AN IDENTITY OF ANDREWS, MULTIPLE INTEGRALS, AND VERY-WELL-POISED HYPERGEOMETRIC SERIES

C. KRATTENTHALER^{†} AND T. RIVOAL

Dedicated to Richard Askey

Abstract. We give a new proof of a theorem of Zudilin that equates a very-well-poised hypergeometric series and a particular multiple integral. This integral generalizes integrals of Vasilenko and Vasilyev which were proposed as tools in the study of the arithmetic behaviour of values of the Riemann zeta function at integers. Our proof is based on limiting cases of a basic hypergeometric identity of Andrews.

1. Introduction

After Apéry’s 1978 proof of the irrationality of ζ(2) and ζ(3) (see [3]), ζ(s) denoting the Riemann zeta function, Beukers [6] gave another proof with the help of his famous integrals

∫_{[0,1]^{2}} \frac{x^{n}(1 − x)^{n}y^{n}(1 − y)^{n}}{(1 − (1 − x)y)^{n+1}} dx dy = α_{n}ζ(2) − β_{n}

and

∫_{[0,1]^{3}} \frac{x^{n}(1 − x)^{n}y^{n}(1 − y)^{n}z^{n}(1 − z)^{n}}{(1 − (1 − (1 − x)y)z)^{n+1}} dx dy dz = 2a_{n}ζ(3) − b_{n}.

Here, n ≥ 0 is an integer, and α_{n}, β_{n}, a_{n}, b_{n} are rational numbers. More precisely, α_{n}, a_{n}, d_{n}^{2}β_{n} and d_{n}^{3}b_{n} are integers, with d_{n} = lcm{1, 2, …, n}.

Extending Vasilenko’s method [17], Vasilyev [18, 19] considered a family of integrals generalizing Beukers’ pattern:

J_{E,n} = ∫_{[0,1]^{E}} \frac{∏_{i=1}^{E} x_{i}^{n}(1 − x_{i})^{n}}{Q_{E}(x_{1}, x_{2}, …, x_{E})^{n+1}} dx_{1} dx_{2} ⋯ dx_{E},  (1.1)

where

Q_{E}(x_{1}, x_{2}, …, x_{E}) = 1 − (⋯ (1 − (1 − x_{E})x_{E−1}) ⋯)x_{1},

and where E is an integer ≥ 2.

He then formulated the following conjecture, which he proved for E = 4, 5 and which is also true for E = 2, 3 because of Beukers’ work.

2000 Mathematics Subject Classification. Primary 33C20; Secondary 11J72.

Key words and phrases. Riemann zeta function, hypergeometric series.

†Research partially supported by the programme “Improving the Human Research Potential” of the European Commission, grant HPRN-CT-2001-00272, “Algebraic Combinatorics in Europe”.

Conjecture 1 (Vasilyev).

(i) For all integers E ≥ 2 and n ≥ 0, there exist rational numbers p_{m,E,n} such that

J_{E,n} = p_{0,E,n} +_{∑_{m=2,...,E} ; m≡E (mod 2)}p_{m,E,n}ζ(m).  (1.2)

(ii) Furthermore, d_{n}^{E}p_{m,E,n} is an integer for all m = 0, 2, 3, …, E.

Part (i) of this conjecture has been proved by Zudilin in [23, Sec. 8], thanks to an unexpected identity between a certain multiple integral J_{m} (generalizing Vasilyev’s ones) and a non-terminating very-well-poised hypergeometric series (see Theorem 1 below). Part (ii) was also proved by Zudilin in [23], except for the coefficient p_{0,E,n}. A sharper version of Part (ii) has been established by the authors in [10] (for m = 0 only up to multiplication by a factor of 2).

The aim of this note is to give a new proof of Zudilin’s identity, which is the content of Theorem 1 in the next section. In fact, our proof shows that, modulo a more or less evident expansion of the Vasilyev-type integral J_{m} as a multiple sum (see Proposition 2 in Section 4), Zudilin’s identity is a limiting case of a thirty year old identity between a terminating multiple sum and a terminating very-well-poised hypergeometric series due to Andrews (see (3.1) below). We believe that this is an interesting observation because attempts to prove Zudilin’s identity by manipulating hypergeometric series directly failed because of convergence problems. Zudilin circumvents these problems by having recourse to a Barnes-type (i.e., contour) integral in place of the very-well-poised hypergeometric series. Our proof shows that there is indeed a “purely hypergeometric” proof (i.e., a proof just using summation and transformation formulas for hypergeometric series), but to be able to accomplish it, one has to go “one level higher in hierarchy,” meaning that one finds a terminating identity “above,” of which the identity which one actually wants to prove is a limiting case.^{1} This identity “above” is Andrews’ identity, and it does indeed have a purely hypergeometric proof (see Section 3 for more information).

As an aside, we mention that, in the recent paper [22], Zlobin shows that the multiple integral J_{m} is also equal to an integral of the same type as those of Sorokin in [15, 16] (see also Fischler [7] for similar results). The latter integral can also be expanded as a multisum, in a manner completely analogous to the way we derive the multisum expansion for J_{m} in the proof of Proposition 2. As a result, one obtains again exactly the right-hand sides of (4.1) and (4.2), respectively. Thus, this provides an alternative proof of Zlobin’s result. As a matter of fact, when this work originated, we went the other way, that is, our starting point was the multisum expansion of Zlobin’s integral, until we realized that, actually, the integral J_{m} admits the same treatment.

1This point is also of interest “philosophically.” There are several proposers (of whom Koornwinder [9] seems to have been the first; see [13, Remark 3.2] and [20, paragraph after the second Eq. (Apery)] for printed versions) of the “conjecture” that above every identity for non-terminating hypergeometric series (which are very often difficult to prove; in particular, the automatic tools described in [11] do not apply) there sits a more general identity for terminating series (which, at least in principle, can be proved automatically), of which the non-terminating identity is a limiting case. Of course, the analyst would object that above every terminating identity there exists an even more general non-terminating identity, of which the terminating one is a special case. Clearly, this dispute is as easy to settle as the dispute about the question of which was first, the hen or the egg …

Zudilin’s identity is recalled in Theorem 1 in the next section. The limiting cases of Andrews’ identity which we need are stated and proved in Proposition 1 in Section 3. One of the lemmas which we need for carrying out these limits generalizes a lemma of Zhao [21] on the convergence of multizeta functions, see the remark after the proof of Lemma 3. The purpose of Section 4 is to relate these identities to the Vasilyev-type integral J_{m}, see Proposition 2. We finally prove Theorem 1 in Section 5.

2. Zudilin’s identity

In order to be able to state Zudilin’s identity, we need to recall the standard notation for (generalized) hypergeometric series,

p+1F_{p} [\begin{matrix}α_{0}, α_{1}, …, α_{p} \\ β_{1}, …, β_{p} \end{matrix}; z] = ∑_{k=0}^{∞} \frac{(α_{0})_{k} (α_{1})_{k} ⋯ (α_{p})_{k}}{k! (β_{1})_{k} ⋯ (β_{p})_{k}} z^{k},

where p ≥ 1, α_{j} ∈ ℂ, β_{j} ∈ ℂ \ ℤ_{≤0} and, by definition, (x)_{0} = 1 and (x)_{ℓ} = x(x + 1) ⋯ (x + ℓ − 1) for ℓ ≥ 1. The series is absolutely convergent for all z ∈ ℂ such that |z| < 1, and also for |z| = 1 provided ℜ𝔢(β_{1} + ⋯ + β_{p}) > ℜ𝔢(α_{0} + α_{1} + ⋯ + α_{p}). Furthermore, it is said to be balanced if α_{0} + ⋯ + α_{p} + 1 = β_{1} + ⋯ + β_{p} and very-well-poised if α_{0} + 1 = α_{1} + β_{1} = ⋯ = α_{p} + β_{p} and α_{1} = \frac{1}{2} α_{0} + 1. See the books [2, 4, 8, 14] for more information on hypergeometric series.

Let z, a_{0}, a_{1}, …, a_{m}, and b_{1}, …, b_{m} be complex numbers such that |z| < 1, ℜ𝔢(b_{i}) > ℜ𝔢(a_{i}) > 0 for all i = 1, 2, …, m, and let us define the Vasilyev-type integral

J_{m} [\begin{matrix}a_{0}, a_{1}, …, a_{m} \\ b_{1}, …, b_{m} \end{matrix}; z] = ∫_{[0,1]^{m}} \frac{∏_{i=1}^{m} x_{i}^{a_{i}−1}(1 − x_{i})^{b_{i}−a_{i}−1}}{(1 − (1 − (⋯ (1 − x_{m})x_{m−1}) ⋯)x_{1}z)^{a_{0}}} dx_{1} dx_{2} ⋯ dx_{m},

(2.1)

which is absolutely convergent under the above conditions. (We will sometimes use the short notation J_{m} for this integral if there is no ambiguity about the parameters.) It is also absolutely convergent for z = 1, provided that we also assume that ℜ𝔢(b_{1} − a_{1}) > ℜ𝔢(a_{0}) if m = 1, respectively ℜ𝔢(b_{1} − a_{1}) ≥ ℜ𝔢(a_{0}) if m > 1. Since previous authors assume more restrictive conditions in the case z = 1 (in particular, restrictions that are not satisfied by Vasilyev’s integrals (1.1)), we sketch the verification of the convergence here for the sake of completeness.

If m = 1, then J_{m} = J_{1} is a beta integral. If m ≥ 2, then, because of

1 − (1 − (⋯ (1 − x_{m})x_{m−1}) ⋯)x_{1} ≥ 1 − x_{1}

and ℜ𝔢(b_{1} − a_{1}) ≥ ℜ𝔢(a_{0}), we have

∫_{ε_{m}}^{1−ε_{m}} … ∫_{ε_{2}}^{1−ε_{2}} ∫_{ε_{1}}^{1−ε_{1}} |\frac{∏_{i=1}^{m} x_{i}^{a_{i}−1}(1 − x_{i})^{b_{i}−a_{i}−1}}{(1 − (1 − (⋯ (1 − x_{m})x_{m−1}) ⋯)x_{1})^{a_{0}}} | dx_{1} dx_{2} ⋯ dx_{m}

≤ ∫_{ε_{m}}^{1−ε_{m}} … ∫_{ε_{2}}^{1−ε_{2}} ∫_{ε_{1}}^{1−ε_{1}} \frac{∏_{i=2}^{m} x_{i}^{ℜ𝔢(a_{i}−1)}(1 − x_{i})^{ℜ𝔢(b_{i}−a_{i}−1)}}{(1 − x_{1} + Xx_{2}x_{1})} dx_{1} dx_{2} ⋯ dx_{m}

≤ ∫_{ε_{m}}^{1−ε_{m}} … ∫_{ε_{2}}^{1−ε_{2}} (∏_{i=2}^{m} x_{i}^{ℜ𝔢(a_{i}−1)}(1 − x_{i})^{ℜ𝔢(b_{i}−a_{i}−1)})

· (−\frac{log(1 − x_{1} + Xx_{2}x_{1})}{1 − Xx_{2}}) |_{x_{1}=ε_{1}}^{1−ε_{1}} dx_{2} ⋯ dx_{m},

for any small ε_{1}, ε_{2}, …, ε_{m} > 0, where we wrote X for 1 − (⋯ (1 − x_{m})x_{m−1}) ⋯)x_{3}. (In case that m = 2, X has to be interpreted as 1.) If we perform the limit ε_{1} → 0, then the right-hand side of this inequality becomes the integral

∫_{ε_{m}}^{1−ε_{m}} … ∫_{ε_{2}}^{1−ε_{2}} (∏_{i=2}^{m} x_{i}^{ℜ𝔢(a_{i}−1)}(1 − x_{i})^{ℜ𝔢(b_{i}−a_{i}−1)}) (−\frac{log(Xx_{2})}{1 − Xx_{2}}) dx_{2} ⋯ dx_{m}. (2.2)

In the integrand, there is no problem as Xx_{2} → 1, since the function log(Xx_{2})/(1 − Xx_{2}) is continuous at Xx_{2} = 1. On the other hand, if we fix η > 0, then for Xx_{2} sufficiently close to 0, we have

| log(Xx_{2})| < (Xx_{2})^{−η} ≤ (1 − x_{3})^{−η}x_{2}^{−η}.

Thus, choosing η = \frac{1}{2} min\{ℜ𝔢(a_{2}), ℜ𝔢(b_{3} − a_{3})\}, we see that the integral in (2.2), and thus the original integral J_{m}, exists.

Theorem 1 (Zudilin). For every integer m ≥ 1, the following identity holds:

J_{m} [\begin{matrix}h_{1}, h_{2}, h_{3}, …, h_{m+1} \\ 1 + h_{0} − h_{3}, 1 + h_{0} − h_{4}, …, 1 + h_{0} − h_{m+2} \end{matrix}; 1]

= \frac{Γ(1 + h_{0}) ∏_{j=3}^{m+1} Γ(h_{j})}{∏_{j=1}^{m+2} Γ(1 + h_{0} − h_{j})} · (∏_{j=1}^{m+1} Γ(1 + h_{0} − h_{j} − h_{j+1}))

× _{m+4}F_{m+3} [\begin{matrix}h_{0}, \frac{1}{2}h_{0} + 1, h_{1}, …, h_{m+2} \\ \frac{1}{2}h_{0}, 1 + h_{0} − h_{1}, …, 1 + h_{0} − h_{m+2} \end{matrix}; (−1)^{m+1}] , (2.3)

provided that 1 + ℜ𝔢(h_{0}) > \frac{2}{m+1} ∑_{j=1}^{m+2} ℜ𝔢(h_{j}), ℜ𝔢(1 + h_{0} − h_{j+1}) > ℜ𝔢(h_{j}) > 0 for j = 2, 3, …, m + 1, and ℜ𝔢(1 + h_{0} − h_{3} − h_{2}) ≥ ℜ𝔢(h_{1}), these conditions ensuring that both sides of (2.3) are well-defined.

In the case of the original integrals J_{E,n} of Vasilyev, the identity in Theorem 1 reads as follows: for any integers n ≥ 0 and E ≥ 2,

J_{E,n} = \frac{n!^{2E+1}(3n + 2)!}{(2n + 1)!^{E+2}} _{E+4}F_{E+3} [\begin{matrix}3n + 2, \frac{3}{2}n + 2, n + 1, …, n + 1 \\ \frac{3}{2}n + 1, 2n + 2, …, 2n + 2 \end{matrix}; (−1)^{E+1}] .

From [5, 12], it follows that such a very-well-poised hypergeometric series gives rise to a decomposition of the shape (1.2).

3. Limiting cases of Andrews’ hypergeometric identity

Let N and s be positive integers, and a, b_{1}, …, b_{s+1}, c_{1}, …, c_{s+1} be complex numbers such that none of 1 + a − b_{j}, 1 + a − c_{j}, j = 1, 2, …, s + 1, and 1 + a + N are non-positive integers.

Andrews’ identity [1, Theorem 4] relates a terminating very-well-poised basic hypergeometric series to a terminating multiple basic hypergeometric series. We shall need here the limiting case of this identity when q → 1, so that the series there reduce to “ordinary” hypergeometric series. That is, we replace a by q^{a}, b_{i} by q^{b_{i}}, c_{i} by q^{c_{i}}, there, and then let q tend to 1. The result can be compactly written in the form

2s+5F_{2s+4} [ \begin{matrix}a, \frac{a}{2} + 1, b_{1}, c_{1}, …, b_{s+1}, c_{s+1}, −N \\ \frac{a}{2}, 1 + a − b_{1}, 1 + a − c_{1}, …, 1 + a − b_{s+1}, 1 + a − c_{s+1}, 1 + a + N \end{matrix} ; 1]

= \frac{(1 + a)_{N} (1 + a − b_{s+1} − c_{s+1})_{N}}{(1 + a − b_{s+1})_{N} (1 + a − c_{s+1})_{N}} ∑_{k_{1},k_{2},...,k_{s}≥0} \frac{(−N)_{k_{1}+···+k_{s}}}{(b_{s+1} + c_{s+1} − a − N)_{k_{1}+···+k_{s}}}

· ∏_{j=1}^{s} \frac{(1 + a − b_{j} − c_{j})_{k_{j}} (b_{j+1})_{k_{1}+···+k_{j}} (c_{j+1})_{k_{1}+···+k_{j}}}{k_{j}! (1 + a − b_{j})_{k_{1}+···+k_{j}} (1 + a − c_{j})_{k_{1}+···+k_{j}}} . (3.1)

The proof in [1] uses Whipple’s transformation between a balanced _{4}F_{3}-series and a verywell-poised _{7}F_{6}-series,

4F_{3}[\begin{matrix}a, b, c, −N \\ e, f, 1 + a + b + c − e − f − N \end{matrix}; 1] = \frac{(−a − b + e + f)_{N} (−a − c + e + f)_{N}}{(−a + e + f)_{N} (−a − b − c + e + f)_{N}}

·_{7}F_{6}[\begin{matrix}−1 − a + e + f, \frac{1}{2} − \frac{a}{2} + \frac{e}{2} + \frac{f}{2}, −a + f, −a + e, b, c, −N \\ −\frac{1}{2} − \frac{a}{2} + \frac{e}{2} + \frac{f}{2}, e, f, −a − b + e + f, −a − c + e + f, −a + e + f + N \end{matrix} ; 1] ,

and the Pfaff–Saalschütz summation in an iterative fashion. In particular, the identity (3.1) reduces to Whipple’s transformation for s = 1.

We prove that the same kind of identity holds for non-terminating hypergeometric series provided the parameters a, b_{j}, and c_{j}, j = 1, 2, …, s + 1, satisfy some further conditions.

Proposition 1. (i) Let s ≥ 1 be an integer, and let a, b_{1}, …, b_{s+1}, c_{1}, …, c_{s+1} be complex numbers such that none of 1 + a − b_{j}, 1 + a − c_{j}, j = 1, 2, …, s + 1, is a non-positive integer. Furthermore, we assume that

ℜ𝔢 ((2s + 1)(a + 1) − 2 ∑_{j=1}^{s+1}(b_{j} + c_{j})) > 0  (3.2)

and

ℜ𝔢 ((1 + a − b_{s+1} − c_{s+1}) + ∑_{j=r}^{s} A_{j}(1 + a − b_{j} − c_{j})) > 0  (3.3)

for all r = 2, 3, …, s + 1 (in the case that r = s + 1, the empty sum ∑_{j=r}^{s} has to be interpreted as 0), for all possible choices of A_{j} = 1 or 2, for j = 2, 3, …, s. Then

2s+4F_{2s+3} [ \begin{matrix}a, \frac{a}{2} + 1, b_{1}, c_{1}, …, b_{s+1}, c_{s+1} \\ \frac{a}{2}, 1 + a − b_{1}, 1 + a − c_{1}, …, 1 + a − b_{s+1}, 1 + a − c_{s+1} \end{matrix} ; −1]

= \frac{Γ(1 + a − b_{s+1}) Γ(1 + a − c_{s+1})}{Γ(1 + a) Γ(1 + a − b_{s+1} − c_{s+1})}

× ∑_{k_{1},k_{2},...,k_{s}≥0} ∏_{j=1}^{s} \frac{(1 + a − b_{j} − c_{j})_{k_{j}} (b_{j+1})_{k_{1}+···+k_{j}} (c_{j+1})_{k_{1}+···+k_{j}}}{k_{j}! (1 + a − b_{j})_{k_{1}+···+k_{j}} (1 + a − c_{j})_{k_{1}+···+k_{j}}} . (3.4)

(ii) Let s ≥ 1 be an integer, and let a, c_{0}, b_{1}, …, b_{s}, c_{1}, …, c_{s} be complex numbers such that none of 1 + a − b_{j}, 1 + a − c_{j}, j = 0, 1, …, s, is a non-positive integer. Furthermore, we assume that

ℜ𝔢 (2s(a + 1) − 2c_{0} − 2 ∑_{j=1}^{s}(b_{j} + c_{j})) > 0,  (3.5)

ℜ𝔢 ((1 + a − b_{s} − c_{s}) + ∑_{j=r}^{s−1} A_{j}(1 + a − b_{j} − c_{j})) > 0  (3.6)

for all r = 2, 3, …, s (in the case that r = s, the empty sum ∑_{j=r}^{s−1} has to be interpreted as 0), and

ℜ𝔢 ((1 + a − c_{0} − b_{1} − c_{1}) + ∑_{j=2}^{s−1} A_{j}(1 + a − b_{j} − c_{j})) > 0,  (3.7)

for all possible choices of A_{j} = 1 or 2, for j = 2, 3, …, s − 1. Then

2s+3F_{2s+2} [ \begin{matrix}a, \frac{a}{2} + 1, c_{0}, b_{1}, c_{1}, …, b_{s}, c_{s} \\ \frac{a}{2}, 1 + a − c_{0}, 1 + a − b_{1}, 1 + a − b_{1}, …, 1 + a − b_{s}, 1 + a − c_{s} \end{matrix} ; 1]

= \frac{Γ(1 + a − b_{s}) Γ(1 + a − c_{s})}{Γ(1 + a) Γ(1 + a − b_{s} − c_{s})} ∑_{k_{1},k_{2},...,k_{s}≥0} \frac{(b_{1})_{k_{1}} (c_{1})_{k_{1}}}{k_{1}! (1 + a − c_{0})_{k_{1}}}

· ∏_{j=2}^{s} \frac{(1 + a − b_{j−1} − c_{j−1})_{k_{j}} (b_{j})_{k_{1}+···+k_{j}} (c_{j})_{k_{1}+···+k_{j}}}{k_{j}! (1 + a − b_{j−1})_{k_{1}+···+k_{j}} (1 + a − c_{j−1})_{k_{1}+···+k_{j}}} . (3.8)

Our proof of this proposition is based on three lemmas, which we state and prove first.

Lemma 1. Let α and β be complex numbers such that β is not a non-positive integer. Then for any non-negative integer k we have

|\frac{Γ(α + k)}{Γ(β + k)}| ≤ D_{1} · (k + 1)^{ℜ𝔢(α−β)},

where D_{1} is a constant which does not depend on k.

Proof. By Stirling’s formula, we have

\frac{Γ(α + k)}{Γ(β + k)} ∼ (k + 1)^{α−β}

as k → ∞. Hence, the claim follows immediately. □

Lemma 2. Let A and B be real numbers such that A + B + 1 < 0, and let C be a nonnegative integer. Then, for any non-negative integer h, we have

∑_{k=0}^{∞}(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ D_{2}(h + 1)^{max{B,A+B+1}} (log(h + 2))^{C+1} ,

where D_{2} is a constant independent of h.

Proof. We split the summation range into the ranges R_{0} = {0, 1, …, 2^{⌈log_{2}(h+1)⌉+1} − h − 1} and

R_{s} = {2^{s} − h, 2^{s} − h + 1, …, 2^{s+1} − h − 1}, s = ⌈log_{2}(h + 1)⌉ + 1, ⌈log_{2}(h + 1)⌉ + 2, …,

where ⌈x⌉ denotes the least integer ≥ x. Since, depending on whether A and B are positive or not, for k ∈ R_{s}, s > 0, we have

(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ (log 2^{s+2})^{C} · \begin{cases}2^{(s+1)(A+B)} & if A, B ≥ 0, \\ 2^{(s−1)A+(s+1)B} & if A < 0, B ≥ 0, \\ 2^{(s+1)A+sB} & if A ≥ 0, B < 0, \\ 2^{(s−1)A+sB} & if A, B < 0, \end{cases}

we infer that

(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ D_{3} · (s + 2)^{C} · 2^{s(A+B)},

with a constant D_{3} which is independent of h. Thus, for the sum over the range {k ≥ 2^{⌈log_{2}(h+1)⌉+1} − h} we have

∑_{k=2^{⌈log_{2}(h+1)⌉+1}−h}^{∞}(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C}

= ∑_{s=⌈log_{2}(h+1)⌉+1}^{∞} ∑_{k∈R_{s}}(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C}

≤ ∑_{s=⌈log_{2}(h+1)⌉+1}^{∞} D_{3} · 2^{s} · (s + 1)_{C+1} · 2^{s(A+B)}

≤ D_{3} · 2^{(⌈log_{2}(h+1)⌉+1)(A+B+1)} \frac{(C + 1)!}{(1 − 2^{A+B+1})^{C+2}}

≤ D_{4} · (h + 1)^{A+B+1},

for a constant D_{4} independent of h.

Now we consider the remaining range, R_{0} = {0, 1, …, 2^{⌈log_{2}(h+1)⌉+1} − h − 1}. For any k ∈ R_{0} we have k ≤ 3h + 3, and therefore

(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ (log(4h + 5))^{C} · \begin{cases}(h + 1)^{B} & if B ≤ 0, \\ (4h + 4)^{B} & if B > 0. \end{cases}

In particular, there is a constant D_{5} independent of k such that

(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ D_{5} (log(h + 2))^{C} (h + 1)^{B}

for all k ∈ R_{0}. Using this fact, we are able to conclude that

∑_{k∈R_{0}}(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C} ≤ D_{5} · (log(h + 2))^{C} (h + 1)^{B} ∑_{k∈R_{0}}(k + 1)^{A}.

Now, if A < −1, then ∑_{k∈R_{0}}(k + 1)^{A} < ζ(−A). If A = −1, then ∑_{k∈R_{0}}(k + 1)^{−1} < log(4h + 4). Finally, if A > −1, then

∑_{k∈R_{0}}(k + 1)^{A} < ∫_{0}^{4h+4} x^{A} dx = \frac{1}{A + 1}(4h + 4)^{A+1}.

In all cases, we obtain that

∑_{k∈R_{0}}(k + 1)^{A}(h + k + 1)^{B} (log(h + k + 2))^{C} < D_{6} (log(h + 2))^{C+1} (h + 1)^{max{B,A+B+1}},

where D_{6} is a constant independent of h.

To conclude the proof of the lemma, the two estimates for the two ranges are combined, and the claimed result follows. □

In the statement of the next lemma, we use the following notation: given two sets S and T, we write S + T for the sum-set {x + y : x ∈ S and y ∈ T }.

Lemma 3. Let E_{j} and F_{j} be real numbers and let Z_{j} denote the set {F_{j}, E_{j} + F_{j} + 1}, j = 1, 2, …, s. If

E_{r} + F_{r} + 1 + max(Z_{r+1} + Z_{r+2} + ⋯ + Z_{s}) < 0  (3.9)

for r = 1, 2, …, s, then the multiple series

∑_{k_{1},...,k_{s}≥0} ∏_{j=1}^{s}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}}  (3.10)

converges.

Proof. By applying Lemma 2 iteratively, we have

∑_{k_{1},...,k_{s}≥0} ∏_{j=1}^{s}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}}

≤ D_{7} ∑_{k_{1},...,k_{s−1}≥0} (∏_{j=1}^{s−1}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}})

· (k_{1} + ⋯ + k_{s−1} + 1)^{max(Z_{s})} log(k_{1} + ⋯ + k_{s−1} + 2)

≤ D_{8} ∑_{k_{1},...,k_{s−2}≥0} (∏_{j=1}^{s−2}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}})

· (k_{1} + ⋯ + k_{s−2} + 1)^{max(Z_{s−1}+Z_{s})} (log(k_{1} + ⋯ + k_{s−2} + 2))^{2} ,

and, after the t-th iteration, 1 ≤ t ≤ s − 1,

∑_{k_{1},...,k_{s}≥0} ∏_{j=1}^{s}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}}

≤ D_{9} ∑_{k_{1},...,k_{s−t}≥0} (∏_{j=1}^{s−t}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}})

· (k_{1} + ⋯ + k_{s−t} + 1)^{max(Z_{s−t+1}+···+Z_{s})} (log(k_{1} + ⋯ + k_{s−t} + 2))^{t} .

To justify these steps, we have to verify that the condition A + B + 1 < 0 in Lemma 2 is satisfied in each iteration. However, this is exactly the condition (3.9) with r replaced by s − t.

Thus, for t = s − 1 we arrive at the estimate

∑_{k_{1},...,k_{s}≥0} ∏_{j=1}^{s}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}}

≤ D_{10} ∑_{k_{1}≥0}(k_{1} + 1)^{E_{1}+F_{1}+max(Z_{2}+···+Z_{s})} (log(k_{1} + 2))^{s−1} .

Since the sum over k_{1} at the right-hand side converges because of (3.9) with r = 1, the claim follows. □

Remark. A careful check of our arguments reveals that, in fact, the conditions in Lemma 3 are optimal, meaning that they describe exactly the domain of convergence of the multiple sum (3.10). This can be seen by verifying that, if condition (3.9) is violated for a particular r, then the subsum

∑_{k_{r},...,k_{s}≥0} ∏_{j=1}^{s}(k_{j} + 1)^{E_{j}} (k_{1} + ⋯ + k_{j} + 1)^{F_{j}}

of (3.10) does not converge. Thus, this lemma generalizes Proposition 1 in [21]. It does at the same time correct that proposition, and it answers the question raised after the (incomplete) proof of the proposition. The question, which is asked there, is to determine the domain of absolute convergence of the multizeta function

ζ(s_{d}, s_{d−1}, …, s_{1}) = ∑_{0<n_{1}<···<n_{d}} \frac{1}{n_{1}^{s_{1}}n_{2}^{s_{2}} ⋯ n_{d}^{s_{d}}} .  (3.11)

Proposition 1 in [21] states that, for all d-tuples (s_{1}, s_{2}, …, s_{d}) with ℜ𝔢(s_{d}) > 1 and ∑_{i=1}^{d} ℜ𝔢(s_{i}) > d, the series ζ(s_{d}, s_{d−1}, …, s_{1}) converges absolutely. (As the case d = 3, s_{1} = 3, s_{2} = −1, s_{3} = 2 shows, these conditions are not sufficient.)

Applying Lemma 3 to

∑_{0<n_{1}≤···≤n_{d}} \frac{1}{n_{1}^{s_{1}}n_{2}^{s_{2}} ⋯ n_{d}^{s_{d}}} = ∑_{k_{1},...,k_{d}≥0} ∏_{j=1}^{d} \frac{1}{(k_{1} + k_{2} + ⋯ + k_{j} + 1)^{s_{j}}}

(that is, one chooses s = d, E_{i} = 0 and F_{i} = − ℜ𝔢(s_{i}), i = 1, 2, …, d, there), it is seen that the domain of absolute convergence of this latter multisum is the set of all d-tuples (s_{1}, s_{2}, …, s_{d}) such that

∑_{i=r}^{d} ℜ𝔢(s_{i}) > d − r + 1,  r = 1, 2, …, d.  (3.12)

Moreover, it is not difficult to see that, for the domain of absolute convergence, it does not matter whether we sum the summand on the right-hand side of (3.11) over 0 < n_{1} < ⋯ < n_{d} or over 0 < n_{1} ≤ ⋯ ≤ n_{d}. Therefore, the domain described by the inequalities (3.12) is at the same time the domain of absolute convergence of ζ(s_{d}, s_{d−1}, …, s_{1}). That is, one has to add the conditions (3.12) for i = 2, …, d − 1 to Zhao’s two conditions to obtain a complete description of the domain of absolute convergence. As a matter of fact, all the arguments given in the proof of Proposition 1 in [21] are correct. However, it is only the case d = 2 which is carried out in detail (in which case there are no missing conditions), and therefore the additional d − 2 conditions are overlooked.

Proof of Proposition 1. (i) We consider first the left-hand side of Andrews’ identity (3.1). We write the hypergeometric series as a sum over k. Let S_{k} denote the k-th summand. Since for N ≥ k > |a| we have

|\frac{(−N)_{k}}{(1 + a + N)_{k}} | ≤ \frac{(N − k + 1) ⋯ (N − 1)N}{(N + 1 − |a|)(N + 2 − |a|) ⋯ (N + k − |a|)} ≤ 1,

and since for k > N we have (−N)_{k} = 0, the modulus of (−N)_{k}/(1 + a + N)_{k} is bounded above by a constant for all k = 0, 1, . . .. Hence, using Lemma 1, we obtain that

|S_{k}| ≤ D_{11} · (k + 1)^{−E−1},

where D_{11} is some constant independent of k, and where E is the left-hand side of (3.2). Since, by (3.2), we have E > 0, the absolutely convergent series ∑_{k=0}^{∞} D_{11} · (k + 1)^{−E−1} dominates the hypergeometric series on the left-hand side of (3.1) term-wise. Thus, by Lebesgue’s dominated convergence theorem, we may perform its limit as N → ∞ termwise. This term-wise limit is exactly the left-hand side of (3.4).

Now we consider the right-hand side of (3.1). We need to temporarily assume that

ℜ𝔢(a − b_{s+1} − c_{s+1}) > 0.  (3.13)

(This is slightly stronger than (3.3) with r = s + 1.) Writing A for a − b_{s+1} − c_{s+1}, for any non-negative integer K ≤ N we have

|\frac{(−N)_{K}}{(b_{s+1} + c_{s+1} − a − N)_{K}} | ≤ \frac{N(N − 1) ⋯ (N − K + 1)}{(N + ℜ𝔢(A))(N + ℜ𝔢(A) − 1) ⋯ (N + ℜ𝔢(A) − K + 1)} ≤ 1,

and since for K > N we have (−N)_{K} = 0, the modulus of (−N)_{K}/(b_{s+1} + c_{s+1} − a − N)_{K} is bounded above by a constant for all K = 0, 1, . . .. Thus, again using Lemma 1, the modulus of the summand indexed by k_{1}, k_{2}, …, k_{s} on the right-hand side of (3.1) is bounded above by

D_{12} ∏_{j=1}^{s}(k_{j} + 1)^{ℜ𝔢(a−b_{j}−c_{j})} (k_{1} + ⋯ + k_{j} + 1)^{ℜ𝔢(b_{j}+c_{j}+b_{j+1}+c_{j+1}−2(a+1))},  (3.14)

for some constant D_{12} independent of the summation indices. Now we apply Lemma 3 with E_{j} = ℜ𝔢(a−b_{j}−c_{j}) and F_{j} = ℜ𝔢(b_{j}+c_{j}+b_{j+1}+c_{j+1}−2(a+1)). This is indeed justified since, for this choice of parameters, the set of conditions (3.9) is exactly the set (3.3). Hence, the sum of the expression (3.14) over all k_{1}, …, k_{s} ≥ 0 converges. Another application of Lebesgue’s dominated convergence theorem then implies that we may perform the limit of the multiple sum on the right-hand side of (3.1) as N → ∞ term-wise. Together with the fact that

lim_{N→+∞} \frac{(1 + a)_{N} (1 + a − b_{s+1} − c_{s+1})_{N}}{(1 + a − b_{s+1})_{N} (1 + a − c_{s+1})_{N}} = \frac{Γ(1 + a − b_{s+1}) Γ(1 + a − c_{s+1})}{Γ(1 + a) Γ(1 + a − b_{s+1} − c_{s+1})},

this establishes the identity (3.4), provided (3.13) holds in addition to the conditions of the statement of the proposition.

We can finally get rid of the restriction (3.13) by analytic continuation. Indeed, by using arguments very similar to those above, one can show that both sides of (3.4) are analytic in the parameters a, b_{1}, …, b_{s+1}, c_{1}, …, c_{s+1} as long as (3.2) and (3.3) are satisfied. In particular, in variation of Lemma 1, one would use the fact that, for fixed complex numbers α and β, there are constants D_{13} and D_{14} such that

D_{13} · (k + 1)^{ℜ𝔢(α−β)} log(k + 2) ≤ |\frac{Γ(x + k) ψ(x + k)}{Γ(β + k)}| ≤ D_{14} · (k + 1)^{ℜ𝔢(α−β)} log(k + 2)

for all non-negative integers k and all complex numbers x in a sufficiently small neighbourhood of α, say for |x − α| < 1. Here, ψ(x) denotes the logarithmic derivative of Γ(x).

(ii) In (3.4), we first shift the parameters to b_{j} → b_{j−1} and c_{j} → c_{j−1}, and then we let b_{0} → +∞. The same kind of argument as above then yields (3.8). □

4. Multisum expansions of the Vasilyev-type integral J_{m}

The link between Andrews’ identity and the Vasilyev-type integrals J_{m} becomes apparent in the next proposition.

Proposition 2. Let z, a_{0}, a_{1}, …, a_{m}, and b_{1}, …, b_{m} be complex numbers such that |z| < 1, ℜ𝔢(a_{0}) > 0, ℜ𝔢(b_{i}) > ℜ𝔢(a_{i}) > 0 for all i = 1, 2, …, m.

(i) If m = 2s ≥ 2 is even, then

J_{m} [\begin{matrix}a_{0}, a_{1}, …, a_{m} \\ b_{1}, …, b_{m} \end{matrix}; z] = ∏_{j=1}^{2s} \frac{Γ(a_{j}) Γ(b_{j} − a_{j})}{Γ(b_{j})}

× ∑_{k_{1},k_{2},...,k_{s}≥0} z^{k_{1}+···+k_{s}} ∏_{j=1}^{s} \frac{(b_{2s−2j+2} − a_{2s−2j+2})_{k_{j}}}{k_{j}!} \frac{(a_{2s−2j+1})_{k_{1}+···+k_{j}}}{(b_{2s−2j+1})_{k_{1}+···+k_{j}}} \frac{(a_{2s−2j})_{k_{1}+···+k_{j}}}{(b_{2s−2j+2})_{k_{1}+···+k_{j}}} . (4.1)

This identity holds also for z = 1 provided ℜ𝔢(b_{1} − a_{1}) ≥ ℜ𝔢(a_{0}), and provided (3.9) holds with E_{j} = ℜ𝔢(b_{2s−2j+2} − a_{2s−2j+2} − 1) and F_{j} = ℜ𝔢(a_{2s−2j} + a_{2s−2j+1} − b_{2s−2j+1} − b_{2s−2j+2}), j = 1, 2, …, s.

(ii) If m = 2s + 1 ≥ 3 is odd, then

J_{m} [\begin{matrix}a_{0}, a_{1}, …, a_{m} \\ b_{1}, …, b_{m} \end{matrix}; z] = ∏_{j=1}^{2s+1} \frac{Γ(a_{j}) Γ(b_{j} − a_{j})}{Γ(b_{j})} · ∑_{k_{1},...,k_{s+1}≥0} z^{k_{1}+···+k_{s}} \frac{(a_{2s+1})_{k_{1}}}{k_{1}!} \frac{(a_{2s})_{k_{1}}}{(b_{2s+1})_{k_{1}}}

× ∏_{j=2}^{s+1} \frac{(b_{2s−2j+4} − a_{2s−2j+4})_{k_{j}}}{k_{j}!} \frac{(a_{2s−2j+3})_{k_{1}+···+k_{j}}}{(b_{2s−2j+3})_{k_{1}+···+k_{j}}} \frac{(a_{2s−2j+2})_{k_{1}+···+k_{j}}}{(b_{2s−2j+4})_{k_{1}+···+k_{j}}} .

(4.2)

This identity holds also for z = 1 provided ℜ𝔢(b_{1} − a_{1}) ≥ ℜ𝔢(a_{0}), and provided (3.9) holds with E_{1} = ℜ𝔢(a_{2s+1} − 1), F_{1} = ℜ𝔢(a_{2s} − b_{2s+1}), E_{j} = ℜ𝔢(b_{2s−2j+4} − a_{2s−2j+4} − 1), and F_{j} = ℜ𝔢(a_{2s−2j+2} + a_{2s−2j+3} − b_{2s−2j+3} − b_{2s−2j+4}), j = 2, 3, …, s + 1.

Proof. For m ≥ 2, we denote by Q_{m}(x_{1}, …, x_{m}; z) the nested expression in the denominator of the integrand in (2.1), that is

Q_{m}(x_{1}, …, x_{m}; z) = 1 − (1 − (⋯ (1 − x_{m})x_{m−1}) ⋯)x_{1}z.

(i) We prove the claim by induction on m. For m = 0, the (empty) integral J_{0} [a_{0}; z]

can be consistently interpreted as 1. In order to do the induction step, we fix m = 2s ≥ 2 and z such that |z| < 1. Then, trivially,

Q_{2s}(x_{1}, …, x_{2s}; z) = Q_{2s−2}(x_{1}, …, x_{2s−2}; z) − zx_{1} ⋯ x_{2s−1}(1 − x_{2s})

= Q_{2s−2}(x_{1}, …, x_{2s−2}; z) (1 − \frac{zx_{1} ⋯ x_{2s−1}(1 − x_{2s})}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)}) ,

where for s = 1 the term Q_{0}(−; z) has to be interpreted as 1. Since for x_{j} ∈ [0, 1], we have

|\frac{zx_{1} ⋯ x_{2s−1}(1 − x_{2s})}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)}| ≤ |z| < 1,

we may apply the binomial theorem to obtain

(1 − \frac{zx_{1} ⋯ x_{2s−1}(1 − x_{2s})}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)})^{−a_{0}} = ∑_{k_{1}=0}^{∞} z^{k_{1}} \frac{Γ(a_{0} + k_{1})}{Γ(a_{0})Γ(k_{1} + 1)} (\frac{x_{1} ⋯ x_{2s−1}(1 − x_{2s})}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)})^{k_{1}} .

Hence

J_{2s} [\begin{matrix}a_{0}, a_{1}, …, a_{2s} \\ b_{1}, …, b_{2s} \end{matrix}; z] = ∫_{[0,1]^{2s}} ∑_{k_{1}=0}^{∞} z^{k_{1}} \frac{Γ(a_{0} + k_{1})}{Γ(a_{0})Γ(k_{1} + 1)}x_{2s−1}^{a_{2s−1}+k_{1}−1}(1 − x_{2s−1})^{b_{2s−1}−a_{2s−1}−1}

· x_{2s}^{a_{2s}−1}(1 − x_{2s})^{b_{2s}−a_{2s}+k_{1}−1} \frac{∏_{j=1}^{2s−2} x_{j}^{a_{j}+k_{1}−1}(1 − x_{j})^{b_{j}−a_{j}−1}}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)^{k_{1}+a_{0}}} dx_{1} ⋯ dx_{2s}.

The conditions on the parameters ensure that the integral

∫_{[0,1]^{2s}} ∑_{k_{1}=0}^{∞} |z^{k_{1}} \frac{Γ(a_{0} + k_{1})}{Γ(a_{0})Γ(k_{1} + 1)}x_{2s−1}^{a_{2s−1}+k_{1}−1}(1 − x_{2s−1})^{b_{2s−1}−a_{2s−1}−1}

· x_{2s}^{a_{2s}−1}(1 − x_{2s})^{b_{2s}−a_{2s}+k_{1}−1} \frac{∏_{j=1}^{2s−2} x_{j}^{a_{j}+k_{1}−1}(1 − x_{j})^{b_{j}−a_{j}−1}}{Q_{2s−2}(x_{1}, …, x_{2s−2}; z)^{a_{0}+k_{1}}} | dx_{1} ⋯ dx_{2s}

is convergent. Thus, we can exchange the integral and the summation, and, using the beta integral evaluation and some straightforward simplifications, we obtain

J_{2s} [\begin{matrix}a_{0}, a_{1}, …, a_{2s} \\ b_{1}, …, b_{2s} \end{matrix}; z] = \frac{Γ(a_{2s}) Γ(a_{2s−1}) Γ(b_{2s} − a_{2s}) Γ(b_{2s−1} − a_{2s−1})}{Γ(b_{2s}) Γ(b_{2s−1})}

· ∑_{k_{1}=0}^{∞} z^{k_{1}} \frac{(b_{2s} − a_{2s})_{k_{1}} (a_{2s−1})_{k_{1}} (a_{0})_{k_{1}}}{k_{1}! (b_{2s})_{k_{1}} (b_{2s−1})_{k_{1}}} J_{2s−2} [\begin{matrix}a_{0} + k_{1}, a_{1} + k_{1}, …, a_{2s−2} + k_{1} \\ b_{1} + k_{1}, …, b_{2s−2} + k_{1} \end{matrix}; z] . (4.3)

If we substitute the induction hypothesis for J_{2s−2}, we arrive exactly at (4.1).

We now perform the limit z → 1: since the conditions on the parameters guarantee that the integral J_{2s} is absolutely convergent for z = 1, dominated convergence implies that one can interchange limit and integral. Similarly, if we put z = 1 in the above multiple sum, then the conditions on the parameters allow us to apply Lemmas 1 and 3 and to conclude that it converges absolutely. Thus, again, dominated convergence implies that we may interchange limit and summation. As a result, Case (i) of Proposition 2 is now completely proved.

(ii) We do not provide all the details for the case where m is odd, m = 2s + 1 ≥ 3, since this case can be treated in a rather similar manner as the case where m is even. A main difference, however, is that, to get started, we use the alternative identity

Q_{2s+1}(x_{1}, …, x_{2s+1}; z) = Q_{2s}(x_{1}, …, x_{2s}; z) (1 − \frac{zx_{1} ⋯ x_{2s+1}}{Q_{2s}(x_{1}, …, x_{2s}; z)}) ,

which implies the expansion

J_{2s+1} [\begin{matrix}a_{0}, a_{1}, …, a_{2s+1} \\ b_{1}, …, b_{2s+1} \end{matrix}; z]

= \frac{Γ(a_{2s+1}) Γ(b_{2s+1} − a_{2s+1})}{Γ(b_{2s+1})} · ∑_{k=0}^{∞} z^{k} \frac{(a_{0})_{k} (a_{2s+1})_{k}}{k! (b_{2s+1})_{k}} J_{2s} [\begin{matrix}a_{0} + k, a_{1} + k, …, a_{2s} + k \\ b_{1} + k, …, b_{2s} + k \end{matrix}; z] . (4.4)

At this point, we substitute the multiple series (4.1) for J_{2s}, and after some simple manipulations we arrive at (4.2). □

Remark. Both of the recursive formulas (4.3) and (4.4) appear already earlier in the article [22] of Zlobin which we mentioned in the Introduction. He used them to express the integrals J_{m} in terms of another family of integrals, like those considered by Sorokin in [15, 16].

5. Proof of Theorem 1

We are now in the position to prove Zudilin’s theorem, by putting together the identities in Proposition 2 and 1. Because of the use of Proposition 2 when z = 1, we shall need to temporarily impose stronger conditions on the parameters than required by the assertion of the theorem. We shall do this without mention. One gets rid of these restrictions at the end by analytic continuation.

Let first m be even, m = 2s. We apply Proposition 2, Eq. (4.1), with a_{j−1} = h_{j}, j = 1, 2, …, 2s + 1, b_{j} = 1 + h_{0} − h_{j+2}, j = 1, 2, …, 2s. Thus, using (4.1), we express the integral on the left-hand side of (2.3) in terms of a multiple sum. If we subsequently apply the identity (3.4) with b_{j} = h_{2s−2j+4}, c_{j} = h_{2s−2j+3} for j = 1, 2, …, s + 1, to the multiple sum, then we arrive at the very-well-poised hypergeometric series on the right-hand side of (2.3).

Similarly, if m is odd, m = 2s+1, then we apply Proposition 2, Eq. (4.2), with a_{j−1} = h_{j}, j = 1, 2, …, 2s + 2, b_{j} = 1 + h_{0} − h_{j+2}, j = 1, 2, …, 2s + 1. Thus, using (4.2), we express the integral on the left-hand side of (2.3) in terms of a multiple sum. If we subsequently apply the identity (3.8) with s replaced by s + 1, b_{j} = h_{2s−2j+4}, j = 1, 2, …, s + 1, c_{j} = h_{2s−2j+3} for j = 0, 1, …, s + 1, to the multiple sum, then we arrive at the very-well-poised hypergeometric series on the right-hand side of (2.3).

Acknowledgement

We thank Wadim Zudilin for an attentive reading of an earlier version of the paper, and, in particular, for several useful suggestions for improvement of the exposition.

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Institut Girard Desargues, Université Claude Bernard Lyon-I, 21, avenue Claude Bernard, F-69622 Villeurbanne Cedex, France

E-mail address: kratt@euler.univ-lyon1.fr

Laboratoire de Mathématiques Nicolas Oresme, CNRS UMR 6139, Université de Caen, BP 5186, 14032 Caen cedex, France

E-mail address: rivoal@math.unicaen.fr