λn = (n − 1)! ds n s =1

and proved that a necessary and sufficient condition for the non-trivial zeros ρ of the 1 Riemann zeta function to lie on the critical line s = + iτ is that λn is non-negative for 2 every positive integer n . Earlier in 1991, Keiper [21] showed that if the Riemann hypothesis is true, then λn > 0 for all n ≥ 1 . Li also showed that (1.5)

⎡ ⎛ 1 ⎞n ⎤ λn = ∑ ⎢1 − ⎜1 − ⎟ ⎥ ρ ⎢ ⎣ ⎝ ρ ⎠ ⎦⎥

Taking logarithms of (1.1) and noting that

(1.6)

s ⎛s⎞ ⎛ s⎞ Γ ⎜ ⎟ = Γ ⎜1 + ⎟ we see that 2 ⎝2⎠ ⎝ 2⎠

⎛ s⎞ s log ξ ( s ) = log Γ ⎜1 + ⎟ − log π + log[( s − 1)ς ( s )] ⎝ 2⎠ 2

We also have the Maclaurin expansion about s = 1 (1.7)

∞

σk

k =1

k

log ξ ( s ) = − log 2 − ∑ (−1) k

( s − 1) k

where the constant term in (1.7) arises because 1 −1 1 lim ξ ( s ) = π 2 Γ(1/ 2) lim[( s − 1)ς ( s )] = s →1 s →1 2 2 We note that the coefficients σ k are defined by

2

σk =

(−1) k +1 d k log ξ ( s ) (k − 1)! ds k s =1

and we see that ∞ d ξ ′( s) log ξ ( s ) = = −∑ (−1) kσ k ( s − 1) k −1 ξ ( s) ds k =1

and comparing this with (1.4) we immediately see that λ1 = σ 1 . Differentiating (1.6) results in

(1.8)

d [( s − 1)ς ( s )] d 1 ⎛ s⎞ 1 log ξ ( s ) = ψ ⎜1 + ⎟ − log π + ds ds 2 ⎝ 2⎠ 2 ( s − 1)ς ( s )

which, referring to (1.4), is clearly relevant in computing λ1 . Before proceeding any further we need to recall some details regarding the Stieltjes constants. 2. A brief survey of the Stieltjes constants

The Stieltjes constants γ n (u ) are the coefficients in the Laurent expansion of the Hurwitz zeta function ς ( s, u ) about s = 1 (2.1)

∞ 1 1 (−1) n = + γ n (u )( s − 1) n ∑ s s − 1 n =0 n ! n =0 ( n + u ) ∞

ς ( s, u ) = ∑

and γ 0 (u ) = −ψ (u ) , where ψ (u ) is the digamma function which is the logarithmic d derivative of the gamma function ψ (u ) = log Γ(u ) . It is easily seen from the definition du of the Hurwitz zeta function that ς ( s,1) = ς ( s ) and accordingly that γ n (1) = γ n . Further details of the Stieltjes constants are contained in [5] to [13] inclusive.

The generalised Euler-Mascheroni constants γ n (or Stieltjes constants) are the coefficients of the Laurent expansion of the Riemann zeta function ς ( s ) about s = 1 (2.2)

∞ 1 (−1) n +∑ ς (s) = γ n ( s − 1) n s − 1 n =0 n !

3

1 ⎤ ⎡ Since lim ⎢ς ( s ) − = γ it is clear that γ 0 = γ . It may be shown, as in [20, p.4], that s →1 s − 1 ⎦⎥ ⎣ (2.3)

⎡ N log n k N log n t ⎤ log n k log n +1 N ⎤ lim dt ⎥ − = −∫ ⎢∑ ⎥ n + 1 ⎦ N →∞ ⎣ k =1 k t ⎣ k =1 k 1 ⎦ ⎡

N

γ n = lim ⎢ ∑ N →∞

where, throughout this paper, we define log 0 1 = 1 . We now refer to the Hasse identity [19] for the Hurwitz zeta function ς ( s, u ) which is valid for all s ∈ C provided Re ( s) ≠ 1 (2.4)

1 ∞ 1 i ⎛ i ⎞ (−1) j ς ( s, u ) = ∑ ∑⎜ ⎟ s − 1 i =0 i + 1 j =0 ⎝ j ⎠ (u + j ) s −1

and we have the well-known limit ∞

(2.5)

lim( s − 1)ς ( s, u ) = ∑ s →1

i =0

∞ 1 i ⎛i⎞ 1 j δ i ,0 = 1 ( 1) − = ∑ ∑ ⎜ ⎟ i + 1 j =0 ⎝ j ⎠ i =0 i + 1

where δ i , j is the Kronecker delta symbol. We see from (2.4) that (2.6)

∞ 1 i ⎛ i ⎞ (−1) j log n +1 (u + j ) d n +1 n +1 [( s − 1)ς ( s, u )] = (−1) ∑ ∑ ⎜ ⎟ (u + j )s −1 ds n +1 i =0 i + 1 j =0 ⎝ j ⎠

and thus (2.7)

∞ d n +1 1 i ⎛i⎞ n +1 n +1 j s s u − = − [( 1) ( , )] ( 1) ς ∑ ∑ ⎜ ⎟(−1) log (u + j ) j ds n +1 i + 1 i =0 j =0 ⎝ ⎠ s =1

We previously showed in [15] that (2.8)

γ n (u ) = −

1 ∞ 1 i ⎛i⎞ j n +1 ∑ ∑ ⎜ ⎟(−1) log (u + j ) n + 1 i =0 i + 1 j =0 ⎝ j ⎠

and comparing this with (2.7) allows us to conclude for n ≥ 0 that (2.9)

d n +1 [( s − 1)ς ( s, u )] = (−1) n (n + 1)γ n (u ) n +1 ds s =1

4

This may also be more directly obtained by differentiating (2.1) p + 1 times where we see that ∞ d p +1 (−1) n [( s − 1) ( s , u )] = ς γ n (u )(n + 1)n(n + 1) ⋅⋅⋅ ( n + 1 − p)( s − 1) n − p ∑ p +1 ds n! n =0 With u = 1 in (2.9) we have d n +1 [( s − 1)ς ( s )] = (−1) n (n + 1)γ n n +1 ds s =1

(2.10)

We now evaluate (1.8) at s = 1 and obtain

λ1 =

d 1 ⎛3⎞ 1 d [log ξ ( s )] = ψ ⎜ ⎟ − log π + [( s − 1)ς ( s )] ds 2 ⎝2⎠ 2 ds s =1 s =1

Using [30, p.20] ⎛ ⎝

1⎞

n −1

1

ψ ⎜ n + ⎟ = −γ − 2 log 2 + 2∑ 2 2k + 1 ⎠

k =0

we see that ⎛3⎞ ⎝ ⎠

ψ ⎜ ⎟ = −γ − 2 log 2 + 2 2

(2.11)

From (2.10) we have d [( s − 1)ς ( s )] = γ 0 = γ ds s =1

(2.12)

and we therefore easily compute the first Li/Keiper constant (2.13)

λ1 =

d 1 ⎛3⎞ 1 1 1 [log ξ ( s )] = ψ ⎜ ⎟ + γ − log π = − log π + γ + 1 − log 2 ds 2 ⎝2⎠ 2 2 2 s =1

1 ⎛3⎞ We note from (2.11) that ψ ⎜ ⎟ is positive; in addition we see that γ − log π > 0 and 2 ⎝2⎠ therefore λ1 is positive. Its approximate value is [5] (2.14)

σ 1 = λ1  0.023...

5

d n n −1 [ s log ξ ( s)] and ds n reference to (1.6) shows that this in turn would require us to compute

However, in order to determine λ n we would need to calculate

d ⎡ ⎤ [( s − 1)ς ( s )] ⎥ d n n −1 d n −1 ⎢ [ s log[( s − 1)ς ( s )] = n −1 ⎢ (n − 1) s n − 2 log[( s − 1)ς ( s ) + s n −1 ds ⎥ ds n ds ⎢ ( s − 1)ς ( s ) ⎥ ⎣ ⎦

It is remarkable that determining the veracity, or otherwise, of the most famous hypothesis in mathematics is so dependent upon obtaining a “manageable” formula for the higher derivatives of the quotient of two functions. Formulae for the Li/Keiper constants have been given by Bombieri and Lagarias [4], Maślanka [26] and Coffey [13], the latter being based on an identity determined by Matsuoka [27]. However, it is not easy to discern particular values of λ n from these representations. It is because of this inherent difficulty that we now proceed to go for second best and derive a recurrence relationship for the λ n constants. 3. A recurrence relation for the Li/Keiper constants

We recall (1.7) above ∞

σk

k =1

k

log ξ ( s ) = − log 2 − ∑ (−1) k

( s − 1) k

and since log ξ ( s) = log ξ (1 − s ) we see that ∞

σk

k =1

k

log ξ ( s ) = − log 2 − ∑

sk

We therefore have from (1.6) (3.1)

∞ σ s ⎛ s⎞ log Γ ⎜1 + ⎟ + log 2 − log π + log[( s − 1)ς ( s)] = −∑ k s k 2 ⎝ 2⎠ k =1 k

and differentiation gives us (3.2)

d [( s − 1)ς ( s )] ∞ σ 1 ⎛ s⎞ 1 ψ ⎜1 + ⎟ − log π + ds = −∑ k ks k −1 2 ⎝ 2⎠ 2 ( s − 1)ς ( s ) k =1 k

6

We now multiply (3.2) across by ( s − 1)ς ( s ) to obtain (3.3) ∞ σ 1 d ⎛ s⎞ 1 ( s − 1)ς ( s )ψ ⎜1 + ⎟ − log π ( s − 1)ς ( s ) + [( s − 1)ς ( s )] = −∑ k ks k −1[( s − 1)ς ( s )] 2 ds ⎝ 2⎠ 2 k =1 k

and with s = 1 we have ∞ σ 1 ⎛3⎞ 1 ψ ⎜ ⎟ − log π + γ = −∑ k k 2 ⎝2⎠ 2 k =1 k

Using ∞

log ξ ( s ) = − log 2 − ∑ k =1

σk k

sk

we see that ∞ σk d n n −1 ξ s s [ log ( )] = − (k + n − 1) ⋅⋅⋅ k s k + n−1 ∑ n ds k =1 k

and hence referring to (1.4) we have ∞ σk 1 d n n −1 1 λn = ξ [ s log ( s )] (k + n − 1) ⋅⋅⋅ k = − ∑ n (n − 1)! ds (n − 1)! k =1 k s =1

We then see that for n ≥ 1 (3.4)

∞

σk

k =1

k

(n − 1)!λ n = −∑

k ⋅⋅⋅ (k + n − 1)

and for example we have

λ1 = −

∞ 1 ∞ σk k = − σ k = σ1 ∑ ∑ 0! k =1 k k =1

λ2 = −

1 ∞ σk ∑ k (k + 1) 1! k =1 k

λ3 = −

1 ∞ σk ∑ k (k + 1)(k + 2) 2! k =1 k

7

We then see that 1 ⎛3⎞ 1 ψ ⎜ ⎟ + γ − log π = λ1 2 ⎝2⎠ 2 which is the same as (2.11). Differentiating equation (3.3) gives us ( s − 1)ς ( s )

d d2 1 (1) ⎛ s ⎞ 1 ⎛ s ⎞ d 1 + + + − − − + 1 1 [( 1) ( )] log [( 1) ( )] [( s − 1)ς ( s )] ψ ψ s ς s π s ς s ⎜ ⎟ ⎜ ⎟ 22 2 ds ds 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠ ds ∞

σk

k =1

k

= −∑

∞ σ d [( s − 1)ς ( s )] − ∑ k k (k − 1) s k − 2 [( s − 1)ς ( s)] ds k =1 k

ks k −1

and with s = 1 we have ∞ ∞ σ 1 (1) ⎛ 3 ⎞ 1 ⎛ 3 ⎞ 1 ψ ⎜ ⎟ + ψ ⎜ ⎟ γ − γ log π − 2γ 1 = −γ ∑ σ k − ∑ k k (k − 1) 4 2 ⎝2⎠ 2 ⎝2⎠ k =1 k =1 k

We have k (k − 1) = k (k + 1) − 2k and hence ∞

∞

σk

k =1

k =1

k

−γ ∑ σ k − ∑

∞

σk

k =1

k

k (k − 1) = −γ ∑

∞

σk

k =1

k

k −∑

∞

σk

k =1

k

k (k + 1) + 2∑

and, therefore, using (3.4), we have (3.5)

1 (1) ⎛ 3 ⎞ 1 ⎛3⎞ 1 ψ ⎜ ⎟ + γ ψ ⎜ ⎟ − γ log π − 2γ 1 = (γ − 2)λ1 + λ 2 4 ⎝2⎠ 2 ⎝2⎠ 2

Since ψ (1 + x ) = ψ ( x ) +

1 we see that x

ψ ( n ) (1 + x ) = ψ ( n ) ( x ) + (−1) n n ! x − n −1 and with x = 1/ 2 we have

ψ ( n ) ( 3 / 2 ) = ψ ( n ) (1/ 2 ) + (−1) n n !2n +1 We have the well-known result from [30]

ψ ( n ) (1/ 2 ) = (−1) n +1 n ![2n +1 − 1]ς (n + 1) 8

k

and hence we have

(

ψ ( n ) ( 3 / 2 ) = (−1) n +1 n ! ⎡⎣ 2n +1 − 1⎤⎦ ς (n + 1) − 2n +1

(3.5.1)

)

In fact we have from [30, p.22]

ψ ( n ) ( x ) = (−1) n +1 n !ς (n + 1, x)

(3.5.2)

and hence we deduce that for n ≥ 1

ψ ( n ) ( 3 / 2 ) = (−1) n +1 cn where cn > 0

(3.5.3)

We thus obtain ⎛3⎞ ⎝ ⎠

ψ (1) ⎜ ⎟ = 3ς (2) − 4 2 We have from (2.11) ⎛3⎞ ⎝ ⎠

ψ ⎜ ⎟ = −γ − 2 log 2 + 2 2 and from (2.13) 1 2

1 2

λ1 = − log π + γ + 1 − log 2 Therefore we have the second Li/Keiper constant 3 4

λ 2 = ς (2) + 1 + γ − γ 2 − 2 log 2 − log π − 2γ 1

(3.6)

Differentiating equation (3.3) n + 1 times using the Leibniz rule gives us ⎛ n + 1⎞ d m 1 s⎞ 1 d n +1 ( n +1− m ) ⎛ ς ψ π [( s − 1) ( s )] 1 + − log [( s − 1)ς ( s )] ∑⎜ ⎟ m ⎜ ⎟ 2n + 2− m ds n +1 ⎝ 2⎠ 2 m = 0 ⎝ m ⎠ ds n +1

(3.7) +

∞ σk d n+2 ς [( s 1) ( s )] − = − ∑ n+2 ds k =1 k

n +1− m ⎛ n + 1⎞ d m k −1 d [ ks ] [( s − 1)ς ( s )] ∑⎜ ⎟ m ds n +1− m m = 0 ⎝ m ⎠ ds n +1

9

First of all, isolating the term for m = 0 and evaluating at s = 1 we have since lim[( s − 1)ς ( s )] = 1 s →1

⎛ n + 1⎞ d m 1 s⎞ ( n +1− m ) ⎛ S1 = ∑ ⎜ 1+ ⎟ ⎟ m [( s − 1)ς ( s)] n + 2− m ψ ⎜ 2 ⎝ 2 ⎠ s =1 m = 0 ⎝ m ⎠ ds n +1

=

1 2n + 2

ψ

( n +1)

m s⎞ 1 ⎛ s ⎞ n +1 ⎛ n + 1⎞ d ( n +1− m ) ⎛ ⎜1 + ⎟ + ∑ ⎜ m ⎟ m [( s − 1)ς ( s )] n + 2− m ψ ⎜1 + ⎟ 2 ⎝ 2 ⎠ m =1 ⎝ ⎝ 2 ⎠ s =1 ⎠ ds

and then rebasing the index to m = p + 1 results in ⎛ ⎝

1

s⎞

⎛ n + 1 ⎞ d p +1

n

⎛ ⎝

1

s⎞

(n− p ) ψ ( n +1) ⎜1 + ⎟ + ∑ ⎜ ⎟ p +1 [( s − 1)ς ( s )] n +1− p ψ ⎜1 + ⎟ n+2 p + 1 ds 2 2 2 2

S1 =

⎠

p =0

⎝

⎠

⎠ s =1

Noting from (2.9) that d p +1 [( s − 1)ς ( s )] = (−1) p ( p + 1)γ p p +1 ds s =1

we obtain S1 =

⎛3⎞

1

n

⎛ n +1⎞

1

⎛3⎞

p (n− p ) ψ ( n +1) ⎜ ⎟ + ∑ ⎜ ⎟(−1) ( p + 1)γ p n +1− p ψ ⎜ ⎟ p + 1 2n + 2 2 2 ⎝ ⎠ p =0 ⎝ ⎝2⎠ ⎠

Carrying out similar operations for the summation on the right-hand side of (3.7), where this time we isolate the term for m = n + 1 , we get ∞

S2 = − ∑ k =1

∞

= −∑ k =1

∞

σk

k =1

k

∞

σk

k =1

k

= −∑ = −∑

n +1− m ⎛ n + 1⎞ d m k −1 d [ ] [( s − 1)ς ( s)] ks ∑⎜ ⎟ k m =0 ⎝ m ⎠ ds m ds n +1− m s =1

σk

n +1

σ k d n +1 k ds

∞

[ks k −1 ] − ∑ n +1 k =1

∞

k (k − 1)...(k − n − 1) − ∑ k =1

∞

k (k − 1)...(k − n − 1) − ∑ k =1

n +1− m ⎛ n + 1⎞ d m k −1 d [ ] [( s − 1)ς ( s)] ks ∑⎜ ⎟ k m =0 ⎝ m ⎠ ds m ds n +1− m s =1

σk

n

⎛ n + 1⎞ d n +1− m ( 1)...( ) [( s − 1)ς ( s)] k k − k − m ∑⎜ ⎟ k m=0 ⎝ m ⎠ ds n +1− m s =1

σk

n

⎛ n + 1⎞ n−m ⎟ k (k − 1)...(k − m)(−1) (n − m + 1)γ n − m k m=0 ⎝ m ⎠

σk

n

∑⎜

10

Therefore, the evaluation of (3.7) at s = 1 results in ⎛3⎞

1

⎛ n +1 ⎞

n

1

⎛3⎞

m ( n−m ) ψ ( n +1) ⎜ ⎟ + ∑ ⎜ ⎟(−1) (m + 1)γ m n +1− m ψ ⎜ ⎟ n+2 2 2 ⎝ 2 ⎠ m=0 ⎝ m + 1⎠ ⎝2⎠

1 − (−1) n (n + 1)γ n log π + (−1) n +1 (n + 2)γ n +1 2 ∞

σk

k =1

k

= −∑

∞

k (k − 1)...(k − n − 1) − ∑ k =1

⎛ n + 1⎞ n−m ⎟ k (k − 1)...(k − m)(−1) (n − m + 1)γ n − m k m=0 ⎝ m ⎠

σk

n

∑⎜

which may be written as (3.8)

⎛3⎞

1

n

⎛ n +1 ⎞

1

⎛3⎞

( n−m ) m ψ ( n +1) ⎜ ⎟ + ∑ ⎜ ⎟(−1) (m + 1)γ m n +1− m ψ ⎜ ⎟ n+2 2 2 ⎝ 2 ⎠ m=0 ⎝ m + 1⎠ ⎝2⎠

1 + (−1) n +1 (n + 1)γ n log π + (−1) n +1 (n + 2)γ n +1 2 ∞

= −∑ k =1

n ∞ ⎛ n + 1⎞ σk n−m k (k − 1)...(k − n − 1) − ∑ ⎜ ( 1) ( n m 1) k (k − 1)...(k − m) γ − − + n−m ∑ ⎟ k m=0 ⎝ m ⎠ k =1 k

σk

The right-hand side of this equation clearly requires a modicum of simplification which we now endeavour to provide. The rising factorial [ x] p is defined by [ x] p = x( x + 1)...( x + p − 1) and it is shown in [2, p.36] that p

[ x] p = ∑ j =1

p ! ⎛ p − 1⎞ ⎜ ⎟ [ x] j j! ⎝ j −1 ⎠

where the falling factorial [ x] j is defined by j

[ x] j = x( x − 1)...( x − j + 1) = ∑ s ( j , i ) x i i =0

and s ( j , i) are known as the Stirling numbers of the first kind. We therefore have

11

p

x( x + 1)...( x + p − 1) = ∑ j =1

p ! ⎛ p − 1⎞ ⎜ ⎟ x( x − 1)...( x − j + 1) j! ⎝ j −1 ⎠

Letting x → − x we see that p

x( x − 1)...( x − p + 1) = (−1) p ∑ j =1

p ! ⎛ p − 1⎞ j ⎜ ⎟ (−1) x( x + 1)...( x + j − 1) j j! ⎝ −1 ⎠

and with x = k we have k (k − 1)...(k − p + 1) = (−1)

(3.9)

p

p

∑ j =1

p ! ⎛ p − 1⎞ j ⎜ ⎟ (−1) k (k + 1)...(k + j − 1) j! ⎝ j −1 ⎠

We then have for the first term in the right-hand side of (3.8) ∞

σk

k =1

k

S3 = −∑

∞ σk (n + 2)! ⎛ n + 1⎞ j k (k + 1)...(k + j − 1) ( − 1) ∑ ⎜ ⎟ j ! ⎝ j − 1⎠ j =1 k =1 k

n+2

k (k − 1)...(k − n − 1) = −(−1)n ∑

Using (3.4) for n ≥ 1 ∞

σk

k =1

k

(n − 1)!λ n = −∑

k ⋅⋅⋅ (k + n − 1)

we may then express S3 in terms of the Li/Keiper constants as (3.10)

n+2 (n + 2)! ⎛ n + 1⎞ (n + 2)! ⎛ n + 1⎞ j n j j λ ( − 1) ( − 1)! = ( − 1) ∑ j ⎜ ⎟ ⎜ ⎟ (−1) λ j j ! ⎝ j − 1⎠ j ⎝ j − 1⎠ j =1 j =1

n+2

S3 = (−1) n ∑

We now consider the second term in the right-hand side of (3.8) n ∞ ⎛ n + 1⎞ σk n−m S4 = −∑ ⎜ ⎟(−1) (n − m + 1)γ n − m ∑ k (k − 1)...(k − m) m=0 ⎝ m ⎠ k =1 k

which on using (3.9) becomes n m +1 ∞ ⎛ n + 1⎞ σk (m + 1)! ⎛ m ⎞ n−m m +1 j n m k (k + 1)...(k + j − 1) γ = −∑ ⎜ ( − 1) ( − + 1) ( − 1) ( − 1) ∑ ∑ n−m ⎟ ⎜ ⎟ j ! ⎝ j − 1⎠ m=0 ⎝ m ⎠ j =1 k =1 k n m +1 ⎛ n + 1⎞ (m + 1)! ⎛ m ⎞ n−m m +1 j = ∑⎜ ( − 1) ( n − m + 1) ( − 1) γ ∑ ⎟ ⎜ ⎟ ( −1) ( j − 1)!λ j n−m j ! ⎝ j − 1⎠ m=0 ⎝ m ⎠ j =1

12

(3.11) n m +1 ⎛ n + 1⎞ (m + 1)! ⎛ m ⎞ n−m m +1 j = ∑⎜ ⎟(−1) (n − m + 1)γ n − m (−1) ∑ ⎜ ⎟ (−1) λ j j ⎝ j − 1⎠ m=0 ⎝ m ⎠ j =1

Using (3.10) and (3.11) we may then write (3.8) as (3.12)

⎛3⎞

1

n

⎛ n +1 ⎞

⎛3⎞

1

( n−m ) m ψ ( n +1) ⎜ ⎟ + ∑ ⎜ ⎟(−1) (m + 1)γ m n +1− m ψ ⎜ ⎟ m + 1 2n + 2 2 2 ⎝ ⎠ m=0 ⎝ ⎝2⎠ ⎠

1 + (−1) n +1 (n + 1)γ n log π + (−1) n +1 (n + 2)γ n +1 2 = (−1)

(n + 2)! ⎛ n + 1⎞ j ⎜ ⎟ (−1) λ j j ⎝ j − 1⎠ j =1

n+2

n

∑

n m +1 ⎛ n + 1⎞ (m + 1)! ⎛ m ⎞ n−m m +1 j n m γ +∑ ⎜ ( − 1) ( − + 1) ( − 1) ∑ n−m ⎟ ⎜ ⎟ (−1) λ j m j − 1 j m =0 ⎝ j =1 ⎠ ⎝ ⎠

After some algebra and using the following elementary binomial identities ⎛ n +1 ⎞ ⎛n⎞ ⎜ ⎟ (m + 1) = (n + 1) ⎜ ⎟ ⎝ m + 1⎠ ⎝m⎠

⎛ n + 1⎞ ⎛ n ⎞ ⎜ ⎟ (n − m + 1) = (m + 1) ⎜ ⎟ ⎝ m ⎠ ⎝ m + 1⎠

1 ⎛ n + 1⎞ 1 ⎛ n + 2⎞ ⎜ ⎟= ⎜ ⎟ j ⎝ j − 1⎠ n + 2 ⎝ j ⎠

1⎛ m ⎞ 1 ⎛ m + 1⎞ ⎜ ⎟= ⎜ ⎟ j ⎝ j − 1⎠ m + 1 ⎝ j ⎠

we have the required recurrence relation (3.13)

1

⎛ 3 ⎞ (n + 1)

ψ ( n +1) ⎜ ⎟ + n +1 n+2 2 ⎝2⎠ 2

n

⎛n⎞

∑ ⎜ m ⎟(−1)

m =0

⎝ ⎠

m

⎛3⎞ ⎝ ⎠

γ m 2mψ ( n − m ) ⎜ ⎟ 2

1 + (−1) n +1 (n + 1)γ n log π + (−1) n +1 (n + 2)γ n +1 2 n m +1 m + 1 ⎛ n + 2⎞ ⎛ n ⎞ ⎛ ⎞ j n +1 j ⎟ (−1) λ j + (−1) ∑ ⎜ ⎟(m + 1)!γ n − m ∑ ⎜ ⎟ (−1) λ j j ⎠ j ⎠ j =1 ⎝ m = 0 ⎝ m + 1⎠ j =1 ⎝

n+ 2

= (−1)n (n + 1)!∑ ⎜

For example with n = 0 we get

13

(3.14)

1 (1) ⎛ 3 ⎞ 1 ⎛3⎞ 1 ψ ⎜ ⎟ + γ ψ ⎜ ⎟ − γ log π − 2γ 1 = −2λ1 + λ 2 + γλ1 4 ⎝2⎠ 2 ⎝2⎠ 2

which was previously derived in (3.5). Using (2.13) this may be expressed as 1 (1) ⎛ 3 ⎞ 2 ψ ⎜ ⎟ − γ − 2γ 1 + 2λ1 = λ 2 4 ⎝2⎠ and referring to (4.4) this becomes 1 (1) ⎛ 3 ⎞ ψ ⎜ ⎟ − η1 + 2λ1 = λ 2 4 ⎝2⎠ ⎛3⎞ and we note that ψ (1) ⎜ ⎟ and η1 are both positive (however at this stage we still need to ⎝2⎠ substitute numerical values of the various constants to determine that λ 2 is nonnegative). □ In what follows we show how it is possible to determine the signs of the first few λn with a modicum of numerical calculations. Using the Jacobi theta function, Coffey [5] showed in 2003 that even derivatives of ξ ( s) are positive for all real values of s , while odd derivatives are positive for s ≥ ½ and negative for s < ½. In particular, ξ ( n ) (1) is positive for n ≥ 1 . This was also proved by Freitas [18] in a different manner in 2005. We note from (1.7) that ∞

σk

k =1

k

ξ ′( s ) = −ξ ( s)∑ (−1)k Since ξ (1) = ξ (0) =

(3.15)

∞

σk

k =1

k

k ( s − 1) k −1 = −ξ ( s )∑

1 this results in 2 1 2

1 2

ξ ′(1) = σ 1 = λ1

and we immediately deduce that both λ1 and σ 1 are positive.

14

k s k −1

1 ∞ 1 σ k = − λ1 which again shows that λ1 is positive and ∑ 2 k =1 2

The (odd) derivative ξ ′(0) = that

∞

∑σ k =1

k

is negative.

We have the second derivative ∞

σk

k =1

k

ξ (2) ( s ) = −ξ ( s)∑ and thus ∞

σk

k =1

k

ξ (2) (1) = −ξ (1)∑

∞

σk

k =1

k

k (k − 1) s k − 2 − ξ (1) ( s)∑ ∞

σk

k =1

k

k (k − 1) − ξ (1) (1)∑

ks k −1

k

Since k (k − 1) = k (k + 1) − 2k we have ∞

σk

k =1

k

ξ (2) (1) = −ξ (1)∑

∞

σk

k =1

k

k (k + 1) + 2ξ (1)∑

∞

σk

k =1

k

k − ξ (1) (1)∑

k

and hence, using (3.4) and (3.15), we obtain (3.16)

1 2

1 2

ξ (2) (1) = λ 2 − λ1 + λ12

Since ξ (2) (1) is positive, we determine that (3.17)

λ 2 > λ1 (2 − λ1 )

and from (2.14) we see that 2 − λ1 > 0. We may therefore conclude that λ 2 is also positive.

Since 1 > λ1 we have λ1 > λ12 and from (3.17) we have

λ 2 > 2λ1 − λ12 > 2λ1 − λ1 and therefore we also have (3.18)

λ 2 > λ1

Continuing as above we have

15

∞

σk

k =1

k

ξ ( s ) = −ξ ( s )∑ (3)

k (k − 1)(k − 2) s

k −3

∞

σk

k =1

k

− 2ξ ( s)∑ (1)

k (k − 1) s

−ξ

k −2

(2)

∞

σk

k =1

k

( s )∑

ks k −1

and using the identity k ( k − 1)(k − 2) = k ( k + 1)(k + 2) − 6k (k + 1) + 6k we obtain ∞

σk

k =1

k

ξ (3) ( s) = −ξ ( s )∑ ∞

σk

k =1

k

− 2ξ (1) ( s )∑

∞

σk

k =1

k

k (k + 1)(k + 2) s k −3 + 6ξ ( s )∑ ∞

σk

k =1

k

∞

σk

k =1

k

k (k + 1) s k − 2 + 4ξ (1) ( s )∑

∞

σk

k =1

k

k s k − 2 − ξ (2) ( s )∑

∞

σk

k =1

k

k (k + 1)(k + 2) + 6ξ (1)∑

∞

σk

k =1

k

−2ξ (1) (1)∑

∞

σk

k =1

k

k (k + 1) + 4ξ (1) (1)∑

∞

σk

k =1

k

k (k + 1) − 6ξ (1)∑ ∞

σk

k =1

k

k − ξ (2) (1)∑

k

Employing (3.16) this becomes

ξ (3) (1) = −

∞ ∞ σk σk 1 ∞ σk ( 1)( 2) 3 ( 1) 3 k k + k + + k k + − k ∑ ∑ ∑ 2 k =1 k k =1 k k =1 k

∞

σk

k =1

k

− λ1 ∑

∞

k (k + 1) + 2λ1 ∑ k =1

σk

∞ σ 1 k − (λ 2 − 2λ1 + λ12 )∑ k k 2 k k =1 k

and we obtain 1 2

ξ (3) (1) = λ 3 − 3λ 2 + 3λ1 + λ1λ 2 − 2λ12 + λ1 (λ 2 − 2λ1 + λ12 ) Since ξ (3) (1) is positive we have ⎛1 ⎝2

⎞ 1 ⎠ 2

λ 3 > 3(λ 2 − λ1 ) − 3λ1 ⎜ λ 2 − λ1 ⎟ − λ13 3 1 = 3(λ 2 − λ1 ) − 3λ1 ( λ 2 − λ1 ) + λ1λ 2 − λ13 2 2

16

σk

k =1

k

ks k −1

In particular we have

ξ (3) (1) = −ξ (1)∑

∞

k (k + 1) s k −3 − 6ξ ( s )∑

k

k s k −3

3 1 = 3(λ 2 − λ1 )(1 − λ1 ) + λ1λ 2 − λ13 2 2 1 = 3(λ 2 − λ1 )(1 − λ1 ) + λ1 (3λ 2 − λ12 ) 2 1 > 3(λ 2 − λ1 )(1 − λ1 ) + λ1 (3λ 2 − λ1 ) > 0 2 and we therefore deduce that λ 3 is positive. It remains to be determined whether or not the above inequality also implies that λ 3 > λ 2 . Using the other version of (1.7) gives us ∞

∞

k =1

k =1

ξ (2) ( s ) = −ξ ( s)∑ (−1) k (k − 1)σ k ( s − 1) k − 2 − ξ (1) ( s)∑ (−1) kσ k ( s − 1) k −1

(3.19)

and for s = 1 we get 1 2

ξ (2) (1) = −ξ (1)σ 2 + ξ (1) (1)σ 1 = [σ 12 − σ 2 ] Hence we see that (3.20)

σ 12 > σ 2

Letting s = 0 in (3.19) results in

ξ (2) (0) = −

∞ σk 1 ∞ σk 1 1 (1) ( 1) ( ) k k − − ξ s k = λ 2 − λ1 + λ12 ∑ ∑ 2 2 k =1 k 2 k =1 k

which is the same as (3.16) since

ξ ( n ) ( s ) = (−1) n ξ ( n ) (1 − s) and therefore

ξ ( n ) (0) = (−1)n ξ ( n ) (1) A further differentiation gives us

17

∞

σk

k =1

k

∞

σk

k =1

k

ξ (3) ( s ) = −ξ ( s )∑ (−1) k − 6ξ ( s )∑ (−1) k

σk

k =1

k

∞

σk

k =1

k

k ( s − 1) k −3 − 2ξ (1) ( s )∑ (−1) k

∞

σk

k =1

k

+ 4ξ (1) ( s )∑ (−1) k

∞

k (k + 1)(k + 2)( s − 1) k −3 + 6ξ ( s )∑ (−1) k

∞

σk

k =1

k

k ( s − 1) k − 2 − ξ ( 2) ( s )∑ (−1) k

k (k + 1) ( s − 1) k −3

k (k + 1)( s − 1) k − 2

k ( s − 1) k −1

It will be seen in (6.2) below that n

⎛n⎞

k =1

⎝ ⎠

1

ξ ( n +1) (1) = (−1) n n !σ n +1 + ∑ ⎜ ⎟ (−1)n − k +1ξ ( k ) (1)(n − k + 1)!σ n − k k 2 □ We now attempt to generalise the above specific results which we obtained for λ1 , λ 2 and λ 3 . It is shown in [22] that (3.22)

d m f ( x) e = e f ( x )Ym ( f (1) ( x), f (2) ( x),..., f ( m ) ( x) ) dx m

where the (exponential) complete Bell polynomials Yn ( x1 ,..., xn ) are defined by Y0 = 1 and for n ≥ 1 (3.23)

Yn ( x1 ,..., xn ) = ∑

π (n)

k

k

1 2 n! ⎛ xn ⎞ ⎛ x1 ⎞ ⎛ x2 ⎞ ⎜ ⎟ ⎜ ⎟ ... ⎜ ⎟ k1 ! k2 !... kn ! ⎝ 1! ⎠ ⎝ 2! ⎠ ⎝ n! ⎠

kn

where the sum is taken over all partitions π (n) of n , i.e. over all sets of integers k j such that k1 + 2k2 + 3k3 + ... + nkn = n

The complete Bell polynomials have integer coefficients and the first five are set out below (Comtet [14, p.307]) (3.24)

Y1 ( x1 ) = x1

18

Y2 ( x1 , x2 ) = x12 + x2

Y3 ( x1 , x2 , x3 ) = x13 + 3x1 x2 + x3 Y4 ( x1 , x2 , x3 , x4 ) = x14 + 6 x12 x2 + 4 x1 x3 + 3x22 + x4 Y5 ( x1 , x2 , x3 , x4 , x5 ) = x15 + 10 x13 x2 + 10 x12 x3 + 15 x1 x22 + 5 x1 x4 + 10 x2 x3 + x5 Suppose that h′( x) = h( x) g ( x) and let f ( x) = log h( x) . We see that f ′( x) =

h′( x) = g ( x) h( x )

and then using (3.22) above we have dm d m log h ( x ) h( x ) = m e = h( x)Ym ( g ( x), g (1) ( x),..., g ( m −1) ( x) ) m dx dx We now write (1.7) as

ξ ′( s ) = ξ ( s) g ( s ) ∞

σk

k =1

k

where g ( s) = −∑ (3.25)

ks k −1 and g (1) = λ1 . We then have

ξ ( m ) ( s ) = ξ ( s)Ym ( g ( s), g (1) ( s),..., g ( m −1) ( s) )

where ∞

σk

k =1

k

g ( r ) ( s ) = −∑ and ∞

σk

k =1

k

g ( r ) (1) = −∑

k (k − 1) ⋅⋅⋅ (k − r ) s k −1− r

k (k − 1) ⋅⋅⋅ (k − r )

Using (3.9) (r + 1)! ⎛ r ⎞ j ⎜ ⎟ (−1) k (k + 1)...(k + j − 1) j ! ⎝ j − 1⎠ j =1

r +1

k (k − 1)...(k − r ) = (−1) r +1 ∑ we have

19

∞ σk (r + 1)! ⎛ r ⎞ j ( 1) − k (k + 1)...(k + j − 1) ∑ ⎜ ⎟ j ! ⎝ j − 1⎠ j =1 k =1 k

r +1

g ( r ) (1) = −(−1) r +1 ∑

and using (3.4) this becomes (r + 1)! ⎛ r ⎞ j ⎜ ⎟ (−1) ( j − 1)!λ j j ! ⎝ j − 1⎠ j =1

r +1

= (−1) r +1 ∑

r +1 λj ⎛ r ⎞ j = (−1) r +1 (r + 1)!∑ ⎜ ⎟ (−1) j j =1 ⎝ j − 1 ⎠

We then have (3.26)

r +1 r + 1 ⎛ ⎞ j g ( r ) (1) = (−1) r +1 r !∑ ⎜ ⎟ (−1) λ j j ⎠ j =1 ⎝

and applying the binomial inversion formula (3.27)

n ⎛n⎞ an = ∑ ⎜ ⎟ (−1) k bk k =0 ⎝ k ⎠

n ⎛n⎞ bn = ∑ ⎜ ⎟ (−1) k ak k =0 ⎝ k ⎠

⇔

to the following expression (where we define λ 0 = g ( −1) (1) = 0 ) (−1) r g ( r −1) (1) r ⎛ r ⎞ = ∑ ⎜ ⎟ (−1) j λ j (r − 1)! j =0 ⎝ j ⎠ we obtain ⎛ r ⎞ g ( j −1) (1)

r

λ j = ∑⎜ ⎟ j =1 ⎝ j ⎠ ( j − 1)!

(3.28)

We have g

( j −1)

∞

σk

k =1

k

( s ) = −∑ (−1) k

k (k − 1) ⋅⋅⋅ (k − j + 1)( s − 1) k − j

which results in g ( j −1) (1) = −( j − 1)!(−1) j σ j

20

and therefore from (3.28) we obtain the well-known result [6] (3.29)

r

⎛r⎞

j =1

⎝ ⎠

λ j = −∑ (−1) j ⎜ ⎟ σ j j

Particular values of (3.26) are g (1) = λ1 g (1) (1) = λ 2 − 2λ1 g (2) (1) = 6λ1 − 6λ 2 + 2λ 3

Using (3.25) we see that 1 2

ξ ( m ) (1) = Ym ( g (1), g (1) (1),..., g ( m −1) (1) ) 1 2

1 2

ξ (1) (1) = Y1 ( g (1) ) = λ1 1 2

ξ (2) (1) = Y2 ( g (1), g (1) (1) ) 1 = [ g 2 (1) + g (1) (1)] 2 1 = [λ12 + λ 2 − 2λ1 ] 2 1 2

ξ (3) (1) = Y3 ( g (1), g (1) (1), g (2) (1) ) and, since Y3 ( x1 , x2 , x3 ) = x13 + 3x1 x2 + x3 , this becomes 1 = [ g 3 (1) + 3g (1) g (1) (1) + g ( 2) (1)] 2 1 2

ξ (3) (1) = [λ13 + 3λ1 (λ 2 − 2λ1 ) + 6λ1 − 6λ 2 + 2λ 3 ]

21

Riordan [28] reports that (3.30)

m m ⎛m⎞ ⎛m⎞ Ym +1 ( x1 ,..., xm +1 ) = ∑ ⎜ ⎟ Ym − k ( x1 ,..., xm − k ) xk +1 = ∑ ⎜ ⎟ Yk ( x1 ,..., xk ) xm − k +1 k =0 ⎝ k ⎠ k =0 ⎝ k ⎠

and for example we have Y2 ( x1 , x2 ) = Y1 ( x1 ) x1 + Y0 x2 = x12 + x2 From (3.25) we see that

ξ ( m +1) ( s) = ξ ( s)Ym +1 ( g ( s ), g (1) ( s ),..., g ( m ) ( s) ) m ⎛m⎞ = ∑ ⎜ ⎟ ξ ( s )Yk g ( s ), g (1) ( s ),..., g ( k −1) ( s) g ( m − k ) ( s) k =0 ⎝ k ⎠

(

)

We therefore have (3.31)

m

⎛m⎞

k =0

⎝ ⎠

ξ ( m +1) ( s ) = ∑ ⎜ ⎟ ξ ( k ) ( s ) g ( m − k ) ( s ) k

which simply corresponds with the Leibniz differentiation rule using ξ ′( s ) = ξ ( s ) g ( s ) . This gives us

ξ (2) ( s ) =ξ ( s ) g (1) ( s ) + ξ (1) ( s ) g ( s ) 1 2

1 2

ξ (2) (1) = [λ 2 − 2λ1 ] + λ12 We therefore have 1 1 [λ 2 − 2λ1 ] + λ12 > 0 2 2 which corresponds with (3.16). □ We note from (1.6) that

22

⎛ s⎞ s log ξ ( s ) = log Γ ⎜1 + ⎟ − log π + log[( s − 1)ς ( s )] ⎝ 2⎠ 2 and therefore g (s) =

ξ ′( s ) 1 ς ′( s ) ⎛ s⎞ 1 = ψ ⎜1 + ⎟ − log π + + s − 1 ς (s) ξ (s) ⎝ 2⎠ 2

We will note from (4.1) that ∞ ς ′( s) 1 + = −∑ηk ( s − 1) k ς ( s) s − 1 k =0

and we therefore obtain (3.32)

g (s) =

∞ ξ ′( s) 1 ⎛ s ⎞ 1 = ψ ⎜ 1 + ⎟ − log π − ∑ηk ( s − 1) k ξ (s) 2 ⎝ 2 ⎠ 2 k =0

and we see that 1 ⎛3⎞ 1 g (1) = g (0) (1) = ψ ⎜ ⎟ − log π − η0 = λ1 2 ⎝2⎠ 2 For r ≥ 1

g ( r ) ( s) =

1 (r ) ⎛ s ⎞ ∞ ψ ⎜1 + ⎟ − ∑ηk k (k − 1) ⋅⋅⋅ (k − r + 1)( s − 1) k − r r +1 2 ⎝ 2 ⎠ k =0

g ( r ) (1) =

1 (r ) ⎛ 3 ⎞ ψ ⎜ ⎟ − r !ηr 2r +1 ⎝2⎠

=

1 (r ) ⎛ 3 ⎞ 1 ψ ⎜ ⎟ − r !ηr − δ r ,0 log π r +1 2 2 ⎝2⎠

where we have introduced the Kronecker delta δ r ,0 to ensure that this equation is also valid for r = 0 r +1 r + 1 ⎛ ⎞ j g ( r ) (1) = (−1) r +1 r !∑ ⎜ ⎟ (−1) λ j j ⎠ j =1 ⎝

This may be written as

23

r +1 r + 1 ⎛ ⎞ 1 (r ) ⎛ 3 ⎞ 1 j (−1) r +1 r !∑ ⎜ ⎟ (−1) λ j = r +1 ψ ⎜ ⎟ − r !η r − δ r ,0 log π j ⎠ 2 2 ⎝2⎠ j =0 ⎝

where we start the summation at j = 0 by defining λ 0 = 0 . ⎛3⎞ We now let r → r − 1 and define η −1 = 0 and ψ ( −1) ⎜ ⎟ = 0 ⎝2⎠ r ⎛r⎞ 1 1 ⎛3⎞ (−1) r (r − 1)!∑ ⎜ ⎟ (−1) j λ j = r ψ ( r −1) ⎜ ⎟ − (r − 1)!ηr −1 − δ r −1,0 log π 2 2 ⎝2⎠ j =0 ⎝ j ⎠

r

⎛r⎞

∑ ⎜ j ⎟ (−1) λ j =0

j

⎝ ⎠

j

⎡ r ⎤ r ⎛3⎞ = (−1) r ⎢ r ψ ( r −1) ⎜ ⎟ − ηr −1 − δ r −1,0 log π ⎥ 2r ! ⎝2⎠ ⎣2 r! ⎦

and applying the binomial inversion formula (3.27) we obtain ⎛r ⎞⎡ j

⎛3⎞

⎛r ⎞⎡ j

⎛3⎞

r

⎤

j

log π ⎥ λ r = ∑ ⎜ ⎟ ⎢ j ψ ( j −1) ⎜ ⎟ − η j −1 − δ j −1,0 2 j! ⎝2⎠ j =0 ⎝ j ⎠ ⎣ 2 j ! ⎦ r

⎤

1 ⎛r⎞

λ r = ∑ ⎜ ⎟ ⎢ j ψ ( j −1) ⎜ ⎟ − η j −1 ⎥ − ⎜ ⎟ log π ⎝2⎠ j =1 ⎝ j ⎠ ⎣ 2 j ! ⎦ 2 ⎝1⎠ r

⎛r⎞

j

⎛3⎞

1⎛r⎞ ⎛ 3⎞

r

⎛r⎞

1 ⎛r⎞

λ r = ∑ ⎜ ⎟ j ψ ( j −1) ⎜ ⎟ + ⎜ ⎟ψ ⎜ ⎟ − ∑ ⎜ ⎟η j −1 − ⎜ ⎟ log π 2 ⎝1⎠ ⎝ 2 ⎠ 2 ⎝ j ⎠ ⎝ 2 ⎠ j =1 ⎝ j ⎠ j =2 ⎝ j ⎠ 2 j !

(3.33)

From (3.5.1) we have

(

ψ ( j −1) ( 3 / 2 ) = (−1) j ( j − 1)! ⎡⎣ 2 j − 1⎤⎦ ς ( j ) − 2 j

)

and we obtain r

⎛r⎞

⎡

1⎤

r

⎛r⎞

1 ⎛r⎞ ⎛ 3⎞

r

⎛r⎞

1 ⎛r⎞

λ r = ∑ ⎜ ⎟ (−1) j ⎢1 − j ⎥ ς ( j ) − ∑ ⎜ ⎟ (−1) j + ⎜ ⎟ψ ⎜ ⎟ − ∑ ⎜ ⎟η j −1 − ⎜ ⎟ log π 2 ⎝ 1 ⎠ ⎝ 2 ⎠ j =1 ⎝ j ⎠ 2 ⎝1⎠ ⎣ 2 ⎦ j =2 ⎝ j ⎠ j =2 ⎝ j ⎠ Using (2.11) ⎛3⎞ ⎝ ⎠

ψ ⎜ ⎟ = −γ − 2 log 2 + 2 2 and noting that

24

r

⎛r⎞

⎛r⎞

r

∑ ⎜ j ⎟ (−1) = ∑ ⎜ j ⎟ (−1) j =2

j

⎝ ⎠

j =0

⎝ ⎠

j

+ r −1 = r −1

we simply obtain another derivation of Coffey’s result [C1] for r ≥ 2 r

⎛r⎞

j =2

⎝ ⎠

⎡ ⎣

1⎤ ⎦

r

⎛r⎞

j =1

⎝ ⎠

1

λ r = ∑ ⎜ ⎟ (−1) j ⎢1 − j ⎥ ς ( j ) + 1 − ∑ ⎜ ⎟η j −1 + 1 − r (γ + 2 log 2 + log π ) j j 2 2

(3.34)

(and we note that the term +1 has been inadvertently omitted in equation (3) in [7] and in equation (12) in [13]). A much simpler derivation of (3.34) is shown below. We see that (3.33) may be written as r

⎛r⎞

⎡1

1

⎤

⎛3⎞

1

⎛3⎞ 1

λr = ∑⎜ ⎟ ψ ( j −1) ⎜ ⎟ − ( j − 1)!η j −1 ⎥ + rγ + rψ ⎜ ⎟ − r log π ⎢ j 2 ⎝2⎠ 2 ⎝2⎠ j = 2 ⎝ j ⎠ ( j − 1)! ⎣ 2 ⎦ r ⎛ r ⎞ 1 ⎡ 1 ( j −1) ⎛ 3 ⎞ ⎤ 1 = ∑⎜ ⎟ − ( j − 1)!η j −1 ⎥ + r (γ − 2 log 2 + 2 − log π ) ψ ⎜ ⎟ ⎢ j ⎝2⎠ j = 2 ⎝ j ⎠ ( j − 1)! ⎣ 2 ⎦ 2

and we note from (3.32) that d 1 1 ⎛ s⎞ ∞ log ξ ( s) + log π = ψ ⎜1 + ⎟ − ∑η k ( s − 1) k ds 2 2 ⎝ 2 ⎠ k =0 Differentiating this j − 1 times gives us dj 1 1 ( j −1) ⎛ s ⎞ ∞ k − j +1 log ( s ) log + = ξ δ π ψ j ,1 ⎜1 + ⎟ − ∑ηk k (k − 1) ⋅⋅⋅ (k − j )( s − 1) j j ds 2 2 ⎝ 2 ⎠ k =0

and evaluating this at s = 1 results in dj 1 1 ⎛3⎞ log ξ ( s ) + δ j ,1 log π = j ψ ( j −1) ⎜ ⎟ − ( j − 1)!η j −1 j ds 2 2 ⎝2⎠ s =1

We then have r

⎛r⎞

1

⎡1

∑ ⎜ j ⎟ ( j − 1)! ⎢⎣ 2 j =2

⎝ ⎠

j

⎤

⎛3⎞

1

⎛3⎞ 1

ψ ( j −1) ⎜ ⎟ − ( j − 1)!η j −1 ⎥ + rγ + rψ ⎜ ⎟ − r log π 2 ⎝2⎠ 2 ⎝2⎠ ⎦

25

r ⎛r⎞ 1 dj 1 = ∑⎜ ⎟ log ξ ( s ) − r (γ + 2 log 2 + log π ) j 2 j = 2 ⎝ j ⎠ ( j − 1)! ds s =1

Applying the Leibniz differentiation rule to (1.4) we see that

λr = =

1 d r r −1 [ s log ξ ( s )] (r − 1)! ds r s =1 r ⎛r⎞ d j d r − j r −1 1 s s log ( ) ξ ∑⎜ ⎟ ds r − j (r − 1)! j =0 ⎝ j ⎠ ds j s =1

r ⎛r⎞ 1 dj = ∑⎜ ⎟ log ξ ( s ) j j = 0 ⎝ j ⎠ ( j − 1)! ds s =1 r ⎛r⎞ 1 dj d = ∑⎜ ⎟ log ξ ( s ) + r log ξ ( s ) j ds j = 2 ⎝ j ⎠ ( j − 1)! ds s =1 s =1 r ⎛ r ⎞ 1 ⎡ 1 ( j −1) ⎛ 3 ⎞ ⎤ 1 = ∑⎜ ⎟ − ( j − 1)!η j −1 ⎥ + r (γ − 2 log 2 + 2 − log π ) ψ ⎜ ⎟ ⎢ j ⎝2⎠ j = 2 ⎝ j ⎠ ( j − 1)! ⎣ 2 ⎦ 2

which corresponds with (3.33) above.

□

We have r

⎛r⎞

1

⎡1

⎤ 1

⎛3⎞

λr = ∑⎜ ⎟ ψ ( j −1) ⎜ ⎟ − ( j − 1)!η j −1 ⎥ + r (γ − 2 log 2 + 2 − log π ) ⎢ j j ⎝2⎠ j = 2 ⎝ ⎠ ( j − 1)! ⎣ 2 ⎦ 2 Using (3.5.2)

ψ ( j −1) ( 3 / 2 ) = (−1) j ( j − 1)!ς ( j ,3 / 2) we may write this as r ⎛r⎞ ⎡ ς ( j ,3 / 2) ⎤ 1 (3.35) λ r = ∑ ⎜ ⎟ (−1) j ⎢ − (−1) j η j −1 ⎥ + r (γ − 2 log 2 + 2 − log π ) j ⎣ 2 ⎦ 2 j =2 ⎝ j ⎠

and it may be possible to simplify the first summation by using the following Hermite integral representation of the Hurwitz zeta function (see for example [31, p.120])

ς ( s, u ) =

∞

u − s u1− s (u + ix) − s − (u − ix) − s + + i∫ dx 2 s −1 0 e 2π x − 1

26

It may be noted that

γ − 2 log 2 + 2 − log π > 0 4. The eta constants

The eta constants ηn are defined by reference to the logarithmic derivative of the Riemann zeta function ∞ d ς ′( s) 1 [log ς ( s)] = =− − ∑η k ( s − 1) k ς ( s) ds s − 1 k =0

(4.1)

s −1 < 3

and we may also note that this is equivalent to ∞ d ς ′( s) 1 log[( s − 1)ς ( s )] = + = −∑ηk ( s − 1) k ς (s) s − 1 ds k =0

(4.2) We then see that

∞ d n +1 s s log[( − 1) ς ( )] = − η k k (k − 1)...(k − n + 1)( s − 1) k − n ∑ ds n +1 k =0

and hence we get (4.3)

lim s →1

d n +1 log[( s − 1)ς ( s )] = −n !ηn ds n +1

We see from (4.2) that ∞ d [( s − 1)ς ( s)] = −∑ηk ( s − 1) k [( s − 1)ς ( s )] ds k =0

and thus d n +1 dn s s [( − 1) ς ( )] = − ds n +1 ds n

∞

∑η k =0

k

( s − 1) k [( s − 1)ς ( s )]

We consider the value at s = 1 d n +1 dn [( s − 1)ς ( s )] = − lim n s →1 ds ds n +1 s =1

∞

∑η k =0

and using (2.10)

27

k

( s − 1) k [( s − 1)ς ( s )]

d n +1 [( s − 1)ς ( s )] = (−1) n (n + 1)γ n ds n +1 s =1

we obtain via the Leibniz differentiation formula n ⎛n⎞ = − n !η n + ∑ ⎜ ⎟ (n − j )!η n − j (−1) j jγ j −1 j =1 ⎝ j ⎠

where we have isolated the first term. n

= − n !η n + n !∑

(−1) j jη n − j γ j −1 j!

j =1

Hence we obtain the recurrence relation (4.4)

n

(−1) (n + 1)γ n = − n !η n + n !∑ n

(−1) j jη n − j γ j −1 j!

j =1

where for n = 0,1, 2 we have respectively:

γ = −η0 2γ 1 = η1 + η0γ = η1 − γ 2 3 γ 2 = −η2 − 3γγ 1 − γ 3 2 Equation (4.4) is equivalent to Coffey’s recurrence relation (4.5) below. Letting u = 1 in (2.1) gives us (−1) n γ n ( s − 1) n +1 ! n n =0 ∞

( s − 1)ς ( s) = 1 + ∑

and defining

g ( s ) = ( s − 1)ς ( s )

L( s) = log[( s − 1)ς ( s)] we have ∞ d (−1) n ( s − 1)ς ( s ) = ∑ γ n (n + 1)( s − 1) n ds n! n=0

28

and thus g (1) (1) = γ 0 = γ g ( k +1) (1) = (−1) k (k + 1)γ k We see that

( s − 1)ς ′( s ) + ς ( s) d log[( s − 1)ς ( s)] = ( s − 1)ς ( s) ds

which may be written as ( s − 1)ς ( s ) L(1) ( s ) = ( s − 1)ς ′( s ) + ς ( s ) =

d [( s − 1)ς ( s )] ds

or g ( s ) L(1) ( s ) = g (1) ( s ) We then have d (n) ⎡ g ( s ) L(1) ( s ) ⎤⎦ = g ( n +1) ( s ) (n) ⎣ ds and applying the Leibniz rule gives us n ⎛n⎞ g ( n +1) ( s ) = ∑ ⎜ ⎟ g ( k ) ( s ) L( n +1− k ) ( s ) k =0 ⎝ k ⎠

Letting s = 1 gives us n ⎛n⎞ (−1) n (n + 1) γ n = g (0) (1) L( n +1) (1) − ∑ ⎜ ⎟ (−1) k −1 k γ k −1 (n − k )!ηn − k k =1 ⎝ k ⎠ n ⎛n⎞ = −n !η n − ∑ ⎜ ⎟ (−1) k k γ k −1 (n − k ) !η n − k k =1 ⎝ k ⎠

(−1) k γ k −1η n − k (k − 1)!

n

= − n !η n − n !∑ k =1

Hence we have n

n !ηn = (−1) n +1 (n + 1) γ n − n !∑ k =1

and reindexing the sum gives us

29

(−1) k γ k −1η n − k (k − 1)!

(4.5)

n −1

n !η n = (−1) n +1 (n + 1) γ n + (−1) n +1 n !∑ k =0

(−1) k +1 γ n − k −1η k (n − k − 1)!

as originally reported by Coffey [5]. A formula for the Stieltjes constants in terms of the (exponential) complete Bell polynomials containing the eta constants η n as the arguments is shown below in equation (6.1). □ Eliminating log ξ ( s) from (1.6) and (1.7) results in ∞ σ ⎛ s⎞ s log Γ ⎜1 + ⎟ − log π + log[( s − 1)ς ( s )] = − log 2 − ∑ (−1) k k ( s − 1) k k ⎝ 2⎠ 2 k =1

and, since ξ ( s ) = ξ (1 − s ) , we have the equivalent representation ∞ σ ⎛ s⎞ s log Γ ⎜1 + ⎟ − log π + log[( s − 1)ς ( s )] = − log 2 − ∑ k s k ⎝ 2⎠ 2 k =1 k

For convenience, we define L( s ) by L( s) = log[( s − 1)ς ( s)] Upon differentiating the above two equations we obtain ∞ 1 ⎛ s⎞ 1 ψ ⎜1 + ⎟ − log π + L(1) ( s ) = −∑ (−1)k σ k ( s − 1)k −1 2 ⎝ 2⎠ 2 k =1 ∞ 1 ⎛ s⎞ 1 ψ ⎜1 + ⎟ − log π + L(1) ( s) = −∑ σ k s k −1 2 ⎝ 2⎠ 2 k =1

For n ≥ 1 we have the higher derivatives ∞ 1 ( n ) ⎛ s ⎞ ( n +1) ψ + + = − 1 L ( s ) (−1) k σ k (k − 1)(k − 2)...( k − n)( s − 1) k −1− n ∑ ⎜ ⎟ 2n +1 2 ⎝ ⎠ k =1 ∞ 1 ( n ) ⎛ s ⎞ ( n +1) ψ σ k (k − 1)(k − 2)...(k − n) s k −1− n + + = − 1 L ( s ) ∑ ⎜ ⎟ 2n +1 2 ⎝ ⎠ k =1

and therefore with s = 1 we obtain

30

1 (n) ψ ( 3 / 2 ) + L( n +1) (1) = −(−1) n +1 n !σ n +1 n +1 2 We have

ψ ( n ) ( 3 / 2 ) = (−1)n +1 n ![2n +1 − 1]ς (n + 1) + (−1) n n !2n +1

and hence we get 1 ⎤ ⎡ (−1) n +1 n ! ⎢1 − n +1 ⎥ ς (n + 1) + (−1) n n !+ L( n +1) (1) = −( −1) n +1 n !σ n +1 ⎣ 2 ⎦ Using L( n +1) (1) = − n !ηn results in ⎡ ⎣

σ n +1 = (−1)n +1ηn − ⎢1 −

1 ⎤ ς (n + 1) + 1 2n +1 ⎥⎦

Letting s = 0 gives us 1 (n) ψ (1) + L( n +1) (0) = −n !σ n +1 n +1 2 and therefore we get L( n +1) (0) = −n !σ n +1 −

1 (−1) n +1 n !ς (n + 1) n +1 2

and substituting for σ n +1 gives us ⎛ 1 ⎤ ⎞ 1 ⎡ = − n !⎜ (−1) n +1η n − ⎢1 − n +1 ⎥ ς ( n + 1) + 1⎟ − n +1 ( −1) n +1 n !ς (n + 1) ⎣ 2 ⎦ ⎝ ⎠ 2 Hence we get for n ≥ 1 1 ⎛ ⎞ L( n +1) (0) = (−1) n n !η n + ⎜1 − n +1 ⎡⎣1 − (−1) n ⎤⎦ ⎟ n !ς (n + 1) − n ! ⎝ 2 ⎠ and with n = 1 we have 1 L(2) (0) = ς (2) − η1 − 1 2 We will also see from (4.6) below that

31

L(2) (0) = −2ς ′′(0) − log 2 (2π ) − 1 and we therefore obtain 1 2

1 4

1 2

ς ′′(0) = η1 − ς (2) − log 2 (2π ) Substituting (4.4) η1 = 2γ 1 + γ 2 we then have an expression for ς ′′(0) 1 2

1 4

1 2

ς ′′(0) = γ 1 + γ 2 − ς (2) − log 2 (2π ) as previously derived by Ramanujan [3] and Apostol [1]. We have ∞

L(1) ( s ) = −∑η k ( s − 1) k k =0

and hence L(1) (1) = −η0 = γ ∞

L( n +1) ( s ) = −∑η k k (k − 1)...(k − n + 1)( s − 1) k − n k =0

Coffey [13] has shown that the sequence (ηn ) has strict sign alteration

ηn = (−1) n +1 ε n where ε n are positive constants and therefore we have L( n +1) (1) = − n !η n = (−1) n n !ε n We therefore note that L(1) (1) is positive and L(2) (1) is negative and the signs strictly alternate thereafter. We see that ∞

∞

k =0

k =0

L(1) (0) = −∑ (−1) k η k = ∑ ε k is positive In fact we have

32

∞

L (0) = −∑ (−1) k η k = log(2π ) − 1 (1)

k =0

∞

∞

k =0

k =0

L( n +1) (0) = −∑ηk k (k − 1)...(k − n + 1)(−1) k − n = (−1) n ∑ ε k k (k − 1)...(k − n + 1) and therefore we note that the signs of L( n +1) (0) also strictly alternate. We see that L(2) ( s ) =

2 ( s − 1)ς ′′( s) + 2ς ′( s ) − ⎡⎣ L(1) ( s ) ⎤⎦ ( s − 1)ς ( s )

Referring to (2.9) we then obtain 2

L(2) (1) = −2γ 1 − ⎡⎣ L(1) (1) ⎤⎦ = −2γ 1 − γ 2 and we therefore obtain (as in 4.4))

η1 = 2γ 1 + γ 2 Since η1 ≥ 0 we deduce that 2γ 1 + γ 2 ≥ 0 Since η2 ≤ 0 we see that 3 − γ 2 − γ 3 − 3γγ 1 ≤ 0 2 We have L(2) (0) = −2ς (2) (0) + 4ς (1) (0) − ⎡⎣ L(1) (0) ⎤⎦

2

= −2ς (2) (0) + 4ς (1) (0) − [ log(2π ) − 1]

2

= −2ς (2) (0) − 2 log(2π ) − [ log(2π ) − 1]

2

and therefore

33

(4.6)

L(2) (0) = −2ς (2) (0) − log 2 (2π ) − 1

5. A formula for the Stieltjes constants in terms of the higher derivatives of the Riemann zeta function ς ( n ) (0) .

We recall Hasse’s formula (2.4) with u = 1 1 n ⎛ n ⎞ (−1) k ( s − 1)ς ( s ) = ( s − 1)ς ( s,1) = ∑ ∑⎜ ⎟ s −1 n = 0 n + 1 k = 0 ⎝ k ⎠ (1 + k ) ∞

and with s → s − 1 we have 1 n ⎛n⎞ k s ∑ ⎜ ⎟(−1) (1 + k ) k 1 + n n =0 k =0 ⎝ ⎠ ∞

sς (1 − s ) = −∑

We refer to the functional equation for the Riemann zeta function (1.2) 2(2π ) − s Γ( s ) cos(π s / 2)ς ( s ) = ς (1 − s ) and, multiplying by s , we see that 1 n ⎛n⎞ k s ∑ ⎜ ⎟(−1) (1 + k ) n =0 n + 1 k =0 ⎝ k ⎠ ∞

f ( s ) = sς (1 − s ) = 2(2π ) − s Γ( s + 1) cos(π s / 2)ς ( s) = −∑ Differentiation results in 1 n ⎛n⎞ s p k ∑ ⎜ ⎟(−1) (1 + k ) log (1 + k ) n =0 n + 1 k =0 ⎝ k ⎠ ∞

f ( p ) ( s ) = −∑

and we have the particular value at s = 0 1 n ⎛n⎞ p k ∑ ⎜ ⎟(−1) log (1 + k ) k n 1 + n =0 k =0 ⎝ ⎠ ∞

f ( p ) (0) = −∑

We recall the following expression for the Stieltjes constants (2.8)

γ p (u ) = −

1 ∞ 1 n ⎛n⎞ k p +1 ∑ ∑ ⎜ ⎟(−1) log (u + k ) p + 1 n =0 n + 1 k =0 ⎝ k ⎠

which gives us

34

γ p = γ p (1) = −

1 ∞ 1 n ⎛n⎞ k p +1 ∑ ∑ ⎜ ⎟(−1) log (1 + k ) p + 1 n =0 n + 1 k =0 ⎝ k ⎠

and we therefore obtain f ( p +1) (0) = ( p + 1)γ p

(5.1) We have

log f ( s ) = log 2 − s log(2π ) + log Γ( s + 1) + log cos(π s / 2) + log ς ( s ) and differentiation results in f ′( s ) π ς ′( s) = − log(2π ) +ψ ( s + 1) − tan(π s / 2) + f (s) 2 ς ( s) We note that f (0) = 2ς (0) = −1 ⎡ ς ′(0) ⎤ f ′(0) = f (0) ⎢ − log(2π ) + ψ (1) + = log(2π ) + γ + 2ς ′(0) ς (0) ⎥⎦ ⎣ and, referring to (5.1), we see that log(2π ) + γ + 2ς ′(0) = γ 0 = γ We then deduce the well-known result (5.2)

1 2

ς ′(0) = − log(2π )

We may write f ′( s ) = g ( s ) f ( s ) where g (s) =

d ⎡log(2π ) − s + log Γ( s + 1) + log cos(π s / 2) + log ς ( s ) ⎤⎦ ds ⎣

= − log(2π ) + ψ ( s + 1) −

π 2

tan(π s / 2) +

35

ς ′( s ) ς ( s)

We note that g (0) = −γ and f ′(0) = γ ⎡ π ς ′( s) ⎤ f ′( s) = ⎢ − log(2π ) + ψ ( s + 1) − tan(π s / 2) + f (s) 2 ς ( s) ⎥⎦ ⎣ ⎡ π ς ′( s) ⎤ f ′′( s ) = ⎢ − log(2π ) +ψ ( s + 1) − tan(π s / 2) + f ′( s) 2 ς ( s ) ⎥⎦ ⎣ 2 2 ⎡ ς ( s)ς ′′( s) − [ς ′( s ) ] ⎤ π⎞ ⎛ 2 + ⎢ψ ′( s + 1) − ⎜ ⎟ sec (π s / 2) + ⎥ f ( s) ς 2 (s) ⎝2⎠ ⎢⎣ ⎥⎦

We then see that 2 2 ⎡ π ⎞ ς (0)ς ′′(0) − [ς ′(0)] ⎤ ⎛ f ′′(0) = −γ − ⎢ψ ′(1) − ⎜ ⎟ + ⎥ ς 2 (0) ⎝2⎠ ⎢⎣ ⎥⎦ 2

We know from (3.5.2) that

ψ ′(1) = ς (2) =

π2 6

and we then have f ′′(0) = −γ 2 +

π2 12

+ 2ς ′′(0) + log 2 (2π )

We also have from (5.1) f ′′(0) = 2γ 1 and we thus obtain (5.3)

1 2

ς ′′(0) = γ 1 + γ 2 −

1 2 1 π − log 2 (2π ) 24 2

This concurs with the result previously obtained by Ramanujan [3] and Apostol [1]. As noted by Coffey [9] for differentiable functions f ( s) and g ( s) such that

36

f ′( s ) = f ( s ) g ( s ) then we have in terms of the (exponential) complete Bell polynomials

(

f ( n ) ( s ) = f ( s )Yn g ( s ), g ′( s ),..., g ( n −1) ( s )

)

where, in the case under review, we have g (s) =

d d 4 ⎡⎣log(2π ) − s + log Γ( s + 1) + log cos(π s / 2) + log ς ( s) ⎤⎦ = ∑ log g p ( s ) ds ds p =1

⎛d 4 d2 f ( n ) ( s ) = f ( s )Yn ⎜ ∑ log g p ( s ), 2 ds ⎝ ds p =1

4

∑ log g p (s),..., p =1

⎞ d n −1 4 log g p ( s ) ⎟ n −1 ∑ ds p =1 ⎠

We have [28] (5.4)

n ⎛n⎞ Yn ( x1 + y1 ,..., xn + yn ) = ∑ ⎜ ⎟ Yn − k ( x1 ,..., xn − k )Yk ( y1 ,..., yk ) k =0 ⎝ k ⎠

and this may be generalised to Bell polynomials with additional arguments as follows n ⎛n⎞ Yn (a1 + b1 + c1 + d1 ,..., an + bn + cn + d n ) = ∑ ⎜ ⎟ Yn − k (a1 ,..., an − k )Yk (b1 + c1 + d1 ,..., bk + ck + d k ) k =0 ⎝ k ⎠ k ⎛k ⎞ Yk (b1 + c1 + d1 ,..., bk + ck + d k ) = ∑ ⎜ ⎟ Yk − j (b1 ,..., bk − j )Y j (c1 + d1 ,..., c j + d j ) j =0 ⎝ j ⎠ j ⎛ j⎞ Y j (c1 + d1 ,..., c j + d j ) = ∑ ⎜ ⎟ Y j −l (c1 ,..., c j −l )Yl (d1 ,..., dl ) l =0 ⎝ l ⎠

and we end up with Yn (a1 + b1 + c1 + d1 ,..., an + bn + cn + d n ) j n k ⎛n⎞ ⎛k ⎞ ⎛ j⎞ = ∑ ⎜ ⎟ Yn − k (a1 ,..., an −k )∑ ⎜ ⎟ Yk − j (b1 ,..., bk − j )∑ ⎜ ⎟ Y j −l (c1 ,..., c j −l )Yl (d1 ,..., dl ) k =0 ⎝ k ⎠ j =0 ⎝ j ⎠ l =0 ⎝ l ⎠

We note from (A.5) in Appendix A that

37

d m h( x) e = e h ( x )Ym ( h (1) ( x), h (2) ( x),..., h ( m ) ( x) ) m dx and letting h( x) → log h( x) we have ⎛d ⎞ dm d m log h ( x ) d2 dm log h ( x ) = = h x e e Y h x h x ( ) log ( ), log ( ),..., log h( x) ⎟ m⎜ 2 m m m dx dx dx dx ⎝ dx ⎠

and we see that ⎛d ⎞ 1 dm d2 dm = h ( x ) Y log h ( x ), log h ( x ),..., log h( x) ⎟ m⎜ m 2 m h( x) dx dx dx ⎝ dx ⎠

We therefore have ⎛d 4 d2 f ( n ) ( s ) = f ( s )Yn ⎜ ∑ log g p ( s ), 2 ds ⎝ ds p =1

=

4

∑ log g p (s),..., p =1

⎞ d n −1 4 log g p ( s ) ⎟ n −1 ∑ ds p =1 ⎠

j n k ⎛ n ⎞ d n−k ⎛ k ⎞ d k− j ⎛ j ⎞ d j −l 1 dl ( ) ( ) ( ) g s g s g s g4 (s) ∑⎜ ⎟ ∑ ⎜ ⎟ k − j 2 ∑ ⎜ ⎟ j −l 3 1 g1 ( s ) g 2 ( s ) g3 ( s ) g 4 ( s ) k =0 ⎝ k ⎠ ds n −k ds l j = 0 ⎝ j ⎠ ds l = 0 ⎝ l ⎠ ds

and with s = 0 this becomes using (5.1) j k ⎛ n ⎞ d n−k ⎛ k ⎞ d k− j ⎛ j ⎞ d j −l dl γ n −1 = 2∑ ⎜ ⎟ n −k g1 (0)∑ ⎜ ⎟ k − j g 2 (0)∑ ⎜ ⎟ j −l g3 (0) l g 4 (0) ds k = 0 ⎝ k ⎠ ds j = 0 ⎝ j ⎠ ds l = 0 ⎝ l ⎠ ds n

where g (s) =

d d 4 ⎡⎣log(2π ) − s + log Γ( s + 1) + log cos(π s / 2) + log ς ( s) ⎤⎦ = ∑ log g p ( s ) ds ds p =1

Having regard to the four components g p ( s ) , the following derivatives are easily computed dm (2π ) − s = (−1) m (2π ) − s log m (2π ) m ds dm (2π ) − s m ds

= (−1) m log m (2π ) s =0

38

dm ⎛πs ⎞ ⎛π ⎞ ⎛ π s mπ ⎞ cos ⎜ ⎟ = ⎜ ⎟ cos ⎜ + ⎟ m ds 2 ⎠ ⎝ 2 ⎠ ⎝2⎠ ⎝ 2 m

dm ⎛πs ⎞ ⎛π ⎞ ⎛ mπ ⎞ cos ⎜ ⎟ = ⎜ ⎟ cos ⎜ ⎟ m ds ⎝ 2 ⎠ s =0 ⎝ 2 ⎠ ⎝ 2 ⎠ m

and we obtain (5.5) j k ⎛n⎞ ⎛k ⎞ ⎛ j ⎞⎛ π ⎞ γ n −1 = 2∑ ⎜ ⎟ Γ ( n − k ) (1)∑ ⎜ ⎟ (−1) k − j log k − j (2π )∑ ⎜ ⎟ ⎜ ⎟ k =0 ⎝ k ⎠ j =0 ⎝ j ⎠ l =0 ⎝ l ⎠ ⎝ 2 ⎠ n

j −l

⎛ ( j − l )π cos ⎜ ⎝ 2

⎞ (l ) ⎟ ς (0) ⎠

This extends the formula previously obtained by Apostol [1]. We note from (A.7) in Appendix A that the derivatives of the gamma function may be expressed in terms of the (exponential) complete Bell polynomials Γ ( m ) (1) = Ym (−γ , x1 ,..., xm −1 ) where x p = (−1) p +1 p !ς ( p + 1) . □ Using (3.3) we see that s ⎤ σ d d ⎡∞ ⎛ s⎞ [( s − 1)ς ( s )] = −( s − 1)ς ( s ) ⎢ ∑ (−1) k k ( s − 1) k + log Γ ⎜1 + ⎟ + log π 2 ⎥ ds ds ⎣ k =1 k ⎝ 2⎠ ⎦

As above we see that k ⎛n⎞ ⎛k ⎞ Yn (a1 + b1 + c1 ,..., an + bn + cn ) = ∑ ⎜ ⎟ Yn − k (a1 ,..., an − k )∑ ⎜ ⎟ Yk − j (b1 ,..., bk − j )Y j (c1 ,..., c j ) k =0 ⎝ k ⎠ j =0 ⎝ j ⎠ n

but it is not clear whether the above methodology will result in an easily manipulated formula. 6. A formula for the Stieltjes constants in terms of the (exponential) complete Bell polynomials containing the eta constants η n as the arguments

We refer to (4.1)

39

∞ ς ′( s) 1 =− − ∑ηk ( s − 1) k ς ( s) s − 1 k =0

and it is easily seen that this may equivalently be written as ∞ d [( s − 1)ς ( s )] = −( s − 1)ς ( s )∑η k ( s − 1) k ds k =0

As noted by Coffey [9] for differentiable functions f ( s) and g ( s) such that f ′( s ) = g ( s ) f ( s ) then we have in terms of the (exponential) complete Bell polynomials

(

f ( n +1) ( s ) = f ( s )Yn +1 g ( s ), g ′( s ),..., g ( n ) ( s )

)

In this particular case we have ∞

f ( s) = ( s − 1)ς ( s) f (1) = 1 and g ( s ) = −∑η k ( s − 1) k g (1) = −η0 = γ k =0

∞

g ( j ) ( s ) = −∑ηk k (k − 1) ⋅⋅⋅ (k − j + 1)( s − 1) k − j k =0

g ( j ) (1) = − j !η j

We recall from (2.10) that d n +1 [( s − 1)ς ( s )] = (−1) n (n + 1)γ n n +1 ds s =1

and we then obtain the relationship for n ≥ 0 (6.1)

(−1) n (n + 1)γ n = Yn +1 ( γ , −1!η1 ,..., −n !η n )

For example with n = 0 and using (3.24) Y1 ( x1 ) = x1 we recover γ 0 = Y1 (γ ) = γ . Similarly with n = 1 we have using Y2 ( x1 , x2 ) = x12 + x2 −2γ 1 = Y2 ( γ , −1!η1 ) = γ 2 − η1

which gives us 40

η1 = γ 2 + 2γ 1 As a third example, using Y3 ( x1 , x2 , x3 ) = x13 + 3x1 x2 + x3 we obtain 3γ 2 = Y3 ( γ , −1!η1 , −2!η2 ) = γ 3 − 3γη1 − 2η 2

2η 2 = γ 3 − 3γη1 − 3γ 2 = γ 3 − 3γ (γ 2 + 2γ 1 ) − 3γ 2 and therefore we have 3γ 2 = −2γ 3 − 6γγ 1 − 2η 2 We may write (6.1) as (−1) n (n + 1)γ n = Yn +1 ( −0!η0 , −1!η1 ,..., − n !ηn )

Using the recurrence relation [28] n n ⎛n⎞ ⎛n⎞ Yn +1 ( x1 ,..., xn +1 ) = ∑ ⎜ ⎟ Yn − k ( x1 ,..., xn − k ) xk +1 = ∑ ⎜ ⎟ Yk ( x1 ,..., xk ) xn − k +1 k =0 ⎝ k ⎠ k =0 ⎝ k ⎠ n ⎛n⎞ = xn +1 + ∑ ⎜ ⎟ Yk ( x1 ,..., xk ) xn − k +1 k =1 ⎝ k ⎠

where xr = −(r − 1)!ηr −1 We see that n ⎛n⎞ (−1) n (n + 1)γ n = −n !η n + ∑ ⎜ ⎟ (−1) k kγ k −1 (n − k )!ηn − k k =1 ⎝ k ⎠

and hence we recover (4.4). From (1.7) we have ∞

σk

k =1

k

ξ ′( s ) = −ξ ( s)∑ (−1) k

k ( s − 1) k −1

∞

σk

k =1

k

and in this case we designate g ( s ) = −∑ (−1) k

41

k ( s − 1) k −1

∞

σk

k =0

k

g ( s ) = −∑ (−1) k ( j)

k (k − 1) ⋅⋅⋅ (k − j )( s − 1) k −1− j

g ( j ) (1) = (−1) j j !σ j +1

ξ ( n ) ( s) = ξ ( s )Yn ( g ( s ), g ′( s ),..., g ( n −1) ( s ) ) 1 2

ξ ( n ) (1) = Yn (σ 1 , −1!σ 2 ,..., (−1)n −1 (n − 1)!σ n ) We have 1 2

ξ ( n +1) (1) = Yn (σ 1 , −1!σ 2 ,..., (−1) n n !σ n +1 ) and using n ⎛n⎞ Yn +1 ( x1 ,..., xn +1 ) = xn +1 + ∑ ⎜ ⎟ Yk ( x1 ,..., xk ) xn − k +1 k =1 ⎝ k ⎠

we obtain (6.2)

1

n

⎛n⎞

k =1

⎝ ⎠

ξ ( n +1) (1) = (−1) n n !σ n +1 + ∑ ⎜ ⎟ (−1) n −k +1ξ ( k ) (1)(n − k + 1)!σ n − k k 2

Appendix A A brief survey of the (exponential) complete Bell polynomials

The (exponential) complete Bell polynomials may be defined by Y0 = 1 and for n ≥ 1 (A.1)

Yn ( x1 ,..., xn ) = ∑

π (n)

k

k

1 2 n! ⎛ xn ⎞ ⎛ x1 ⎞ ⎛ x2 ⎞ ⎜ ⎟ ⎜ ⎟ ... ⎜ ⎟ k1 ! k2 !... kn ! ⎝ 1! ⎠ ⎝ 2! ⎠ ⎝ n! ⎠

kn

where the sum is taken over all partitions π (n) of n , i.e. over all sets of integers k j such that (A.2)

k1 + 2k2 + 3k3 + ... + nkn = n

The complete Bell polynomials have integer coefficients and, by way of illustration, the first six are set out below [14, p.307]

42

Y1 ( x1 ) = x1

(A.3)

Y2 ( x1 , x2 ) = x12 + x2 Y3 ( x1 , x2 , x3 ) = x13 + 3x1 x2 + x3 Y4 ( x1 , x2 , x3 , x4 ) = x14 + 6 x12 x2 + 4 x1 x3 + 3 x22 + x4 Y5 ( x1 , x2 , x3 , x4 , x5 ) = x15 + 10 x13 x2 + 10 x12 x3 + 15 x1 x22 + 5 x1 x4 + 10 x2 x3 + x5 Y6 ( x1 , x2 , x3 , x4 , x5 , x6 ) = x16 + 6 x1 x5 + 15 x2 x4 + 10 x23 + 15 x12 x4 + 15 x23 + 60 x1 x2 x3 +20 x13 x3 + 45 x12 x22 + 15 x14 x1 + x6 The complete Bell polynomials are also given by the exponential generating function in Comtet’s book [14, p.134] ∞ ∞ ⎛ ∞ tj ⎞ tn tn exp ⎜ ∑ x j ⎟ = 1 + ∑ Yn ( x1 ,..., xn ) = ∑ Yn ( x1 ,..., xn ) n ! n =0 n! n =1 ⎝ j =1 j ! ⎠

(A.4)

Let us now consider a function f ( x) which has a Taylor series expansion around x : we have ⎛ ∞ ⎛ ∞ tj ⎞ tj ⎞ e f ( x +t ) = exp ⎜ ∑ f ( j ) ( x) ⎟ = e f ( x ) exp ⎜ ∑ f ( j ) ( x) ⎟ j!⎠ j!⎠ ⎝ j =0 ⎝ j =1 ∞ ⎧ tn ⎫ = e f ( x ) ⎨1 + ∑ Yn ( f (1) ( x), f (2) ( x),..., f ( n ) ( x) ) ⎬ n !⎭ ⎩ n =1

We see that d m f ( x ) ∂ m f ( x +t ) ∂ m f ( x +t ) e = e = e m dx m ∂x m ∂ t t =0 t =0

and we therefore obtain (as noted by Kölbig [22] and Coffey [9]) (A.5)

d m f ( x) e = e f ( x )Ym ( f (1) ( x), f (2) ( x),..., f ( m ) ( x) ) m dx

Differentiating (A.5) we see that

43

d Ym ( f (1) ( x), f (2) ( x),..., f ( m ) ( x) ) = Ym +1 ( f (1) ( x), f (2) ( x),..., f ( m +1) ( x) ) dx − f (1) ( x)Ym ( f (1) ( x), f (2) ( x),..., f ( m ) ( x) ) As an example of (A.5), letting f ( x) = log Γ( x) we obtain (A.6)

d m log Γ ( x ) = Γ ( m ) ( x) = Γ( x)Ym (ψ ( x),ψ (1) ( x),...,ψ ( m −1) ( x) ) e m dx ∞

= ∫ t x −1e −t log m t dt 0

and since [30, p.22]

ψ ( p ) ( x) = (−1) p +1 p !ς ( p + 1, x) we may express Γ ( m ) ( x) in terms of ψ ( x) and the Hurwitz zeta functions. In particular, Coffey [9] notes that (A.7)

Γ ( m ) (1) = Ym (−γ , x1 ,..., xm−1 )

where x p = (−1) p +1 p !ς ( p + 1) . Values of Γ ( m ) (1) are reported in [30, p.265] for m ≤ 10 and the first three are Γ (1) (1) = −γ Γ (2) (1) = ς (2) + γ 2 Γ (3) (1) = −[2ς (3) + 3γς (2) + γ 3 ] The general form is m

Γ ( m ) (1) = (−1) m ∑ ε mj j =1

where ε mj are positive constants. □

44

The following is extracted from an interesting series of papers written by Snowden [29, p.68]. Let us consider the function f ( x) with the following Maclaurin expansion ∞

(A.8)

bn n x n =1 n

log f ( x) = b0 + ∑

and we wish to determine the coefficients an such that ∞

(A.9)

f ( x) = ∑ an x n n =0

By differentiating (A.8) and multiplying the two power series, we get n

(A.10)

nan = ∑ bk an − k k =1

Upon examination of this recurrence relation Snowden reports it is easy to see that (A.11)

n !an = a0 [b1 , −b2 , b3 ,..., (−1) n +1 bn ]

where the symbol [c1 , c2 , c3 ,..., cn ] is defined as the n × n determinant c1

c2

c3

c4 c3 (n − 1) c1 c2 c2 0 (n − 2) c1 0 0 (n − 3) c1 #

#

#

0

0

0

.

.

. . .

. . .

.

cn . cn −1 . cn − 2 . cn −3

# # # # 0 0 0 1

# c1

Since log f (0) = log a 0 = b0 we have (A.12)

∞ ⎡ xn ⎤ f ( x) = eb0 ⎢1 + ∑ [b1 , −b2 , b3 ,..., (−1) n +1 bn ] ⎥ n! ⎦ ⎣ n =1

Multiplying (A.9) by α it is easily seen that (after correcting the misprint in Snowden’s paper)

45

(A.13)

∞ ⎡ xn ⎤ f α ( x) = eα b0 ⎢1 + ∑ [α b1 , −α b2 , α b3 ,..., (−1) n +1α bn ] ⎥ n! ⎦ ⎣ n =1

and, in particular, with α = −1 we obtain (A.14)

∞ ⎡ xn ⎤ n + − − − 1 [ , , ,..., ( 1) ] b b b b n ⎢ ∑ 1 2 3 ⎥ n! ⎦ ⎣ n =1

1 = e −b0 f ( x)

Differentiating (A.13) with respect to α would give us an expression for f α ( x) log f ( x) . Differentiating (A.8) gives us ∞

f ′( x) = f ( x)∑ bk x k −1 = f ( x) g ( x) k =1

and hence

(

f ( n ) ( x) = f ( x)Yn g ( x),..., g ( n −1) ( x)

)

Alternatively, differentiating (A.9) results in ∞

f ( n ) ( x) = ∑ ak k (k − 1) ⋅⋅⋅ (k − n + 1) x k − n k =0

and letting x = 0 gives us (A.15)

n !an = eb0 Yn ( b1 ,1!b2 ,..., (n − 1)!bn )

From (A.11) we then see that (A.16)

Yn ( b1 ,1!b2 ,..., (n − 1)!bn ) = [b1 , −b2 , b3 ,..., ( −1) n +1 bn ]

and more generally (A.17)

⎡x xn ⎤ x Yn ( x1 ,..., xn ) = ⎢ 1 , − 2 ,..., (−1) n +1 (n − 1)!⎥⎦ ⎣ 0! 1!

We now write (A.13) as ∞ ⎡ xn ⎤ f α ( x) = eα b0 ⎢1 + ∑ Dn (α ) ⎥ n! ⎦ ⎣ n =1

46

where for convenience we have designated Dn (α ) by Dn (α ) = [α b1 , −α b2 , α b3 ,..., (−1) n +1α bn ] Letting α = 0 we see that ∞

1 = 1 + ∑ Dn (0) n =1

xn n!

and therefore we have Dn (0) = 0 which is only to be expected since all of the elements of the top row of the defining determinant are equal to zero. Differentiation with respect to α results in ∞

f α ( x) log f ( x) = eα b0 ∑ Dn′ (α ) n =1

xn + b0 eα b0 n!

∞ ⎡ xn ⎤ + D 1 ( α ) ⎢ ∑ n ⎥ n! ⎦ ⎣ n =1

Letting α = 0 gives us ∞

log f ( x) = ∑ Dn′ (0) n =1

∞ ⎡ xn xn ⎤ + b0 ⎢1 + ∑ Dn (0) ⎥ n! n! ⎦ ⎣ n =1

∞

= b0 + ∑ Dn′ (0) n =1

xn n!

We started out with (A.8) ∞

bn n x n =1 n

log f ( x) = b0 + ∑

and, hence equating coefficients in the two power series, we obtain Dn′ (0) = (n − 1)!bn We have

47

∞ dm α n(n − 1) ⋅⋅⋅ (n − m + 1) x n − m α b0 f ( x ) = e D ( ) α ∑ n dx m n! n =1 ∞ d dm α n(n − 1) ⋅⋅⋅ (n − m + 1) x n − m α b0 ′ f ( x ) = e D ( ) α ∑ n dα dx m n! n =1 ∞

+b0 eα b0 ∑ Dn (α ) n =1

n(n − 1) ⋅⋅⋅ (n − m + 1) x n − m n!

and hence ∞ d dm α n(n − 1) ⋅⋅⋅ (n − m + 1) f ( x ) = Dn′ (0) ∑ m dα dx n! n =1 x =1,α = 0

We then have ∞ bn d dm α ( ) = f x n(n − 1) ⋅⋅⋅ (n − m + 1) ∑ m dα dx n n = 1 x =1,α = 0

which is equivalent to ∞ bn dm f x n(n − 1) ⋅⋅⋅ (n − m + 1) log ( ) = ∑ m ds n =1 n x =1

REFERENCES

[1]

T.M. Apostol, Formulas for Higher Derivatives of the Riemann Zeta Function. Math. of Comp., 169, 223-232, 1985.

[2] V.K. Balakrishnan, Schaum’s Outline of Theory and Problems of Combinatorics. McGraw-Hill Inc, 1995. [3]

B.C. Berndt, Chapter eight of Ramanujan’s Second Notebook. J. Reine Agnew. Math, Vol. 338, 1-55, 1983. http://www.digizeitschriften.de/no_cache/home/open-access/nach-zeitschriftentiteln/

[4]

E. Bombieri and J.C. Lagarias, Complements to Li’s criterion for the Riemann hypothesis. J. Number Theory 77, 274-287 (1999). [PostScript] [Pdf]

[5] M.W. Coffey, Relations and positivity results for the derivatives of the Riemann ξ function. J. Comput. Appl. Math., 166, 525-534 (2004)

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[6]

M.W. Coffey, Toward verification of the Riemann Hypothesis: Application of the Li criterion.. Math. Phys. Analysis and Geometry, 8, 211-255, 2005. arXiv:math-ph/0505052 [ps, pdf, other], 2005.

[7]

M.W. Coffey, Polygamma theory, the Li/Keiper constants and validity of the Riemann Hypothesis. arXiv:math-ph/0507042 [ps, pdf, other], 2005. [8] M.W. Coffey, New results on the Stieltjes constants: Asymptotic and exact evaluation. J. Math. Anal. Appl., 317 (2006) 603-612. arXiv:math-ph/0506061 [ps, pdf, other] [9] M.W. Coffey, A set of identities for a class of alternating binomial sums arising in computing applications. 2006. arXiv:math-ph/0608049v1 [10] M.W. Coffey, New summation relations for the Stieltjes constants Proc. R. Soc. A ,462, 2563-2573, 2006. [11] M.W. Coffey, The Stieltjes constants, their relation to the η j coefficients, and representation of the Hurwitz zeta function. arXiv:math-ph/0612086 [ps, pdf, other], 2007. [12] M.W. Coffey, Series of zeta values, the Stieltjes constants, and a sum Sγ (n) . arXiv:0706.0345 [ps, pdf, other], 2007. [13] M.W. Coffey, New results concerning power series expansions of the Riemann xi function and the Li/Keiper constants. (preprint) 2007. [14] L. Comtet, Advanced Combinatorics, Reidel, Dordrecht, 1974.

[15] D.F. Connon, Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers. Volume II(b), 2007. arXiv:0710.4024 [pdf]

[16] D.F. Connon, Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers. Volume VI, 2007. arXiv:0710.4032 [pdf]

[17] D.F. Connon, Some applications of the Stieltjes constants. arXiv:0901.2083 [pdf], 2009. [18] P. Freitas, A Li-type criterion for zero-free half-planes of Riemann's zeta function. arXiv:math/0507368 [ps, pdf, other], 2005. [19] H. Hasse, Ein Summierungsverfahren für Die Riemannsche ς - Reithe. Math. Z.32, 458-464, 1930. http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D23956&p=462

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[20] A. Ivić, The Riemann Zeta- Function: Theory and Applications. Dover Publications Inc, 2003. [21] J.B. Keiper, power series expansions of Riemann’s ξ function. Math. Comp.58, 765-773 (1992). [22] K.S. Kölbig, The complete Bell polynomials for certain arguments in terms of Stirling numbers of the first kind. J. Comput. Appl. Math. 51 (1994) 113-116. Also available electronically at: A relation between the Bell polynomials at certain arguments and a Pochhammer symbol. CERN/Computing and Networks Division, CN/93/2, 1993. http://doc.cern.ch/archive/electronic/other/preprints//CM-P/CM-P00065731.pdf [23] J.C. Lagarias, On a positivity property of the Riemann xi function. Acta Arithmetica, 89 (1999), No. 3, 217-234 . [PostScript] [Pdf] [24] X.-J. Li, The positivity of a sequence of numbers and the Riemann Hypothesis. J. Number Theory, 65, 325-333, 1997. [25] K. Maślanka, Effective method of computing Li’s coefficients and their properties. math.NT/0402168 [abs, ps, pdf, other] [26] K. Maślanka, An explicit formula relating Stieltjes constants and Li’s numbers. math.NT/0406312 [abs, ps, pdf, other] [27] Y. Matsuoka, A note on the relation between generalized Euler constants and the zeros of the Riemann zeta function. J. Fac. Shinshu Univ. 53, 81-82, 1985. [28] J. Riordan, An introduction to combinatorial analysis. Wiley, 1958. [29] A. Snowden, Collection of Mathematical Articles. 2003. http://www.math.umd.edu/~asnowden/math-cont/dorfman.pdf [30] H.M. Srivastava and J. Choi, Series Associated with the Zeta and Related Functions. Kluwer Academic Publishers, Dordrecht, the Netherlands, 2001. [31] Z.X. Wang and D.R. Guo, Special Functions. World Scientific Publishing Co Pte Ltd, Singapore, 1989. Donal F. Connon Elmhurst Dundle Road Matfield Kent TN12 7HD [email protected]

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