ON LINEAR FORMS WITH COEFFICIENTS IN ℕ ζ(1 + ℕ)

The detailed version, Part 1 (INTRODUCTION)

Leonid A. Gutnik

In memory of Professor Alexander Buchstab

Foreword

This is the first part of the detailed version of my work “On linear forms with coefficients in ℕζ(1 + ℕ),” [8]. The goal of this work is to give a complete proof of the following theorem.

Theorem. Let

ϕ_{i}(x_{1}, x_{2}) = ∑_{k=1}^{2}(2ik − i − k + 2) ζ(i + k + 1) x_{k}, i = 1, 2.

For any d ∈ ℝ, let ∥d∥ be the distance between d and ℤ; let γ = 43, 464412. There is a positive constant c such that

∥ϕ_{1}(x_{1}, x_{2})∥ + ∥ϕ_{2}(x_{1}, x_{2})∥ ≥ c(|x_{1}| + |x_{2}|)^{−γ},

where x_{1} ∈ ℤ, x_{2} ∈ ℤ, |x_{1}| + |x_{2}| > 0.

Remark. The lattices L_{1}, L_{2} in ℝ^{n} are said to be incommensurable, if L_{1}∩L_{2} = {0}. The qualitative part of the theorem asserts that lattice generated by the vectors f_{1} = (2 ζ(3), 3 ζ(4)) and f_{2} = (3 ζ(4), 6 ζ(5)) and the lattice ℤ^{2} (generated by the vectors (1, 0) and (0, 1)) are incommensurable.

Corollary. If p ∈ ℚ, q ∈ ℚ, and p^{2} + q^{2} > 0, then

{2p ζ(3) + 3q ζ(4), 3p ζ(4) + 6q ζ(5)} ⊈ ℚ,

and therefore

\{\frac{ζ(3 + 2k)}{ζ(4)}, \frac{12 ζ(3)ζ(5) − 9 ζ(4)^{2}}{ζ(4)} \} ⊈ ℚ, k = 0, 1.

Typeset by AMS-TEX

In the first part, I give a short survey of the C. S. Mejer functions, and then define and compute my auxiliary functions. In what follows, given a ∈ ℂ and M ⊆ ℂ, M ≠ ∅, let

a + M = M + a = {w ∈ ℂ: w = a + z, z ∈ M}.

Acknowledgment. I am obliged to Professor A.G.Aleksandrov and to Professor B.Z.Moroz for their help during the preparation of the preprint, for correcting my English, and for their attention to my work.

§ 1. Short survey of the Mejer functions

In this work, I use the functions introduced and studied by C. S. Mejer in a long series of papers published during the decade 1936 – 1946. The functions of Mejer can be defined as follows ([9], ch. 5). Let (1.1)

G_{p,q}^{(m,n)} (z| \begin{matrix}a_{1}, & …, & a_{p} \\ a_{1}, & …, & a_{p} \end{matrix} ) = \frac{1}{2πi} ∫_{L} \frac{z^{s} ∏_{j=1}^{m} Γ(b_{j} − s) ∏_{j=1}^{n} Γ(1 − a_{j} + s)}{∏_{j=m+1}^{q} Γ(1 − b_{j} + s) ∏_{j=n+1}^{p} Γ(a_{j} − s)} ds,

where an empty product is, by definition, equal to 1, 0 ≤ m ≤ q, 0 ≤ n ≤ p; the parameters a_{j} ∈ ℂ, j = 1, …, p, and b_{k} ∈ ℂ, k = 1, …, q, are chosen in such a way that none of the poles of the function Γ(b_{k} − s), k = 1, …, m, is equal to a pole of one of the functions Γ(1 − a_{j} + s), j = 1, …, n. There are 3 possibilities to choose the curve L.

(A) First, the curve L = L_{0} may be chosen to pass from −i∞ to +i∞ in such way that for any k = 1, …, m all the poles of the function Γ(b_{k} − s) lie to the right of it and for any j = 1, …, n all the poles of the function Γ(1 − a_{j} + s) lie to the left of it. The integral (1.1) is convergent in either of the following two cases:

(A1) | arg(z)| < (m + n − \frac{p}{2} − \frac{q}{2})π;

(A2) | arg(z)| ≤ (m + n − \frac{p}{2} − \frac{q}{2})π and \frac{p−q}{2} + Re Δ^{∗} < −1, where

(1.2)  Δ^{∗} = ∑_{k=1}^{q} b_{k} − ∑_{k=1}^{q} a_{j}.

(B) Second, the curve L = L_{1} may be chosen to pass from +i∞ to +i∞, encircling each of the poles of the functions Γ(b_{k} −s), k = 1, …, m, in the negative direction, but not including any of the poles of the functions Γ(1 − a_{j} + s), j = 1, …, n. The integral (1.1) is convergent in each of the following three cases:

(B1) p < q;

(B2) 1 ≤ p ≤ q and |z| < 1;

(B3) 1 ≤ p ≤ q, |z| ≤ 1 and Re Δ^{∗} < −1.

(C) Third, the curve L = L_{2} may be chosen to pass from −∞ to −∞, encircling each of the poles of the functions Γ(1−a_{j}+s), j = 1, …, n, in the positive direction, but not including any of the poles of the functions Γ(b_{k} − s), k = 1, …, m. The integral (1.1) is convergent in each of the following 3 cases:

(C1) q < p;

(C2) 1 ≤ q ≤ p and |z| > 1;

(C3) 1 ≤ q ≤ p, |z| ≥ 1 and Re Δ^{∗} < −1.

If both conditions (A) and (B) (respectively (A) and (C)) are satisfied, then the result does not depend on whether the curve L is defined as in (A) or as in (B) (respectively in (C)).

Let g denote the integrand of the integral (1.1), let G denote the integral (1.1) with L = L_{k}, where k = 1, 2, and let S_{k} be the set of all the unremovable singularities of g encircled by L_{k}. If one of the conditions (B1) – (B3) (respectively (C1) – (C3)) holds for k = 1 (respectively for k = 2), then

(1.3)  G = (−1)^{k} ∑_{s∈S_{k}} Res (g; s),

where Res (g; s) stands for the residue of the function g at the point s. Let us prove these assertions. Let σ_{0} > 0, τ_{0} > 0, and let

(1.4)  min(σ_{0}, τ_{0}) ≥ 1 + ∑_{j=0}^{p} 2(|a_{j}| + 1) + ∑_{k=0}^{q} 2(|b_{k}| + 1).

The curve L_{0} may be chosen in such a way that, except for a compact piece, it coincides with a part of the imaginary axis | Im s| ≥ τ_{0}; the curves L_{1} and L_{2} may be chosen so that, except for a compact part, the curve L_{1} coincides with the union of the two rays σ ±iτ_{0}, σ ≥ σ_{0}, and the curve L_{2}, except for a compact part, coincides with the union of the two rays σ ± iτ_{0}, σ ≤ −σ_{0}.

Let r_{0} ≥ σ_{0} + τ_{0} be chosen big enough, so that the specified compact parts of L_{0}, L_{1} and L_{2} lie inside the disk K_{0} = {s ∈ ℂ: |s| ≤ r_{0}}.

Let u ≥ 0, η(u) = ∫_{0}^{u}([t] − t + \frac{1}{2})dt; then 0 ≤ η(u) ≤ 1/8. Let

\overline{D}_{0} = \{s ∈ ℂ: |s| ≥ r_{0}\}\\{s ∈ ℂ: Re s < 0, | Im s| < τ_{0}\}.

If z = x + iy, |y| ≥ τ_{0}, then the integral in the complex Stirling formula ([10], ch. 4)

log Γ(z) = (z − \frac{1}{2}) log z − z + \frac{1}{2} log(2π) + ∫_{0}^{∞} \frac{η(u)}{(u + z)^{2}} du

can be estimated as follows:

| ∫_{0}^{∞} \frac{η(u)}{(u + z)^{2}} du| ≤ ∫_{0}^{∞}|\frac{η(u)}{(u + z)^{2}} |du =

∫_{0}^{∞}|\frac{η(u)}{(u + x)^{2} + y^{2}} |du < ∫_{−∞}^{+∞}|\frac{η(u)}{(u + x)^{2} + y^{2}} |du < \frac{π}{8τ_{0}} = O(1).

If z = x + iy and x ≥ σ_{0}, then

| ∫_{0}^{∞} \frac{η(u)}{(u + z)^{2}} du| ≤ ∫_{0}^{∞}|\frac{η(u)}{(u + z)^{2}} |du =

∫_{0}^{∞}|\frac{η(u)}{(u + x)^{2} + y^{2}} |du ≤ ∫_{0}^{∞}|\frac{η(u)}{(u + x)^{2}} |du < \frac{1}{8σ_{0}} = O(1).

Therefore

(1.5)  log Γ(s + c) = (s + c − \frac{1}{2})(log s − 1) + O(1)

in \overline{D}_{0}, for any constant c with 2|c| ≤ min\{σ_{0}, τ_{0}\}.

Let s = σ + iτ = |s| e^{iψ} with |ψ| ≤ π, ϕ = arg z, then

(1.6)  Re log Γ(s + c) = (σ + Re (c) − \frac{1}{2})(log |s| − 1) − τψ + O(1).

One may write g(s) = g_{1}(s)g_{2}(s) with

(1.7)  g_{1}(s) = z^{s} ∏_{j=1}^{p} Γ(1 − a_{j} + s)/ ∏_{j=1}^{q} Γ(1 − b_{j} + s),

(1.8)  g_{2}(s) = ∏_{j=n+1}^{p} \frac{sin(π(a_{j} − s))}{π} ∏_{j=1}^{m} \frac{π}{sin(π(b_{j} − s))}.

Since τψ ≥ 0, it follows from (1.6) that

(1.9)  g_{1}(s) = O(1) · |z|^{σ}e^{|ϕ||τ|}(|s|/e)^{((p−q)(σ+ \frac{1}{2} )+Re Δ^{∗})}e^{−τ(p−q)ψ} =
= O(1) · |z|^{σ}(|s|/e)^{((p−q)(σ+ \frac{1}{2} )+Re Δ^{∗})}e^{|τ|(p−q)( \frac{π}{2} −|ψ|)}e^{|τ|(ϕ−(p−q) \frac{π}{2} )}

in \overline{D}_{0}. In view of our choice of τ_{0}, it follows that

| sin(π(a_{j} − s))| = O(1) · e^{π|τ|}, | sin(π(b_{k} − s))|^{−1} = O(1) · e^{−π|τ|}

for s ∈ \overline{D}_{0}, j = 1, …, p and k = 1, …, q, and consequently

(1.10)  g_{2}(s) = O(1) · e^{π(p−m−n)|τ|} .

Therefore, according to (1.9) and (1.10) the following equality (1.11)

|g(s)| = O(1) · |z|^{σ}e^{|ϕ||τ|}(|s|/e)^{((p−q)(σ+ \frac{1}{2} )+Re Δ^{∗})}e^{−|τ|(p−q)|ψ|}e^{π(p−m−n)|τ|} =

= O(1) · |z|^{σ}(|s|/e)^{((p−q)(σ+ \frac{1}{2} )+Re Δ^{∗})} · e^{((p−q)( \frac{π}{2} −|ψ|)+|ϕ|+π( \frac{p+q}{2} −m−n))|τ|}

holds in \overline{D}_{0}. According to (1.11), on the part of the curve L_{0} lying on the rays σ = 0, |τ| ≥ τ_{0} we have the equality

|g(s)| = O(1) · (|τ|/e)^{( \frac{p−q}{2} +Re Δ^{∗})}e^{|τ|(|ϕ|+π( \frac{p+q}{2} −m−n))};

if the condition (A1) holds, then

|g(s)| = |τ|^{O(1)}e^{|τ|(|ϕ|+π( \frac{p+q}{2} −m−n))},

where |τ| ≥ τ_{0}, and the integral (1.1) is convergent; if the condition (A2) holds, then

|g(s)| = O(1) · (|τ|/e)^{( \frac{p−q}{2} +Re Δ^{∗})},

where |τ| ≥ τ_{0}, and the integral (1.1) is convergent. Let the number σ_{0} be chosen so that

(1.12)  σ_{0} > 3 + |\frac{p − q}{2} + Re Δ^{∗}|.

If one of the condition (B), (C) holds, then

(1.13)  (p − q)σ = −|(p − q)||σ|,

(1.14)  (p − q)(\frac{π}{2} − |ψ|) = −|p − q||\frac{π}{2} − |ψ||,

(1.15)  σ(log |z|) = −|σ|| log |z||

in \overline{D}_{0}, and (1.9), (1.11) takes respectively the form

(1.16)  |g_{1}(s)| = O(1) · e^{−|σ|| log |z||}(|s|/e)^{−|σ||p−q|+ \frac{p−q}{2} +Re Δ^{∗}}×
e^{−|p−q|| \frac{π}{2} −|ψ|||τ|}e^{(|ϕ|−(p−q) \frac{π}{2} )|τ|}
(1.17)  |g(s)| = O(1) · e^{−|σ|| log |z||}(|s|/e)^{−|σ||p−q|+ \frac{p−q}{2} +Re Δ^{∗}}×
e^{(−|p−q|| \frac{π}{2} −|ψ||+|ϕ|+π( \frac{p+q}{2} −m−n))|τ|}.

If one of conditions (B1), (C1) holds and |σ| ≥ σ_{0}, then |p − q| ≥ 1 and, in view of (1.12) and (1.17), we have

(1.18)  |g_{1}(s)| = O(1) · e^{−|σ|| log |z||}(|s|/e)^{−3}e^{−|p−q|| \frac{π}{2} −|ψ||}e^{(|ϕ|−(p−q) \frac{π}{2} )|τ|},

(1.19) |g(s)| = O(1) · e^{−|σ|| log |z||}(|s|/e)^{−3}e^{(−|p−q|| \frac{π}{2} −|ψ||+|ϕ|+π( \frac{p+q}{2} −m−n))|τ|}

in \overline{D}_{0}. Therefore if one of conditions (B1), (C1) holds, then on the rays τ = τ_{0}, |σ| ≥ σ_{0} we have the equality

(1.20)  |g(s)| = O(1) · (|σ|/e)^{−3},

and the integral (1.1) is convergent; if one of conditions (B2), (C2) holds, then on the rays τ = τ_{0}, |σ| ≥ σ_{0} with | log |z|| > 0 we have the equality

(1.21)  |g(s)| = |σ|^{O(1)} · e^{−|σ|| log |z||},

and the integral (1.1) is convergent. Finally, if one of condition (B3), (C3) holds, then, in view of (1.17),

(1.22)  |g(s)| = O(1) · |σ|^{Re Δ^{∗}}

on the rays τ = τ_{0}, |σ| > σ_{0}, and the integral (1.1) is convergent. For any r ≥ r_{0} and k = 0, 1, 2 let us denote by L_{r,k} the part of the curve L_{k} contained in the disk K_{r} = {s ∈ ℂ: |s| ≤ r} and directed as the curve L_{k}. Let further C_{r,k,j}, where k = 1, 2, j = 0, 1, be the smallest counter-clockwise directed arc of the circle |s| = r connecting the point ir(−1)^{j} with the point iτ_{0}(−1)^{j} − (−1)^{k}\sqrt{r^{2} − τ_{0}^{2}}. Finally, let L_{r,k}^{∗}, where k = 1, 2, be the positively directed curve consisting of the curves L_{0,r}, L_{k,r}, C_{r,k,0} and C_{r,k,1}. Clearly,

(1.23)  ∫_{L_{k}} g(s) = lim_{r→∞} ∫_{L_{r,k}} g(s)ds

for k = 0, 1, 2. The function g(s) has no singular points in the domain bounded by the curve L_{k}^{∗}. Therefore

(1.24) ∫_{L_{r,k}}g(s)ds − (−1)^{k} ∫_{C_{r,k,0}}g(s)ds − ∫_{L_{r,0}}g(s)ds − (−1)^{k} ∫_{C_{r,k,1}}g(s)ds = ∫_{L_{k}^{∗}} g(s)ds = 0 .

For k = 1, 2 and j = 0, 1, let Č_{r,k,j} be the counter-clockwise directed part of the arc C_{r,k,j} lying in the strip |σ| ≤ σ_{0}, and let Ĉ_{r,k,j} be the counter-clockwise directed part of the arc C_{r,k,j} lying outside this strip. If p ≠ q and one of conditions (B1), (C1) holds, then according to (1.19)

(1.25)  |g(s)| = O(1) · (r/e)^{−3}

on the Ĉ_{r,k,j} for k = 1, 2 and j = 0, 1, and, because the length of the arc Ĉ_{r,k,j} is not bigger than πr/2, it follows that

(1.26)  lim_{r→∞} ∫_{Ĉ_{r,k,j}} g(s)ds = 0.

Let γ = 4^{−1} min{||ϕ| + π(p + q)/2 − m − n|, | log |z||}. In view of (1.17), if p = q and one of conditions (B), (C) holds, then

(1.27)  |g(s)| = O(1) · e^{−γr}(r/e)^{Re Δ^{∗}}

on the Ĉ_{r,k,j} for k = 1, 2 and j = 0, 1; now, if the condition (A) is satisfied, then the equality γ = 0 implies the inequality Re (Δ^{∗}) < −1, and because the length of the arc Ĉ_{r,k,j} is not bigger than πr/2, relation (1.26) still holds. Moreover,

|τ| = \sqrt{r^{2} − σ_{0}^{2}} = r − O(1), |ψ| = \frac{π}{2} − O(arcsin(σ_{0}/r)) = \frac{π}{2} − O(1/r)

on the Č_{r,k,j} for k = 1, 2, j = 0, 1. Therefore if one of conditions (B), (C) holds, then according to (1.17)

(1.28)  |g(s)| = O(1) · (r/e)^{\frac{p−q}{2} +Re Δ^{∗}}e^{(|ϕ|+π( \frac{p+q}{2} −m−n))|τ|} =
O(1) · (r/e)^{\frac{p−q}{2} +Re Δ^{∗}}e^{(|ϕ|+π( \frac{p+q}{2} −m−n))r}

on the curve Č_{r,k,j}. Consequently, if the condition (A1) and one of the condition (B), (C) hold, then

(1.29)  lim_{r→∞} ∫_{Č_{r,k,j}}g(s)ds = lim_{r→∞} O(1) · e^{(|ϕ|+π( \frac{p+q}{2} −m−n))r}r^{O(1)} = 0

because in this case |ϕ| + π(\frac{p+q}{2} − m − n) < 0. If the condition (A2) holds, then

(1.30)  lim_{r→∞} ∫_{Č_{r,k,j}}g(s)ds = lim_{r→∞} O(1) · r^{\frac{p−q}{2} +Re Δ^{∗}+1} = 0.

We have thus proved that if both of the convergence conditions (A) and (B) hold, then

∫_{Ľ_{0}} g(s)ds = ∫_{L_{1}} g(s)ds;

and if both of the convergence conditions (A) and (C) hold, then

∫_{Ľ_{0}} g(s)ds = ∫_{L_{2}} g(s)ds.

Now I shall prove the equality (1.3). Let us choose σ_{1} so that it is bigger than r_{0} and lies outside of the countable set

⋃_{μ=0}^{1} ⋃_{j=1}^{p} ((−1)^{μ} Re (a_{j}) + ℤ) ⋃ ⋃_{ν=0}^{1} ⋃_{k=1}^{q} ((−1)^{ν} Re (b_{k}) + ℤ).

For k = 1, 2 and n ∈ ℕ−1, let r(n) = \sqrt{τ_{0}^{2} + (σ_{1} + n)^{2}}, let c_{n,k} denote the segment

[−(−1)^{k}(σ_{1} + n) − iτ_{0}, −(−1)^{k}(σ_{1} + n) + iτ_{0}],

and let L̃_{n,k} be the counter-clockwise directed curve consisting of the curve L_{r(n),k} and the segment c_{n,k}; finally, let S̃_{n,k} denote the set of all the unremovable singularities encircled by the curve L̃_{n,k}. It is clear that

\frac{1}{2πi} ∫ ; S̃_{n,k} g(s)ds = ∑_{s∈S̃_{n,k}} Res (g; s),

where k = 1, 2. The equality (1.3) will be proved if we establish that for each k = 1, 2 the equality

(1.31)  lim_{n→∞} ∫_{c_{n,k}} g(s)ds = 0

holds. First, consider the case k = 1. In view of the choice of σ_{1}, the function g_{2}(s) is continuous on the segment c_{0,k}, and therefore there is a constant M_{0} with |g_{2}(s)| ≤ M_{0} for s ∈ c_{0,k}, where k = 1, 2. Since the period of the function |g_{2}(s)| is equal to one, we have

(1.32)  |g_{2}(s)| ≤ M_{0}

for s ∈ c_{n,k}, k = 1, 2, and n ∈ ℕ. Let us first consider the case k = 1. If condition (B) holds, then (1.31) follows from (1.16). Suppose now that condition (C) is satisfied. Write

(1.33)  g_{1}(s) = g_{3}(s)g_{4}(s)

with

(1.34)  g_{3}(s) = z^{s} ∏_{j=1}^{q} Γ(b_{j} − s)/ ∏_{j=1}^{p} Γ(a_{j} − s),

(1.35)  g_{4}(s) = ( ∏_{j=1}^{p} \frac{sin(π(a_{j} − s))}{π} ∏_{j=1}^{q} \frac{π}{sin(π(b_{j} − s))})^{−1}.

Since the period of the function |g_{4}(s)| is equal to one, there is a constant M_{2} such that

(1.36)  |g_{4}(s)| ≤ M_{2}

for s ∈ c_{n,2}, n ∈ ℕ − 1, in view of the choice of σ_{1}. Because |τ| ≤ τ_{0} if s ∈ c_{n,2} and −c_{n,2} lies in \overline{D}_{0}, n ∈ ℕ − 1, it follows from (1.6) that

g_{1}(s) = O(1)·|z|^{σ}(|s|/e)^{∑_{j=1}^{q} Re (b_{j}−σ− \frac{1}{2} )− ∑_{j=1}^{p} Re (b_{j}−σ− \frac{1}{2} )} =

= O(1)·|z|^{σ}(|s|/e)^{((p−q)(σ+ \frac{1}{2} )+Re Δ^{∗})}

with |s| ≥ −σ = σ_{1} + n. Therefore the equality (1.31) holds also for k = 2. This proves (1.3).

§ 2. My auxiliary functions

I shall work with the set Ω^{∗} consisting of all the points z ∈ ℂ for which

(2.1)  |z| ≥ 1 and − \frac{3π}{2} < arg(z) ≤ \frac{π}{2} .

Thus log(−z) = log z − iπ for Re (z) > 0. Let Δ ∈ ℕ, 1 < Δ, δ_{0} = 1/Δ,

γ_{1} = (1 − δ_{0})/(1 + δ_{0}), d_{l} = Δ + (−1)^{l}, l = 1, 2.

To introduce the first of my auxiliary function f_{1}(z, ν), I use the auxiliary set

Ω^{(h)} = \{z ∈ ℂ: |z| ≤ 1, −\frac{3π}{2} < ϕ = arg(z) ≤ \frac{π}{2} \}.

I shall prove that, for each ν ∈ ℕ, the function f_{1}(z, ν) belongs to ℚ[z]; therefore using the principle of analytic continuation we may regard it as being defined in ℂ, and consequently, in Ω^{∗}. For ν ∈ ℕ, let

(2.2)  f_{1}(z, ν) =

= −(−1)^{ν(Δ+1)}G_{6,6}^{(1,3)} (z| \begin{matrix}−νd_{1}, & −νd_{1}, & −νd_{1}, & 1 + νd_{2}, & 1 + νd_{2}, & 1 + νd_{2} \\ 0, & 0, & 0, & ν, & ν, & ν \end{matrix} )

= −(−1)^{ν(Δ+1)} \frac{1}{2πi} ∫_{L_{1}} g_{6,6}^{(1,3)}(s)ds,

where

g_{6,6}^{(1,3)}(s) = z^{s}Γ(−s)Γ(1 + s)^{−2}(Γ(1 + νd_{1} + s)/(Γ(1 − ν + s)Γ(1 + νd_{2} − s)))^{3},

and the curve L_{1} passes from +∞ to +∞ encircling the set ℕ − 1 in the negative direction, but not including any point of the set −ℕ. Here p = q = 6, m = 1, n = 3, a_{1} = a_{2} = a_{3} = −νd_{1}, a_{4} = a_{5} = a_{6} = 1 + νd_{2}, b_{1} = b_{2} = b_{3} = 0, b_{4} = b_{5} = b_{6} = ν, Δ^{∗} = −3ν − 3 and, since we take |z| ≤ 1, convergence conditions (B2) and (B3) hold. To compute the function f_{1}(z, ν), we use formula (1.3) and the well-known formula

(2.3)  Γ(s) = Γ(s + l) ∏_{k=1}^{l}(s + l − k)^{−1}

with l ∈ ℕ. The set of unremovable singular points of the function g_{6,6}^{(1,3)}(s), which are encircled by the curve L_{1}, consists of the points s = ν, …, νd_{2}, all these points are poles of the first order, and, for k = 0, …, νΔ, the following equality holds:

Res (g_{6,6}^{(1,3)}; ν + k) = (−z)^{ν+k}((ν + k)!)^{−3}((νΔ + k)!)^{3}(k!)^{−3}((νΔ − k)!)^{−3} =

= (−z)^{ν+k}((νd_{1})!/(νΔ)!)^{3}\binom{νΔ}{k}^{3}\binom{νΔ + k}{νd_{1}}^{3}.

The function f_{1}(z, ν) is equal to a finite sum

(2.4)  f_{1}(z, ν) = ((νd_{1})!/(νΔ)!)^{3}z^{ν}(−1)^{νΔ} ∑_{k=0}^{νΔ}(−z)^{k}\binom{νΔ}{k}^{3}\binom{νΔ + k}{νd_{1}}^{3}.

Therefore, as it has been already remarked, using the principle of analytic continuation we may regard it as being defined in ℂ and consequently in Ω^{∗}.

Let Ω_{0}^{∗} = {z ∈ Ω^{∗}: Re (z) > 0}. If z ∈ Ω_{0}^{∗}, then

−\frac{3π}{2} < ϕ = arg(−z) = arg(z) − π < −\frac{π}{2} ,

and therefore −z ∈ Ω^{∗}. Now, let me introduce my second auxiliary function defined for z ∈ Ω_{0}^{∗}.

Let

f_{2}(z, ν) = −(−1)^{νΔ} \frac{1}{2πi} ∫_{L_{2}} g_{6,6}^{(4,3)}(s)ds =

= −(−1)^{νΔ}G_{6,6}^{(4,3)} (−z| \begin{matrix}−νd_{1}, & −νd_{1}, & −νd_{1}, & 1 + νd_{2}, & 1 + νd_{2}, & 1 + νd_{2} \\ 0, & 0, & 0, & ν, & nu, & ν \end{matrix} ) ,

where z ∈ Ω_{0}^{∗}, ν ∈ ℕ, and

g_{6,6}^{(4,3)} = g_{6,6}^{4,3)}(s) = (−z)^{s}Γ(−s)^{3}Γ(−s+ν)Γ(1−ν+s)^{−2}Γ(1+νd_{1}+s)^{3}Γ(1+νd_{2}−s)^{−3},

and the curve L_{2} passes from −∞ to −∞, encircling the set −ℕ in the positive direction but no point in the set ℕ − 1. Here p = q = 6, m = 4, n = 3, a_{1} = a_{2} = a_{3} = −νd_{1}, a_{4} = a_{5} = a_{6} = 1 + νd_{2}, b_{1} = b_{2} = b_{3} = 0, b_{4} = b_{5} = b_{6} = ν, Δ^{∗} = −3ν − 3; since now | − z| ≥ 1, convergence conditions (C2) and (C3) are satisfied. To compute the function f_{2}(z, ν), we use formula (1.3). The set of all the unremovable singular points of the function g_{6,6}^{(4,3)}(s), encircled by the curve L_{2},

12 LEONID A. GUTNIK

consists of the points s = −1 − νd_{1} − k with k ∈ ℕ − 1; each of these points is a pole of the first order. Therefore making use of (2.3) one obtains

Res (g_{6,6}^{(4,3)}; −1 − νd_{1} − k) =

= (−z)^{−1−νd_{1}−k}((νd_{1} + k)!)^{3}((νΔ + k)!)^{3}(−1)^{k}(k!)^{−3}((1 + 2νΔ + k)!)^{−3} =

= (−1)^{1+νd_{1}}z^{−(1+νd_{1}+k)} (\frac{∏_{j=1}^{νΔ−ν}(1 + νΔ − ν + k − j)}{∏_{j=0}^{νΔ}(1 + νΔ + k + j)})^{3},

(2.5)  f_{2}(z, ν) = ∑_{k=0}^{+∞} z^{−(1+νd_{1}+k)} (\frac{∏_{j=1}^{νΔ−ν}(1 + νΔ − ν + k − j)}{∏_{j=0}^{νΔ}(1 + νΔ + k + j)})^{3}.

Let a ∈ ℕ − 1, b ∈ ℕ + a, and

(2.6) R(a; b; t) = \frac{b!}{(b − a)!} ∏_{κ=a+1}^{b}(t − κ) ∏_{κ=0}^{b} \frac{1}{(t + κ)},  R_{0}(t; ν) = R(ν; νΔ; t).

Let t = 1 + νΔ + k with k ∈ ℕ − 1; in view of (2.5), it follows that

(2.7)  f_{2}(z, ν)((νΔ)!/(νd_{1})!)^{3} = (−1)^{ν} ∑_{t=νΔ+1}^{∞} R_{0}(t; ν)^{3}z^{−t+ν}.

Since R_{0}(t; ν) = 0 for t = ν + 1, …, νΔ, we have

(2.8)  f_{2}(z, ν)((νΔ)!/(νd_{1})!)^{3} = (−1)^{ν} ∑_{t=ν+1}^{∞} R_{0}(t; ν)^{3}z^{−t+ν}.

Let

f_{3} = (−1)^{ν(Δ+1)} \frac{1}{2πi} ∫_{L_{2}} g_{6,6}^{(5,3)}(s)ds =

(−1)^{ν(Δ+1)}G_{6,6}^{(5,3)} (z| \begin{matrix}−νd_{1}, & −νd_{1}, & −νd_{1}, & 1 + νd_{2}, & 1 + νd_{2}, & 1 + νd_{2} \\ 0, & 0, & 0, & ν, & ν, & ν \end{matrix} ) ,

where z ∈ Ω_{0}^{∗}, ν ∈ ℕ, and

g_{6,6}^{(5,3)} = g_{6,6}^{(5,3)}(s) = z^{s}Γ(−s)^{3}Γ(ν −s)^{2}Γ(1−ν +s)^{−1}Γ(1+νd_{1} +s)^{3}Γ(1+νd_{2} −s)^{−3}.

Here p = q = 6, m = 5, n = 3, a_{1} = a_{2} = a_{3} = −νd_{1}, a_{4} = a_{5} = a_{6} = 1 + νd_{2}, b_{1} = b_{2} = b_{3} = 0, b_{4} = b_{5} = b_{6} = ν, Δ^{∗} = −3ν − 3; convergence conditions (C2) and (C3) are satisfied since now |z| ≥ 1. The set of all the unremovable singular points of the function g_{6,6}^{(5,3)}(s), encircled by the curve L_{2}, consists of the points s = −1 − νd_{1} − k with k ∈ ℕ − 1; each of these points is a pole of the second order. Therefore

Res (g_{6,6}^{(5,3)}; −νd_{1} − 1 − k) = lim_{s→−νd_{1}−1−k} \frac{∂}{∂s}((s + νd_{1} + 1 + k)^{2}g_{6,6}^{(5,3)}),

where k ∈ ℕ − 1.

Let s = −νd_{1} − 1 − k + u and

H_{1}(u) = g_{6,6}^{(5,3)}(−νd_{1} − 1 − k + u) = z^{−νd_{1}−1−k+u} Γ(νd_{1} + 1 + k − u)^{3}×

Γ(νΔ + 1 + k − u)^{2}Γ(−νΔ − k + u)^{−1}Γ(−k + u)^{3}Γ(2νΔ + 2 + k − u)^{−3} =

= z^{−νd_{1}−1−k+u} Γ(νd_{1} + 1 + k − u)^{3}Γ(νΔ + 1 + k − u)^{3}Γ(νΔ + 1 + k − u)^{−1}×

Γ(−νΔ − k + u)^{−1}Γ(1 + k − u)^{3}Γ(−k + u)^{3}Γ(1 + k − u)^{−3}Γ(2 + 2νΔ + k − u)^{−3} =

= (π/ sin(πu))^{2}(−1)^{νΔ}z^{−νd_{1}−1−k+u} Γ(νd_{1} + 1 + k − u)^{3}×

Γ(νΔ + 1 + k − u)^{3}Γ(1 + k − u)^{−3}Γ(2 + 2νΔ + k − u)^{−3} = (π/ sin(πu))^{2}H^{∗}(u),

where

H^{∗}(u) = (−1)^{νΔ} z^{−νd_{1}−1−k+u}×

Γ(νd_{1} + 1 + k − u)^{3}Γ(1 + k − u)^{−3}Γ(1 + νΔ + k − u)^{3}Γ(2 + 2νΔ + k − u)^{−3} =

= (−1)^{νΔ}z^{−T +ν}Γ(T)^{3}Γ(T + 1 + νΔ)^{−3}Γ(T − νΔ)^{−3}Γ(T − ν)^{3} =

= (−1)^{νΔ}z^{−T +ν}( ∏_{κ=ν+1}^{νΔ}(T − κ))^{3}( ∏_{κ=0}^{k+νΔ}(T + κ)^{−1})^{3} =

= (−1)^{νΔ}z^{−T +ν}R_{0}(T; ν)^{3}((νΔ)!/(νd_{1})!)^{−3}

and T = νΔ + 1 + k − u. Therefore

(−1)^{νd_{2}}((νΔ)!/(νd_{1})!)^{3} Res (g_{6,6}^{(5,3)}; −1 − νd_{1} − k) =

= (−1)^{ν}(z^{−T +ν}R_{0}(T; ν)^{3} log z − \frac{∂}{∂T} R_{0}(T; ν)^{3})|_{T =1+νΔ+k}

because (πu/(sin(πu))^{2} is an even function. Thus

f_{3}(z, ν) = f_{2}(z, ν) log z − (−1)^{ν}((νΔ)!/(νd_{1})!)^{−3} · ∑_{t=νΔ+1}^{∞} z^{−t+ν} \frac{∂}{∂t}R_{0}(t; ν)^{3};

since R_{0}(t; ν)^{3} has zeros of the third order in the points t = 1, …, νΔ, it follows that

(2.9) f_{3}(z, ν) = f_{2}(z, ν) log z − (−1)^{ν}((νΔ)!/(νd_{1})!)^{−3} ∑_{t=1+ν}^{∞} z^{−t+ν} \frac{∂}{∂t}R_{0}(t; ν)^{3}.

Let

(2.10)  f_{4}(z, ν) = −((νΔ)!/(νd_{1})!)^{−3}(−z)^{ν} ∑_{t=νΔ+1}^{∞} z^{−t} \frac{∂}{∂t}R_{0}(t; ν)^{3};

then

(2.11)  f_{3}(z, ν) = f_{2}(z, ν) log z + f_{4}(z, ν).

Let

f_{5}^{∨}(z, ν) = −\frac{(−1)^{νΔ}}{2πi} ∫_{L_{2}} g_{6,6}^{(6,3)}(s)ds =

= −(−1)^{νΔ} G_{6,6}^{(6,3)} (−z| \begin{matrix}−νd_{1}, & −νd_{1}, & −νd_{1}, & 1 + νd_{2}, & 1 + νd_{2}, & 1 + νd_{2} \\ 0, & 0, & 0, & ν, & ν, & ν \end{matrix} ) ,

where z ∈ Ω_{0}^{∗}, ν ∈ ℕ, and

g_{6,6}^{(6,3)} = g_{6,6}^{(6,3)}(s) = (−z)^{s}Γ(−s)^{3}Γ(ν − s)^{3}Γ(1 + νd_{1} + s)^{3}Γ(1 + νd_{2} − s)^{−3}.

Here p = q = 6, m = 6, n = 3, a_{1} = a_{2} = a_{3} = −νd_{1}, a_{4} = a_{5} = a_{6} = 1 + νd_{2}, b_{1} = b_{2} = b_{3} = 0, b_{4} = b_{5} = b_{6} = ν, Δ^{∗} = −3ν − 3; since −z ∈ Ω^{∗}, each of convergence conditions (C2) and (C3) is satisfied. The set of all the unremovable singular points of the function g_{6,6}^{(5,3)}(s), encircled by the curve L_{2}, consists of the points s = −1 − νd_{1} − k with k ∈ ℕ − 1; each of these points is a pole of the third order. Therefore

Res (g_{6,6}^{(6,3)}; −νd_{1} − 1 − k) = lim_{s→−νd_{1}−1−k} \frac{1}{2}(\frac{∂}{∂s})^{2}((s + νd_{1} + 1 + k)^{2}g_{6,6}^{(5,3)}),

where k ∈ ℕ − 1.

Let s = −νd_{1} − 1 − k + u, and

H_{2}(u) = g_{6,6}^{(6,3)}(−νd_{1} − 1 − k + u) =

= (−z)^{−νd_{1}−1−k+u}Γ(−k+u)^{3}Γ(νd_{1}+1+k−u)^{3}Γ(νΔ+1+k−u)^{3}Γ(2+2νΔ+k−u)^{−3}

= (−1)^{k}(−z)^{−νd_{1}−1−k+u}Γ(1 + νΔ + k − u)^{3}Γ(2 + 2νΔ + k − u)^{−3}×
× Γ(νd_{1} + 1 + k − u)^{3}Γ(1 + k − u)^{−3}(π/ sin(πu))^{3} = (π/ sin(πu))^{3}H_{2}^{∗}(u),

where

H_{2}^{∗}(u) = (−1)^{k}(−z)^{−T +ν}R_{0}(T; ν)^{3}((νΔ)!/(νd_{1})!)^{−3}

and T = νΔ + 1 + k − u, t = νΔ + 1 + k, as before.

Therefore

−(−1)^{νΔ}((νΔ)!/(νd_{1})!)^{3} Res (g_{6,6}^{(6,3)}; −1 − νd_{1} − k) =

= −(−1)^{νΔ+k}(\frac{1}{2}((log(−z))^{2}(−z)^{−t+ν}R_{0}(t; ν)^{3} − (−z)^{−t+ν} log(−z)\frac{∂}{∂T} R_{0}(t; ν)^{3}+

+\frac{1}{2}((−z)^{−t+ν}(\frac{∂}{∂T} )^{2}R_{0}(t; ν)^{3} + π^{2}(−z)^{−t+ν}R_{0}(t; ν)^{3}))

because (πu/ sin(πu))^{3} = 1 + \frac{1}{2}(πu)^{2} + ….

Consequently,

(2.12)  f_{5}^{∨}(z, ν) = \frac{1}{2}f_{2}(z, ν)((log(−z))^{2} + π^{2}) + f_{4}(z, ν) log(−z)+

+\frac{1}{2}(−z)^{ν}((νΔ)!/(νd_{1})!)^{−3} ∑_{t=νΔ+1}^{∞} z^{−t}(\frac{∂}{∂T} )^{2}R_{0}(t; ν)^{3} =

= \frac{1}{2} f_{2}(z, ν)(log z)^{2} + f_{4}(z, ν) log z+

+\frac{1}{2}(−z)^{ν}((νΔ)!/(νd_{1})!)^{−3} ∑_{t=νΔ+1}^{∞} z^{−t}(\frac{∂}{∂T} )^{2}R_{0}(t; ν)^{3} − πif_{3}(z, ν),

where z ∈ Ω_{0}^{∗}.

Since R_{0}(t; ν)^{3} has zeros of the third order at the points t = 1 + ν, …, νΔ, it follows that

f_{5}^{∨}(z, ν) = \frac{1}{2}f_{2}(z, ν)(log z)^{2} + f_{4}(z, ν) log z +

16 LEONID A. GUTNIK

+\frac{1}{2}(−z)^{ν}((νΔ)!/(νd_{1})!)^{−3} ∑_{t=ν+1}^{∞} z^{−t}(\frac{∂}{∂T} )^{2}R_{0}(t; ν)^{3} − iπf_{3}(z, ν).

Let

(2.13)  f_{6}(z, ν) = \frac{1}{2}(−z)^{ν}((νΔ)!/(νd_{1})!)^{−3} ∑_{t=1}^{∞} z^{−t}(\frac{∂}{∂T} )^{2}R_{0}(t; ν)^{3},

f_{5}(z, ν) = f_{6}(z, ν) + \frac{1}{2}f_{2}(z, ν)(log z)^{2} + f_{4}(z, ν) log z,

where z ∈ Ω_{0}^{∗}. Then

(2.14)  f_{5}(z, ν) = f_{5}^{∨}(z, ν) + iπf_{3}(z, ν),

where z ∈ Ω_{0}^{∗}. Let further

(2.15) f_{j}^{∗}(z, ν) = ((νΔ)!/(νd_{1})!)^{3}f_{j}(z, ν), j = 1, …, 6.

Expanding function R_{0}(t; ν)^{3} into partial fractions, we obtain

R_{0}(t; ν)^{3} = ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(t + k)^{−3} + ∑_{k=0}^{νΔ} β_{ν,k}^{∗}(t + k)^{−2} + ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}(t + k)^{−3}

with

(2.16)  α_{ν,k}^{∗} = (−1)^{ν+νΔ+k}\binom{νΔ}{k}^{3}\binom{νΔ + k}{νΔ − ν}^{3},

(2.17)

β_{ν,k}^{∗} = lim_{t→−k} \frac{∂}{∂t}(R_{0}(t; ν)^{3}(t + k)^{3}) = α_{ν,k}^{∗}(3(− ∑_{κ=ν+k+1}^{νΔ+k}κ^{−1} − ∑_{κ=1}^{νΔ−k} κ^{−1} + ∑_{κ=1}^{k} κ^{−1})),

(2.18)  2γ_{ν,k}^{∗} = α_{ν,k}^{∗}(3(− ∑_{κ=ν+k+1}^{νΔ+k}κ^{−1} − ∑_{κ=1}^{νΔ−k} κ^{−1} + ∑_{κ=1}^{k} κ^{−1}))^{2}−

− α_{ν,k}^{∗}(3( ∑_{κ=ν+k+1}^{νΔ+k}κ^{−2} − ∑_{κ=1}^{νΔ−k} κ^{−2} − ∑_{κ=1}^{k} κ^{−2})),

ON LINEAR FORMS WITH COEFFICIENTS IN ℕ ζ(1 + ℕ) 17

where k = 0, …, νΔ. It follows from (2.4), (2.7), (2.10), (2.13) and (2.15) - (2.18) that

(2.19)  f_{2}^{∗}(z, ν) = (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t}R_{0}(t; ν)^{3} =

= (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(t + k)^{−3} + (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} β_{ν,k}^{∗}(t + k)^{−2} +

+ (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}(t + k)^{−1} = (−z)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−3} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} β_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−2} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−1} =

= α^{∗}(z; ν)L_{3}(z^{−1}) + β^{∗}(z; ν)L_{2}(z^{−1}) + γ^{∗}(z; ν)(− log(1 − 1/z)) − ϕ^{∗}(z; ν) with

(2.20)  L_{n}(z) = ∑_{t=1}^{+∞} z^{t}t^{−n},
(2.21)  α^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}z^{k} = f_{1}^{∗}(z; ν),
(2.22)  β^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} β_{ν,k}^{∗}z^{k}
(2.23)  γ^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k},
(2.24)  ϕ^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−3} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} β_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−2} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−1} =

= (−z)^{ν} ∑_{t=1}^{ν} z^{−t}(t^{−3}α^{∗}(z; ν) + t^{−2}β^{∗}(z; ν) + t^{−1}γ^{∗}(z; ν)) +

+ (−z)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−3} + (−z)^{ν} ∑_{k=0}^{νΔ} β_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−2} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−1};

(2.25)  f_{4}^{∗}(z, ν) = −(−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} \frac{∂}{∂t}R_{0}(t; ν)^{3} =

= (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} 3α_{ν,k}^{∗}(t + k)^{−4} + (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} 2β_{ν,k}^{∗}(t + k)^{−3} +

+ (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}(t + k)^{−2} = (−z)^{ν} ∑_{k=0}^{νΔ} 3α_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−4} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 2β_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−3} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−2} =

= 3α^{∗}(z; ν)L_{4}(z^{−1}) + 2β^{∗}(z; ν)L_{3}(z^{−1}) + γ^{∗}(z; ν)L_{2}(z)^{−1} − ψ^{∗}(z; ν),

where

(2.26)  ψ^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} 3α_{ν,k}^{∗} z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−4} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 2β_{ν,k}^{∗} z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−3} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗} z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−2} =

= (−z)^{ν} ∑_{t=1}^{ν} z^{−t}(3t^{−4}α^{∗}(z; ν) + 2t^{−3}β^{∗}(z; ν) + t^{−2}γ^{∗}(z; ν)) +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 3α_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−4} + (−z)^{ν} ∑_{k=0}^{νΔ} 2β_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−3} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−2};

(2.27)  f_{6}^{∗}(z, ν) = \frac{1}{2}(−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t}(\frac{∂}{∂t})^{2}R_{0}(t; ν)^{3} =

= (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} 6α_{ν,k}^{∗}(t + k)^{−5} + (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} 3β_{ν,k}^{∗}(t + k)^{−4} +

+ (−z)^{ν} ∑_{t=1+ν}^{+∞} z^{−t} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}(t + k)^{−3} = (−z)^{ν} ∑_{k=0}^{νΔ} 6α_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−5} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 3β_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−4} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν+k}^{+∞} z^{−t}t^{−3} =

= 6α^{∗}(z; ν)L_{5}(z^{−1}) + 3β^{∗}(z; ν)L_{4}(z^{−1}) + γ^{∗}(z; ν)L_{3}(z^{−1}) − ξ^{∗}(z; ν),

and

(2.28)  ξ^{∗}(z; ν) = (−z)^{ν} ∑_{k=0}^{νΔ} 6α_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−5} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 3β_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−4} + (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1}^{k+ν} z^{−t}t^{−3} =

= (−z)^{ν} ∑_{t=1}^{ν} z^{−t}(6t^{−5}α^{∗}(z; ν) + 3t^{−4}β^{∗}(z; ν) + t^{−3}γ^{∗}(z; ν)) +

+ (−z)^{ν} ∑_{k=0}^{νΔ} 6α_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−5} + (−z)^{ν} ∑_{k=0}^{νΔ} 3β_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−4} +

+ (−z)^{ν} ∑_{k=0}^{νΔ} γ_{ν,k}^{∗}z^{k} ∑_{t=1+ν}^{k+ν} z^{−t}t^{−3}.

References

[1] Hata M., Legendre type polynomials and irrationality measures, J. Reine Angew. Math. 407 (1990), 99 − 105.

[2] Gutnik L., On a measure of the irrationality of dilogarithms at rational points, VINITI 4345-84 (1984), 1 − 74.

[3] , On the rank over ℚ of some real matrices, VINITI 5736-84 (1984), 1 − 29.

[4] , To the question of the smallness of some linear forms, VINITI 2413-B93 (1993), 1 − 94.

[5] , About linear forms, whose coefficients are ℚ-proportional to the number log 2, and the values of ζ(s) for integer s, VINITI 3258-B96 (1996), 1 − 70.

[6] , The lower estimate for some linear forms with coefficients ℕ ζ(1 + ℕ), VINITI 3072-B97 (1997), 1 − 77.

[7] , About the behaviour of solutions of difference equations of Poincaré type, VINITI 3384-B97 (1997), 1 − 41.

[8] , On linear forms with coefficients in ℕ ζ(1+ℕ), Max-Plank-Institut für Mathematik, Bonn, Preprint Series 3 (2000), 1 − 13.

[9] Bateman H. and Erdélyi A., Higher transcendental functions, New-York – Toronto – London, Mc. Grow-Hill Book Company, Inc., 1953.

[10] Titchmarsh E. C., The Theory of Functions, Oxford University Press, 1939.

Moscow State Institute of Electronics and Mathematics Bolshoi Trekhsvyatitelskij pereulok, 1-3/12, str. 8 Moscow 109028 Russian Federation E-mail: gutnik@gutnik.mccme.ru