A Series Representation for Riemann's Zeta Function and some Interesting Identities that Follow
aa r X i v : . [ m a t h . C A ] S e p A Series Representation for Riemann’s Zeta Function and some Interesting Identities that Follow.Michael Milgram Consulting Physicist, Geometrics Unlimited, Ltd.Box 1484, Deep River, Ont. Canada. K0J 1P0Sept. 9, 2020MSC classes: 11M06, 11M26, 11M35, 11M99, 26A09, 30B40, 30E20, 33C20, 33B20, 33B99Keywords: Riemann Zeta Function, Dirichlet Eta function, alternating Zeta function, incompleteZeta function, Generalized Exponential Integral, Bernoulli numbers, Euler numbers, Harmonicnumbers, infinite series, evaluation of integrals
Abstract
Using Cauchy’s Integral Theorem as a basis, what may be a new series representation forDirichlet’s function η ( s ), and hence Riemann’s function ζ ( s ), is obtained in terms of theExponential Integral function E s ( iκ ) of complex argument. From this basis, infinite sums areevaluated, unusual integrals are reduced to known functions and interesting identities areunearthed. The incomplete functions ζ ± ( s ) and η ± ( s ) are defined and shown to be intimatelyrelated to some of these interesting integrals. An identity relating Euler, Bernouli and Harmonicnumbers is developed. It is demonstrated that a known simple integral with complex endpointscan be utilized to evaluate a large number of different integrals, by choosing varying pathsbetween the endpoints. In a recent paper [1], I have developed an integral equation for Riemann’s function ξ ( s ), based onLeClair’s series [2] involving the Generalized Exponential Integral. Glasser, in a subsequent paper[3], has shown that this integral equation is equivalent to the standard statement of Cauchy’sIntegral Theorem applied to ξ ( s ), from which he closed the circle [3, Eq. (13)] by re-derivingLeClair’s original series representation. This raises the question of what will be found if Cauchy’sIntegral Theorem is applied to other functions related to Riemann’s function ζ ( s ), notably theso-called Dirichlet function η ( s ) (sometimes called the alternating zeta function), workingbackwards to a series representation? That question is the motivation for this work.Section 2 gives some preliminaries and develops the reduction of Cauchy’s Integral Theorem toyield a series representation of η ( s ) in terms of (generalized) Integro-Exponential functions [4] E s ( κ ) (a.k.a Incomplete Gamma functions - see (4.1)). Although it is well-known that ζ ( s ) can beexpressed [2], [5] in terms of such functions in various ways, (see Appendix B) the form derivedhere differs substantially from these since it involves a series that indexes half-integer values of theimaginary argument of E s ( iκ ) and does not possess generic s ↔ (1 − s ) symmetry. Severalsubsections explore variations of this series, revealing some interesting identities, not all of whichpertain to either ζ ( s ) or η ( s ). Section 3 presents a brief analysis of these results as they pertain to ζ ′ ( s ). In Section 4, the series representation is converted into an integral representation whoseendpoints are complex, and in a number of subsections this representation is reduced to a realintegral representation by choosing a number of different paths between the endpoints in the [email protected] • The main result upon which this paper is based (see (2.24)):(2 s − ζ ( s ) = − π s − sin ( πs/ ⌊ ( n − / ⌋ X j =0 Γ (2 j + 1 − s ) E (2 j )Γ (2 j + 1)+ π s − n e iπ n/
2Γ ( s − n ) cos ( πs/ ∞ X k =0 E n − s ( − iπ ( k + 1 / − n E n − s ( iπ ( k + 1 / k + 1 / n • An unusual recursion relation for Bernoulli numbers (see (2.32)): B N +2 = Γ (3 + 2 N )2 N (2 N +2 − N − X k =0 k (cid:0) − k +2 (cid:1) B k +2 Γ (1 − k + 2 N ) Γ (2 k + 3) + 1 + N N +1 (2 N +2 − , N ≥ , • The sum of an infinite series (see (2.35)): π p − m ( − m e − iπp/ Γ (1 + p ) ∞ X k =0 ( − p E − p ( iπ ( k + 1 / E − p ( − iπ ( k + 1 / k + 1 / − p +2 m = − m X j = ⌊− p/ m +1 / ⌋ E (2 j )Γ (2 m − j + 1) Γ (2 j + 1) . (1.1) • ζ (2 m + 1) expressed in terms of E ( κ ) (see (2.39)): (cid:0) m +1 − (cid:1) ζ (2 m + 1) = − ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (2 j + 1) Γ (1 − j + 2 m )+ iπ ∞ X k =0 E ( iπ ( k + 1 / − E ( − iπ ( k + 1 / k + 1 / m +1 . • Bernoulli, Euler and Harmonic numbers are related (see (2.46)): m − X j =0 E (2 j ) H m − j Γ (1 − j + 2 m ) Γ (2 j + 1) = m X k =1 k (cid:0) k − (cid:1) B k (1 − k + 2 m ) Γ (2 − k + 2 m ) Γ (2 k + 1) − m Γ (1 + 2 m )2 For integer p > m, m < c m < m + 1, a contour integral representation for ζ (2 m + 1) (see(E.16)): (cid:0) m +1 − (cid:1) ζ (2 m + 1) = i ( − p π p +1 Z c m + i ∞ c m − i ∞ Γ (2 p − t ) π t sin ( π t ) ∞ X k =0 ( − k ( k + 1 / t − p − m − d t − ( − m π m p − − m X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) + m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1) . • A divergent, asymptotic series representation for the alternating Hurwitz Zeta Function η (1 , ( t + 1) /
2) (see (2.64)): η (1 , ( t + 1) / ∼ m X k =0 E (2 k ) t k +1 ( t → ∞ ) , ∀ m ≥ . (1.2) • A representation for the lower incomplete Zeta function ζ − ( s ) (see (4.10)): ζ − ( s ) = 2 s − Z π/ − π/ e − π sin( θ ) / sin ( π cos ( θ ) / θ ( s − π sin ( θ )) − cos ( π cos ( θ )) d θ . • An interesting integral with 0 ≤ n (see (4.17)): Z π/ − π/ sin (( π/
2) cos ( θ ) + (2 n − θ ) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ( − n B n π n (cid:0) − n − (cid:1) Γ (2 n + 1) • An interesting integral with 2 n > Z π/ − π/ sin (( π/
2) cos ( θ ) − (2 n + 1) θ ) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 0 • An integral interesting on its own merits (see (4.29)): Z π/ − π/ cos ( θ ) sin (( π/
2) cos ( θ )) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 1 / π / , (1.3) • A theorem for p >
0, (see (D.2)):( − p Γ (2 p ) E p ( − iπ ( k + 1 / E p ( iπ ( k + 1 / π ( k + 1 / p − = 2 ( − k p − X j =1 ( − j Γ (2 j )( π ( k + 1 / j − π (k + 1 / − π/ • A hypergeometric identity (see (D.10)): F ( 12 − p ; 12 , − p ; − π ( k + 1 / − p +1 ( π ( k + 1 / p Γ (2 p − F ( 12 ; 32 ,
32 ; − π ( k + 1 / − k + p ( π ( k + 1 / p − Γ (2 p − p − X j =1 Γ (2 j ) ( − j ( π ( k + 1 / j A series representation for ζ ( s ) By way of review, the standard definition for Dirichlet’s function η ( s ) (also called the alternatingzeta function) is given by η ( s ) = ∞ X k =1 ( − k k − s (2.1)valid for all s , related to ζ ( s ) by η ( s ) = (1 − − s ) ζ ( s ) . (2.2)Another well-known representation is(2 s − ζ ( s ) = ∞ X k =0 / ( k + 1 / s , ℜ ( s ) > . (2.3)The Cauchy Integral Theorem for η ( s ), with an additional factor 2 − s embedded for convenience,is 2 − s (cid:0) − − s (cid:1) ζ ( s ) = 12 πi I (cid:0) − − v (cid:1) ζ ( v ) 2 − v s − v d v (2.4)where the contour of integration encloses the non-negative real axis as well as the pole at the point v = s in a clockwise direction. Notice that there is no singularity at v = 1 because of a cancellationbetween the first two numerator terms of the integrand at v = 1. Let v := v + 1 in (2.4) giving η ( s ) = (cid:0) − − s (cid:1) ζ ( s ) = 2 s − πi I (1 − − v ) ζ ( v + 1) 2 − v s − v − v (2.5)where the contour now crosses the real axis at v < min( − , ℜ ( s )). Apply the slightly rewrittenform of the well-known functional equation ζ ( v + 1) = ζ ( − v ) 2 v +1 π v cos ( π v/
2) Γ ( − v ) (2.6)to obtain, with the same contour, η ( s ) = (cid:0) − − s (cid:1) ζ ( s ) = 2 s − iπ I (1 − − v ) ζ ( − v ) Γ ( − v ) π v cos ( π v/ s − v − v . (2.7)Notice that the product ζ ( − v ) Γ( − v ) is non-singular on the non-negative real v − axis. Substitutethe cosine term in (2.7) as a sum of exponentials giving η ( s ) = (cid:0) − − s (cid:1) ζ ( s ) = 2 s − πi I (1 − − v ) ζ ( − v ) ( iπ ) v Γ ( − v ) s − v − v + { ( iπ ) v → ( − iπ ) v } . (2.8)where each exponential has been written ase ± iπv/ = ( ± i ) v . (2.9)Now apply (2.3) and interchange the summation and integration because for ℜ ( v ) < − − s (cid:0) − − s (cid:1) ζ ( s ) = − πi ∞ X k =0 I (cid:0) iπ (cid:0) k + (cid:1)(cid:1) v Γ ( − v ) s − v − v ! − πi ∞ X k =0 (cid:18)I ( − iπ ( k + 1 / v Γ ( − v ) s − v − v (cid:19) (2.10)Because the contour lies to the left of the point v = s −
1, each of the contour integrals in (2.10) isrecognizable as a contour integral representation of the Generalized Integro-Exponential function[4, Eq. (2.6a)] E s ( z ) = 12 πi I Γ( − v ) z v ( s − − v ) d v (2.11)4here the integration contour encloses the pole at v = s − v − axis in aclockwise direction. Another alternate (defining) representation is E s ( iπ κ ) ≡ Z ∞ v − s e − iπκv d v . (2.12)When (2.11) is substituted into (2.10) we find the (slowly converging - see Section 2.1) seriesrepresentation ζ ( s ) = 2 s − − s − ∞ X k =0 ( E s ( iπ ( k + 1 / E s ( − iπ ( k + 1 / . (2.13) Remark:
Because it has never been proven ([10, Section 2.13]) that the functional equation (2.6) uniquely defines ζ ( s ) (and therefore η ( s )), up to the point in (2.10) where (2.3) is applied, (2.8) isvalid for any function (that could have been labelled ζ ( s ) arbitrarily) which might obey (2.6) inany of its forms. Although (2.13) converges slowly numerically, it does converge. To test this observation, define t ± k ( s ) = E s ( iπ ( k + 1 / ± E s ( − iπ ( k + 1 / k → ∞ t + k ( s ) ∼ ( − k (cid:18) − π k + 1 π k + 1 π k (cid:18) −
12 + 2 s ( s + 1) π (cid:19)(cid:19) + O (cid:0) k − (cid:1) . (2.15)Applying Gauss’ test gives (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t + k +1 ( s ) t + k ( s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∼ − k + 32 k + O (cid:0) k − (cid:1) (2.16)indicating ambiguity. However, from (2.15), we see that, for finite values of s , | t + k ( s ) | → t − k ( s ) we have t − k ( s ) ∼ is ( − k π (cid:18) − k + 2 k + 1 k (cid:18) −
32 + 2 ( s + 1) ( s + 2) π (cid:19)(cid:19) (2.17)and, with respect to Gauss’ test, (cid:12)(cid:12)(cid:12)(cid:12) t − k +1 ( s ) t − k ( s ) (cid:12)(cid:12)(cid:12)(cid:12) = 1 − k + O (cid:0) k − (cid:1) (2.18)which proves convergence.In a later Section, we will consider sums involving t ± k ( s + p ) / ( k + 1 / p where p >
0. Testing suchsums for p := 2 p , gives (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t + k +1 ( s + 2 p ) t + k ( s + 2 p ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∼ − pk , (2.19)proving absolute convergence by Gauss’ test, and for p := 2 p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − k +1 ( s + 2 p − t − k ( s + 2 p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∼ − pk , (2.20)again proving absolute convergence by the same test.5 .2 Recursion The recursion rule for E s ( z ) is E s ( z ) = zE s − ( z ) − e − z − s s = 1 , (2.21)leading, after the use of forward recursion, to η ( s ) = (1 − − s ) ζ ( s ) = 2 s − isπ ∞ X k =0 E s +1 ( − iπ ( k + 1 / − E s +1 ( iπ ( k + 1 / k + 1 / s − (2.22)with improved convergence properties (see (2.19)). The general form of (2.22) can be obtainedsimilarly by induction, giving for any n ≥ η ( s ) = (cid:0) − − s (cid:1) ζ ( s ) = 2 s Γ ( s ) ⌊ ( n − / ⌋ X j =0 ( − j Γ ( s + 2 j ) π j +1 ∞ X k =0 ( − k ( k + 1 / j +1 − s − Γ ( n + s )Γ ( s ) (cid:18) − π (cid:19) n e iπ n/ ∞ X k =0 (cid:18) E n + s ( − iπ ( k + 1 / − n E n + s ( iπ ( k + 1 / k + 1 / n (cid:19) . (2.23)As n increases in (2.23), the two terms composing the right-hand side converge to each other, sonumerically significant error occurs at large values of n , although the convergence rate of thesecond infinite sum improves also. An equivalent form may be obtained by replacing s := 1 − s and applying the functional equation (2.6) to obtain(2 s − ζ ( s ) = − π s − sin ( πs/ ⌊ ( n − / ⌋ X j =0 Γ (2 j + 1 − s ) E (2 j )Γ (2 j + 1)+ π s − n e iπ n/
2Γ ( s − n ) cos ( πs/ ∞ X k =0 E n − s ( − iπ ( k + 1 / − n E n − s ( iπ ( k + 1 / k + 1 / n (2.24)In (2.24), the infinite inner sum appearing in the first term of (2.23) has been replaced by aknown result in terms of Euler numbers E (2 j ) - see (C.27). The application of the result (2.23) isfairly straightforward for both even and odd values of n if s = ± m . These exceptional cases areinvestigated in the following subsections. (2.23) with n even or odd yields a recursion formula for Bernoulli numbers Consider the case n := 2 n . Application of the reversed logic presented in Section 2 applied to(2.23) along with (2.12) leads to ∞ X k =0 E s +2 n ( − iπ ( k + 1 / E s +2 n ( iπ ( k + 1 / k + 1 / n = 2 Z ∞ ∞ X k =0 v − s − n cos ( π ( k + 1 / v )( k + 1 / n d v . (2.25)Expand the integration Z ∞ ... d v = Z ∞ ... d v − Z ... d v (2.26)interchange the (convergent) sum and integral, and consider the first integration term on the rightof (2.26) 6 X k =0 Z ∞ v − s − n cos ( π ( k + 1 / v )( k + 1 / n d v = ( − n π s +2 n − ∞ X k =0 ( k + 1 / s − Γ (1 − s − n ) sin ( πs/ − n π s +2 n − (cid:0) − s − (cid:1) ζ (1 − s ) Γ (1 − s − n ) sin ( πs/ π n (cid:0) − s − (cid:1) ζ ( s ) ( − n Γ ( s )2 s Γ ( s + 2 n ) . (2.27)The first equality in (2.27) follows from the application of (C.13) with − < −ℜ ( s ) − n <
0, thesecond equality follows from the first by the principle of analytic continuation using (2.3) with ℜ ( s ) <
0, and the third equality follows from simple identities and invocation of the functionalequation (2.6) with v := v − Z v − s − n ∞ X k =0 cos ( π ( k + 1 / v )( k + 1 / n d v = (2 π ) n ( − n n ) Z v − s − n E n − ( v/
2) d v = ( − n +1 (2 π ) n n − X k =0 (cid:0) − k − − k +2 n (cid:1) B n − k Γ ( k + 1) Γ (1 − k + 2 n ) ( − s + 2 n − k ) (2.28)The second equality in (2.28) arises from the definition (C.18) followed by straightforwardintegration with − s − n > −
1. Now apply (2.25), (2.26), (2.27) and (2.28) into (2.23), along withthe use of (C.27) to identify the last (inner) sum of (2.23) in terms of Euler numbers, to arrive atthe identity n − X j =0 Γ (2 j + s ) E (2 j )Γ (2 j + 1) = 2 n Γ ( s + 2 n ) n − X k =0 (cid:0) − k − n − k (cid:1) B n − k Γ ( k + 1) Γ (1 − k + 2 n ) (2 n − s − k ) . (2.29)This result can be used to obtain an interesting recursion relation among Bernoulli numbers B n due to the presence of the arbitrary parameter s . To solve (2.29), first let n := N , then let n := N + 1 and subtract, giving E (2 N )Γ (2 N + 1) = − N N − X k =0 B N − k (cid:0) − k − − k +2 N (cid:1) Γ ( k + 1) Γ ( − k + 2 N + 1) ( − s + 2 N − k )+ Γ ( s + 2 N + 2) 2 N +2 Γ ( s + 2 N ) N +1 X k =0 B − k +2 N (cid:0) − k − − k +2 N (cid:1) Γ ( k + 1) Γ (3 − k + 2 N ) (1 + s + 2 N − k ) , (2.30)which, after shifting the index of the second sum by two, reorganizing, and recognizing that B N +1 = 0 (except define B = − / E (2 N )Γ (2 N + 1) = X ( N ) s + Y ( N ) , (2.31)where X ( N ) and Y ( N ) are complicated functions only of the variable N . Because the left-handside of (2.31) does not depend on s , it must be that X ( N ) = 0, and when X ( N ) = 0 is simplified,one obtains B N +2 = Γ (3 + 2 N )2 N (2 N +2 − N − X k =0 k (cid:0) − k +2 (cid:1) B k +2 Γ (1 − k + 2 N ) Γ (2 k + 3) + 1 + N N +1 (2 N +2 − , N ≥ , (2.32)7ne of a large number of possible recursive definitions of Bernoulli numbers B N (e.g. [11, Eqs. (4)and (6)], [12] and compare with [13, Eq. 24.5.8]). This result can alternatively be derived from(C.19) using the fact that E (2 n + 1) = 0 as well as from a convolution type of identity forBernoulli polynomials at x = 1 / x = 1 /
4, the latter (privately) communicated by C. Vignat,Tulane University. The remaining term Y ( N ) of (2.31) eventually simplifies to reproduce (C.19).By following steps similar to those given above, the same results emerge for the case n := 2 n + 1. (2.24) with s = 2 m The evaluation of (2.24) with either s = 2 m or s = 2 m + 1 requires that limits be taken. Considerthe case s = 2 m . Substitute s = 2 m + ǫ in (2.24) and evaluate the limit ǫ → (cid:0) − m (cid:1) ζ (2 m ) = ( − m π m m − X j =0 E (2 j )Γ (2 m − j ) Γ (2 j + 1) , (2.33)a result equivalent to [13, Eq. 2.4.15], valid for all non-negative values of the arbitrary parameter n . Also, note both [14, Eq. (3.1)] as well as [13, Eq. 24.5.4] where we find the related identity m X j =0 E (2 j )Γ (2 m − j + 1) Γ (2 j + 1) = 0 . (2.34) (2.24) with s = 2 m + 1 , n < m + 1 , and m ≥ For the case s = 2 m + 1, again set s = 2 m + 1 + ǫ and evaluate the limit ǫ → n < m + 1.In this case, on the right-hand side, a term of order ǫ − arises, the coefficient of which mustvanish (because the left-hand side is not divergent) giving rise to an identity as follows. Let n = 2 m − p with 0 ≤ p ≤ m and use (2.34) to find π p − m ( − m e − iπp/ Γ (1 + p ) ∞ X k =0 ( − p E − p ( iπ ( k + 1 / E − p ( − iπ ( k + 1 / k + 1 / − p +2 m = − m X j = ⌊− p/ m +1 / ⌋ E (2 j )Γ (2 m − j + 1) Γ (2 j + 1) . (2.35)Notice that if p = 2 r or p = 2 r + 1, the right-hand side of (2.35) is invariant for constant values of m and therefore so is the left-hand side, leading to some interesting identities. Suppose p = 2 m , inwhich case, the right-hand side of (2.35) vanishes by (2.34), giving ∞ X k =0 E − m ( iπ ( k + 1 / E − m ( − iπ ( k + 1 / . (2.36)Further, if p = 2 m −
1, by extending the lower limit of the finite sum in (2.35), we similarly find ∞ X k =0 E − m ( iπ ( k + 1 / − E − m ( − iπ ( k + 1 / k + 1 / iπ m , (2.37)and, if p = 2 m − ∞ X k =0 E − m ( iπ ( k + 1 / E − m ( − iπ ( k + 1 / k + 1 / = − π Γ (2 m − m + 1) (2.38) Remark.
Replacing E − p ( ± iπ ( k + 1 / s = − m in (2.13) with recourse to (2.36) yields8he well-know result ζ ( − m ) = 0 - the trivial zeros. Finally, although (2.37) and (2.38), related by p = 2 r and p = 2 r + 1 with r = m − p -dependent factors have been transposed to the right-handside for the sake of uniformity of presentation.Consider now the terms associated with ǫ . The analysis is long and complicated, involving thederivative of E s ( z ) with respect to its order s (see [4]) as well as considerable simplification, all ofwhich employ identities listed in [4, Eqs. (2.1), (2.20c), and (2.21)]. As before, the result isindependent of the parameter n , so, a simple instance of the final result with n = 0 is (cid:0) m +1 − (cid:1) ζ (2 m + 1) = − ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (2 j + 1) Γ (1 − j + 2 m )+ iπ ∞ X k =0 E ( iπ ( k + 1 / − E ( − iπ ( k + 1 / k + 1 / m +1 . (2.39)For other values of n , the equivalent of (2.39) reduces to a tautology. Notice that both sides of(2.39) are singular if m = 0 because of (2.37). (2.24) with s = 2 m + 1 and n > m For the same case ( s = 2 m + 1) this time with n > m , in order to repeat the same procedure asabove, it is necessary to split the finite sum in (2.24) into two parts: 0 ≤ j ≤ m and m + 1 ≤ j ≤ ⌊ ( n − / ⌋ . The numerator and denominator of the first of these sums both vanishwhen s → m + 1 so a limit must be evaluated, givinglim s → m +1 m X j =0 E (2 j )Γ ( s − j ) Γ (2 j + 1)cos ( π s/ = ( − m π m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1) . (2.40)Set n = 2 m + p, p >
0, and after some simplification (2.24) becomes (cid:0) m +1 − (cid:1) ζ (2 m + 1) = ( − p Γ ( p ) e iπp/ π p ∞ X k =0 E p ( − iπ ( k + 1 / − p E p ( iπ ( k + 1 / k + 1 / p +2 m − ( − m π m ⌊ p/ m +1 / ⌋− X j = m +1 Γ (2 j − m ) E (2 j )Γ (2 j + 1) − ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1)(2.41)Notice that the left-hand side of (2.41) is independent of p and therefore so must be theright-hand side. An interesting identity emerges from (2.41) by applying (2.12) to the first term on its right-handside with p = 1 and identifying the sine integral Si as follows Z ∞ sin ( π ( j + 1 / v ) v d v = − Si ( π ( j + 1 / π/ π/ − Z sin ( π ( j + 1 / v ) v d v . (2.42)Each of the terms on the right-hand side of the second equality of (2.42) can be summed over theindex j according to (2.35) with recourse to (2.3), (C.16), (C.17) and straightforward integration,9n that order ( Remark: the term corresponding to j = 0 vanishes because, for m > B m +1 = 0), to eventually yield the identity m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1) = m − X j =0 (cid:0) j − j +1 (cid:1) B j +1 (2 m − j ) Γ (1 − j + 2 m ) Γ ( j + 2) . (2.43)Also see (2.39). Since ψ (1) = − γ , (2.43) is intriguing because it suggests the existence of analgebraic relationship between the Euler-Mascheroni constant γ , whose rationality status isunknown, and the irrational number π embedded in the Bernoulli numbers B j +1 . This is not thecase however. Proof:
From its recursion property ψ (1 − j + 2 m ) = m − j − X k =0 m − j − k ) − γ , (2.44)substitute (2.44) into (2.43), reverse the inner sum and separate all terms containing γ to find m X j =0 E (2 j ) m − j − X k =0 k + 1Γ (1 − j + 2 m ) Γ (2 j + 1) − γ m X j =0 E (2 j )Γ (1 − j + 2 m ) Γ (2 j + 1)= m − X k =0 (cid:0) k − k +1 (cid:1) B k +1 (2 m − k ) Γ (1 − k + 2 m ) Γ ( k + 2) . (2.45)Note that the coefficient of the γ term in (2.45) vanishes by (2.34) and therefore, effectively, γ isnot embedded in (2.43). QED
From (2.45), after setting k := 2 k + 1 on the right-hand side we are left with an interestingidentity that relates Euler, Bernoulli and Harmonic numbers: m − X j =0 E (2 j ) H m − j Γ (1 − j + 2 m ) Γ (2 j + 1) = m X k =1 k (cid:0) k − (cid:1) B k (1 − k + 2 m ) Γ (2 − k + 2 m ) Γ (2 k + 1) − m Γ (1 + 2 m )(2.46)This result can be used to obtain a recursive formula for E (2 j ) by inserting (C.21) into (2.46) aftersetting m := m + 1, and then considering the final term ( k = m ) in each sum to eventually obtain E (2 m ) = − m − X k =0 E (2 k )Γ (2 k + 1) m − k X j =0 Γ (1 + 2 m )Γ (2 m + 2 − k − j ) (2 j + 1) Γ (2 + 2 j ) (2.47)+4 Γ (1 + 2 m ) m − X k =0 E (2 k ) H m − k +2 Γ (3 − k + 2 m ) Γ (2 k + 1) + 1(1 + 2 m ) ( m + 1) . (2.48)Further, by substituting the integral representation of H m − j (see [15, first entry, with t → − x ]) into the left-hand side of (2.46), inverting the summation and integration operators,making a change of integration variables ( x → x ) and utilizing [13, Eq. 24.2.10] to evaluate thesum, we arrive at2 m Γ (2 m + 1) Z E (2 m, u/ u d u = − m − X j =0 E (2 j ) H m − j Γ (2 j + 1) Γ (2 m − j + 1) m > , (2.49)10n addition to the collection of similar integrals evaluated in [16, Section 10]. Remark:
Theright-hand side of (2.46) can be also be reduced to the left-hand side of (2.49), thereby providingan alternative derivation of (2.46) - (C. Vignat, Toulane University, private communication.) ζ (2 m + 1) Further interesting results can be found by considering (2.41) with n = 2 m + p as before, thensetting p := 2 p giving (cid:0) m +1 − (cid:1) ζ (2 m + 1) + ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1)= ( − p Γ (2 p ) π p ∞ X k =0 E p ( − iπ ( k + 1 / E p ( iπ ( k + 1 / k + 1 / p +2 m − ( − m π m p − X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) . (2.50)The result (2.50) can be analyzed with recourse to Meijer’s G-function; since the analysis issomewhat lengthy the details have been relegated to Appendix (E). With reference to thatAppendix, a simple change of variables in (E.16), eventually yields (cid:0) m +1 − (cid:1) ζ (2 m + 1) = A ( m ) − ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1) . (2.51)where A ( m ) ≡ π Z ∞−∞ Γ (1 − iv ) π iv P ∞ k =0 ( − k ( k + 1 / − m − iv cosh ( π v ) d v (2.52)The sum in the integrand of A ( m ) has a known integral representation [17, Eq. 3.523(3)]: ∞ X k =0 ( − k ( k + 1 / m +2 − iv = 2 m +1 − iv Γ (2 m + 2 − iv ) Z ∞ t m +1 − iv cosh ( t ) d t . (2.53)Applying (2.53) to (2.52) and reordering the integration operators gives A ( m ) = 2 m +1 π Z ∞ t m +1 cosh ( t ) Z ∞−∞ Γ (1 − iv )Γ (2 m + 2 − iv ) cosh ( π v ) (cid:16) π t (cid:17) iv d v d t . (2.54)Since the arguments of the Γ function ratio in (2.54) differ by an integer, this ratio can profitablybe written as a β function (not Dirichlet’s Beta function), which in turn has a well-known integralrepresentation, givingΓ (1 − iv )Γ (2 m + 2 − iv ) = β (1 − iv, m + 1)Γ (2 m + 1) = 1Γ (2 m + 1) Z x − iv (1 − x ) m d x. (2.55)Substitute (2.55) into (2.54) and again reorder the integral operations to obtain A ( m ) = 2 m +1 π Γ (2 m + 1) Z ∞ t m +1 cosh ( t ) Z (1 − x ) m Z ∞−∞ π v ) (cid:16) π xt (cid:17) iv d v d x d t . (2.56)The innermost integration in (2.56) can be found from [17, Eq. 3.981(3)] by converting theintegral into the range (0 , ∞ ), writing (cid:0) π xt (cid:1) iv in the form of complex trigonometric functionsand applying symmetry to the result, eventually giving Z ∞−∞ π v ) (cid:16) π xt (cid:17) iv d v = 4 π xt x t + π . (2.57)Thus, after rescaling the t − integration by π , (2.56) becomes A ( m ) = 2 m +3 π m Γ (2 m + 1) Z ∞ t m +2 cosh ( π t ) Z (1 − x ) m x t x + 1 d x d t . (2.58)11 .7.1 Reduction of (2.58)The inner integral in (2.58) is listed in [17, Eq. 3.254(1)], giving Z (1 − x ) m x t x + 1 d x = Γ (2 m + 1) F (1 , , /
2; 2 + m, / m ; − t )Γ (2 m + 3) , (2.59)in which case, we have A ( m ) = 2 m π m Γ (3 + 2 m ) Z ∞ t m +23 F (1 , , /
2; 2 + m, / m ; − t )cosh ( π t ) d t . (2.60)This result can be verified by direct integration for at least the cases m = 1 and m = 2. See [14,Section 6]. (2.58) and an asymptotic divergent series for Hurwitz’alternating Zeta function Since both integrals composing (2.58) are convergent at both endpoints with m >
0, rather thandirectly evaluating (2.58) as in the previous section, interchange the order of integration giving A ( m ) = 2 m +3 π m Γ (2 m + 1) Z (1 − x ) m x Z ∞ t m +2 (4 t x + 1) cosh ( π t ) d t d x . (2.61)From [14, Eq. (3.10)] with a = 1 /x , the inner integral can be evaluated to obtain: Z ∞ t m +2 (4 t + 1 /x ) cosh ( π t ) d t = ( − m m +3 x m Γ (2 m + 2) m X j =0 ( − j (2 x ) j E (cid:0) j, x + (cid:1) Γ ( j + 2) Γ (2 m − j + 1) − x (cid:18) ψ (cid:18) x + 34 (cid:19) − ψ (cid:18) x + 14 (cid:19)(cid:19)(cid:21) . (2.62)Substitute (C.22) into (2.62), apply the Lemma (C.23), substitute into (2.61) with a change ofvariables x → /t and, noting that E (2 k + 1) = 0, eventually obtain A ( m ) = − ( − m π m m + 1) Z ∞ ( t − m " ψ (cid:18) t (cid:19) − ψ (cid:18) t (cid:19) − m X k =0 E (2 k ) t k +1 d t . (2.63)The bracketed term in (2.63) must asymptotically decrease faster than O ( t − m − ) since theoriginal integral was convergent at x = 0, and nothing has happened to alter that state of affairs;thus the transformed integrand must be of O ( t − − ǫ ) at its upper limit. From this we infer that,for t → ∞ , ψ (cid:18) t (cid:19) − ψ (cid:18) t (cid:19) ∼ m X k =0 E (2 k ) t k +1 (2.64)and since the left-hand side of (2.64) is independent of m , this must be true for all values of m >
0, in which case (2.64) is equivalent to a divergent asymptotic series representation of theleft-hand side, that is ψ (cid:18) t (cid:19) − ψ (cid:18) t (cid:19) ∼ ∞ X k =0 E (2 k ) t k +1 . (2.65)This result that can be easily verified numerically by expanding the left-hand side asymptoticallyusing either the Mathematica [18] or Maple [19] “Series” command, to any order desired. Notethe identity ([20, Eq. 22(5)]) ψ (cid:18) t (cid:19) − ψ (cid:18) t (cid:19) = 2 η (1 , ( t + 1) /
2) (2.66)12here η (1 , ( t + 1) /
2) is the alternating Hurwitz Zeta function (a.k.a the Lerch transcendent -Φ(1 / , s, a ) - see [21]). Thus (2.65) and (2.66) together give an infinite (asymptotic) divergentseries representation of a special case of the alternating Hurwitz Zeta function to all inverse ordersof t - see (1.2). Also see [22, Theorem 1a]. The derivative ∂∂s E s ( z ) defines the “Generalized Integro-Exponential function” as follows E js ( z ) ≡ ( − j Γ( j + 1) ∂ j ∂s j E s ( z ) j ≥ . (3.1)Two important properties [4, Eqs. (2.4) and (2.8)] are the recursion E js ( z ) = zE js − ( z ) − E j − s ( z )1 − s , s = 1 , j ≥ , (3.2)and the limiting case E − s ( z ) = exp( − z ) . (3.3)Thus after straightforward differentiation of (2.13) we havedd s η ( s ) = ln (2) η ( s ) + 2 s − ∞ X k =0 (cid:0) E s ( − iπ ( k + 1 / E s ( iπ ( k + 1 / (cid:1) (3.4)which can alternatively be written as (cid:0) − − s (cid:1) ζ ′ ( s ) = ln (2) (cid:0) − − s (cid:1) ζ ( s ) + 2 s − ∞ X k =0 (cid:0) E s ( − iπ ( k + 1 / E s ( iπ ( k + 1 / (cid:1) . (3.5)Again using forward recursion, from (2.22), (3.2) and (3.5) we find the equivalent form of (2.22),that is (cid:0) − − s (cid:1) ζ ′ ( s ) = 2 s is π ∞ X k =0 E s +1 ( iπ ( k + 1 / − E s +1 ( − iπ ( k + 1 / k + 1 / (cid:18) ln (2) (cid:0) − − s (cid:1) + 1 − − s s (cid:19) ζ ( s ) − s − s . (3.6)For the record, the general form corresponding to (2.23) with n ≥ (cid:0) − − s (cid:1) ζ ′ ( s ) = 2 s − Γ ( s ) ⌊ n/ − / ⌋ X j =0 ( ψ (2 j + s ) − ψ ( s + n )) Γ (2 j + s ) E (2 j )Γ (2 j + 1)+ 2 s − e iπn/ ( − n Γ ( s + n )Γ ( s ) π n ∞ X k =0 E s + n ( − iπ ( k + 1 / − n E + n ( iπ ( k + 1 / k + 1 / n + (cid:0)(cid:0) − − s (cid:1) ln (2) + ( ψ ( s + n ) − ψ ( s )) (cid:0) − − s (cid:1)(cid:1) ζ ( s ) . (3.7) The functions E s ( z ) appearing in (2.12) are related to incomplete (upper) Gamma functions via E s ( z ) = z − s Γ(1 − s, z ) . (4.1)13ith reference to (4.1), a useful integral representation [23, page 42] isΓ ( s, ± iκ ) = e ± isπ/ Z ∞ κ e ∓ iv v s − d v (4.2)or, with κ >
0, and a change of integration variablesΓ ( s, ± iκ ) = Γ ( s, κ ) + κ s Z ± i e − κ v v s − d v . (4.3)Keeping (2.13) in mind, we havee isπ/ Γ ( s, − iκ ) + e − isπ/ Γ ( s, iκ ) = e isπ/ κ s Z − i e − κ v v s − d v + e − isπ/ κ s Z i e − κ v v s − d v + 2 cos ( sπ/
2) Γ ( s, κ ) , (4.4)which, with (4.1), gives E s ( iκ ) + E s ( − iκ ) = i e − isπ/ Z − i e − κv v − s d v − i e isπ/ Z i e − κv v − s d v + 2 sin ( sπ/ E s ( κ ) (4.5)after setting s → − s . Further, the identification κ = π ( k + 1 / ∞ X k =0 e − π ( k +1 / v = 12 sinh ( π v/
2) (4.6)gives ∞ X k =0 E s ( iπ ( k + 1 / E s ( − iπ ( k + 1 / i e − isπ/ Z − i v − s sinh ( π v/
2) d v − i e isπ/ Z i v − s sinh ( π v/
2) d v + 2 sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / . (4.7)Thus, from (2.13) and (4.7) we have2 − s − s − ζ ( s ) = i e − isπ/ Z − i v − s sinh ( π v/
2) d v + i e isπ/ Z i v − s sinh ( π v/
2) d v + 2 sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / . (4.8)Notice that the structure of ζ ( s ) is fundamentally different for s = 2 n or s = 2 n + 1 because of thecoefficient sin( πs/
2) in (4.8). At this point, there are a number of interesting possible choices forthe integration paths in (4.8), each of which leads to different identities. See Figure 1. v = e iθ A simple choice suggests a change of variables v = e iθ with − π/ < θ < π/
2, equivalent tochoosing a circular path in the right-hand complex v -plane centred at the origin passing throughthe limiting points of the integrals in (4.7) (see Figure 1). With this choice, and followingsimplification (specifically, reverse one integral (i.e. θ → π/ − θ ) and utilize symmetry under θ → − θ ), (2.13) becomes 14 igure (1) Four possible paths of integration from − i to i . ζ ( s ) (cid:0) − − s (cid:1) s = 2 − s η ( s ) = − sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) + θ ( s − π sin ( θ )) − cos ( π cos ( θ )) d θ . (4.9)Also, see [23, Ex. 3.2, page 43]. From Appendix A, we thus have another representation for thelower incomplete function ζ − ( s ) (see (C.4)). ζ − ( s ) = 2 s − Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) + θ ( s − π sin ( θ )) − cos ( π cos ( θ )) d θ (4.10)valid for all s . Corollary.
In (4.9), replace s → s + 1 and concurrently replace s → − s , add the two results and simplify toeventually obtain Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ )) cos ( s θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 2 − s − (cid:0) − − s (cid:1) ζ − (1 + s ) + 2 s − (1 − s ) ζ − (1 − s )(4.11)= 2 − s − η − (1 + s ) + 2 s − η − (1 − s ) . (4.12)Similarly, by subtracting, we identify Z π/ − π/ e − ( π/
2) sin( θ ) cos (( π/
2) cos ( θ )) sin ( s θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 2 − s − (cid:0) − − s (cid:1) ζ − (1 + s ) − s − (1 − s ) ζ − (1 − s )(4.13)= 2 − s − η − (1 + s ) − s − η − (1 − s ) (4.14)15 emark: The following subsections list some special cases that may be worthy of specialattention. It is obvious that variations of all of the following can be obtained by simpletransformations of integration variables (e.g. θ → π/ − θ etc.) In the case that s = 1, take the limit of the left-hand side of (4.9) and eventually obtain Z π/ − π/ sin (( π/
2) cos ( θ )) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ln (2) + 2 ∞ X k =0 E ( π ( k + 1 / Z π/ sin (( π/
2) cos ( θ )) cosh (( π/
2) sin ( θ ))cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 12 ln (2) + ∞ X k =0 E ( π ( k + 1 / . (4.16)The sum in both of these can be identified from (C.7) with s = 1, noting that η (1) does not sharethe singularity of ζ (1). s = ± n and s = 1 ± n The case s = 0 , ± , ± . . . leads to some interesting identifications from the well-knownidentification ([13, Eq. 25.6.2]) of ζ (2 n ) in terms of Bernoulli numbers B n .If 0 ≤ s = 2 n then Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) + (2 n − θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ( − n B n π n (cid:0) − n − (cid:1) Γ (2 n + 1) . (4.17)In the above, if n = 0, we have Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 1 . (4.18)If s = − n <
0, then Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − (2 n + 1) θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 0 . (4.19)If 1 ≤ s = 2 n + 1 then Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) + 2 n θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 4 − n (cid:0) − − n (cid:1) ζ (2 n + 1)+ 2 ( − n ∞ X k =0 E n +1 ( π ( k + 1 / , (4.20)and, in the above, if n = 0, by taking a limit, we obtain Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ))cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ln (2) + 2 ∞ X k =0 E ( π ( k + 1 / . (4.21)Finally, if s = − n + 1 < Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − θ n )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 2 n − B n (cid:0) n − (cid:1) n + 2 ( − n ∞ X k =0 E − n ( π ( k + 1 / . (4.22)16n (4.22), recall that for m = 1 , . . . E − m ( κ ) = Γ ( m + 1) e − κ κ m +1 m X j =0 κ j j ! = Γ(1 + m, κ ) /κ m +1 , (4.23)which, together with (C.33) and (C.34) yields an alternate form for (4.22) Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − n θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 2 n (cid:0) − n (cid:1) ζ (1 − n )+ 2 ( − n Γ (2 n ) π n − X j =1 j +1 Li j +1 (cid:0) e − π/ (cid:1) − Li j +1 (e − π )Γ (2 n − j ) π j + 2 ( − n π ln (cid:18) sinh ( π/ π/ − (cid:19) . (4.24)Adding and subtracting then yields the following identities Z π/ − π/ e − ( π/
2) sin( θ ) cos (2 n θ ) sin (( π/
2) cos ( θ ) − θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = Z π/ − π/ e − ( π/
2) sin( θ ) sin (2 n θ ) cos (( π/
2) cos ( θ ) − θ )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ( − n B n π n (cid:0) − n − (cid:1)
2Γ (2 n + 1) , (4.25) Z π/ − π/ e − ( π/
2) sin( θ ) cos (2 n θ ) sin (( π/
2) cos ( θ ))cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ζ (2 n + 1) 2 − n − (cid:0) − − n (cid:1) + ( − n ∞ X k =0 E n +1 ( π ( k + 1 / E − n ( π ( k + 1 / B n n − (cid:0) n − (cid:1) n , (4.26)and Z π/ − π/ e − ( π/
2) sin( θ ) sin (2 n θ ) cos (( π/
2) cos ( θ ))cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = ζ (2 n + 1) 2 − n − (cid:0) − − n (cid:1) + ( − n ∞ X k =0 E n +1 ( π ( k + 1 / − E − n ( π ( k + 1 / − B n n − (cid:0) n − (cid:1) n . (4.27) By adding and subtracting cases corresponding to s = 0 and s = 1, after some simplification weobtain Z π/ − π/ e − ( π/
2) sin( θ ) sin ( θ/
2) cos (( π/
2) cos ( θ ) − θ/ π sin ( θ )) − cos ( π cos ( θ )) d θ = ln (2) / − / ∞ X k =0 E ( π ( k + 1 / . (4.28)Similarly, adding and subtracting the cases s=0 and s=2 along with some elementarytrigonometric identities, respectively yields Z π/ − π/ cos ( θ ) sin (( π/
2) cos ( θ )) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 1 / π / , (4.29)and Z π/ − π/ sin ( θ ) cos (( π/
2) cos ( θ )) e − ( π/
2) sin( θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = − / π / . (4.30)17 .1.4 Special case s=1/2 The case s = 1 / Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − θ/ π sin ( θ )) − cos ( π cos ( θ )) d θ = √ (cid:16) − √ (cid:17) ζ (1 /
2) + ∞ X k =0 E / ( π ( k + 1 / ! = √ (cid:16) − √ (cid:17) ζ − (1 /
2) (4.31)the second equality arising by virtue of (C.4). An alternate representation for ζ (1 /
2) can be foundby employing the identity ([13, Eq. 7.11.3]) E / ( z ) = r πz (cid:0) − erf (cid:0) √ z (cid:1)(cid:1) . (4.32)From the integral representation of the error function ([13, Eq. 7.2.1]) applied to (2.13) after somesimplification, we find ζ (1 /
2) = 11 − √ ∞ X k =0 (cid:18) √ Z cos (cid:0) π ( k + 1 / t (cid:1) d t − / p k + 1 / (cid:19) (4.33)leading to a series representation ζ (1 /
2) = 11 − √ ∞ X k =0 F c (cid:0) √ k + 1 (cid:1) − p ( k + 1 / . (4.34)where F c is the Fresnel C function ([13, Eq. 7.2.7]). From its asymptotic series ([19]), it can beshown that the terms of the alternating series (4.34) converge as O ( k − ) and the series istherefore convergent providing that all its terms are grouped as shown.Finally, differentiating (4.9) with respect to s , setting s = 1 / Z π/ − π/ e − ( π/ θ ) cos (( π/
2) cos ( θ ) − θ/ π sin ( θ )) − cos ( π cos ( θ )) θ d θ = −√ ∞ X k =0 E / ( π ( k + 1 / (cid:16)(cid:16) √ / (cid:17) ln (2) + ( γ + π/ π )) (cid:16) √ / − (cid:17) (cid:17) ζ (1 / − π (cid:16) √ / − (cid:17) ζ + (1 / An interesting set of results can be obtained by setting s = 1 / it in (4.9) and identifying E / it ( iπκ ) by its integral representation (2.12). Define J ( t ) ≡ Z π/ − π/ e − ( π/
2) sin( θ ) cos (( π/
2) cos ( θ ) − θ/
2) sinh ( θ t )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ (4.36) J ( t ) ≡ Z π/ − π/ e − ( π/
2) sin( θ ) sin (( π/
2) cos ( θ ) − θ/
2) cosh ( θ t )cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ (4.37) J ( t ) ≡ Z ∞ sin ( t ln ( v )) √ v sinh ( π v/
2) d v (4.38) J ( t ) ≡ Z ∞ cos ( t ln ( v )) √ v sinh ( π v/
2) d v . (4.39)In (4.9), split the resulting identities into real and imaginary components, and solve for J ( t ) and J ( t ) to yield 18 ( t ) = H ( t ) sin ( t ln (2)) ζ R (1 / it ) + (2 − H ( t ) cos ( t ln (2))) ζ I (1 / it ) − J ( t ) cosh ( π t/ √ J ( t ) sinh ( π t/ √ J ( t ) = (2 + H ( t ) cos ( t ln (2))) ζ R (1 / it ) + H ( t ) sin ( t ln (2)) ζ I (1 / it )+ J ( t ) sinh ( π t/ √ J ( t ) cosh ( π t/ √ H ( t ) ≡ t ln (2)) − √ . (4.42)From the definition (C.3), each of the integrals J ( t ) and J ( t ) can be connected with the real andimaginary parts of the upper incomplete function ζ + ( s ) as follows: ζ + R (1 / it ) = J ( t ) Q ( t ) + J ( t ) Q ( t ) (4.43) ζ + I (1 / it ) = − J ( t ) Q ( t ) + J ( t ) Q ( t ) , (4.44)where Q ( t ) = − P ( t ) cosh ( π t/ / − P ( t ) sinh ( π t/ / Q ( t ) = − P ( t ) cosh ( π t/ / P ( t ) sinh ( π t/ / , (4.46)and P ( t ) = (cid:0) √ t ln (2)) − (cid:1) sin ( t ln (2))2 √ t ln (2)) − P ( t ) = 2 √ t ln (2))) − √ − cos ( t ln (2))2 √ t ln (2)) − . (4.48)Inverting (4.43) and (4.44) identifies the integrals J ( t ) and J ( t ) in terms of the real andimaginary components of the upper incomplete function ζ + (1 / it ) as follows: J ( t ) = 2 √ (cid:0) − t ln (2)) + 3 √ (cid:1) (cid:0) Q ( t ) ζ + R (1 / it ) − Q ( t ) ζ + I (1 / it ) (cid:1) cosh ( π t ) (4.49) J ( t ) = 2 √ (cid:0) − t ln (2)) + 3 √ (cid:1) (cid:0) Q ( t ) ζ + I (1 / it ) + Q ( t ) ζ + R (1 / it ) (cid:1) cosh ( π t ) . (4.50)Further simplification is possible by utilizing (4.49), (4.50) and (C.6) in (4.40) and (4.41) toeliminate J ( t ) and J ( t ). After considerable (Maple) simplification we arrive at the identities J ( t ) = H ( t ) sin ( t ln (2)) ζ − R (1 / it ) + (2 − H ( t ) cos ( t ln (2))) ζ − I (1 / it ) (4.51) J ( t ) = (2 − H ( t ) cos ( t ln (2))) ζ − R (1 / it ) − H ( t ) sin ( t ln (2)) ζ − I (1 / it ) . (4.52)Alternatively, inversion of (4.51) and (4.52) produces an integral representation for eachcomponent of the lower incomplete function ζ − (1 / it ). Specifically ζ − R (1 / it ) = − ( H ( t ) cos ( t ln (2)) − J ( t )6 − √ t ln (2)) + H ( t ) J ( t ) sin ( t ln (2))6 − √ t ln (2)) (4.53) ζ − I (1 / it ) = − ( H ( t ) cos ( t ln (2)) − J ( t )6 − √ t ln (2)) − H ( t ) J ( t ) sin ( t ln (2))6 − √ t ln (2)) . (4.54) Remark:
As opposed to the representation (C.4) that does not converge when s = 1 / t = 0 reduces (4.53) to the convergentrepresentation (4.31). In other words, (4.53) and (4.54) are convergent representations of theincomplete function ζ − ( s ) on the critical line s = 1 / it .19 .2 Integration path B: v = 1 ± i − e iθ A second choice for the integration path in (4.8) corresponds to integrating over circular pathscentred at ( i, ± i ) for each of the two integrals in (4.7) respectively (see Figure 1). After somesimplification, this yields the identity ζ ( s ) (cid:0) − s − (cid:1) s − = 2 sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / Z π/ (3 − θ ) − θ )) − s/ (cid:0) cos( πB/ − sA − θ ) e πC/ − cos( πB/ sA + θ ) e − πC/ (cid:1) cosh ( π C ) − cos ( πB ) d θ, (4.55)where C = cos( θ ) − B = sin( θ ) − A = arctan( C/B ) . To simplify special cases, it is worth noting the following identitiessin ( A ) = 1 − cos ( θ ) p − θ ) − θ ) (4.56)cos ( A ) = 1 − sin ( θ ) p − θ ) − θ ) (4.57)arctan( z ) + arctan(1 /z ) = ± π/ , (4.58)the sign of the latter chosen according to whether S ( z ) > S ( z ) < S ( z ) = 1 if 0 < ℜ ( z ) or ℜ ( z ) = 0 and 0 < ℑ ( z ) S ( z ) = -1 if 0 > ℜ ( z ) or ℜ ( z ) = 0 and 0 > ℑ ( z ) . (4.59)The simplest of the special cases, corresponding to s = 0, gives Z π/ sin ( − π/ θ ) + θ ) e π/ − θ )) cosh ( π ( − θ ))) + cos ( π sin ( θ )) + sin ( π/ θ ) + θ ) e − π/ − θ )) cosh ( π ( − θ ))) + cos ( π sin ( θ )) d θ = 1 , (4.60)which, after the transformation θ → π/ − θ alternatively becomes Z π/ cos ( π/ θ ) + θ ) e π/ − θ )) + cos ( − π/ θ ) + θ ) e − π/ − θ )) cosh ( π (sin ( θ ) − π cos ( θ )) d θ = 1 . (4.61)For comparison, see (4.17) with n = 0, which, with B = 1, can be rewritten Z π/ sin ( π/ θ ) − θ ) e − π/ θ ) + sin ( π/ θ ) + θ ) e π/ θ ) cosh ( π sin ( θ )) − cos ( π cos ( θ )) d θ = 1 (4.62)or Z π/ − cos ( π/ θ ) + θ ) e − π/ θ ) + cos ( − π/ θ ) + θ ) e π/ θ ) cosh ( π cos ( θ )) − cos ( π sin ( θ )) d θ = 1 . (4.63)Any of the above can also be rewritten in a different form after the transformation sin( θ ) = v ,giving for example the following variant of (4.17): Z sinh (cid:0) π/ √ − v (cid:1) cos ( π v/ (cid:0) π √ − v (cid:1) − cos ( π v ) + v cosh (cid:0) π/ √ − v (cid:1) sin ( π v/ √ − v (cid:0) cosh (cid:0) π √ − v (cid:1) − cos ( π v ) (cid:1) d v = 1 / . (4.64)20 .3 Integration path C: Two straight lines: v ± := ± i + (1 ∓ i ) v By joining the integration endpoints ( ± i,
0) and (0 ,
1) with straight lines, (see Figure 1), (4.7)becomes ζ ( s ) (cid:0) − − s (cid:1) s = − sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / Z (cid:20) cos ( πv/
2) cosh ( πv/
2) + sin ( πv/
2) sinh ( π v/ π v ) + cosh ( π v ) cos( s arctan( vv − − cos ( π v/
2) cosh ( πv/ − sin ( πv/
2) sinh ( πv/ π v ) + cosh ( π v ) sin( s arctan( vv − (cid:21) (cid:16) ( v − + v (cid:17) − s/ d v . (4.65)The case s = 0 yields the identity Z cosh ( π v/ π v/
2) + sinh ( π v/
2) sin ( π v/ π v ) + cos ( π v ) d v = 1 / , (4.66)which can also be rewritten in the interesting form Z cosh ( π v/
2) cos ( π v/ ( π v/ − sin ( π v/
2) d v = 1 − Z sinh ( π v/
2) sin ( π v/ ( π v/ − sin ( π v/
2) d t . (4.67)
Remarks:
After considerable execution time, the computer program Mathematica [18]successfully evaluates (4.66). Also, by unknown means, both Glasser (private communication) andMathematica obtain the even more interesting variation: Z cosh ( π v/
2) cos ( π v/ ( π v/ − sin ( π v/
2) d v = 12 + 1 π log (coth( π/ . (4.68)After subtracting (4.66), the case s = − Z (cid:0) v − v (cid:1) cos ( π v/
2) cosh ( π v/ − sinh ( π v/
2) sin ( π v/ v cosh ( π v ) + cos ( π v ) d v = − / . (4.69)Similarly, the case s = 2 produces the identity Z cos ( π v/
2) cosh ( π v/ (cid:0) v − (cid:1) − (cid:0) v − v + 1 (cid:1) sin ( π v/
2) sinh ( π v/ π v ) + cos ( π v )) (2 v − v + 1) d v = − π / , (4.70)and, again after taking (4.66) into consideration, the case s = 1 gives Z v ( v −
1) cos ( π v/
2) cosh ( π v/
2) + sinh ( π v/
2) sin ( π v/ v (cosh ( π v ) + cos ( π v )) (2 v − v + 1) d v = (4.71) − / ∞ X k =0 E ( π ( k + 1 / − ln (2) / / (cid:0) v − v + 1 (cid:1) = 0 on the real line. In this case, the integration endpoints of (4.7) are joined by straight lines (see Figure 1) in thefollowing order: (0 , − i ), (1 , − i ), (1 , , + i ) and (0 , + i ) and, after the appropriate change of21ariables and simplification, the following identity emerges: ζ ( s ) (cid:0) − − s (cid:1) = − s sin ( sπ/ ∞ X k =0 E s ( π ( k + 1 / s − Z (cid:18) cos ( − π v/ s arctan (1 /v )) e π/ − cos ( π v/ s arctan (1 /v )) e − π/ cosh ( π ) − cos ( π v )+ cos ( s arctan ( v ))cosh ( π v/ (cid:19) (cid:0) v + 1 (cid:1) − s/ d v . (4.73)As before, many special cases are embedded in (4.73). In the case s = 0, the simplified forminvolves integrals all of which are amenable to analysis by both Mathematica and Maple [18], [19]and so this case is not discussed here. Similar but simpler integrals can be found in [17, Sections3.531 and 3.532]. What is believed to be a new series representation of Riemann’s zeta function has been developed,from which many novel identities flowed, most of which appear to have no obvious application. Itis suggested that further consideration of the integrals J ( t ) and J ( t ) in (4.53) and (4.54) mayshed light on the behaviour of ζ − (1 / it ) in the asymptotic limit t → ∞ . The author thanks Larry Glasser for making him aware of Russell’s 1876 work [6] and ChristopheVignat for pointing out (2.49).
References [1] Michael Milgram. An integral equation for Riemanns Zeta function and its approximatesolution.
Abstract and Applied Analysis , /2020/1832982(1832982), May 2020.https://doi.org/10.1155/2020/1832982.[2] Andre LeClair. An electrostatic depiction of the validity of the Riemann Hypothesis and aformula for the N-th zero at large N.
Int. J. Mod. Phys. , A28:1350151, 2013. Also availablefrom http://arxiv.org/abs/1305.2613v3.[3] M. L. Glasser. A note on the Riemann ξ -function, 2019. available fromhttps://arxiv.org/abs/1901.07011.[4] Milgram M.S. The Generalized Integro-Exponential Function. Mathematics of Computation
Applied Mathematical Sciences , 3(60):2973–2984, 2009. alsoavailable from: https://core.ac.uk/download/pdf/228177306.pdf.[6] W.H.L. Russell. On certain integrals.
Proc. Royal Soc. London , 25(176), 1876.[7] Tewodros Amdeberhan and Victor Moll. A dozen integrals: Russell-style, 2008. availablefrom https://arxiv.org/abs/0808.2692.[8] Mark W. Coffey. Generalizations of Russell-style integrals. June 2018.https://arxiv.org/abs/1806.07962. 229] Mark W. Coffey. Integrals in Gradshteyn and Ryzhik: Hyperbolic and Trigonometricfunctions, 2018. https://arxiv.org/abs/1803.00632.[10] E.C. Titchmarsh and D.R Heath-Brown.
The Theory of the Riemann Zeta-Function
Hacettepe Journal ofMathematics and Statistics , 42(4):319–329, 2013.[13] F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark, editors.
NIST Handbook ofMathematical Functions . Cambridge University Press, New York, NY, 2010. Print companionto [24].[14] Michael Milgram. Some additions to a family of integrals related to Hurwitz’ zeta function.2020.[15] Wolfram Research. HarmonicNumber. Retrieved fromhttps://functions.wolfram.com/GammaBetaErf/HarmonicNumber/07/01/01/0001/.[16] V.H. Moll and C. Vignat. Integrals involving Bernoulli and Euler polynomials.
SCIENTIA,Series A, Mathematical Sciences , 30:55–78, 2020.[17] I.S. Gradshteyn and I.M. Ryzhik.
Tables of Integrals, Series and Products, corrected andenlarged Edition . Academic Press, 1980.[18] Wolfram Research, Champagne, Illinois.
Mathematica, version 12 , 2019.[19] Maplesoft, a division of Waterloo Maple Inc.
Maple. [20] H.M. Srivastava and Junesang Choi.
Zeta and q-Zeta Functions and Associated Series andIntegrals . Elsevier, 32 Jamestown Rd.,London, NW1 7BY, first edition, 2012.[21] K.S. Williams and N-Y Zhang. Special values of the Lerch Zeta function and the evaluationof certain intgrals.
Proceedings of the American Mathematical Society , 119(1), 1993.[22] J. M. Borwein, P. B. Borwein, and K. Dilcher. Pi, Euler numbers, and asymptoticexpansions.
The American Mathematical Monthly
Asymptotics and Special Functions . Academic Press, 1974.[24] NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/, Release 1.0.9 of2014-08-29.[25] Dan Romik. The Taylor coefficients of the Jacobi theta constant θ . arXiv e-prints , July2018. arXiv:1807.06130.[26] M.S. Milgram. Integral and series representations of Riemann’s Zeta function, Dirichlet’s Etafunction and a medley of related results. Journal of Mathematics, Article ID 181724 , 2013.http://dx.doi.org/10.1155/2013/181724.[27] A. Magnus W. Oberhettinger F. Erdelyi and Tricome F.G.
Higher Transcendental Functions ,volume 1. McGraw-Hill, 1953.[28] Eldon R Hansen.
A Table of Series and Products . Prentice-Hall Inc., Englewood Cliffs, N.J.,1975.[29] E.W. Weisstein. Dirichlet Beta Function. Mathworld – a Wolfram Web Resource., Dec. 2016.Retrieved from http://mathworld.wolfram.com/DirichletBetaFunction.html.2330] E.W. Weisstein. Log Gamma function. mathworld – a wolfram web resource., 2005.Retrieved from https://mathworld.wolfram.com/LerchTranscendent.html.[31] Luke Y.
The special functions and their approximations, Volume I , volume 53 of
Mathematics in Science and Engineering . Academic Press, 1969.[32] Jos´e Luis L´opez and Nico M. Temme. ”Large degree asymptotics of generalized Bernoulli andEuler polynomials”.
Journal of Mathematical Analysis and Applications , 363(1):197 – 208,2010. https://doi.org/10.1016/j.jmaa.2009.08.034.
Appendices
A Notation
Throughout, the symbols j, k, m, n, p, r, N are positive integers, non-zero except as noted. E s ( z ) isthe (generalized) exponential integral, E ( n ) is an Euler number, E ( n, z ) is an Euler polynomial, B n is a Bernoulli number, H n is a harmonic number, erf is the error function, γ is theEuler-Mascheroni constant and Si ( z ) is the sine integral (D.1). The floor function (“greatestinteger less than”) is symbolized as usual by ⌊ ... ⌋ . Subscripted ζ R ( s ) and ζ I ( s ) refer to the realand imaginary components of ζ ( s ) or its incomplete forms. Other symbols are defined where theyfirst appear. Any summation where the lower summation limit exceeds the upper summationlimit is identically zero. At times, a complicated computer simplification has been used [19] toreduce an expression; such use is indicated by the notation “(Maple)”. B ζ ( s ) represented by various sums involving E s ( z ) In the literature, a number of representations are known in which ζ ( s ) or its associate ξ ( s ) areexpressed in terms of exponential integral functions. • From Paris [5] (where λ is a parameter satisfying | arg( λ ) | ≤ π/
2) we have, with modifiednotation, ζ ( s ) = ( π λ ) s/ Γ ( s/ ∞ X n =1 E − s/ (cid:0) π n λ (cid:1) + 1 √ λ ∞ X n =1 E s/ / (cid:18) π n λ (cid:19) + 1 √ λ ( s − − s ! . (B.1) • From [2] we have the representation ξ ( s ) = π ( s − ∞ X n =1 n E − s/ (cid:0) π n (cid:1) − π s ∞ X n =1 n E ( s − / (cid:0) π n (cid:1) + 4 π ∞ X n =1 n e − π n , (B.2)where ξ ( s ) ≡ ( s − π − s/ ζ ( s ) Γ (1 + s/
2) (B.3)and where it was later shown ([25, Eq.(3)]) that4 π ∞ X n =1 n e − π n = π /
2Γ (3 / . (B.4)24 The incomplete functions ζ ± ( s ) and η ± ( s ) , definitions andLemmas In this Appendix, various results needed in the text are either defined, proven, or collected fromthe literature. • The sum ∞ X k =0 E s ( π ( k + 1 / ∞ X k =0 E s ( π ( k + 1 / Z ∞ v − s sinh ( π v/
2) d v . (C.1)From [26, Eq.(29)] we have the representation ζ ( s ) = 2 s − sin ( π s/ − s − Z ∞ v − s sinh ( π v/
2) d v σ < , (C.2)after a simple change of variables. Define the upper and lower incomplete functions ζ ± ( s ) as ζ + ( s ) = 2 s − sin ( π s/ − s − Z ∞ v − s sinh ( π v/
2) d v , (C.3)and ζ − ( s ) = 2 s − sin ( π s/ − s − Z v − s sinh ( π v/
2) d v σ < , (C.4)and similarly for the alternating function η ∓ ( s ) ≡ (1 − − s ) ζ ∓ ( s ) (C.5)so that ζ ( s ) = ζ − ( s ) + ζ + ( s ) (C.6)and similarly for η ( s ). Thus, (C.1) becomes ∞ X k =0 E s ( π ( k + 1 / Z ∞ v − s sinh ( π v/
2) d v = ζ + ( s ) (cid:0) − s − (cid:1) s sin ( π s/
2) = − η + ( s )2 s sin ( π s/ . (C.7) Remark:
By writing (C.4) as follows ζ − ( s ) = 2 s − sin ( π s/ − s − (cid:18)Z (cid:18) v − s sinh ( π v/ − v − s − π d v (cid:19) − π s (cid:19) σ < , (C.8)we find ζ − (0) = − / ζ + (0) = 0 . (C.10)Similarly, by expanding (C.2) about s = 1, from the (leading) term of order ( s − − weidentify Z ∞ (cid:18) v sinh ( π v/ − π v (cid:19) d v = − ln (2) (C.11)and from the term of order ( s − with known coefficient γ we find Z ∞ ln ( v ) v (cid:18) π v/ − π v (cid:19) d v = ( γ − / . (C.12)25 From Erdelyi [27, Eq. 1.5.1(37)] with − < −ℜ ( s ) − n < Z ∞ v − s − n cos ( π ( k + 1 / v ) d v = ( π ( k + 1 / s +2 n − Γ ( − s − n + 1) ( − n sin ( π s/ . (C.13) • From [4, Eq. (2.21) with j = 0]: E − p ( ± iπ ( j + 1 / p + 1) e ∓ iπ ( j +1 / ( ± i π ( j + 1 / p +1 p X k =0 ( ± iπ ( j + 1 / k Γ ( k + 1) . (C.14) • From Hansen [28, Eqs. (17.4.10) and (14.4.10) respectively] or NIST [13, Eqs. (24.8.4) and(24.8.5)] with 0 ≤ v/ ≤ ∞ X k =0 cos ( π ( k + 1 / v )( k + 1 / n = (2 π ) n ( − n E (2 n − , v/ n ) (C.15)and ∞ X k =0 sin ( π ( k + 1 / v )( k + 1 / n +1 = (2 π ) n +1 ( − n E (2 n, v/ n + 1) (C.16)where E (2 n − , v/
2) represents the Euler polynomial, defined (from [19, FunctionAdvisor])in terms of Bernoulli numbers B k by E ( n, z ) = Γ ( n + 1) n X k =0 z k B − k + n Γ ( k + 1) Γ (2 − k + n ) (cid:0) − − k + n (cid:1) (C.17)which, reversed, becomes E ( n, z ) = Γ ( n + 1) n X k =0 (cid:0) − k (cid:1) z n − k B k +1 Γ (1 − k + n ) Γ (2 + k ) , (C.18)reducing to E ( n ) = Γ ( n + 1) n X k =0 (cid:0) k +1 − k (cid:1) B k +1 Γ (1 − k + n ) Γ ( k + 2) , (C.19)if z = 1 / E ( n ) = 2 n E ( n, / . (C.20)The following useful identity [13, Eq. (24.4.15] inverts the above: B n = Γ (2 n + 1)2 n (2 n − n − X k =0 E (2 k )Γ (2 k + 1) Γ (2 n − k ) . (C.21) • From [13, Eq. 24.2.10] E (cid:18) j, x + 12 (cid:19) = j X k =0 (cid:18) jk (cid:19) E ( k )2 k (cid:18) x (cid:19) j − k , (C.22)we have the following lemma Lemma:
For m > t ≥
1, 26 (2 m + 2) m X j =0 j X k =0 ( − j E ( k ) t − k Γ ( k + 1) Γ ( j − k + 1) ( j + 1) Γ (2 m − j + 1) = m X k =0 ( − k t − k E ( k ) . (C.23) Proof:
Reorder the series in (C.23) giving m X j =0 j X k =0 ( − j E ( k ) t − k Γ ( k + 1) Γ ( j − k + 1) ( j + 1) Γ (2 m − j + 1)= m X k =0 ( − k t − k E ( k )Γ ( k + 1) m − k X j =0 ( − j Γ ( j + 1) ( j + k + 1) Γ (2 m − j − k + 1) . (C.24)The inner sum can be identified m − k X j =0 ( − j Γ ( j + 1) ( j + k + 1) Γ (2 m − j − k + 1) = F ( k + 1 , − m + k ; k + 2; 1)( k + 1) Γ (2 m + 1 − k ) (C.25)and the hypergeometric function evaluates to F ( k + 1 , − m + k ; k + 2; 1) = Γ ( k + 2) Γ (2 m + 1 − k )Γ (2 m + 2) , (C.26)all of which is easily seen to reduce to (C.23). Q.E.D. • From [29], a sum appearing frequently in the text, is identified by any of the following: ∞ X k =0 ( − k ( k + 1 / n +1 = ψ (2 n, / − ψ (2 n, / n + 1) 2 n +1 = β (2 n + 1) 2 n +1 = ( − n π n +1 E (2 n )2Γ (2 n + 1)(C.27)where β ( x ) is the Dirichlet Beta Function and ψ ( n, x ) is the polygamma function. • From Mathematica [18] we have the following sum ∞ X j =0 ψ (2 j + 1) ( − j z − j j Γ (2 j + 1) = − (cid:18) Ci (cid:18) √ z (cid:19) + ln ( z )2 (cid:19) cos (cid:18) √ z (cid:19) + ln (2) cosh (cid:18) √− z (cid:19) − sin (cid:18) √ z (cid:19) Si (cid:18) √ z (cid:19) (C.28)where Ci and Si are the cosine and sine integrals respectively (see [13, Eq. 6.2.9 and 6.2.16]).In the case that z = − /π / ( k + 1 / , (C.28) reduces to ∞ X j =0 ψ (2 j + 1) ( − j ( k + 1 / j π j Γ (2 j + 1) = − ( − k Si ( π ( k + 1 / . (C.29)27 Taking (4.23) into account, the series appearing in (4.22) with n >
0, can be written ∞ X k =0 E − n ( π ( k + 1 / n ) π n − X j =0 π − j e − π/ Γ (2 n − j ) ∞ X k =0 e − π k ( k + 1 / j +1 , (C.30)and from the standard definition [30] of the Lerchphi transcendent Φ, we have ∞ X k =0 e − π k ( k + 1 / j +1 = Φ (cid:0) e − π , j + 1 , / (cid:1) (C.31)which can be rewritten in terms of generic polylog functions Li j ( x ) using the identitiesΦ ( z, s, a ) = Φ (cid:0) z , s, a/ (cid:1) + z Φ (cid:0) z , s, a/ / (cid:1) s (C.32)with a = 1 and [30, Eq. (6)], eventually, with n >
0, yielding the identity ∞ X k =0 E − n ( π ( k + 1 / n ) π n − X j =0 Li j +1 (cid:0) e − π/ (cid:1) j +1 Γ (2 n − j ) π j − n − X j =0 Li j +1 (e − π )Γ (2 n − j ) π j . (C.33) Remark:
Relevant to the above, note both the identity Li ( x ) = − ln (1 − x ) (C.34)and the connection to the Bose-Einstein distribution (see [30, Eq. (13)]). D Appendix C - a Theorem
In the main text, we are interested in combinations of the exponential integral function ofcomplex conjugate argument. Here we present a proof of the following identity, where k ∈ , . . . , p ∈ , . . . and Si(z) is the sine integral:Si(z) = Z z0 sin(t)t dt . (D.1) Theorem D.1.
For integer p ≥ − p Γ (2 p ) E p ( − iπ ( k + 1 / E p ( iπ ( k + 1 / π ( k + 1 / p − = 2 ( − k p − X j =1 ( − j Γ (2 j )( π ( k + 1 / j − π (k + 1 / − π/
2) (D.2)
Proof:
Begin by converting each of the functions E p and Si to a (convergent) integral representationusing (2.12) and (D.1) respectively, giving an equivalent form of (D.2) Z ∞ v − p cos (( k + 1 / vπ ) d v = ( − p ( π ( k + 1 / p − Γ (2 p ) − Z sin (( k + 1 / vπ ) v d v + ( − k p − X j =1 ( − j Γ (2 j )( k + 1 / j π j + π/ (D.3)28fter integrating once by parts, the left-hand side of (D.3) becomes Z ∞ v − p cos (( k + 1 / vπ ) d v = − ( − k π ( k + 1 /
2) + 2 pπ ( k + 1 / Z ∞ sin (( k + 1 / vπ ) v − − p d v . (D.4)We proceed by induction. Set p = 1 in (D.3), integrate again by parts, complete the incompleteintegral by utilizing Z ∞ sin( t ) t dt = π π ( k + 1 /
2) Si ( π ( k + 1 / − − k π ( k + 1 /
2) + 2 ( − k − π ( k + 1 / π ( k + 1 / π ( k + 1 /
2) Si ( π ( k + 1 / − π ( k + 1 /
2) (D.6)which is obviously true. Having shown that it is true for p = 1, use (D.4) to posit that theequivalent general form of (D.2) is true for any value of p : Z ∞ sin (( k + 1 / vπ ) v − − p d v = − ( − p ( π ( k + 1 / p Γ (2 p + 1) Z sin (( k + 1 / vπ ) v d v + ( − k + p ( π ( k + 1 / p Γ (2 p + 1) p − X j =1 ( − j Γ (2 j )( π ( k + 1 / j + ( − p π π ( k + 1 / p Γ (2 p + 1) + ( − k p ; (D.7)then transform (D.7) by setting p → p + 1. Integrating by parts converts the left-hand side of thetransformed (D.7) as follows: Z ∞ sin (( k + 1 / vπ ) v − − p d v = − ( k + 1 / π R ∞ sin (( k + 1 / vπ ) v − − p d v p + 1) (1 + p ) + ( − k p . (D.8)Similarly, convert the transformed sum within (D.7) as follows p X j =1 ( − j Γ (2 j )( k + 1 / j π j = p − X j =1 ( − j Γ (2 j )( k + 1 / j π j + ( − p Γ (2 p )( k + 1 / p π p , (D.9)substitute both (D.8) and (D.9) into the transformed (D.7), simplify (Maple) and it will be seenthat, under the change p → p + 1, the transformed (D.7) reduces to itself. By the logic ofinduction, (D.7) is true and therefore so is (D.2). QEDRemark:
Since integration-by-parts underlies the recursion rule (2.21), the proof of (D.2) couldhave been alternatively obtained by the p − fold application of (2.21).Since each of the integrals in (D.7) can be expressed as a hypergeometric function (see [31, Eqs.6.2(5) and 6.2(6)]) we find Corollary F ( 12 − p ; 12 , − p ; − π ( k + 1 / − p +1 ( π ( k + 1 / p Γ (2 p − F ( 12 ; 32 ,
32 ; − π ( k + 1 / − k + p ( π ( k + 1 / p − Γ (2 p − p − X j =1 Γ (2 j ) ( − j ( π ( k + 1 / j . (D.10)29 A lengthy derivation
This Appendix contains the details of a lengthy analysis based on (2.50). With reference to thatidentity, and to reiterate: both sides are independent of the integer p >
0. Consider the final termin (2.50): − ( − m π m p − X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) . (E.1)For large values of n (and therefore p ), an asymptotic approximation to the Euler numbers E ( n )can be extracted by setting z = 1 / N terms) the following expression [32, Eq.(3.6)] for the Euler polynomial E ( n, z ) E ( n, z ) = 4 Γ ( n + 1) ∞ X k =0 sin ((2 k + 1) π z − π n/ k + 1) π ) n +1 , (E.2)along with the definition (C.20) giving E (2 j + 2 + 2 m ) ≈ − j + 3 + 2 m ) ( − j + m π j +3+2 m N X k =0 ( − k ( k + 1 / j +3+2 m (E.3)where N = 1 , , . . . . In the case lim N = ∞ , (E.3) is equivalent to (C.27) and the approximationbecomes an equality. In general, and for later use in the derivation of (E.16), note that2 p − X j =1 ( − p − j Γ (2 p − j ) π j ∞ X k =0 ( − k ( k + 1 / p +2 m − j +1 = π ( − m π p +2 m p − X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) . (E.4)For the moment however, we shall treat (E.3) as an approximation - in fact it is a very good onenumerically even for small values of N . When (E.3) is substituted into (E.1), both sums are finiteand can therefore be interchanged, yielding( − m π m p − X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) ≈ − π N X k =0 ( − k ( k + 1 / m +3 p − X j =0 Γ (2 j + 2) ( − j ( k + 1 / j π j . (E.5)The inner sum of (E.5) can be formally identified in terms of hypergeometric functions as follows p − X j =0 Γ (2 j + 2) ( − j ( k + 1 / j π j = F (1 , , /
2; ; − π ( k + 1 / )+ Γ (2 p ) ( − p ( k + 1 / p − π p − F (1 , p, / p ; ; − π ( k + 1 / ) . (E.6)The first F ( ... ; . ) in (E.6) represents the sum on the left-hand side corresponding to p → ∞ , andthe second F ( ... ; . ) represents a correction term for finite values of p . Although rigorouslydivergent when written in series representation, each of these functions can have meaning assignedto it through recourse to Meijer’s G-function, and contour integration. Formally, for both x and q complex, (see [31, Eq. 5.3(1) ]) 30 F (1 , q, / q ; ; x ) = 2 q √ π Γ (2 q ) G , , (cid:18) − x (cid:12)(cid:12)(cid:12) , − q, / − q ;0; (cid:19) = 2 q √ π Γ (2 q ) G , , (cid:18) − x (cid:12)(cid:12)(cid:12) , / q, q ; (cid:19) , (E.7)and we are particularly interested in the case that x = − / ( π ( k + 1 / ) for both general integervalues of q as well as q = 1 in particular. From the theory of G-functions, the final term in (E.7)can be identified (see [31, Eq. 5.2(7)]) G , , (cid:18) − x (cid:12)(cid:12)(cid:12) , / q, q ; (cid:19) = − √ π Γ ( − q )2 q x F (1; 2 − q, / − q ; x − )+ π / csc ( π q ) (cid:0) − x − (cid:1) q F ( ; 1 / x − ) − π / (cid:0) − x − (cid:1) / q sec ( π q ) F ( ; 3 / x − ) , (E.8)a result that is valid for | x | > q . Remark:
The act of generalizing integer p → q , evaluating the result for general values of q andthen reducing the result employing q → p as a limit, is known as “regularization”. The functionsin (E.8) are rooted in confluent hypergeometric series (related to Lommel functions) that possessan infinite radius of convergence. This is significant because the identification x = − / ( π ( k + 1 / ) will eventually result in a series over the index ( k + 1 / .We now take the limit q → p , and, recognizing that both k and p are integers, each of the termsin (E.8) can be simplified; thus (E.7) becomes F (1 , p, / p ; ; − π ( k + 1 / ) = π ( k + 1 / Γ (2 p ) p − X j =0 Γ (2 p − j −
2) ( k + 1 / j π j ( − j − π p ( − p ( k + 1 / p Γ (2 p ) ∞ X j =0 ψ (2 j + 1) ( − j ( k + 1 / j π j Γ (2 j + 1) − ( − p + k π p +1 ( k + 1 / p p ) , (E.9)which, with recourse to (C.29) can be rewritten asΓ (2 p ) π p F (1 , p, / p ; ; − π ( k + 1 / ) = − ( − p ( k + 1 / p p − X j =1 ( − j Γ (2 j ) π − j ( k + 1 / j + ( − p + k ( k + 1 / p Si ( π ( k + 1 / − π ( − p + k ( k + 1 / p / Remark:
As noted previously, (E.9) must vanish as p → ∞ consistent with the commondenominator factor Γ(2 p ).) Since p > p = 1 in (E.9), to identify F (1 , , /
2; ; − π ( k + 1 / ) = π ( k + 1 / ∞ X j =0 ψ (2 j + 1) ( − j ( k + 1 / j π j Γ (2 j + 1) + ( − k π , (E.11)a convergent series, which, with the help of (C.29) - and true for any choice of integer p - becomes F (1 , , /
2; ; − k + 1 / π ) = − π ( − k ( k + 1 / (Si ( π ( k + 1 / − π/ . (E.12)31 Remark:
Inserting (E.9) and (E.12) into (E.6) yields a trivial identity, reinforcing confidencethat the somewhat unusual analysis methods employed thus far are valid.)Comparing (E.10) and (D.2) now gives E p ( − iπ ( k + 1 / E p ( iπ ( k + 1 / − − k π ( k + 1 / F (1 , p, / p ; ; − π ( k + 1 / ) , (E.13)and with reference to (E.7) and [31, Eq. (5.2(1)], the right-hand side of (E.13) can be rewritten asa well-defined contour integral: F (1 , p, / p ; ; − π ( k + 1 / ) = − i p − π / Γ (2 p ) Z c + i ∞ c − i ∞ Γ (1 − t ) Γ (1 / p − t ) Γ ( p − t ) Γ ( t ) ( π/ t ( k + 1 / t d t (E.14)where 0 < c < m < c < m + 1, (denoted by c m ) and, with p > m , by adding the appropriate residues we obtain E p ( − iπ ( k + 1 / E p ( iπ ( k + 1 / k + 1 / p +2 m = i ( − k π Γ (2 p ) Z c m + i ∞ c m − i ∞ Γ (2 p − t ) π t sin ( π t ) ( k + 1 / − t +2 p +2 m +1 d t + 2 ( − k π Γ (2 p ) ( k + 1 / p +2 m − j +1 m X j =1 ( − j π j Γ (2 p − j ) . (E.15)Now, employing (E.4) and (C.27), substitute (E.15) into (2.50), split one of the sums to cancel,and reverse a remaining sum, giving (cid:0) m +1 − (cid:1) ζ (2 m + 1) = i ( − p π p +1 Z c m + i ∞ c m − i ∞ Γ (2 p − v ) π v sin ( π v ) ∞ X k =0 ( − k ( k + 1 / v − p − m − d v − ( − m π m p − − m X j =0 Γ (2 j + 2) E (2 j + 2 + 2 m )Γ (2 j + 3 + 2 m ) − ( − m π m m X j =0 E (2 j ) ψ (1 − j + 2 m )Γ (1 − j + 2 m ) Γ (2 j + 1) . (E.16)As noted earlier, the sum of the first two terms on the right-hand side of (E.16) is independent of p . One possibility now is to choose c m = m + 1 / p = mm