Proceedings of the Steklov Institute of Mathematics, Vol. 218, 1997, pp. 340–351. Translated from Trudy Matematicheskogo Instituta imeni V. A. Steklova, Vol. 218, 1997, pp. 343–353. Original Russian Text Copyright \begin{matrix}c \\ ◯ \end{matrix} 1997 by Pavlov. English Translation Copyright \begin{matrix}c \\ ◯ \end{matrix} by ���� �����/Interperiodica Publishing (Russia).

Meromorphic Continuation of Dirichlet Series Connected with the Distribution of Numbers Modulo 1

A. I. Pavlov

Received February, 1997

In 1921 Hecke proved the following statement in his work [1]. Let α = \sqrt{D}, where the integer D ≥ 1 is a square-free number under the conditions D ≢ 1 (mod 4). Then the series

F (s) = ∑_{n=1}^{∞} \frac{\{αn\} − 1/2}{n^{s}},

where { · } is the fractional part of the number, which converges for Re s > 1, can be analytically continued to the whole plane of the complex variable s, except for the points s_{q,r} = −2q+2πir/ ln η, where q = 0, 1, 2, …, r = 0, ±1, ±2, …, η is the principal unit of the field 𝕂(\sqrt{D}), where 𝕂 is the field of rational numbers. At the points s_{q,r} the function F (s) has simple poles.

In 1923 Hardy and Littlewood, in their work [2], generalized Hecke’s result to the case of any quadratic irrationality α.

Let M be the set of all functions which are analytic in the domain Re s > 1 and which admit a meromorphic continuation to the whole plane of the variables s. We shall prove here the following statements.

Theorem 1. If the integer q ≥ 2 and P (x) is an arbitrary polynomial, then

∑_{n=1}^{∞} \frac{P (\{n^{1/q}\})}{n^{s}} ∈ M.

Theorem 2. If the integer q ≥ 2 and P (x) is an arbitrary polynomial, then

∑_{n=1}^{∞} \frac{P (\{log_{q} n\})}{n^{s}} ∈ M.

Remark 1. In Theorems 1 and 2, in the case of P (x) ≡ x, the arrangement of poles of the corresponding function is indicated.

Theorem 3. If the integer q ≥ 2, then there exists an infinite set of irrational α such that

∑_{n=1}^{∞} \frac{\{αq^{n}\}}{n^{s}} ∈ M.

Remark 2. For each α from Theorem 3 we can indicate the arrangement of poles of the corresponding functions.

Proof of Theorem 1. We shall begin with a special case which is interesting in itself since much (if not all) can be calculated here to the end. According to the definition of the integral part for Re s > 3/2 we have

∑_{n=1}^{∞} \frac{\{\sqrt{n}\}}{n^{s}} = ∑_{k=1}^{∞} ∑_{k^{2}≤n<(k+1)^{2}} \frac{\sqrt{n} − k}{n^{s}} = ζ(s − \frac{1}{2}) − ∑_{k=1}^{∞} k( ∑_{n=k^{2}}^{∞} n^{−s} − ∑_{n=(k+1)^{2}} n^{−s})

= ζ(s − \frac{1}{2}) − f_{1}(s),

where

f_{1}(s) = ∑_{k=1}^{∞} ∑_{n=k^{2}}^{∞} n^{−s}.  (1)

Using for Re s > 0 the relation

n^{−s} = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1}e^{−nx} dx  (2)

(here and in the sequel Γ(s) is the gamma-function), we get from (1) the relation

f_{1}(s) = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{ψ(x/π)}{1 − e^{−x}} dx,

where

ψ(x) = ∑_{k=1}^{∞} e^{−πxk^{2}}.

Let

f_{2}(s) = \frac{1}{Γ(s)} ∫_{0}^{1} x^{s−1} \frac{ψ(x/π)}{1 − e^{−x}} dx,  f_{3}(s) = \frac{1}{Γ(s)} ∫_{1}^{∞}x^{s−1} \frac{ψ(x/π)}{1 − e^{−x}} dx.  (3)

Then f_{1}(s) = f_{2}(s) + f_{3}(s), and, as it is easy to see, the function f_{3}(s) is entire. We know that

∑_{k=−∞}^{∞} e^{−πxk^{2}} = \frac{1}{\sqrt{x}} ∑_{k=−∞}^{∞} e^{−πk^{2}/x}

for x > 0. Consequently, for x > 0 we have

ψ(x) = \frac{1}{\sqrt{x}}ψ(\frac{1}{x}) + \frac{1}{2}(\frac{1}{\sqrt{x}} − 1).  (4)

From (3) and (4) we obtain

f_{2}(s) = \frac{1}{2Γ(s)} ∫_{0}^{1} \frac{x^{s−1}}{1 − e^{−x}} (\sqrt{\frac{π}{x}} − 1) dx + f_{4}(s),  (5)

where

f_{4}(s) = \frac{\sqrt{π}}{Γ(s)} ∫_{0}^{1} x^{s−3/2} \frac{ψ(π/x)}{1 − e^{−x}} dx

and the function f_{4}(s) is entire. It is easy to see that

∫_{0}^{1} \frac{x^{s−1}}{1 − e^{−x}} (\sqrt{\frac{π}{x}} − 1) dx = ∫_{0}^{1}(\sqrt{π}x^{s−3/2} − x^{s−1}) dx + ∫_{0}^{1} \frac{\sqrt{π}x^{s−3/2} − x^{s−1}}{e^{x} − 1} dx

= \frac{\sqrt{π}}{s − 1/2} − \frac{1}{s} + ∫_{0}^{∞} \frac{\sqrt{π}x^{s−3/2} − x^{s−1}}{e^{x} − 1} dx − f_{5}(s),  (6)

where

f_{5}(s) = ∫_{1}^{∞} \frac{\sqrt{π}x^{s−3/2} − x^{s−1}}{e^{x} − 1} dx,

and the function f_{5}(s) is entire. Hence, using (5), (6), and the relation

ζ(s)Γ(s) = ∫_{0}^{∞} \frac{x^{s−1}}{e^{x} − 1} dx,  Re s > 1,  (7)

we obtain

f_{2}(s) = \frac{1}{Γ(s)}(\frac{\sqrt{π}}{2s − 1} − \frac{1}{2s}) + \frac{1}{2Γ(s)}(\sqrt{π}Γ(s − \frac{1}{2})ζ(s − \frac{1}{2}) − Γ(s)ζ(s)) − \frac{f_{5}(s)}{2Γ(s)}. (8)

We know that the function 1/Γ(s) is entire. Therefore, using relations (3) and (8), we obtain for Re s > 3/2 the relation

∑_{n=1}^{∞} \frac{\{\sqrt{n}\}}{n^{s}} = (1 − \frac{\sqrt{π}Γ(s − 1/2)}{2Γ(s)})ζ(s − \frac{1}{2}) + \frac{ζ(s)}{2} − \frac{\sqrt{π}}{(2s − 1)Γ(s)} + Φ(s),  (9)

where the funtion Φ(s) is entire. It is easy to see that the right-hand side of relation (9) is analytic function throughout the plane of the complex variable s, except for the simple poles at the points 1, 1/2, and 1/2 − k, where k is an odd positive integer.

Let us now consider the general case. Obviously, it suffices to prove Theorem 1 for the series

∑_{n=1}^{∞} \frac{\{n^{1/q}\}^{m}}{n^{s}} ,

where the integer m ≥ 1. Since

\{n^{1/q}\} = (n^{1/q} − [n^{1/q}])^{m} = ∑_{ν=0}^{m}(−1)^{ν}\binom{m}{ν}[n^{1/q}]^{ν}n^{(m−ν)/q},

MEROMORPHIC CONTINUATION OF DIRICHLET SERIES 343

it follows that

∑_{n=1}^{∞} \frac{\{n^{1/q}\}^{m}}{n^{s}} = ∑_{ν=0}^{m}(−1)^{ν}\binom{m}{ν} ∑_{n=1}^{∞}[n^{1/q}]^{ν}n^{−(s−(m−ν)/q)}.

This means that everything reduces to the analytic continuation of the series

f_{q,m}(s) = ∑_{n=1}^{∞}[n^{1/q}]^{m}n^{−s},

where the integers q ≥ 2 and m ≥ 1. However, for Re s > 1 + m/q we have

f_{q,m}(s) = ∑_{k=1}^{∞} ∑_{k^{q}≤n<(k+1)^{q}}[n^{1/q}]n^{−s} = ∑_{k=1}^{∞} k^{m} ∑_{k^{q}≤n<(k+1)^{q}} n^{−s}

= ∑_{k=1}^{∞}(k^{m} − (k − 1)^{m}) ∑_{n=k^{q}}^{∞} n^{−s} = ∑_{ν+1}^{m}(−1)^{ν+1}\binom{m}{ν} ∑_{k=1}^{∞} k^{m−ν} ∑_{n=k^{q}}^{∞} n^{−s}.

Therefore, in order to prove Theorem 1, it suffices, for any integer ν ≥ 0, to continue meromorphically the series

φ_{q,ν}(s) = ∑_{k=1}^{∞} k^{ν} ∑_{n=k^{q}} n^{−s}

to the whole plane of the variable s. Using relation (2), for any integer N ≥ 1 we obtain

∑_{n=N}^{∞} n^{−s} = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{e^{−Nx}}{1 − e^{−x}} dx.

Therefore

φ_{q,ν}(s) = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{ψ_{q,ν}(x)}{1 − e^{−x}} dx,

where

ψ_{q,ν}(x) = ∑_{k=1}^{∞} k^{ν}e^{−xk^{q}}, x > 0.

Let

φ_{q,ν}^{(1)} = \frac{1}{Γ(s)} ∫_{0}^{1} x^{s−1} \frac{ψ_{q,ν}(x)}{1 − e^{−x}} dx, φ_{q,ν}^{(2)} = \frac{1}{Γ(s)} ∫_{1}^{∞}x^{s−1} \frac{ψ_{q,ν}(x)}{1 − e^{−x}} dx.

Then φ_{q,ν}(s) = φ_{q,ν}^{(1)}(s) + φ_{q,ν}^{(2)}(s), and φ_{q,ν}^{(2)} is an entire function of s. It is known [3, p. 83] that for x → 0, x > 0, and for any fixed integer N ≥ 1 the relation

ψ_{q,ν}(x) = \frac{1}{q}Γ(\frac{ν + 1}{q})x^{−(ν+1)/q} + ∑_{k=0}^{N} c_{k}x^{k} + O(x^{N+1/2})

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

is satisfied, where the constants c_{k} do not depend on x under the sign O either. Consequently, for any integer N ≥ 1 we have

φ_{q,ν}^{(1)} = \frac{1}{qΓ(s)}Γ(\frac{ν + 1}{q}) ∫_{0}^{1} \frac{x^{s−(ν+1)/q}}{1 − e^{−x}} dx + \frac{1}{Γ(s)} ∑_{k=0}^{N} c_{k} ∫_{0}^{1} \frac{x^{s+k−1}}{1 − e^{−x}} dx + R_{N}^{(1)}(s),  (10)

where the function R_{N}^{(1)}(s) is analytic for Re s > −(N + 1/2). It is easy to see that for any real α the relation

∫_{0}^{1} \frac{x^{s+α−1}}{1 − e^{−x}} dx = \frac{1}{s + α} + ∫_{0}^{∞} \frac{x^{s+α−1}}{e^{x} − 1} dx − g_{α}(s),

where

g_{α}(s) = ∫_{1}^{∞} \frac{x^{s+α−1}}{e^{x} − 1} dx

is entire function of s for any α, holds true for any Re s > −α. Consequently, using relation (7), we obtain

∫_{0}^{1} \frac{x^{s+α−1}}{1 − e^{−x}} dx = \frac{1}{s + α} + Γ(s + α)ζ(s + α) − g_{α}(s).  (11)

Returning to the function φ_{q,ν}(s), with the aid of relations (10) and (11), for Re s > max(−(1+ν)/q, −(N + 1/2)) we arrive at the relation

φ_{q,ν}(s) = \frac{1}{q}Γ(\frac{1 + ν}{q} )(s − \frac{1 + ν}{q} )^{−1} + \frac{1}{q}Γ(\frac{1 + ν}{q} )\frac{Γ(s − (ν + 1)/q)ζ(s − (ν + 1)/q)}{Γ(s)}

+ \frac{1}{Γ(s)} ∑_{k=0}^{N} c_{k}(\frac{1}{s + k} + Γ(s + k)ζ(s + k)) + G_{N}(s),  (12)

where the function G_{N}(s) is analytic for Re s > −(N + 1/2). Since N is an arbitrary positive integer, it follows from (12) that the function φ_{q,ν}(s) is analytic for any complex s, except for the simple poles at the points (ν + 1)/q and (ν + 1)/q − k, where k is an odd positive integer, and −k, where k is a nonnegative integer.

Proof of Theorem 2. We shall begin with a special case where P (x) ≡ x. Then, for Re s > 1, we have

∑_{n=1}^{∞} \frac{\{log_{q} n\}}{n^{s}} = ∑_{n=1}^{∞} \frac{log_{q} n − [log_{q} n]}{n^{s}} = − log_{q} e × ζ′(s) − f_{6}(s),

where

f_{6}(s) = ∑_{n=1}^{∞} \frac{[log_{q} n]}{n^{s}} = ∑_{k=0}^{∞} ∑_{q^{k}≤n<q^{k+1}} kn^{−s} = ∑_{k=1}^{∞} ∑_{n=q^{k}}^{∞} n^{−s}.

Using relation (2), we get, for Re s > 1, the relation

f_{6}(s) = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{ψ_{q}(x)}{1 − e^{−x}} dx,

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

where

ψ_{q}(x) = ∑_{k=1}^{∞} e^{−xq^{k}}.

We can show (see the lemma below) that the relation

ψ_{q}(x) = −\frac{ln x}{ln q} + c_{0} + ∑_{k=1}^{N} \frac{(−1)^{k+1}q^{k}}{(q^{k} − 1)k!} x^{k} + \frac{1}{ln q} ∑_{k=−∞}^{∞} x^{γ_{k}}Γ(γ_{k}) + O(x^{N+1/2}),  (13)

where γ_{k} = 2πik/ ln q and c_{0} is a constant dependent only on q, is satisfied for any integer N ≥ 1 when x > 0 and x → 0. We shall show that the infinite series from the right-hand side of relation (13) converges. Hence, if

f_{7}(s) = \frac{1}{Γ(s)} ∫_{0}^{1} x^{s−1} \frac{ψ_{q}(x)}{1 − e^{−x}} dx, f_{8}(s) = \frac{1}{Γ(s)} ∫_{1}^{∞}x^{s−1} \frac{ψ_{q}(x)}{1 − e^{−x}} dx,

then f_{6}(s) = f_{7}(s) + f_{8}(s), the function f_{8}(s) is entire, and according to (13), the relation

f_{7}(s) = −\frac{1}{Γ(s) ln q} ∫_{0}^{1} \frac{x^{s−1} ln x}{1 − e^{−x}} dx + \frac{c_{0}}{Γ(s)} ∫_{0}^{1} \frac{x^{s−1}}{1 − e^{−x}} dx + \frac{1}{Γ(s)} ∑_{k=1}^{N} c_{k} ∫_{0}^{1} \frac{x^{s+k−1}}{1 − e^{−x}} dx

+ \frac{1}{Γ(s)} ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} \frac{Γ(γ_{k})}{ln q} ∫_{0}^{1} \frac{x^{s−γ_{k}−1}}{1 − e^{−x}} + R_{N}^{(2)}(s),

where c_{k} = (−1)^{k+1}q^{k}/((q^{k} − 1)k!) and the function R_{N}^{(2)}(s) is analytic for Re s > −(N + 1/2), holds true. Since the relations

∫_{0}^{1} x^{s−1} ln x dx = −\frac{1}{s^{2}} , \frac{1}{Γ(s)} ∫_{0}^{∞} \frac{x^{s−1} ln x}{e^{x} − 1} dx = ζ′(s) + \frac{Γ′(s)}{Γ(s)}ζ(s)

are satisfied for Re s > 1, we find by analogy that

f_{7}(s) = \frac{1}{s^{2}Γ(s) ln q} − \frac{1}{ln q} (ζ′(s) + \frac{Γ′(s)}{Γ(s)}ζ(s)) + \frac{c_{0}}{sΓ(s)} + c_{0}ζ(s) + \frac{1}{Γ(s)} ∑_{k=1}^{N} \frac{c_{k}}{s + k}

+ \frac{1}{Γ(s)} ∑_{k=1}^{N} c_{k}Γ(s + k)ζ(s + k) + \frac{1}{Γ(s)} ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} \frac{Γ(γ_{k})}{(s + γ_{k}) ln q}

+ \frac{1}{Γ(s) ln q} ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} Γ(γ_{k})Γ(s − γ_{k})ζ(s − γ_{k}) + R_{N}^{(3)}(s),

where the function R_{N}^{(3)}(s) is analytic for Re s > −(N + 1/2). It follows from the last relation that the function f_{7}(s) can be analytically continued to the whole plane of the variable s, except for the simple poles at the points 1, 0, −1, −2, … and at the points −m + γ_{k}, where k = ±1, ±2, … and m = 0, −1, −3, −5, ….

Before passing to the general case, we shall prove the following statement.

Lemma. If the integer ν ≥ 0 and the real q > 1, then the relation

ψ_{ν,q}(x) =^{def} ∑_{k=1}^{∞} k^{ν}e^{−xq^{k}} = ∑_{k=0}^{ν+1} c_{k}(ln x)^{k} + ∑_{k=1}^{N} \frac{(−1)^{k+1}q^{k}}{(q^{k} − 1)k!} x^{k}

+ ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} x^{−γ_{k}} ∑_{μ=0}^{ν}(− ln x)^{μ} ∑_{m=1+μ}^{1+ν} a_{ν,μ,m}Γ^{(m−1−μ)}(γ_{k}) + O(x^{N+1/2}),

where γ_{k} = 2πik/ ln q and the constants c_{k} and a_{ν,μ,m} do not depend on x, is satisfied for any positive integer N ≥ 1 when x > 0 and x → 0.

Proof. It is known [4, p. 50] that if |z| < 1 and the integer ν ≥ 0, then

∑_{k=1}^{∞} k^{ν}z^{k} = ∑_{m=1}^{ν+1} c_{ν,m}(\frac{z}{z − 1})^{m}.

Consequently, for Re w > 0 we have

∑_{k=1}^{∞} k^{ν}q^{−kw} = ∑_{m=1}^{ν+1} c_{ν,m}(q^{w} − 1)^{−m}.

Therefore, using the relation

e^{−x} = \frac{1}{2πi} ∫_{σ−i∞}^{σ+i∞}x^{−w}Γ(w) dw, σ > 0,

we get, for x > 0, the relation

ψ_{ν,q}(x) = ∑_{m=1}^{ν+1} c_{ν,m} \frac{1}{2πi} ∫_{σ−i∞}^{σ+i∞}\frac{Γ(w)}{x^{w}(q^{w} − 1)^{m}} dw.  (14)

The function Γ(w)/(x^{w}(q^{w} − 1)^{m}) of w has the following singularities:

a pole of order m + 1 at the point w = 0 with the residue (ln q)^{−m} \frac{(− ln x)^{m}}{m!} ; poles of order m at the points w = γ_{k}, where the integer k ≠ 0, with the residues

\frac{(ln q)^{−m}}{(m − 1)!}(\frac{Γ(w)}{x^{w}})^{(m−1)}|_{w=γ_{k}};

simple poles at the points w = −k, where the integer k ≥ 1, with the residue \frac{(−1)^{k+1}(qx)^{k}}{(q^{k} − 1)k!} . Therefore, replacing the straight line Re w =σ in relation (14) by the straight line Re w =−(N+1/2), what we can do if we use the relation

ln |Γ(x + iy)| = −\frac{π}{2} |y| + (x − \frac{1}{2}) ln |y| + O(1),  (15)
PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

for the fixed x and |y| → ∞, we obtain from (14), for x > 0 and x → 0, the relation

ψ_{ν,q}(x) = ∑_{k=0}^{ν+1} c_{k}(ln x)^{k} + ∑_{k=1}^{N} \frac{(−1)^{k+1}q^{k}}{(q^{k} − 1)k!} x^{k}

+ ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} x^{−γ_{k}} ∑_{m=1}^{ν+1} \frac{c_{ν,m}}{(m − 1)!(ln q)^{m}} ∑_{μ=0}^{m−1} \binom{m − 1}{μ}Γ^{(μ)}(γ_{k})(− ln x)^{m−1−μ} + O(x^{N+1/2}). (16)

Let us show that for 0 ≤ μ ≤ ν the series

∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} x^{−γ_{k}}Γ^{(μ)}(γ_{k})  (17)

converges. Indeed, relation (15) gives the estimate |Γ(iy)| = O(exp\{−\frac{π}{3} |y|\}), where y is real and |y| → ∞. It is known that

\frac{Γ′(z)}{Γ(z)} = γ − \frac{1}{z} + ∑_{n=1}^{∞} \frac{z}{n(z + n)},

where γ is Euler’s constant. Consequently, by induction on μ ≥ 0, we obtain

|Γ^{(μ)}(iy)| = O(exp\{−\frac{π}{4} |y|\}), |y| → ∞.

Therefore the convergence of series (17) follows from the definition of γ_{k}. Let

C_{ν,m} = \frac{c_{ν,m}}{(m − 1)!(ln q)^{m}} .

Then

∑_{m=1}^{ν+1} C_{ν,m} ∑_{μ=0}^{m−1} \binom{m − 1}{μ}Γ^{(μ)}(γ_{k})(− ln x)^{m−1−μ} =∑_{μ=0}^{ν}(−1)^{μ}(ln x)^{μ} ∑_{m=1+μ}^{ν+1} C_{ν,m}\binom{m − 1}{μ}Γ^{(m−1−μ)}(γ_{k}).

Setting

C_{ν,m}\binom{m − 1}{μ} = a_{ν,μ,m},

we get the statement of the lemma.

Passing to the general case of Theorem 2, we note that it suffices to prove the theorem for the case P (x) ≡ x^{m}, where the integer m ≥ 1. For Re s > 1 the relation

f_{8}(s) =^{def} ∑_{n=1}^{∞} \frac{\{log_{q} n\}^{m}}{n^{s}} = ∑_{n=1}^{∞} n^{−s}(log_{q} n − [log_{q} n])^{m} = ∑_{n=1}^{∞} n^{−s} ∑_{μ=0}^{m} \binom{m}{μ}(−1)^{μ}[log_{q} n]^{μ}(log_{q} n)^{m−μ}.

is satisfied.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

348 PAVLOV

It is easy to see that if

f_{μ,q}(s) = ∑_{n=1}^{∞}[log_{q} n]^{μ}n^{−s},

then

f_{8}(s) = ∑_{μ=0}^{m}(−1)^{m}\binom{m}{μ}(ln q)^{μ−m}(f_{μ,q}(s))^{(m−μ)}

Consequently, it is sufficient to continue analytically the series f_{μ,q}(s) for every integer μ ≥ 1. We have

f_{μ,q}(s) = ∑_{k=0}^{∞} ∑_{q^{k}≤n<q^{k+1}} k^{μ}n^{−s} = ∑_{k=1}^{∞}(k^{μ} − (k + 1)^{μ}) ∑_{n=q^{k}}^{∞} n^{−s} = ∑_{ν=0}^{μ−1}(−1)^{μ−ν+1}\binom{μ}{ν} ∑_{k=1}^{∞} k^{ν} ∑_{n=q^{k}} n^{−s}.

Therefore, in order to prove Theorem 2, it suffices to continue meromorphically the series

φ_{ν,q}(s) = ∑_{k=1}^{∞} k^{ν} ∑_{n=q^{k}} n^{−s}

for any integer ν ≥ 0.

Using relation (2), we obtain, for Re s > 1, the relation

φ_{ν,q}(s) = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{ψ_{ν,q}(x)}{1 − e^{−x}} dx,

where

ψ_{ν,q}(x) = ∑_{k=1}^{∞} k^{ν}e^{−xq^{k}}.

Let

φ_{ν,q}^{(1)}(s) = \frac{1}{Γ(s)} ∫_{0}^{1} x^{s−1} \frac{ψ_{ν,q}(x)}{1 − e^{−x}} dx, φ_{ν,q}^{(2)}(s) = \frac{1}{Γ(s)} ∫_{1}^{∞}x^{s−1} \frac{ψ_{ν,q}(x)}{1 − e^{−x}} dx.

Then φ_{ν,q}(s) = φ_{ν,q}^{(1)}(s) + φ_{ν,q}^{(2)}(s), and the function φ_{ν,q}^{(2)}(s) is entire. Using the lemma, we obtain

φ_{ν,q}^{(1)}(s) = ∑_{k=0}^{∞} c_{k} \frac{1}{Γ(s)} ∫_{0}^{1} \frac{x^{s−1}(ln x)^{k}}{1 − e^{−x}} dx + ∑_{k=1}^{N} \frac{(−1)^{k+1}q^{k}}{(q^{k} − 1)k!Γ(s)} ∫_{0}^{1} \frac{x^{s+k−1}}{1 − e^{−x}} dx

+ ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} ∑_{μ=0}^{ν}(−1)^{μ} ∑_{m=1+μ}^{1+ν} a_{ν,μ,m}Γ^{(m−1−μ)}(γ_{k})\frac{1}{Γ(s)} ∫_{0}^{1} \frac{x^{s−γ_{k}−1}(ln x)^{μ}}{1 − e^{−x}} dx + R_{N}^{(2)}(s),

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

where the function R_{N}^{(2)}(s) is analytic for Re s > −(N + 1/2). If Re s > 1, then the relations

\frac{1}{Γ(s)} ∫_{0}^{1} \frac{x^{s+k−1}}{1 − e^{−x}} dx = \frac{1}{(s + k)Γ(s)} + \frac{Γ(s + k)ζ(s + k)}{Γ(s)} − \frac{1}{Γ(s)} ∫_{1}^{∞} \frac{x^{s+k−1}}{e^{x} − 1} dx,

\frac{1}{Γ(s)} ∫_{0}^{1} \frac{x^{s−1}(ln x)^{k}}{1 − e^{−x}} dx = \frac{(−1)^{k}k!}{s^{k+1}Γ(s)} + \frac{(Γ(s)ζ(s))^{(k)}}{Γ(s)} − \frac{1}{Γ(s)} ∫_{1}^{∞} \frac{x^{s−1}(ln x)^{k}}{e^{x} − 1} dx,

\frac{1}{Γ(s)} ∫_{0}^{1} \frac{x^{s−γ_{k}−1}(ln x)^{μ}}{1 − e^{−x}} dx = \frac{(−1)^{μ}μ!}{(s − γ_{k})^{μ+1}Γ(s)} + \frac{(Γ(s − γ_{k})ζ(s − γ_{k}))^{(μ)}}{Γ(s)}

− \frac{1}{Γ(s)} ∫_{1}^{∞} \frac{x^{s−γ_{k}−1}(ln x)^{μ}}{e^{x} − 1} dx.

(18)

hold true. Note that the third term on the right-hand side of each of relations (18) is an entire function of s. Therefore, using relations (18), we get from (16) the relation

φ_{ν,q}^{(1)}(s) = \frac{1}{Γ(s)} ∑_{k=0}^{N} \frac{(−1)^{k}k!c_{k}}{s^{k+1}} + \frac{1}{Γ(s)} ∑_{k=0}^{N} c_{k}^{(1)}(Γ(s)ζ(s))^{(k)} + \frac{1}{Γ(s)} ∑_{k=1}^{N} \frac{c_{k}^{(2)}}{s + k}

+ \frac{1}{Γ(s)} ∑_{k=1}^{N} c_{k}^{(2)}Γ(s + k)ζ(s + k) + \frac{1}{Γ(s)} ∑_{μ=0}^{ν} μ! ∑_{m=1+μ}^{1+ν} a_{ν,μ,m} ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} \frac{Γ^{(m−1−μ)}(γ_{k})}{(s − γ_{k})^{μ+1}}

+ \frac{1}{Γ(s)} ∑_{μ=0}^{ν}(−1)^{μ} ∑_{m=1+μ}^{1+ν} a_{ν,μ,m} ∑_{\begin{matrix}k=−∞ \\ k≠0 \end{matrix}}^{∞} Γ^{(m−1−μ)}(γ_{k})(Γ(s − γ_{k})ζ(s − γ_{k}))^{(μ)} + R_{N}^{(2)}(s) + Φ_{1}(s), (19)

where c_{k}^{(1)} = c_{k}, c_{k}^{(2)} = (−1)^{k+1}q^{k}/((q^{k} − 1)k!) and Φ_{1}(s) is an entire function. The right-hand side of relation (19) is analytic for Re s > −(N + 1/2), except for the points 0, −1, −2, …, −N and the points γ_{k} − m and γ_{k}, where k = ±1, ±2, … and m is an odd number. Since N is an arbitrary positive integer, this completes the proof of Theorem 2.

Proof of Theorem 3. Since

\{αq^{n}\} = q^{n} ∑_{k=n+1}^{∞} \frac{a_{k}}{q^{k}} ,

we have, for Re s > 1, the relation

f_{9}(s) =^{def} ∑_{n=1}^{∞} \frac{\{αq^{n}\}}{n^{s}} = ∑_{n=1}^{∞} n^{−s}q^{n} ∑_{k=n+1}^{∞} \frac{a_{k}}{q^{k}} = ∑_{k=2}^{∞} \frac{a_{k}}{q^{k}} ∑_{n=1}^{k−1} q^{n}n^{−s}.

Using relation (2), we obtain

∑_{n=1}^{k−1} q^{n}n^{−s} = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1}(∑_{n=1}^{k−1}(qe^{−x})^{n}) dx.

Consequently, for Re s > 1 and α_{1} = {αq} we have

f_{9}(s) = \frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{α_{1}e^{−x} − ∑_{k=2}^{∞} a_{k}e^{−kx}}{1 − qe^{−x}} dx.  (20)

We assume that for x = ln q the fraction on the right-hand side of relation (20) is equal to the number

∑_{k=2}^{∞} \frac{a_{k}(k − 1)}{q^{k}}.

Let a_{k_{0}+k} = 1 for fixed integers r ≥ 2 and k_{0} ≥ 2 if k = m^{r} for a certain integer m ≥ 1, and a_{k_{0}+k} = 0 if k ≠ m^{r} for any integer m ≥ 1. Then

∑_{k=2}^{∞} a_{k}e^{−kx} = ∑_{k=2}^{k_{0}} a_{k}e^{−kx} + (∑_{m=1}^{∞} e^{−xm^{r}})e^{−xk_{0}}.

This means that from (20) we have the relation

f_{9}(s) = −\frac{1}{Γ(s)} ∫_{0}^{∞}x^{s−1} \frac{ψ_{r}(x)e^{−k_{0}x}}{1 − qe^{−x}} dx − \frac{1}{Γ(s)} ∑_{k=2}^{k_{0}} a_{k} ∫_{0}^{∞} \frac{x^{s−1}e^{−kx}}{1 − qe^{−x}} dx + \frac{α_{1}}{Γ(s)} ∫_{0}^{∞} \frac{x^{s−1}e^{−x}}{1 − qe^{−x}} dx,

where

ψ_{r}(x) = ∑_{m=1}^{∞} e^{−xm^{r}}.

Let x_{q} = ln q/2. Then

f_{9}(s) = \frac{α_{1}}{Γ(s)} ∫_{0}^{x_{q}} \frac{x^{s−1}e^{−x}}{1 − qe^{−x}} dx − \frac{1}{Γ(s)} ∫_{0}^{x_{q}}x^{s−1} \frac{ψ_{r}(x)e^{−k_{0}x}}{1 − qe^{−x}} dx − \frac{1}{Γ(s)} ∑_{k=2}^{k_{0}} a_{k} ∫_{0}^{x_{q}} \frac{x^{s−1}e^{−kx}}{1 − qe^{−x}} dx + Φ_{2}(s),

(21)

where Φ_{2}(s) is an entire function. For any k > 0 the function

\frac{1}{Γ(s)} ∫_{0}^{x_{q}} \frac{x^{s−1}e^{−kx}}{1 − qe^{−x}} dx

is entire. This follows from the relation (the integer m ≥ 1)

\frac{1}{Γ(s)} ∫_{0}^{x_{q}} \frac{x^{s−1}e^{−kx}}{1 − qe^{−x}} dx = \frac{(−1)^{m}}{s(s + 1) ⋯ (s + m − 1)Γ(s)} ∫_{0}^{x_{q}}x^{s+m−1}(\frac{e^{−kx}}{1 − qe^{−x}} )^{(m)} dx + Ψ_{k}(s),

in which Ψ_{k}(s) is an entire function and which can be easily proved by means of integration by parts. Using for x > 0 and x → 0 the asymptotic relation

ψ_{r}(x) = \frac{1}{r}Γ(\frac{1}{r})x^{−1/r} + ∑_{k=0}^{N} c_{k}x^{k} + O(x^{N+1/2}),

where the constants c_{k} do not depend on x and the integer N ≥ 1 is fixed, we obtain

\frac{1}{Γ(s)} ∫_{0}^{x_{q}}x^{s−1} \frac{ψ_{r}(x)e^{−k_{0}x}}{1 − qe^{−x}} dx = \frac{1}{r} \frac{Γ(1/r)}{Γ(s)} ∫_{0}^{x_{q}} \frac{x^{s−1/r−1}e^{−k_{0}x}}{1 − qe^{−x}} dx

+ \frac{1}{Γ(s)} ∑_{k=0}^{N} c_{k} ∫_{0}^{x_{q}} \frac{x^{s+k−1}e^{−k_{0}x}}{1 − qe^{−x}} dx + R_{N}^{(3)}(s),  (22)

where the function R_{N}^{(3)}(s) is analytic for Re s > −(N + 1/2). By means of integration by parts it is easy to prove that the first term on the right-hand side of relation (22) can be analytically continued to the whole plane of the variable s, except for the points s = 1/r − m, where m is a nonnegative integer, and the second term is an entire function. Since N is a positive integer, it suffices to use relations (21) and (22) in order to complete the proof of Theorem 3.

ACKNOWLEDGMENTS

This work was supported by the Russian Foundation for Fundamental Research (project no. 97-01-00721).

REFERENCES

1. Hecke, E., Über analytische Funktionen und die Verteilung von Zahleu mod eins, Hamburg Math. Abh., 1921, vol. 1, pp. 54–76.

2. Hardy, G. H. and Littlewood, J. E., Some Problems of Diophantine Approximation. The Analytic Character of Sum of a Dirichlet’s Series Considered by Hecke, Hamburg Math. Abh., 1923, vol. 3, pp. 57–68.

3. Gel’fond, A. O., Vychety i ikh prilozheniya (Residues and their Applications), Moscow: Nauka, 1966.

4. Privalov, I. I., Granichnye svoistva analiticheskikh funktsii (Boundary Properties of Analytic Functions), Moscow: Gostekhizdat, 1950.

Translated by I. Aleksandrova