aa r X i v : . [ m a t h . C A ] D ec ASYMPTOTICS OF CERTAIN q -SERIES RUIMING ZHANG
Abstract.
In this work we study complete asymptotic expansions for theq-series P ∞ n =1 1 n b q n a and P ∞ n =1 σ α ( n ) n b q n a in the scale function (log q ) n as q → − , where a > , q ∈ (0 , , b, α ∈ C and σ α ( n ) is the divisor function σ α ( n ) = P d | n d α . Preliminaries
In this work we study complete asymptotic expansions for the q-series P ∞ n =1 1 n b q n a and P ∞ n =1 σ α ( n ) n b q n a in the scale function (log q ) n as q → − , where a > , q ∈ (0 , , b, α ∈ C and σ α ( n ) is the divisor function σ α ( n ) = P d | n d α . Unlike methodsused [3, 4], our method does not apply Fourier transform or the modular properties,it can not give P ∞ n =1 1 n b q n a a complete asymptotic expansion in exponential scaleswhen a = 2 and b is an even integer. However, this shortcoming can be overcomeby applying the functional equations for the corresponding zeta functions which areequivalent to the symmetry x → /x .The Euler gamma function is defined by(1.1) Γ ( z ) = Z ∞ e − x x z − dx, ℜ ( z ) > , and its analytic continuation is given by(1.2) Γ ( z ) = Z ∞ e − x x z − dx + ∞ X n =0 ( − n n ! ( n + z ) , z ∈ C \ N . Let a, b ∈ R and a < b , it is known that [1, 5, 6](1.3) Γ( σ + it ) = O (cid:16) e − π | t | / | t | σ − / (cid:17) , t ∈ R as t → ±∞ , uniformly with respect to σ ∈ [ a, b ] . The digamma function is definedby(1.4) ψ ( z ) = Γ ′ ( z )Γ( z ) , z ∈ C and the Euler’s constant is(1.5) γ = − ψ (1) ≈ . . Mathematics Subject Classification.
Key words and phrases. q-series; divisor functions; asymptotics.The work is supported by the National Natural Science Foundation of China grants No.11371294 and No. 11771355. q -SERIES 2 The Riemann zeta function ζ ( s ) is defined by(1.6) ζ ( s ) = ∞ X n =1 n s , ℜ ( s ) > , then its analytic continuation, which is also denoted as ζ ( s ) , is an meromorphicfunction that has a simple pole at with residue . The meromorphic function ζ ( s ) satisfies the functional equation [1, 2, 5, 6](1.7) ζ ( s ) = 2 s π s − Γ(1 − s ) sin (cid:16) πs (cid:17) ζ (1 − s ) . For α, β ∈ R and α ≤ σ ≤ β , it is known that [6](1.8) ζ ( σ + it ) = O (cid:16) | t | | α | +1 / (cid:17) as t → ±∞ , uniformly with respect to σ . The Stieltjes constants γ n are thecoefficients in the Laurent expansion,(1.9) ζ ( s ) = 1 s − ∞ X n =0 ( − n n ! γ n ( s − n , where γ = γ and γ ≈ − . . Moreover, the Glaisher’s constant A ≈ . is defined as(1.10) log A = 112 − ζ ′ ( − . The Bernoulli numbers B n are defined by(1.11) ze z − ∞ X n =0 B n n ! z n , | z | < π. Then(1.12) B = 1 , B = − , B n − = 0 , n ∈ N . By (1.7) we get(1.13) ζ ( − n ) = 0 , ζ (1 − n ) = − B n n , n ∈ N . The function σ α ( n ) for α ∈ C is defined as the sum of the α -th powers of thepositive divisors of n , [2, 5](1.14) σ α ( n ) = X d | n d α , where d | n stands for " d divides n ". We also use the notations d ( n ) = σ and σ ( n ) = σ ( n ) . It is known that [2, 5](1.15) ∞ X n =1 σ α ( n ) n s = ζ ( s ) ζ ( s − α ) , ℜ ( s ) > max { , ℜ ( α ) + 1 } and(1.16) ∞ X n =1 d ( n ) n s = ζ ( s ) ℜ ( s ) > . SYMPTOTICS OF CERTAIN q -SERIES 3 Main Results
Theorem 1.
Given a positive integer k , let a j ∈ N , b j ∈ C for all j satisfying ≤ j ≤ k .If (2.1) k Y j =1 ζ ( a j s + b j ) = ∞ X n =1 f k ( n ) n s , ℜ ( s ) > max ≤ j ≤ k (cid:26) − ℜ ( b j ) a j (cid:27) , where ζ ( s ) is the Riemann zeta function, then for all x, a > , b ∈ C and c > satisfying c > max ≤ j ≤ k n −ℜ ( b j + b ) aa j o we have (2.2) πi Z c + i ∞ c − i ∞ Γ( s ) k Y j =1 ζ ( aa j s + b j + b ) dsx s = ∞ X n =1 f k ( n ) n b e − n a x . Furthermore, ∞ X n =1 f k ( n ) n b e − n a x = X j Residue (cid:26) g ( s ) , s = 1 − b − b j aa j (cid:27) + X n Residue { g ( s ) , s = − n } (2.3) as x → + , where the first sum is over all the distinct pairs a j b j j = 1 , . . . , k whilethe last sum is over all nonnegative integers n such that (2.4) − n = 1 − b − b j aa j , j = 1 , . . . , k. Proof.
For ℜ ( s ) > max ≤ j ≤ k n −ℜ ( b j + b ) aa j o , since each factor of Q kj =1 ζ ( aa j s + b j + b ) is an absolute convergent Dirichlet series, then the product itself is also an absoluteconvergent Dirichlet series. Let s be any complex number satisfying σ = ℜ ( s ) > max ≤ j ≤ k (cid:26) − ℜ ( b j + b ) aa j (cid:27) , then by the theory of Dirichlet series we know the partial sums P n ≤ x f k ( n ) n as b areabsolutely and uniformly bounded for all x > . Let N be a large positive integerand M = ∞ X n =1 | f k ( n ) | n aσ + ℜ ( b ) , s = s + 1 , a = N − , b = N in Lemma 2 of section 11.6 in [2] to get | f k ( N ) | ≤ M N σ . Hence,(2.5) ∞ X n =1 | f k ( N ) | n ℜ ( b ) e − n a x < ∞ , a, x > , b ∈ C . By the inverse Mellin transform of (1.1) we get(2.6) πi Z c + i ∞ c − i ∞ Γ( s ) dsx s = e − x for all x, c > .Let g ( s ) = Γ( s ) k Y j =1 ζ ( aa j s + b j + b ) x − s , SYMPTOTICS OF CERTAIN q -SERIES 4 then for any positive c satisfying c > max ≤ j ≤ k n −ℜ ( b j + b ) aa j o , by (2.6) we get πi Z c + i ∞ c − i ∞ g ( s ) ds = 12 πi Z c + i ∞ c − i ∞ Γ( s ) ∞ X n =1 f k ( n ) n as + b ! dsx s (2.7) = ∞ X n =1 f k ( n ) n b πi Z c + i ∞ c − i ∞ Γ( s ) ds ( n a x ) s = ∞ X n =1 f k ( n ) n b e − n a x , where we have applied (2.5) and the Fubini’s theorem to exchange the order ofsummation and integration.Since ζ ( s ) has a simple pole at s = 1 and Γ( s ) has simple poles at all non-positiveintegers, then all the possible poles of the meromorphic function g ( s ) are s = 1 − b − b j aa j , j = 1 , . . . , k and all non-positive integers. Let N ∈ N and M ∈ R such that N > max ≤ j ≤ k (cid:26) | b + b j | aa j (cid:27) + 1 , M > max ≤ j ≤ k (cid:26) | b + b j | aa j (cid:27) , we integrate g ( s ) over the rectangular contour R ( M, N ) with vertices, c − iM, c + iM, − N −
12 + iM, − N − − iM. Then by Cauchy’s theorem we have(2.8) Z R ( M,N ) g ( s ) ds πi = X j Residue (cid:26) g ( s ) , s = 1 − b − b j aa j (cid:27) + X n Residue { g ( s ) , s = − n } , where the first sum is over all the distinct pairs from a j b j , j = 1 , . . . , k whereasthe last sum is over all n satisfying ≤ n ≤ N and (2.4).On the other hand, we also have(2.9) Z R ( M,N ) g ( s ) ds πi = (Z c + iMc − iM − Z − N +12 + iM − N +12 − iM ) g ( s ) ds πi + (Z c − iM − N +12 − iM − Z c + iM − N +12 + iM ) g ( s ) ds πi Fix N and x , by (1.3) and (1.8), since the integrands of the last two integrals havethe estimate g ( s ) = O (cid:16) e − ( π/ − ǫ ) M (cid:17) , M → ∞ , where ǫ is an arbitrary positive number such that < ǫ < π , then the last twointegrals have limit as M → ∞ . Then by taking limit M → ∞ in (2.9) and (2.8)we get πi Z c + i ∞ c − i ∞ g ( s ) ds = 12 πi Z − N +12 + i ∞− N +12 − i ∞ g ( s ) ds (2.10) + X j Residue (cid:26) g ( s ) , s = 1 − b − b j aa j (cid:27) + X n Residue { g ( s ) , s = − n } , where the summations are the same as in (2.8). SYMPTOTICS OF CERTAIN q -SERIES 5 Since (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) πi Z − N +12 + i ∞− N +12 − i ∞ g ( s ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ x N +1 / π Z − N +12 + i ∞− N +12 − i ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( s ) k Y j =1 ζ ( aa j s + b j + b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt, then again by (1.3) and (1.8) we get(2.11) πi Z − N +12 + i ∞− N +12 − i ∞ g ( s ) ds = o (cid:0) x N (cid:1) as x → . Then by (2.10) and (2.11) we get(2.12) πi Z c + i ∞ c − i ∞ g ( s ) ds = X j Residue (cid:26) g ( s ) , s = 1 − b − b j aa j (cid:27) + X n Residue { g ( s ) , s = − n } as x → , where the first sum is over all the distinct pairs from a j b j , j = 1 , . . . , k while the last sum is over all nonnegative integers n satisfying (2.4). Finally, (2.2)is obtained by combining (2.7) and (2.12). (cid:3) Corollary 2.
Let a, x > , b ∈ C . If b = 1 + an, n ∈ N ∪ { } , then (2.13) ∞ X n =1 e − n a x n b = x b − a a Γ (cid:18) − ba (cid:19) + ∞ X n =0 ( − x ) n n ! ζ ( b − an ) as x → .If there exists a n ∈ N ∪ { } such that b = 1 + an , then ∞ X n =1 e − n a x n b = ( − x ) n ( γa + ψ ( n + 1) − log( x )) an ! + ∞ X n = 0 n = n , ( − x ) n n ! ζ ( b − an ) (2.14) as x → . Proof.
When b = 1 + an, n ∈ N ∪ { } , the integrand Γ( s ) ζ ( as + b ) x s is meromorphicand has the following simple poles s = 1 − ba , , − , − , . . . with residuesResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = 1 − ba (cid:27) = x ( b − /a a Γ (cid:18) − ba (cid:19) Residue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) . Then (2.13) is obtained by applying Theorem 1.When b = 1 + an for some nonnegative integer n , then Γ( s ) ζ ( as + b ) x s = Γ( s ) ζ ( a ( s + n ) + 1) x s has a double pole at − n with residueResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − n (cid:27) = ( − n x n ( γa + ψ ( n + 1) − log( x )) an ! , SYMPTOTICS OF CERTAIN q -SERIES 6 all the other nonpositive integers are simple poles with residues,Residue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − n = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) . Then by Theorem 1 we have ∞ X n =1 e − n a x n b = ( − x ) n ( γa + ψ ( n + 1) − log( x )) an ! + ∞ X n = 0 an = ( b − , ( − x ) n n ! ζ ( b − an ) as x → . (cid:3) Example 3.
When a = 2 , b = 0 we have(2.15) ∞ X n =1 e − n x = 12 √ πx + ∞ X n =0 ( − x ) n n ! ζ ( − n ) = 12 √ πx as x → , which means the error term is better than any x n . When a = 2 , b = 1 ,the double pole happens at n = 0 , then(2.16) ∞ X n =1 e − n x n = γ − log( x )2 − ∞ X n =1 B n ( − x ) n n !(2 n ) as x → . When a = 2 , b = − , then n + 1 = − has no nonnegative integersolutions. Thus,(2.17) ∞ X n =1 ne − n x = 12 x + 12 x ∞ X n =1 B n ( − x ) n n ! , or(2.18) ∞ X n = −∞ | n | e − n x = 1 x + 1 x ∞ X n =1 B n ( − x ) n n ! as x → . Corollary 4.
For all x, a > , b ∈ C and c > −ℜ ( b ) a we have (2.19) πi Z c + i ∞ c − i ∞ Γ( s ) ζ ( as + b ) dsx s = ∞ X n =1 d ( n ) n b e − n a x . Furthermore, if an = b − for all nonnegative integers n , then (2.20) ∞ X n =1 d ( n ) n b e − n a x = x b − a Γ (cid:0) − ba (cid:1) (cid:0) ψ (cid:0) − ba (cid:1) + 2 γa − log( x ) (cid:1) a + ∞ X n =0 ( − x ) n n ! ζ ( b − an ) as x → + . SYMPTOTICS OF CERTAIN q -SERIES 7 If am = b − for certain nonnegative m , then ∞ X n =1 d ( n ) n b e − n a x = ( − x ) m a m ! (cid:8) ( aγ ) − a γ + (2 aγ − log x ) ψ ( m + 1) − aγ log x (2.21) + ψ ( m + 1) − ψ (1) ( m + 1) + log ( x )2 + π (cid:27) + ∞ X n = 0 n = m ( − x ) n n ! ζ ( b − an ) , as x → + . Proof.
When an = b − for all nonnegative integers n , then the meromorphicfunction Γ( s ) ζ ( as + b ) x − s has a double pole at s = (1 − b ) /a with residueResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = 1 − ba (cid:27) = x b − a Γ (cid:0) − ba (cid:1) (cid:0) ψ (cid:0) − ba (cid:1) + 2 γa − log( x ) (cid:1) a and simple poles at all nonpositive integers n ∈ N with residueResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) . Then by Theorem 1 we get ∞ X n =1 d ( n ) n b e − n a x = x b − a Γ (cid:0) − ba (cid:1) (cid:0) ψ (cid:0) − ba (cid:1) + 2 γa − log( x ) (cid:1) a + ∞ X n =0 ( − x ) n n ! ζ ( b − an ) as x → .When α = 0 , am = b − for certain nonnegative integer m , then the meromor-phic function Γ( s ) ζ ( as + b ) x − s has a triple pole at s = − m with residueResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − m (cid:27) = ( − x ) m a m ! (cid:8) ( aγ ) − a γ + (2 aγ − log x ) ψ ( m + 1) − aγ log x + ψ ( m + 1) − ψ (1) ( m + 1) + log ( x )2 + π (cid:27) . It has simple poles at all other nonpositive integers with residueResidue (cid:26) Γ( s ) ζ ( as + b ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) . Then by Theorem 1 we get ∞ X n =1 d ( n ) n b e − n a x = ( − x ) m a m ! (cid:8) ( aγ ) − a γ + (2 aγ − log x ) ψ ( m + 1) − aγ log x + ψ ( m + 1) − ψ (1) ( m + 1) + log ( x )2 + π (cid:27) + ∞ X n = 0 n = m ( − x ) n n ! ζ ( b − an ) as x → + . (cid:3) SYMPTOTICS OF CERTAIN q -SERIES 8 Example 5.
Let a = 2 , b = 2 , then by (2.20) to get(2.22) ∞ X n =1 d ( n ) n e − n x = √ xπ (cid:0) log x − ψ ( − ) − γ (cid:1) π as x → + , the remainder here is better than any x n . Let a = 2 , b = 1 , then by(2.21) to get ∞ X n =1 d ( n ) n e − n x = 6 log x −
45 log x + 6 γ + π − γ
12 + 14 ∞ X n =1 B n ( − x ) n n ( n + 1)! (2.23)as x → + . Corollary 6.
Let α ∈ C and α = 0 , then for all x, a > , b ∈ C and c > max ≤ j ≤ k { −ℜ ( b ) , −ℜ ( b − α ) } a we have (2.24) πi Z c + i ∞ c − i ∞ Γ( s ) ζ ( as + b ) ζ ( as + b − α ) dsx s = ∞ X n =1 σ α ( n ) n b e − n a x . Furthermore, if an = b − and an = b − − α for all nonnegative integers n ∈ N ,then ∞ X n =1 σ α ( n ) n b e − n a x = ∞ X n =0 ( − x ) n n ! ζ ( b − an ) ζ ( b − an − α ) (2.25) + x ( b − /a a Γ (cid:18) − ba (cid:19) ζ (1 − α ) + x ( b − − α ) /a a Γ (cid:18) − b + αa (cid:19) ζ (1 + α ) as x → + .If am = b − for certain nonnegative integer m and an = b − − α for allnonnegative integers n ∈ N , then ∞ X n =1 σ α ( n ) n b e − n a x = ( − x ) m ( aζ ′ (1 − α ) + ζ (1 − α )( γa + ψ ( m + 1) − log( x ))) am ! (2.26) + x m − α/a a Γ (cid:18) α − maa (cid:19) ζ (1 + α ) + ∞ X n = 0 n = m ( − x ) n n ! ζ (1 − a ( n − m )) ζ (1 − α − a ( n − m )) as x → + .If an = b − for all nonnegative integers n ∈ N and am = b − − α for certain m ∈ N , then ∞ X n =1 σ α ( n ) n b e − n a x = x m + α/a a Γ (cid:18) − am + αa (cid:19) ζ (1 − α ) (2.27) + ( − x ) m ( aζ ′ (1 + α ) + ζ (1 + α )( γa + ψ ( m + 1) − log( x ))) am !+ ∞ X n = 0 n = m ( − x ) n n ! ζ (1 + α − a ( n − m )) ζ (1 − a ( n − m )) as x → + . SYMPTOTICS OF CERTAIN q -SERIES 9 If am = b − and α = a ( m − ℓ ) for certain nonnegative integers m, ℓ ∈ N with m = ℓ , then ∞ X n =1 σ α ( n ) n b e − n a x = ∞ X n = 0 n = m, ℓ ( − x ) n n ! ζ (1 − a ( n − m )) ζ (1 − a ( n − ℓ )) (2.28) + ( − x ) m ( aζ ′ (1 − a ( m − ℓ )) + ζ (1 − a ( m − ℓ ))( γa + ψ ( m + 1) − log( x ))) am !+ ( − x ) ℓ ( aζ ′ (1 + a ( m − ℓ )) + ζ (1 + a ( m − ℓ ))( γa + ψ ( ℓ + 1) − log( x ))) aℓ ! as x → + .Proof. When α = 0 , an = b − and an = b − − α for all nonnegative integers n ∈ N , the meromorphic function Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x − s has simple polesat − ba , − b + αa , , − , − , . . . with residuesResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = 1 − ba (cid:27) = x ( b − /a a Γ (cid:18) − ba (cid:19) ζ (1 − α ) , Residue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = 1 − b + αa (cid:27) = x ( b − − α ) /a a Γ (cid:18) − b + αa (cid:19) ζ (1 + α ) andResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) ζ ( b − an − α ) for n ∈ N . Then by Theorem 1 we get ∞ X n =1 σ α ( n ) n b e − n a x = ∞ X n =0 ( − x ) n n ! ζ ( b − an ) ζ ( b − an − α )+ x ( b − /a a Γ (cid:18) − ba (cid:19) ζ (1 − α ) + x ( b − − α ) /a a Γ (cid:18) − b + αa (cid:19) ζ (1 + α ) as x → + .When am = b − for certain nonnegative integer m and an = b − − α for allnonnegative integers n ∈ N , then the meromorphic function Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x − s has a double pole at s = − m with residueResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − m (cid:27) = ( − x ) m ( aζ ′ (1 − α ) + ζ (1 − α )( γa + ψ ( m + 1) − log( x ))) am ! , and a simple pole at s = − m + αa with residueResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − m + αa (cid:27) = x m − α/a a Γ (cid:18) α − maa (cid:19) ζ (1 + α ) SYMPTOTICS OF CERTAIN q -SERIES 10 and simple poles at all nonpositive integers other than − m with residuesResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) ζ ( b − α − an ) . Hence, ∞ X n =1 σ α ( n ) n b e − n a x = ( − x ) m ( aζ ′ (1 − α ) + ζ (1 − α )( γa + ψ ( m + 1) − log( x ))) am !+ x m − α/a a Γ (cid:18) α − maa (cid:19) ζ (1 + α ) + ∞ X n = 0 n = m ( − x ) n n ! ζ ( b − an )) ζ ( b − α − an ) as x → + .When an = b − for all nonnegative integers n ∈ N and am = b − − α forcertain m ∈ N , then the meromorphic function Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x − s hasa double simple pole s = − m with residueResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − m (cid:27) = ( − x ) m ( aζ ′ (1 + α ) + ζ (1 + α )( γa + ψ ( m + 1) − log( x ))) am ! and a simple pole s = − m − αa with residueResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − m − αa (cid:27) = x m + α/a a Γ (cid:18) − am + αa (cid:19) ζ (1 − α ) and simple poles at all nonpositive integers other than − m with residueResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − n (cid:27) = ( − x ) n n ! ζ ( b − an ) ζ ( b − α − an ) . Thus, ∞ X n =1 σ α ( n ) n b e − n a x = ( − x ) m ( aζ ′ (1 + α ) + ζ (1 + α )( γa + ψ ( m + 1) − log( x ))) am !+ x m + α/a a Γ (cid:18) − am + αa (cid:19) ζ (1 − α ) + ∞ X n = 0 n = m ( − x ) n n ! ζ ( b − an ) ζ ( b − α − an ) as x → + .When am = b − and α = a ( m − ℓ ) for certain nonnegative integers m, ℓ ∈ N with m = ℓ , then the meromorphic function Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x − s has twodouble poles at s = − m and s = − ℓ with residuesResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − m (cid:27) = ( − x ) m ( aζ ′ (1 − a ( m − ℓ )) + ζ (1 − a ( m − ℓ ))( γa + ψ ( m + 1) − log( x ))) am ! SYMPTOTICS OF CERTAIN q -SERIES 11 and Residue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − ℓ (cid:27) = ( − x ) ℓ ( aζ ′ (1 + a ( m − ℓ )) + ζ (1 + a ( m − ℓ ))( γa + ψ ( ℓ + 1) − log( x ))) aℓ ! respectively. It has simple poles at all other nonpositive integers other than − m, − ℓ with residuesResidue (cid:26) Γ( s ) ζ ( as + b ) ζ ( as + b − α ) x s , s = − n (cid:27) = ( − x ) n n ! ζ (1 − a ( n − m )) ζ (1 − a ( n − ℓ )) . Then, ∞ X n =1 σ α ( n ) n b e − n a x = ∞ X n = 0 n = m, ℓ ( − x ) n n ! ζ (1 − a ( n − m )) ζ (1 − a ( n − ℓ ))+ ( − x ) m ( aζ ′ (1 − a ( m − ℓ )) + ζ (1 − a ( m − ℓ ))( γa + ψ ( m + 1) − log( x ))) am !+ ( − x ) ℓ ( aζ ′ (1 + a ( m − ℓ )) + ζ (1 + a ( m − ℓ ))( γa + ψ ( ℓ + 1) − log( x ))) aℓ ! as x → + . (cid:3) Example 7.
When a = 2 , α = 1 , b = , by (2.25) we get(2.29) ∞ X n =1 σ ( n ) √ n e − n x = π (cid:18) (cid:19) x − − Γ (cid:18) (cid:19) x − + ∞ X n =0 ( − x ) n n ! ζ (cid:18) − n (cid:19) ζ (cid:18) − − n (cid:19) as x → + . When a = 2 , α = 1 , b = 1 , then m = 0 in (2.26). Then ∞ X n =1 σ ( n ) n e − n x = π / √ x + log x − log √ π − γ , as x → + , it implies that the difference between two sides of the above formulais smaller than any x n . Let a = 2 , b = 1 , α = − in (2.28), then m = 0 , ℓ = 1 .Then, ∞ X n =1 σ − ( n ) n e − n x = − ζ (3)2 log x + 2 ζ ′ (3) + ζ (3) γ − x log x
24+ 24 log A + γ + 124 x + 14 ∞ X n =2 B n − B n ( n − n ( − x ) n n ! as x → + . References [1] G. E. Andrews, R. Askey and R. Roy,
Special Functions,
Cambridge University Press, Cam-bridge, 1999.[2] T. M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics,Springer-Verlag, New York-Heidelberg, 1976.[3] B. C. Berndt and B. Kim, Asymptotic Expansions of Certain Partial Theta Functions, Pro-ceedings of AMS, Volume 139, Number 11, November 2011, 3779–3788
SYMPTOTICS OF CERTAIN q -SERIES 12 [4] K. Bringmann, A. Folsom and A. Milas, Asymptotic behavior of partial and false theta func-tions arising from Jacobi forms and regularized characters, Journal of Mathematical Physics58, 011702 (2017); ; doi: 10.1063/1.4973634.[5] Nist DLMF, http://dlmf.nist.gov/ [6] Hans Rademacher, Topics in Analytic Number Theory , Springer-Verlag, Berlin, 1973
College of Science, Northwest A&F University, Yangling, Shaanxi 712100, P. R.China.
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