aa r X i v : . [ m a t h . C A ] J a n Asymptotics of a Mathieu-Gaussian series
R. B. Paris
Division of Computing and Mathematics,Abertay University, Dundee DD1 1HG, UK
Abstract
We consider the asymptotic expansion of the functional series S µ,γ ( a ; λ ) = ∞ X n =1 n γ e − λn /a ( n + a ) µ for real values of the parameters γ , λ > µ ≥ | a | → ∞ in the sector | arg a | < π/ γ the expansion is of algebraic type with terms involving the Riemannzeta function and a terminating confluent hypergeometric function. Of principal interestin this study is the case corresponding to even integer values of γ , where the algebraic-type expansion consists of a finite number of terms together with a contribution comprisingan infinite sequence of increasingly subdominant exponentially small expansions. Thissituation is analogous to the well-known Poisson-Jacobi formula corresponding to the case µ = γ = 0. Numerical examples are provided to illustrate the accuracy of these expansions. Mathematics subject classification (2010):
Keywords:
Mathieu series, asymptotic expansions, exponential asymptotics, Mellin trans-form method
1. Introduction
The functional series ∞ X n =1 n ( n + a ) µ (1.1)in the case µ = 2 was introduced by Mathieu in his 1890 book [2] dealing with the elasticity ofsolid bodies. The asymptotic expansion for large a of more general functional series has beendiscussed in [4] and [11]. More recently, Gerhold and Tomovski [1] extended the asymptotic studyof such Mathieu series by introducing in (1.1) the factor z n , where | z | ≤
1. From this result theywere able to deduce, in particular, the large- a expansions of the trigonometric Mathieu series ∞ X n =1 n sin nx ( n + a ) µ , ∞ X n =1 n cos nx ( n + a ) µ . Subsequently, the above trigonometric series were generalised to include the oscillatory Besselfunctions J ν ( x ) and Y ν ( x ) with argument proportional to n/a , and their large- a asymptoticsdetermined in [6]. In addition, this last study also considered the inclusion of the modified Besselfunction K ν ( x ) of similar argument, which contains the decaying exponential as a special case.The asymptotic expansion we consider in this paper is the Mathieu series coupled with aGaussian exponential of the form S µ,γ ( a ; λ ) := ∞ X n =1 n γ e − λn /a ( n + a ) µ ( µ ≥ , λ >
0) (1.2)1
R. B. Paris for | a | → ∞ in the sector | arg a | < π/
4. It will be supposed throughout that γ is real, althoughthe analysis is easily modified to incorporate complex γ . We shall employ the Mellin transformapproach used in [4, 6, 11], where our interest will be primarily concerned with even integervalues of γ (positive or negative). We shall find that the asymptotic expansion of S µ,γ ( a ; λ )with these parameter values for large complex a in the sector | arg a | < π/ γ . This is also found to be the casewhen λ = 0 in (1.2); see [5] for details. A well-known related series corresponding to µ = γ = 0is the Poisson-Jacobi formula [10, p. 124] S , ( a ; λ ) = ∞ X n =1 e − λn /a = a r πλ −
12 + a r πλ ∞ X n =1 e − π n a /λ . (1.3)This sum is also seen to consist of a finite algebraic contribution together with an infinite sumof exponentially small terms when | a | → ∞ in the sector | arg a | < π/ σ and t , we have the estimatesΓ( σ ± it ) = O ( t σ − e − πt ) , | ζ ( σ ± it ) | = O ( t Ω( σ ) log α t ) ( t → + ∞ ) , (1.4)where Ω( σ ) = 0 ( σ > − σ (0 ≤ σ ≤ − σ ( σ <
0) and α = 1 (0 ≤ σ ≤ α = 0otherwise [9, p. 95]. The zeta function ζ ( s ) has a simple pole of unit residue at s = 1 and theevaluations for positive integer kζ (0) = − , ζ ( − k ) = 0 , ζ (2 k ) = (2 π ) k k )! | B k | ( k ≥ , (1.5)where B k are the Bernoulli numbers. Finally, we have the well-known functional relation satisfiedby ζ ( s ) given by [3, p. 603] ζ ( s ) = 2 s π s − ζ (1 − s )Γ(1 − s ) sin πs. (1.6)
2. An integral representation
The generalised Mathieu series defined in (1.2) can be written as S µ,γ ( a ; λ ) = a − δ ∞ X n =1 h ( n/a ) , h ( x ) := x γ e − λx (1 + x ) µ , δ := 2 µ − γ, (2.1)where the parameter δ is real and λ >
0. We employ a Mellin transform approach as discussedin [7, Section 4.1.1]. The Mellin transform of h ( x ) is H ( s ) = R ∞ x s − h ( x ) dx , where [3, (13.4.4)] H ( s ) = Z ∞ x γ + s − e − λx (1 + x ) µ dx = Γ( γ + s ) U ( γ + s , γ + s − µ, λ ) (2.2)in the half-plane ℜ ( s ) > − γ , with U ( a, b, z ) being the confluent hypergeometric function of thesecond kind. The transform H ( s ) can be represented alternatively in the form H ( s ) = {H ( s ) + H ( s ) } , (2.3) symptotics of a Mathieu-Gaussian series H ( s ) = Γ( γ + s )Γ( µ − γ + s )Γ( µ ) F ( γ + s ; 1 + γ + s − µ ; λ ) , (2.4) H ( s ) = λ µ − ( γ + s ) / Γ( γ + s − µ ) F ( µ ; 1 − γ + s + µ ; λ ) . (2.5)Using the Mellin inversion theorem (see, for example, [7, p. 118]), we find S µ,γ ( a ; λ ) = a − δ πi ∞ X n =1 Z c + ∞ ic −∞ i H ( s )( n/a ) − s ds = a − δ πi Z c + ∞ ic −∞ i H ( s ) ζ ( s ) a s ds, (2.6)where ζ ( s ) is the Riemann zeta function and c > max { , − γ } . The inversion of the order ofsummation and integration is justified by absolute convergence provided c satisfies this condition.From the estimates in (1.4) and the fact that the confluent hypergeometric functions appearingin H ( s ) and H ( s ) are both O (1) as ℑ ( s ) → ±∞ , the integral in (2.6) then defines S µ,γ ( a ; λ )for complex a in the sector | arg a | < π/
4. The integration path in (2.6) lies to the right of thesimple pole of ζ ( s ) at s = 1 and the poles of Γ(( γ + s ) /
2) at s = − γ − k ( k = 0 , , , . . . ), thesebeing the only poles of the integrand since the U function in (2.2) has no poles; see AppendixA for a demonstration of this fact.We consider the integral in (2.6) taken round the rectangular contour with vertices at c ± iT and − c ′ ± iT , where c ′ > T >
0. The contribution from the upper and lower sides ofthe rectangle s = σ ± iT , − c ′ ≤ σ ≤ c , vanishes as T → ∞ provided | arg a | < π/
4, sincefrom (1.4), the modulus of the integrand is controlled by O ( T Ω( σ )+( σ − δ − / log T e − ∆ T ), where∆ = π/ − | arg a | . Displacement of the integration path to the left over the pole at s = 1 andthose of H ( s ) at s = − k − γ (when µ >
0) then yields S µ,γ ( a ; λ ) − a − δ H (1) ∼ a − µ ∞ X k =0 ( − ) k ( µ ) k a k ζ ( − k − γ ) F ( − k ; 1 − µ − k ; λ ) , (2.7)where H (1) = 12 Γ( γ ) U ( γ , γ − µ, λ ) . (2.8)When µ = 0, we have H ( s ) ≡ H ( s ) at s = − k − γ yield S ,γ ( a ; λ ) − (cid:18) a λ (cid:19) ( γ +1) / Γ( γ + 12 ) ∼ ∞ X k =0 ( − λ ) k k ! a k ζ ( − k − γ ) , (2.9)where we have used the fact that U ( α, α, x ) = x − α . Both expansions (2.7) and (2.9) holdas | a | → ∞ in | arg a | < π/
4. The hypergeometric functions appearing in the sum in (2.7) arepolynomials in λ of degree k when µ is non-integer; for positive integer µ they can be expressedby Kummer’s theorem as e λ multiplied by a polynomial in λ of degree µ − µ > γ . If γ equals an odd negativeinteger, the pole at s = 1 has both a double pole contribution (resulting from H ( s )) and asimple pole contribution (resulting from H ( s )). An example of the expansion when γ = − γ = 0 in (2.9) the expansion for S , ( a ; λ )correctly reduces to the first two terms in the Poisson-Jacobi formula (1.3), but does not accountfor the exponentially small contribution. Further consideration of this case is discussed at theend of Section 3.Finally, we note that when γ = 2 p is an even integer, there will be a finite number ofpoles of the integrand of the sequence s = − k − p on account of the trivial zeros of ζ ( s ) at s = − , − , . . . . This results in the number of terms in the asymptotic series in (2.7) and (2.9) The F function appearing in H ( s ) can be written as e λ F (1 − µ ; 1 + ( γ + s ) − µ ; − λ ) by Kummer’stransformation [3, p. 325]. R. B. Paris being either finite or zero. This situation is the main subject of this paper. We shall show that,in addition to a finite algebraic contribution, there is a sequence of increasingly subdominantexponentially small terms in the large- a limit. This is analogous to the exponentially smallcontribution appearing on the right-hand side of the Poisson-Jacobi formula (1.3).
3. The exponentially small contribution to S µ,γ ( a ; λ ) when γ = 2 p Let γ = 2 p be an even integer and µ ≥
0. Then the quantity δ defined in (2.1) is δ = 2( µ − p ).The number of poles of the sequence s = − k − p is finite (when p ≤
0) or zero (when p ≥ ζ ( s ). Then we have upon displacement of the integration path S µ,γ ( a ; λ ) = a − δ H (1) + H µ,γ ( a ; λ ) + J ( a ; λ ) , (3.1)where H µ,γ ( a ; λ ) = p ≥ − a − δ ( p = 0) a − δ P | p | k =0 ζ (2 k ) R k ( µ, | p | ) a k ( p ≤ − . (3.2)The quantity R k ( µ, | p | ) denotes the residue of H ( s ) at s = 2 k , 0 ≤ k ≤ | p | when p ≤ − R k ( µ, | p | ) = ( − ) q − k ( q − k )! U ( k − q, k − q − µ, λ ) ( q = − p, ≤ k ≤ q ) . Routine calculations show that when p = − , −
2, for example, we have R ( µ,
1) = − ( µ + λ ) , R ( µ,
1) = 1 ,R ( µ,
2) = µ (1 + µ ) + µλ + λ , R ( µ,
2) = − ( µ + λ ) , R ( µ,
2) = 1 . The integral J ( a ; λ ) appearing in (3.1) is defined by J ( a ; λ ) = a − δ πi Z c + ∞ ic −∞ i H ( − s ) ζ ( − s ) a − s ds ( c > , where we have replaced s by − s . Use of the functional relation for ζ ( s ) in (1.6) followed byexpansion of ζ (1 + s ) (permissible since c >
0) leads to J ( a ; λ ) = − a − δ πi Z c + ∞ ic −∞ i H ( − s ) ζ (1 + s )Γ(1 + s ) sin πsπ (2 πa ) − s ds = − a − δ π X k ≥ k · πi Z c + ∞ ic −∞ i H ( − s )Γ(1 + s ) sin πs (2 πka ) − s ds. From (2.2) and an application of Kummer’s transformation [3, (13.2.40)] we have H ( − s ) = Γ( p − s ) λ µ − p + s/ U ( µ, µ − p + s, λ ) , whence J ( a ; λ ) = ( − ) p a − δ λ p − µ X k ≥ k · πi Z c + ∞ ic −∞ i − s Γ(1 + s )Γ(1 − p + s ) X − s/ k U ( µ, µ − p + s, λ ) ds, (3.3)where X k := π k a λ . (3.4)We first consider the case p = 0 , , , . . . . Since there are no poles of the integrand in (3.3)in ℜ ( s ) > | s | is symptotics of a Mathieu-Gaussian series − s √ π Γ(1 + s )Γ(1 − p + s ) = Γ( + s )Γ(1 + s )Γ(1 − p + s ) = p X j =0 ( − ) j c j Γ( s + ϑ − j ) , (3.5)where ϑ = p + . The coefficients are given explicitly by c j = 1 j ! ( − p ) j ( − p + ) j = ( − p ) j j j ! . (3.6)We observe that c j = 0 for j > p so that the above sum of gamma functions terminates and sois exact . Substitution of the expansion (3.5) in (3.3), combined with the integral representation[3, (13.4.4)] U ( µ, µ − p + s, λ ) = 1Γ( µ ) Z ∞ e − λt t µ − (1 + t ) − p − s/ dt ( µ > , then shows that, provided µ > J ( a ; λ ) = ( − ) p a − δ λ p − µ √ π Γ( µ ) X k ≥ k p X j =0 ( − ) j c j × Z ∞ e − λt t µ − (1 + t ) p (cid:18) πi Z c + ∞ ic −∞ i Γ( s + ϑ − j ) (cid:18) X k t (cid:19) − s/ ds (cid:19) dt. The inner integral appearing in J ( a ; λ ) can be evaluated by making use of the well-knownresult 12 πi Z L Γ( s + α ) z − s ds = z α e − z ( | arg z | < π ) , (3.7)where L is a path parallel to the imaginary s -axis lying to the right of all the poles of Γ( s + α );see, for example, [7, Section 3.3.1]. Evaluation of the inner integral when | arg a | < π/ exact result J ( a ; λ ) = ( − ) p π µ e λ Γ( µ ) (cid:18) λπa (cid:19) µ − p − / X k ≥ k p e − πka p X j =0 ( − ) j c j (cid:18) λπ k a (cid:19) j I jk , where I jk = Z ∞ t µ − e − ψ ( t ) (1 + t ) p − j +1 / dt, ψ ( t ) := λ (1 + t ) + X k t − πka. (3.8)In the special case µ = 0, we have from (3.3) (since U (0 , b, z ) = 1) that J ( a ; λ ) = ( − ) p a p √ πλ p X k ≥ k p X j =0 ( − ) j c j · πi Z c + ∞ ic −∞ i Γ( + ϑ − j ) X − s/ k ds. Evaluation of the integral by means of (3.7) then produces J ( a ; λ ) = ( − ) p (cid:18) λπa (cid:19) − p − / X k ≥ k p e − π k a /λ p X j =0 ( − ) j c j (cid:18) λπ k a (cid:19) j ( µ = 0) . Then we have the following theorem:
R. B. Paris
Theorem 1 . For µ ≥ , λ > , δ = 2 µ − γ and γ = 2 p , where p is a non-negative integer, wehave when | arg a | < π/ S µ,γ ( a ; λ ) = a − δ H (1) − a − δ δ p + J ( a ; λ ) , where H (1) is defined in (2.8) and δ p is the Kronecker symbol. The exponentially small contri-bution J ( a ; λ ) is given exactly by the double sums J ( a ; λ ) = ( − ) p π µ e λ Γ( µ ) (cid:18) λπa (cid:19) µ − p − / X k ≥ k p e − πka p X j =0 ( − ) j c j (cid:18) λπ k a (cid:19) j I jk ( µ > , (3.9) and J ( a ; λ ) = ( − ) p (cid:18) λπa (cid:19) − p − / X k ≥ k p e − π k a /λ p X j =0 ( − ) j c j (cid:18) λπ k a (cid:19) j ( µ = 0) , (3.10) where the coefficients c j = ( − p ) j / (2 j j !) and the integrals I jk are defined in (3.8). Remark 1.
When µ = p = 0, we find from (2.8), (3.1) and (3.10) (since U ( , , λ ) = λ − / )the result S , ( a ; λ ) = a r πλ −
12 + a r πλ X k ≥ e − π k a /λ , which is the Poisson-Jacobi formula stated in (1.3).When p = 1, we have U ( , , λ ) = λ − / and S , ( a ; λ ) = a √ π λ / − (cid:18) πa λ (cid:19) / X k ≥ (cid:18) k − λ π a (cid:19) e − π k a /λ . We observe that this last case can also be obtained by differentiation of the Poisson-Jacobiformula with respect to λ , since S , p ( a ; λ ) = ( − ) p a p ∂ p ∂λ p S , ( a ; λ ) ( p ≥ . Remark 2.
When p = − , − , . . . , the expansion (3.5) does not terminate and becomes aninverse factorial expansion. Then we have the exponentially small contribution given by J ( a ; λ ) ∼ ( − ) p π µ e λ Γ( µ ) (cid:18) λπa (cid:19) µ − p − / X k ≥ k p e − πka ∞ X j =0 ( − ) j c j (cid:18) λπ k a (cid:19) j I jk ( µ >
0) (3.11)as | a | → ∞ in | arg a | < π/
4. Alternative form of expansion for positive integer values of µ Let µ = m be a positive integer and γ = 2 p . We split H ( s ) into its two constituent parts givenby (2.3) and write J ( a ; λ ) = J ( a ; λ ) + J ( a ; λ ) in an obvious manner. J ( a ; λ ) . Then we have J ( a ; λ ) = − a − δ π X k ≥ k · πi Z c + ∞ ic −∞ i H ( − s )Γ(1 + s ) sin πs (2 πka ) − s ds, symptotics of a Mathieu-Gaussian series H ( − s ) = ( − ) m − p πe λ Γ( m ) sin πs Γ( p − s )Γ(1 − m + p − s ) F (1 − m ; 1 − m + p − s ; − λ )= ( − ) m − p πe λ Γ( m ) sin πs m − X r =0 (1 − m ) r ( − λ ) r r ! Γ( p − s )Γ(1 − m + p − s + r )= ( − ) p − πe λ Γ( m ) sin πs m − X r =0 (1 − m ) r λ r r ! Γ( m − p − r + s )Γ(1 − p + s ) . If we make the change of summation index r → m − − ℓ and use the fact that (1 − m ) m − − ℓ =( − ) m − ℓ Γ( m ) /ℓ !, we find H ( − s ) = ( − ) m − p πe λ sin πs m − X ℓ =0 A ℓ (1 − p + s ) ℓ , A ℓ := ( − ) ℓ λ m − − ℓ ( m − − ℓ )! ℓ ! . (4.1)The Pochhammer symbol appearing in (4.1) can be written in the form(1 − p + s ) ℓ = ℓ X r =0 B rℓ ( s + 1) r , where B = 1 , B = − p, B = B = ( − p ) , B = − p, B = ,B = ( − p ) , B = (5 − p + 4 p ) , B = (1 − p ) , B = , . . . . (4.2)Then we obtain J ( a ; λ ) = ( − ) k − p − e λ a − δ m − X ℓ =0 ℓ X r =0 A ℓ B rℓ X k ≥ k · πi Z c + ∞ ic −∞ i Γ(1 + s + r )(2 πka ) − s ds. (4.3)The integrals appearing in J ( a ; λ ) can be evaluated by (3.7) to produce the final exact result J ( a ; λ ) = ( − ) m − p − a p +1 πe λ m − X r =0 m − X ℓ = r A ℓ B rℓ σ r ( a )(2 πa ) r a m − r (4.4)for positive integer m . Here we have defined the sums σ r ( a ) := e πa X k ≥ k r e − πka , (4.5)which have the evaluations for 0 ≤ r ≤ σ ( a ) = e πa πa , σ ( a ) = e πa πa , σ ( a ) = e πa cosh πa πa ,σ ( a ) = e πa (2 + cosh πa )8 sinh πa , . . . . Note that J ( a ; λ ) ≡ m = 0, since H ( s ) vanishes for these values. J ( a ; λ ) . We have J ( a ; λ ) = − a − δ π X k ≥ k · πi Z c + ∞ ic −∞ i H ( − s )Γ(1 + s ) sin πs (2 πka ) − s ds, R. B. Paris where from (2.5) H ( s ) = ( − ) m − p − πλ m − p − s/ sin πs F ( m ; 1 + m − p + s ; λ )Γ(1 + m − p + s )= ( − ) m − p − πλ m − p − s/ sin πs ∞ X r =0 ( m ) r λ r r !Γ(1+ m − p + r + s ) . Then we obtain J ( a ; λ ) = ( − λ ) m − p a − δ √ π X k ≥ k ∞ X r =0 ( m ) r λ r r ! K kr , (4.6)where K kr = √ π πi Z c + ∞ ic −∞ i − s Γ(1 + s )Γ(1+ m − p + r + s ) X − s/ k ds (4.7)with X k defined in (3.4). Since there no poles of the integrand in ℜ ( s ) > | s | is everywhere large onthe new path. The quotient of gamma functions in the integrand can then be expanded in amanner similar to that in (3.5) to find [7, p. 53]2 − s √ π Γ(1 + s )Γ(1+ m − p + r + s ) = Γ( + s )Γ(1 + s )Γ(1+ m − p + r + s )= M − X j =0 ( − ) j ˆ c j ( r )Γ( + ϑ − j ) + ρ M ( s )Γ( s + ϑ − M ) , where M is a positive integer, ϑ = + p − m − r and ρ M ( s ) = O (1) as | s | → ∞ in | arg s | < π .The coefficients are given explicitly byˆ c j ( r ) = 1 j ! ( m − p + r ) j ( m − p + r + ) j = (2 m − p + 2 r ) j j j ! . (4.8)When m = r = 0, the coefficients ˆ c j ( r ) reduce to c j in (3.6). Then K kr = M − X j =0 ( − ) r c j ( r ) · πi Z c + ∞ ic −∞ i Γ( s + ϑ − j ) X − s/ k ds + R M,r = X ϑk e − X k (cid:26) M − X j =0 ( − ) j c j ( r ) X − jk + O ( X − Mk ) (cid:27) (4.9)by (3.7), where the remainder term R M,r = 14 πi Z c + ∞ ic −∞ i ρ M ( s )Γ( s + ϑ − j ) X − s/ k ds = O ( X ϑ − Mk e − X k )as | a | → ∞ in | arg a | < π/ J ( a ; λ ) ∼ ( − ) m − p (cid:18) λπa (cid:19) δ − / X k ≥ e − π k a /λ k δ ∞ X r =0 ∞ X j =0 ( − ) j c j ( r ) ( m ) r λ r r ! (cid:18) λπ k a (cid:19) r + j = ( − ) m − p (cid:18) λπa (cid:19) δ − / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r + δ , (4.10) symptotics of a Mathieu-Gaussian series r and j has been summed ‘diagonally’ (see [8, p. 58]) and thecoefficients C r are given by C r ≡ C r ( m, p, λ ) = r X n =0 ( − λ ) n ( m ) n n ! ˆ c r − n ( n ) . Use of (4.8) shows that the C r may be expressed in terms of a terminating F hypergeometricseries: C r = r X n =0 ( − λ ) r ( m ) n n !( r − n )! 2 n − r (2 m − p ) r (2 m − p ) n = (2 m − p ) r r r ! r X n =0 ( − r ) n ( m ) n λ n n !( m − p ) n ( m − p + ) n = (2 m − p ) r r r ! F ( − r, m ; m − p, m − p + ; λ ) , (4.11)where we have made use of the results ( a +2 n ) r − n = ( a ) r / ( a ) n and (2 a ) n = 2 n ( a ) n ( a + ) n .Then we have the following theorem: Theorem 2 . Let µ = m and γ = 2 p , where m ≥ , p are integers, and δ = 2( m − p ) with λ > .Then we have the representation S µ,γ ( a ; λ ) = a − δ H (1) + H µ,γ ( a ; λ ) + J ( a ; λ ) , (4.12) where H (1) and H µ,γ ( a ; λ ) are defined in (2.8) and (3.2). The exponentially small contribution J ( a ; λ ) has the expansion J ( a ; λ ) + ( − ) m − p πa p +1 e λ − πa m − X r =0 m − X ℓ = r A ℓ B rℓ σ r ( a )(2 π ) r a m − r ∼ ( − ) m − p (cid:18) λπa (cid:19) δ − / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r + δ (4.13) as | a | → ∞ in | arg a | < π/ . The coefficients A ℓ , B rℓ and C r are defined in (4.1), (4.2) and(4.11) and the functions σ r ( a ) are given by (4.5). When m = 0 the double sum on the left-handside of (4.13) vanishes.
5. Examples
We present some examples of the expansion of S µ,γ ( a ; λ ) stated in Theorem 2. Example 1.
Let µ = 0 and γ = 2 p . Then from (4.12) and (4.13) we have, for p = 1 , , . . . , S , p ( a ; λ ) = ∞ X n =1 n p e − λn /a ∼ (cid:18) λa (cid:19) − p − / Γ( p + )+( − ) p (cid:18) λπa (cid:19) − p − / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r − p (5.1)as | a | → ∞ in | arg a | < π/
4, where from (4.11) C r = ( − p ) r / (2 r r !). The case p = 0 is coveredin Remark 1.In the case p ≤ − p = − q and note that H ( s ) = λ q − s/ Γ( s − q ) when µ = 0. Theresidue of H ( s ) at s = 2 q − k ( k = 0 , , , . . . ) is given by ( − ) k /k !. Then S , − q ( a ; λ ) = ∞ X n =1 e − λn /a n q ∼ (cid:18) λa (cid:19) q − / Γ( − q ) + q X k =0 ( − ) k k ! (cid:18) λa (cid:19) k ζ (2 q − k )0 R. B. Paris +( − ) q (cid:18) λπa (cid:19) q − / ∞ X r =0 ( − ) r (2 q ) r r r ! (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k q +2 r (5.2)as | a | → ∞ in | arg a | < π/ Example 2.
Let µ = 1 and γ = 0 (so that δ = 2). Then we have S , ( a ; λ ) = ∞ X n =1 e − λn /a n + a = πe λ a erfc √ λ − a + J ( a ; λ ) , (5.3)since [3, (13.6.8)] H (1) = √ π U ( , , λ ) = πe λ erfc √ λ, where erfc is the complementary error function. From (4.13), the exponentially small contribu-tion is J ( a ; λ ) − πe λ a e − πa sinh πa ∼ (cid:18) λπa (cid:19) / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r +2 (5.4)as | a | → ∞ in | arg a | < π/
4, where from (4.11) the coefficients C r are given by C r = 2Γ( r + ) √ π F ( − r ; ; λ ) . The first few C r are therefore C = 1 , C = − λ, C = − λ + λ ,C = − λ + λ − λ . We remark that the case µ = 1, γ = 2 can be obtained directly from Theorem 2, but alsofollows from the Poisson-Jacobi formula (1.3), together with (5.3) and (5.4), since S , ( a ; λ ) = ∞ X n =1 n e − λn /a n + a = ∞ X n =1 (cid:18) − a n + a (cid:19) e − λn /a = S , ( a ; λ ) − a S , ( a ; λ ) . Example 3.
Let µ = 2 and γ = 0 (so that δ = 4). Then we have S , ( a ; λ ) = ∞ X n =1 e − λn /a ( n + a ) = √ π a U ( , − , λ ) − a + J ( a ; λ ) . (5.5)From (4.1) and (4.2), the coefficients A = λ , A = − B = 1, B = B = , so that J ( a ; λ ) = − πe λ − πa a sinh πa (cid:26) λ − − πae πa πa (cid:27) and J ( a ; λ ) ∼ (cid:18) λπa (cid:19) / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r +4 . From (4.11) the coefficients C r are given by C r = 4( r + 1)3 √ π Γ( r + ) F ( − r ; ; λ ) symptotics of a Mathieu-Gaussian series C = 1 , C = 5 − λ, C = − λ + 3 λ ,C = − λ + 54 λ − λ . Then the exponentially small contribution is J ( a ; λ ) + πe λ − πa a sinh πa (cid:26) λ − − πae πa πa (cid:27) ∼ (cid:18) λπa (cid:19) / ∞ X r =0 ( − ) r C r (cid:18) λπ a (cid:19) r X k ≥ e − π k a /λ k r +4 (5.6)as | a | → ∞ in | arg a | < π/ µ = 2 and p = 2 , S , ( a ; λ ) = S , ( a ; λ ) − a S , ( a ; λ ) S , ( a ; λ ) = S , ( a ; λ ) − a S , ( a ; λ ) − a S , ( a ; λ ) .
6. Numerical results and concluding remarks
The expansion of the exponentially small contribution (when γ = 2 p ) given in Theorem 1 is exactfor µ ≥
0. It is possible to employ an asymptotic expansion for the integrals I jk , but this wouldnecessarliy introduce an error. However, we have evaluated these integrals to high numericalprecision and have thereby verified the expansion (3.10) of J ( a ; λ ) for several parameter valuesto 50 decimal precision.We present some numerical examples of the large- a expansion of S µ,γ ( a ; λ ) given in Theorem2 to demonstrate the accuracy of our results. We subtract from the sum S µ,γ ( a ; λ ) the finiteterms appearing in (4.12) by definingˆ S µ,γ ( a ; λ ) := S µ,γ ( a ; λ ) − { a − δ H (1) + H µ,γ ( a ; λ ) + J ( a ; λ ) } (6.1)and comparing it with the exponentially small asymptotic expansion J ( a ; λ ) in (4.10). Westress that the contribution J ( a ; λ ) in (4.4) is an exact result when µ is an integer. In Table 1we show the values of the absolute relative error in the high-precision computation of ˆ S µ,γ ( a ; λ )from (1.1) using the asymptotic expansion for J ( a ; λ ) for different truncation index r . Thevalues of µ and γ chosen correspond to the examples given in Section 5. The final entry ineach column gives the value of ˆ S µ,γ ( a ; λ ). It is seen that the exponentially small contributionto S µ,γ ( a ; λ ) when γ is an even integer agrees well with the expansion given in Theorem 2.It is worth mentioning that J ( a ; λ ) given in Theorems 1 and 2 appears to comprise two differ-ent types of exponentially small terms, namely exp( − πka ) in Theorem 1 and both exp( − πka )and exp( − π k a /λ ), k ≥ I jk ap-pearing in Theorem 1 reveals that they also contain the more subdominant terms exp( − π k a /λ ).To see this we consider e − πka I jk = e − πka Z ∞ t µ − e − ψ ( t ) (1 + t ) β dt, β = 2 p − j + . The phase function ψ ( t ) in (3.8) has a saddle point at t = t s , where 1+ t s = p X k /λ = πka/λ and ψ ( t s ) = 0, ψ ′′ ( t s ) = 2 X k / (1+ t s ) = 2 λ / ( πka ). For large complex a in the sector | arg a | < π/ t = φ = π − a R. B. Paris
Table 1:
The absolute relative error in the computation of ˆ S µ,γ ( a ; λ ) from (6.1) for different µ , γ and truncationindex r in the asymptotic expansion J ( a ; λ ) when λ = 2 and a = 3. r µ = 0 , γ = − µ = 1 , γ = 0 µ = 2 , γ = 00 3 . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × − . × −
10 9 . × − . × − . × −
15 3 . × − . × − . × −
20 4 . × − . × − . × − ˆ S µ,γ − . × − − . × − +4 . × − to the singularity at t = − t s to infinityin ℜ ( t ) >
0. The contribution from the saddle is controlled by2 e − πka r π ψ ′′ ( t s ) t µ − s (1 + t s ) β = r πλ (cid:18) πkaλ (cid:19) µ − β − / e − πka while that from the neighbourhood of the origin is approximately e − X k + iµφ Z ∞ e −| X k | τ τ µ − dτ = O ( X − µk e − X k ) , which produces the more subdominant exponential terms.Finally, we note that the alternating version of (1.1) can be expressed in terms of S µ,γ ( a ; λ )since ∞ X n =1 ( − ) n − n γ ( n + a ) µ e − λn /a = S µ,γ ( a ; λ ) − − δ S µ,γ ( a ; λ ) . Application of Theorems 1 and 2 then enables the large- a expansion of the alternating series tobe determined. Appendix A: The pole structure of H ( s )The function H ( s ) defined in (2.3), (2.4) and (2.5) has pokes at s = − k − γ , k = 0 , , , . . . and apparent poles at s = ± k + δ , δ = 2 µ − γ . We shall show in this appendix that H ( s ) isregular at these last points. We have H ( s ) = πG ( s )2 sin π ( µ − γ + s ) , where G ( s ) := Γ( γ + s )Γ( µ ) F ( γ + s ; 1+ γ + s − µ ; λ ) − λ µ − ( γ + s ) / F ( µ ; 1 − γ + s + µ ; λ ) . Here F denotes the normalised confluent hypergeometric function defined by F ( a ; b, z ) = 1Γ( b ) F ( a ; b ; z ) , which is defined for all values of the parameter b . symptotics of a Mathieu-Gaussian series s k = 2 k + δ so that ( γ + s k ) = k + µ . Then G ( s k ) = ( µ ) k F ( µ + k ; k + 1; λ ) − λ − k ∞ X r = k ( µ ) r λ r r !Γ(1 + r − k )= ( µ ) k F ( µ + k ; k + 1; λ ) − ∞ X r =0 ( µ ) r + k λ r r !Γ(1 + k + r )= ( µ ) k F ( µ + k ; k + 1; λ ) − ( µ ) k ∞ X r =0 ( µ + k ) r λ r r !Γ(1 + k + r ) ≡ . Hence H ( s ) is regular at s k = 2 k + δ .A similar argument when s k = − k + δ shows that G ( s k ) = Γ( µ − k )Γ( µ ) ∞ X r = k ( µ − k ) r λ r r !Γ(1 − k + r ) − λ k F ( µ ; 1 + k ; λ )= Γ( µ − k )Γ( µ ) ∞ X r =0 ( µ − k ) r + k λ r + k r !Γ(1 + k + r ) − λ k F ( µ ; 1 + k ; λ )= λ k ∞ X r =0 ( µ ) r λ r r !Γ(1 + k + r ) − λ k F ( µ ; 1 + k ; λ ) ≡ , so that H ( s ) is also regular at the points s k = − k + δ . Appendix B: The expansion in the case γ = − a expansion of S µ,γ ( a ; λ ) given in (2.7) in the special case γ = − s = 1 is a double pole. We set s = 1 + ǫ , with ǫ → H ( s ) = Γ( ǫ )Γ( µ − ǫ )2Γ( µ ) F ( ǫ ; 1 − µ + ǫ ; λ )= 1 ǫ (cid:26) ǫ ( ψ (1) − ψ ( µ )) + O ( ǫ ) (cid:27) F ( ǫ ; 1 − µ + ǫ ; λ ) , where F ( ǫ ; 1 − µ + ǫ ; λ ) = 1 + ǫλ − µ ) (cid:26) λ (2 − µ )2! + 2! λ (2 − µ )
3! + · · · (cid:27) + O ( ǫ )= 1 + ǫλ − µ ) F (1 ,
1; 2 , − µ ; λ ) + O ( ǫ ) . Using the fact that ζ (1 + ǫ ) = ǫ − (1 + γ E ǫ + O ( ǫ )) and ψ (1) = − γ E , where γ E is the Euler-Mascheroni constant, we obtain the residue resulting from H ( s ) at s = 1 given by a (cid:26) log a + 12 γ E − ψ ( µ ) + λ − µ ) F (1 ,
1; 2 , − µ ; λ ) (cid:27) ( µ = 1 , , . . . ) . The residue resulting from H ( s ) is a H (1) = 12 λ µ Γ(1 − µ ) F ( µ ; 1 + µ ; λ ) ( µ = 1 , , . . . ) . Hence, provided µ = 1 , , . . . , S µ, − ( a ; λ ) = ∞ X n =1 e − λn /a n ( n + a ) µ ∼ a − δ (cid:26) log a + 12 γ E − ψ ( µ ) + λ − µ ) F (1 ,
1; 2 , − µ ; λ ) (cid:27) R. B. Paris + a − δ ∞ X k =1 ( − ) k ( µ ) k k ! ζ (1 − k ) F ( − k ; 1 − µ − k ; λ ) a − k (B.1)as a → ∞ in | arg a | < π/ µ is a positive integer a limiting process is required. To illustrate, we consider onlythe case µ = 1. We find that H (1 + ǫ ) = e λ /ǫ and H (1 + ǫ ) = − λǫ (1 − ǫ ) Γ(1 + ǫ ) F (1; 2 − ǫ ; λ )= − λǫ (cid:26) ǫ (1 − γ E − log λ ) + O ( ǫ ) (cid:27) F (1; 2 − ǫ ; λ ) , where F (1; 2 − ǫ ; λ ) = 1+ λ (cid:18) ǫ (cid:19) + λ · (cid:18) ǫ ǫ (cid:19) + · · · + O ( ǫ )= F (1; 2; λ ) + ǫ (cid:26) λ λ · (cid:18)
12 + 13 (cid:19) + λ · · (cid:18)
12 + 13 + 14 (cid:19) + · · · (cid:27) + O ( ǫ )= e λ − λ + ǫ ∞ X n =1 λ n τ ( n )(2) n + O ( ǫ ) , τ ( n ) := n X r =1 r + 1 . Then we obtain the expansion S , − ( a ; λ ) = ∞ X n =1 e − λn /a n ( n + a ) ∼ a − δ (cid:26) e λ (log λ + γ E −
1) + log ( a /λ ) + γ E + 1 − λ ∞ X n =1 λ n τ ( n )(2) n (cid:27) + a − δ ∞ X k =1 ( − ) k ( µ ) k k ! ζ (1 − k ) e k ( λ ) a − k (B.2)as a → ∞ in | arg a | < π/
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