Multiple zeta values, Padé approximation and Vasilyev's conjecture
aa r X i v : . [ m a t h . N T ] S e p MULTIPLE ZETA VALUES, PAD ´E APPROXIMATION ANDVASILYEV’S CONJECTURE
S. FISCHLER AND T. RIVOAL
Abstract.
Sorokin gave in 1996 a new proof that π is transcendental. It is based ona simultaneous Pad´e approximation problem involving certain multiple polylogarithms,which evaluated at the point 1 are multiple zeta values equal to powers of π . In this paperwe construct a Pad´e approximation problem of the same flavour, and prove that it hasa unique solution up to proportionality. At the point 1, this provides a rational linearcombination of 1 and multiple zeta values in an extended sense that turn out to be valuesof the Riemann ζ function at odd integers. As an application, we obtain a new proofof Vasilyev’s conjecture for any odd weight, concerning the explicit evaluation of certainhypergeometric multiple integrals; it was first proved by Zudilin in 2003. Introduction
The goal of this paper is to provide a completely new proof of Vasilyev’s conjecturefor any odd weight d ≥ d ≥ n ≥ J d,n := Z [0 , d Q dj =1 x nj (1 − x j ) n Q d ( x , . . . , x d ) n +1 d x j ∈ Q + Q ζ (2 + e d ) + Q ζ (4 + e d ) + · · · + Q ζ ( d ) (1.1)where e d = 0 if d is even, e d = 1 otherwise, and Q ( x ) := 1 − x , Q d ( x , . . . , x d ) : = 1 − Q d − ( x , . . . , x d − ) x d , d ≥
2= 1 − (1 − ( · · · − (1 − x ) x · · · ) x d − ) x d . This conjecture was already known to be true for d = 2 and d = 3, since Beukers [3]used these integrals to get new and quick versions of Ap´ery’s proofs [1] of the irrationalityof ζ (2) and ζ (3). Vasilyev himself proved his conjecture in the cases d = 4 and d = 5,results which in fact led him to the conjecture. The first complete proof was given byZudilin [18] who showed that J d,n is equal to a very-well-poised hypergeometric serieswhose value was already known to be in Q + Q ζ (2 + e d ) + Q ζ (4 + e d ) + · · · + Q ζ ( d ). Twoother proofs of Vasilyev’s conjecture were subsequently found, one by Zlobin [16] (directattack) and another indirect one by Krattenthaler-Rivoal [10] (limiting case of Andrews’ Date : November 1, 2018.2010
Mathematics Subject Classification.
Primary 11M32, 41A21; Secondary 11J72, 33C60.
Key words and phrases.
Multiple zeta value, Pad´e approximation, Hypergeometric integrals,Polylogarithm. hypergeometric identity, in the spirit of Zudilin). The fourth one, given in the presentpaper, is completely different since it relies on solving a simultaneous Pad´e approximationproblem involving multiple polylogarithms.To state this problem we need some notations. Given any finite word σ built on a(possibly infinite) alphabet { a, b, . . . } , we denote by { σ } j := σσ · · · σ the concatenation j times of σ . By convention, { σ } = ∅ . We will use two alphabets, namely N ∗ = { , , . . . } and { ℓ, s } . We consider multiple polylogarithms in the following extended sense:Li a a ··· a p − b b ··· b p ( z ) := X k & k & ··· & k p ≥ z k k b k b · · · k b p p (1.2)where | z | < b j ∈ N ∗ and a j ∈ { ℓ, s } for all j . For j = 1 , . . . , p −
1, the symbol & ∈ { >, ≥} in k j & k j +1 is determined by the following rule: it is set to > if a j = s , andto ≥ if a j = ℓ . In this way, s stands for a strict inequality, and ℓ for a large one. If a j = s for any j we obtain the usual multiple polylogarithm Li b b ··· b p ( z ); if a j = ℓ for any j weobtain the variant denoted by La b b ··· b p ( z ) in [4] and by Le b b ··· b p ( z ) by Ulanski˘ı and Zlobin.Sorokin used in [14] the functions Li { sℓ } r { } r +1 (1 − x ) and Li { ℓs } r − ℓ { } r (1 − x ), which he denotedrespectively by ε r ( x ) and ϕ r ( x ). In this paper, all multiple polylogarithms Li a a ··· a p − b b ··· b p ( z )will be considered for z ∈ C \ [1 , ∞ ) using analytic continuation. As usual, the integer p in (1.2) is called the depth , and b + · · · + b p is the weight .Our main result is the explicit resolution of the following simultaneous Pad´e approxi-mation problem. Given integers n, r ≥
0, we want to find polynomials A ρ,r,n ( z ), B ρ,r,n ( z ), C ρ,r,n ( z ), D r,n ( z ) ∈ C [ z ], for 0 ≤ ρ ≤ r , all of degree at most n , such that S r,n ( z ) := r X ρ =0 (cid:20) A ρ,r,n ( z ) Li { ℓs } ρ ℓ { } ρ +1 (cid:16) z (cid:17) + B ρ,r,n ( z ) Li { ℓs } ρ ℓ { } ρ +2 (cid:16) z (cid:17) + C ρ,r,n ( z ) Li { sℓ } ρ { } ρ +1 (cid:16) z (cid:17)(cid:21) + D r,n ( z ) = O (cid:16) z ( r +1)( n +1) (cid:17) U j,r,n ( z ) := r X ρ = j A ρ,r,n ( z ) Li { ℓ } r − ρ { } r − ρ (1 − z ) + B j,r,n ( z ) = O (cid:0) (1 − z ) n +1 (cid:1) , j = 0 , . . . , rV j,r,n ( z ) := r X ρ = j A ρ,r,n ( z ) Li { ℓ } r − ρ { } r − ρ +1 (1 − z ) + C j,r,n ( z ) = O (cid:0) (1 − z ) n +1 (cid:1) , j = 0 , . . . , r. We will denote by P r,n this Pad´e approximation problem. The various symbols O havethe following meaning. The function S r,n ( z ) is obviously analytic at z = ∞ and we ask itsorder there to be at least ( r + 1)( n + 1). Similarly, the functions U j,r,n ( z ) and V j,r,n ( z ) areanalytic at z = 1 and we ask their orders there to be at least n + 1. This is a mixed Pad´eapproximation problem, namely in between type I problems and type II problems. Similarmixed Pad´e approximation problems often occur in the Diophantine theory of (multiple)zeta values; see for instance [8, 13, 14]. The problem P r,n can be trivially converted into a linear algebra problem: it amountsto solving a system of (3 r + 4)( n + 1) − r + 4)( n + 1) unknowns (thecoefficients of the polynomials). Hence, there is at least one non identically zero solution.Our main theorem shows that the solution is unique up to a multiplicative constant. Theorem 1.
For any integers n, r ≥ , the function S r,n ( z ) in P r,n is given by the followinghypergeometric integral (up to a multiplicative constant), which converges for any z ∈ C \ [0 , : S r,n ( z ) = ( − n +1 z ( r +1)( n +1) × Z [0 , r +3 u ( r +1)( n +1) − (1 − u ) n r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) ( z − u u v · · · u j − v j − u j ) n +1 ( z − u u v · · · u j v j ) n +1 (cid:1) d u d v . (1.3)For r = 0, the problem P ,n and the integral for S ,n ( z ) exactly match those consideredby Sorokin in [13], from which he deduced a new proof of Ap´ery’s theorem. However, ourderivation of the integral for S ,n ( z ) is different from Sorokin’s.For any r ≥
0, the integral representation (1.3) provides a new proof of Vasilyev’sconjecture, by taking z = 1 (see § ζ (2 r + 1) (see [2, 11]) bysolving a Sorokin-type Pad´e problem involving multiple polylogarithms as in Theorem 1,as Sorokin did [13] for Ap´ery’s theorem (see § π , which relies onthe resolution of a simultaneous Pad´e approximation problem involving certain multiplepolylogarithms (see § { s } r − { } r (1) = π r (2 r +1)! for anyinteger r ≥ S r,n ( z ) can be used to get explicit expression of the polynomials, all ofwhich obviously have rational coefficients. This can be done by various means, for instanceone can convert the integral into the series S r,n ( z ) = n ! X k ≥···≥ k r +1 ≥ ( k − k + 1) n ( k − k + 1) n · · · ( k r − k r +1 + 1) n ( k r +1 − n ) nr Y j =0 (cid:0) ( k j + ( r − j + 1)( n + 1)) e j n +1 ( k j +1 + ( r − j )( n + 1)) n +1 (cid:1) z k + r ( n +1) (where e = 2 and e j = 1 for j ≥
1) and then use the algorithm described in [4].The paper is organised as follows. In §
2, we deduce Vasilyev’s conjecture for odd valuesof d from Theorem 1. In §
3, we present a few tools needed for the proof of Theorem 1, in particular an iterative construction of hypergeometric multiple integrals. In §
4, weprove an important representation formula for multiple polylogarithms and derive a fewconsequences useful in the resolution of P r,n . Section 5, devoted to the proof of Theorem 1,is decomposed in many steps. The first two steps show how to reduce the problem P r,n toSorokin’s problem for π (recalled in § § ζ (2 r + 1) are irrational [2, 11].2. New proof of Vasilyev’s conjecture for odd weights
To deduce Vasilyev’s conjecture from Theorem 1, we first define (when b ≥
2) extendedmultiple zeta values by ζ a a ··· a p − b b ··· b p := Li a a ··· a p − b b ··· b p (1) = X k & k & ··· & k p ≥ k b k b · · · k b p p (2.1)with the same definition for the symbols & as in Eq. (1.2). In particular, when a j = s forall j , we have the usual multiple zeta values ζ { s } p − b b ··· b p = ζ ( b , b , . . . , b p ).Then we remark that the Pad´e conditions for the functions U j,r,n ( z ) and V j,r,n ( z ) in P r,n ensure that all polynomials B j,r,n ( z ) and C j,r,n ( z ) vanish at z = 1 ( j = 0 , . . . , r ). Sincemultiple polylogarithms have (at most) a logarithmic singularity at z = 1, this implies thatwhen we take the limit z → − n +1 Z [0 , r +3 u ( r +1)( n +1) − (1 − u ) n r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) (1 − u u v · · · u j − v j − u j ) n +1 (1 − u u v · · · u j v j ) n +1 (cid:1) d u d v = r X ρ =0 A ρ,r,n (1) ζ { ℓs } ρ ℓ { } ρ +1 + D r,n (1)where A ρ,r,n (1) and D r,n (1) are rational numbers. Moreover, it is proved in [6, Corollaire 8]that this multiple integral is equal to J r +3 ,n for any integer r ≥ § { s } r − { } r (1) = π r (2 r +1)! for Sorokinin [14]. Proposition 1.
For any integer k ≥ , we have ζ { ℓs } k − ℓ { } k − = ζ { ℓ } k { } k = 2 ζ (2 k + 1) . (2.2) Proof.
The second equality in (2.2) is due to Zlobin [17]. To prove the first equality, whichwe haven’t found in the literature, we use the representation of (extended) multiple zeta values as Chen iterated integrals. Indeed, we have ζ { ℓs } k − ℓ { } k − = Z { ≤ x k +1 ≤···≤ x ≤ } d x x x (1 − x )(1 − x ) x (1 − x )(1 − x ) · · · x k (1 − x k )(1 − x k +1 )= Z { ≤ y k +1 ≤···≤ y ≤ } d y y y (1 − y ) y y (1 − y ) y · · · y k (1 − y k )(1 − y k +1 ) = ζ { ℓ } k { } k , where we have made the change of variables x j = 1 − y k +2 − j , j = 1 , . . . , k + 1. (cid:3) General results on multiple polylogarithms
We gather in this section various results, useful in the proof of Theorem 1 but whichmay also be of independent interest.3.1.
Differentiation rules for multiple polylogarithms.
In this section, we describehow to differentiate a multiple polylogarithm. To begin with, we state formulas of whichthe proofs are straightforward; we will use them without further mentions. The letter a denotes a finite word built on the alphabet { ℓ, s } , the letter b a finite word built on thealphabet N ∗ , and t any integer ≥ z Li ( z ) = 11 − z , dd z h Li (cid:16) z (cid:17)i = 1 z (1 − z ) , dd z Li ℓ a b ( z ) = 1 z (1 − z ) Li ab ( z ) , dd z h Li ℓ a b (cid:16) z (cid:17)i = 11 − z Li ab (cid:16) z (cid:17) , dd z Li ℓ a t b ( z ) = 1 z Li ℓ a ( t − b ( z ) , dd z h Li ℓ a t b (cid:16) z (cid:17)i = − z Li ℓ a ( t − b (cid:16) z (cid:17) , dd z Li s a b ( z ) = 11 − z Li ab ( z ) , dd z h Li s a b (cid:16) z (cid:17)i = 1 z (1 − z ) Li ab (cid:16) z (cid:17) , dd z Li s a t b ( z ) = 1 z Li s a ( t − b ( z ) , dd z h Li s a t b (cid:16) z (cid:17)i = − z Li s a ( t − b (cid:16) z (cid:17) . We now state a general lemma, whose proof can be done by induction using the formulasabove.
Lemma 1.
Let d, n ≥ , and A ( z ) ∈ C [ z ] be a polynomial of degree ≤ d . Then we have d n +1 d z n +1 (cid:0) A ( z ) Li a a ··· a p − b b ··· b p ( z ) (cid:1) = p +1 X i =0 b i X b ′ =1 b A i,b ′ ( z ) z n +1 (1 − z ) n +1 Li a i a i +1 ··· a p − b ′ b i +1 b i +2 ··· b p ( z ) for some polynomials b A i,b ′ ( z ) of degree ≤ d + n + 1 ; here we let b p +1 = 1 so that in the sumthere is one term corresponding to i = p + 1 , and the associated polylogarithm is equal to 1. It is not difficult to see that in this lemma, each polynomial b A i,b ′ ( z ) depends only on b ,. . . , b i − , a , . . . , a i − , and b i − b ′ . However we won’t use this remark in the present paper.Using the above relations in the same way, an analogous lemma yields polynomials b A ′ i,b ′ ( z ) of degree ≤ d + n + 1 such thatd n +1 d z n +1 (cid:0) A ( z ) Li a a ··· a p − b b ··· b p (1 /z ) (cid:1) = p +1 X i =0 b i X b ′ =1 b A ′ i,b ′ ( z ) z n +1 (1 − z ) n +1 Li a i a i +1 ··· a p − b ′ b i +1 b i +2 ··· b p (1 /z ) . To take advantage of vanishing conditions like the ones on U j,r,n ( z ) and V j,r,n ( z ) in thePad´e problem P r,n , the following lemma is very useful. Lemma 2.
Let n ′ ≥ , and g ( z ) be a function holomorphic at z = 1 , such that g ( z ) = O (cid:0) ( z − n +1 (cid:1) as z → . Then we have d n +1 d z n +1 (cid:0) g ( z ) Li a a ··· a p − b b ··· b p ( z ) (cid:1) = p +1 X i =0 b i X b ′ =1 h i,b ′ ( z ) Li a i a i +1 ··· a p − b ′ b i +1 b i +2 ··· b p ( z ) for some functions h i,b ′ ( z ) holomorphic at z = 1 . As in Lemma 1, we let b p +1 = 1 so thatin the sum there is one term corresponding to i = p + 1 , and the associated polylogarithmis equal to 1. In other words, no pole appears at z = 1 if g vanishes to order at least n + 1 at thispoint (since polylogarithms have at most a logarithmic divergence at 1).3.2. An integral operator.
Sorokin solved several Pad´e approximation problems involv-ing multiple polylogarithms (see [13] and [14], amongst other papers), which always led tohypergeometric multiple integrals. We define now an integral operator intimately relatedto his approach (and therefore also to Theorem 1).Given integers a, b, n ≥ F ( z ), we let H n +1 a,b ( F )( z ) = ( − n +1 z n +1 − a Z u a + b − n − (1 − u ) n ( u − z ) b F (cid:16) zu (cid:17) d u. (3.1)The assumptions on F and the properties of the function H n +1 a,b ( F ) defined in this way aredetailed in the following lemma. Lemma 3.
Let F ( z ) be holomorphic on C \ [0 , and at z = ∞ ; denote by ω ≥ its orderof vanishing at ∞ . Given a, b, n ≥ , let ω ′ = ω + a + b − n − and assume that ω ′ ≥ .Then H n +1 a,b ( F ) is holomorphic on C \ [0 , and at z = ∞ ; its order of vanishing at ∞ is exactly ω ′ . Moreover, ( i ) Letting R = H n +1 a,b ( F ) , we have F ( z ) = 1 n ! z a (1 − z ) b R ( n +1) ( z ) . (3.2)( ii ) If R ( z ) is a function holomorphic on C \ [0 , and at z = ∞ such that R ( ∞ ) = 0 and Eq. (3.2) holds, then R = H n +1 a,b ( F ) . We shall apply this lemma in two cases: either F ( ∞ ) = 0 and a + b ≥ n + 1, or F isthe constant function F ( z ) = 1 and a + b ≥ n + 2. In both cases we have ω ′ ≥
1, so that H n +1 a,b ( F ) is holomorphic on C \ [0 ,
1] and at z = ∞ , and H n +1 a,b ( F )( ∞ ) = 0. Proof.
Let G ( z ) = z ω F ( z ); then G ( z ) is holomorphic on C \ [0 ,
1] and at ∞ , with G ( ∞ ) = 0.By definition of ω ′ we have H n +1 a,b ( F )( z ) = ( − n +1 z − ω ′ Z u ω ′ − (1 − u ) n ( uz − b G (cid:16) zu (cid:17) d u. Since ω ′ ≥ u/z = 1 for any u ∈ [0 ,
1] (since z ∈ C \ [0 , H n +1 a,b ( F ) is holomorphic on C \ [0 ,
1] and at z = ∞ . It has order equal to ω ′ at ∞ because G ( ∞ ) = 0.To prove ( i ) and ( ii ), we perform the change of variable x = z/u and deduce H n +1 a,b ( F )( z ) = ( − n +1 Z ∞ z ( x − z ) n x a (1 − x ) b F ( x )d x. Then assertions ( i ) and ( ii ) follow immediately from the following lemma, obtained fromthe arguments given in [12, p. 60]. (cid:3) Lemma 4.
Let
R, S be functions analytic on a neighborhood of ∞ , with R ( ∞ ) = 0 . Then: n ! R ( n +1) ( z ) = S ( z ) ⇐⇒ R ( z ) = ( − n +1 Z ∞ z ( x − z ) n S ( x )d x. For Diophantine applications the value H n +1 a,b ( F )(1) is often the most interesting one;conditions for this value to exist are given by the following lemma, whose proof is straight-forward. Lemma 5.
Assume that b ≤ n + 1 and F ( z ) has (at most) a power of logarithm divergenceas z → , with z ∈ C \ [0 , . Then H n +1 a,b ( F )( z ) has also (at most) a power of logarithmdivergence as z → , with z ∈ C \ [0 , .Moreover, if in addition b ≤ n then H n +1 a,b ( F )( z ) has a finite limit as z → , with z ∈ C \ [0 , , and this limit is given by taking z = 1 in the integral representation ofEq. (3.1) , which is then convergent. In Pad´e approximation problems with multiple polylogarithms, multiple integrals appearby applying successively integral operators H n +1 a,b with various parameters. We shall write H n +1 a,b H n ′ +1 a ′ ,b ′ for H n +1 a,b ◦ H n ′ +1 a ′ ,b ′ , so that H n +1 a,b H n ′ +1 a ′ ,b ′ ( F ) = H n +1 a,b ( H n ′ +1 a ′ ,b ′ ( F )). We shall considerin § § H n +1 a ,b H n +1 a ,b · · · H n p +1 a p ,b p ( ) , where the a j , b j , n j are non-negative integers and denotes the function equal to 1 on C \ [0 , C \ [0 ,
1] and at z = ∞ , and Lemma 5 plays the analogous role for the behaviour at z = 1. In the proof of Theorem 1 we shall use the following result which describes the behaviourof this integral operator under the change of variable z − z . Lemma 6.
For any integers a j , b j , n j , j = 1 , . . . , p such that H n +1 a ,b H n +1 a ,b · · · H n p +1 a p ,b p ( ) isholomorphic on C \ [0 , and at ∞ , we have H n +1 a ,b H n +1 a ,b · · · H n p +1 a p ,b p ( )(1 − z ) = ( − p + n + n + ··· + n p H n +1 b ,a H n +1 b ,a · · · H n p +1 b p ,a p ( )( z ) for all z ∈ C \ [0 , .Proof. This is a consequence of the following fact. Given f ( z ), we set f ∂ ( z ) := f (1 − z ).Then R ( z ) = H n +1 a,b ( S )( z ) ⇐⇒ R ∂ ( z ) = ( − n +1 H n +1 b,a ( S ∂ )( z ) . This equivalence results from Lemma 3: S ( z ) = 1 n ! z a (1 − z ) b R ( n +1) ( z ) ⇐⇒ S (1 − z ) = ( − n +1 n ! z b (1 − z ) a (cid:0) R (1 − z ) (cid:1) ( n +1) . (cid:3) Functional linear independence of polylogarithms.
The extended multiple po-lylogarithms introduced in the introduction are very useful to state and prove our result,but they are not really new functions: they are linear combinations over Z of usual multiplepolylogarithms (corresponding to α = . . . = α p − = s in (1.2)). This follows from thefollowing elementary relation (which is the starting point of [4]):Li a ··· a j − ℓa j +1 ··· a p − b b ··· b p ( z ) = Li a ··· a j − sa j +1 ··· a p − b b ··· b p ( z ) + Li a ··· a j − a j +1 ··· a p − b ··· b j − b ′ b j +2 ··· b p ( z ) (3.3)where b ′ = b j + b j +1 .In the proof of Theorem 1 we shall use the following result. Lemma 7.
For any k , let a k be a word on the alphabet { ℓ, s } of length k − , with a = a = ∅ . Then the polylogarithms Li a k { } k (1 /z ) , for k ≥ , are linearly independent over thefield M of functions meromorphic at 1.Proof. To begin with, let us consider for any p ≥ F p of all functions analytic on C \ [0 ,
1] that can be written as P pi =0 h i ( z )(log(1 − z )) i where h ( z ), . . . , h p ( z ) are functionsholomorphic on C \ [0 ,
1] and at z = 1. Of course all functions holomorphic on C \ [0 , z = 1 belong to F , and Li (1 /z ) = − log(1 − z ) belongs to F . We claim that forany p ≥
0, for any α , . . . , α p − ∈ { ℓ, s } and any b , . . . , b p ≥
1, we haveLi a ··· a p − b b ··· b p (1 /z ) ∈ F p . Let us prove this claim by induction on the weight b + · · · + b p . We have already noticedthat it holds if b + · · · + b p ≤
1. Now remark that if f is analytic on C \ [0 ,
1] and g ∈ F p are such that f ′ ( z ) = − z g ( z ) then f ∈ F p , because F p is stable under primitivationand products with functions holomorphic at 1. On the other hand, if f ′ ( z ) = − z g ( z ) or f ′ ( z ) = z (1 − z ) g ( z ) then f ∈ F p +1 . Using the differentiation rules for polylogarithms statedat the beginning of § Now assume that for some k ≥ a k { } k (1 /z ) is a linear combination over M of the Li a j { } j (1 /z ) for 0 ≤ j ≤ k −
1. Using the claim this implies Li a k { } k (1 /z ) ∈ F k − .Now applying Eq. (3.3) as many times as needed one can write Li a k { } k (1 /z ) − Li { s } k − { } k (1 /z )as a Z -linear combination of extended multiple polylogarithms of depth k −
1; applyingthe claim again proves that Li { s } k − { } k (1 /z ) = ( − k (cid:0) log(1 − z ) (cid:1) k belongs to F k − (thisidentity belongs to the folklore and is readily proved by induction and differentiation).But this provides a non-trivial linear relation, with coefficients holomorphic at 1, betweenpowers of the function log(1 − z ). This is impossible since log( z ) is transcendental overthe field of functions meromorphic at the origin. This contradiction concludes the proof ofLemma 7. (cid:3) Weight functions of multiple polylogarithms
In this section we study the weight functions of multiple polylogarithms and computesome of them. This part is at the heart of the proof of Theorem 1, since weights obey thesame derivation rules as the corresponding polylogarithms (see below).If b = ∅ , Li a ∅ ( z ) = 1 / (1 − z ) and none of the considerations below apply. From now on,we consider non-empty words b . It is well-known that usual multiple polylogarithms Li a b ( z )(with a = ss · · · s ) can be analytically continued to the cut plane C \ [1 , + ∞ ). They vanishat z = 0 and their growth as z → ∞ is at most a power of log( z ), with 0 < arg( z ) < π .Moreover, the function defined on the cut bylim y → [Li ss ··· s b ( x + iy ) − Li ss ··· s b ( x − iy )]is C ∞ on (1 , + ∞ ) with at most a (power of) logarithm singularity at x = 1 and x = ∞ .All these properties also hold for Li a b ( z ) for any word a because such functions are simplylinear combinations with rational coefficients of the Li ss ··· s b ( z ) (using repeatedly Eq. (3.3)above).As an (important) application, we prove the following lemma. Lemma 8.
For any fixed z ∈ C \ [0 , , any a and any b = ∅ , we have Li a b (cid:18) z (cid:19) = Z ω a b ( x ) z − x d x, (4.1) where ω a b ( x ) := 12 iπ lim y → (cid:20) Li a b (cid:18) x + iy (cid:19) − Li a b (cid:18) x − iy (cid:19)(cid:21) ∈ L ([0 , . (4.2) The weight function ω a b ( x ) is C ∞ on (0 , , with at most (power of ) logarithm singularitiesat x = 0 and x = 1 .Proof. For any fixed z ∈ C \ [1 , + ∞ ), let us consider the Cauchy representation formulaLi a b ( z ) = z iπ Z C Li a b ( t ) t ( t − z ) d t, where C is any simple closed curve surrounding z and not crossing the cut [1 , + ∞ ). Wecan deform C to a simple closed curve defined as follows: given ε > R > | z | < R ), we glue together two straightlines [1 + iε, R + iε ], [1 − iε + R, R − iε ], asemi-circle of center 1 and diameter [1 − iε, iε ] and an arc of circle of center 0 passingthrough R + iε and R − iε (both arcs not crossing [1 , + ∞ )). The analytic properties ofLi a b ( z ) are such that we can let ε → R → ∞ to get the representationLi a b ( z ) = z Z ∞ ω a b (1 /t ) t ( t − z ) d t = z Z ω a b ( x )1 − zx d x (by letting x = 1 /t ) , where ω a b ( x ) is defined by (4.2). We obtain (4.1) by changing z to 1 /z . (cid:3) (This proof is not specific to multiple polylogarithms. Such weighted integral represen-tations are known as Stieltjes representations; see [9, p. 591, Theorem 12.10d].)We note two important consequences of the expression (4.2) for ω a b ( x ). To begin with,if dd z (cid:20) Li a b (cid:18) z (cid:19)(cid:21) = R ( z ) Li a ′ b ′ (cid:18) z (cid:19) , then dd x ω a b ( x ) = R ( x ) ω a ′ b ′ ( x )where the function R ( z ) is one of − z , − z and 1 z (1 − z ) (see § weights obey the same derivation rules as the corresponding polylogarithms. This observa-tion will be crucial in § a b (1) is finite,then ω a b (1) = 0. Lemma 9.
For any x ∈ (0 , and any integer k ≥ , we have ω { ℓs } k − ℓ { } k ( x ) = Li { sℓ } k − { } k − ( x ) , (4.3) ω { sℓ } k { } k +1 ( x ) = Li { ℓs } k − ℓ { } k ( x ) , (4.4) and ω { ℓs } k ℓ { } k +1 ( x ) = k X j =0 Li { ℓ } j { } j (1 − x ) Li { sℓ } k − j { } k − j +1 ( x )+ k +1 X j =1 Li { ℓ } j − { } j (1 − x ) Li { ℓs } k − j ℓ { } k − j +2 ( x ) (4.5)= − Li { sℓ } k { } k ( x ) + Li { ℓ } k { } k +1 (1) . (4.6) Proof.
Equations (4.3) and (4.4) are readily checked by expanding z − x = P ∞ n =0 x n z n +1 inthe integral (4.1). To prove (4.5), we remark that both sides differentiate to the samefunction − x ω { ℓs } k ℓ { } k +2 ( x ) = − x Li { sℓ } k { } k +1 ( x ), since all functions but this precise one are killedby telescoping when differentiating the right hand side of (4.5). It follows that the functionson both sides of (4.5) differ only by a constant. This constant must be 0 because bothsides vanish at x = 1 (see the remark just before Lemma 9). The same argument yieldsalso ω { ℓs } k ℓ { } k +1 ( x ) = − Z x Li { sℓ } k { } k +1 ( x )d x = − Li { sℓ } k { } k ( x ) + C k for some constant C k . This constant is seen to be equal to Li { ℓ } k { } k +1 (1) by taking x = 0in (4.5). This proves (4.6), and concludes the proof of Lemma 9. (cid:3) In the setting of the Pad´e problem P r,n , we define the function P r,n ( z ) = r X ρ =0 (cid:20) A ρ,r,n ( z ) ω { ℓs } ρ ℓ { } ρ +1 ( z ) + B ρ,r,n ( z ) ω { ℓs } ρ ℓ { } ρ +2 ( z ) + C ρ,r,n ( z ) ω { sℓ } ρ { } ρ +1 ( z ) (cid:21) obtained from S r,n by replacing every polylogarithm with its weight (see Lemma 11 below).By (4.3), (4.4) and (4.6), this function P r,n is analytic on the disk | z | <
1, with a (powerof) logarithm singularity at z = 1. In particular, it is in L ([0 , U j,r,n ( z ) and V j,r,n ( z ).As in the rest of the paper, we continue analytically all polylogarithms to C \ [1 , + ∞ ). Lemma 10.
For any z ∈ C \ [1 , + ∞ ) , P r,n ( z ) = r X j =0 (cid:20) U j,r,n ( z ) Li { sℓ } j { } j +1 ( z ) + V j,r,n ( z ) Li { ℓs } j − ℓ { } j ( z ) (cid:21) . We conclude this section with the precise connection between P r,n ( z ) and S r,n ( z ). Lemma 11.
In the setting of the Pad´e problem P r,n , for any z ∈ C \ [0 , we have S r,n ( z ) = Z P r,n ( x ) z − x d x. Proof.
By definition of S r,n ( z ) and Lemma 8, for any z ∈ C \ [0 ,
1] we have S r,n ( z ) = r X ρ =0 (cid:20) A ρ,r,n ( z ) Z ω { ℓs } ρ ℓ { } ρ +1 ( x ) z − x d x + B ρ,r,n ( z ) Z ω { ℓs } ρ ℓ { } ρ +2 ( x ) z − x d x + C ρ,r,n ( z ) Z ω { sℓ } ρ { } ρ +1 ( x ) z − x d x (cid:21) + D r,n ( z )= Z P r,n ( x ) z − x d x + r X ρ =0 Z (cid:20) A ρ,r,n ( z ) − A ρ,r,n ( x ) z − x ω { ℓs } ρ ℓ { } ρ +1 ( x )+ B ρ,r,n ( z ) − B ρ,r,n ( x ) z − x ω { ℓs } ρ ℓ { } ρ +2 ( x ) + C ρ,r,n ( z ) − C ρ,r,n ( x ) z − x ω { sℓ } ρ { } ρ +1 ( x ) (cid:21) d x + D r,n ( z ) . Hence, S r,n ( z ) = Z P r,n ( x ) z − x d x + polynomial( z ) . (4.7)But, as z → ∞ , S r,n ( z ) = O (1 /z ) and R P r,n ( x ) z − x d x → P r,n ( x ) ∈ L ([0 , (cid:3) Resolution of the Pad´e problem P r,n In this section we prove Theorem 1, using the tools of §§ S r,n ( z ) of the Pad´e problem P r,n , we apply the differential operator z n +1 n ! (cid:0) dd z (cid:1) n +1 and provein §§ Q r,n and stated in § § Q r,n is nothingbut Sorokin’s problem [14] for π , denoted by R r,n , up to a change of variable z − z .Since Sorokin has proved that R r,n has a unique solution up to proportionality, the sameresult holds for Q r,n and P r,n .To conclude the proof of Theorem 1, we deduce in §§ S r,n ( z ) from Sorokin’s integral representation of the solution of R r,n , usingthe integral operator introduced in § First reduction.
Let S r,n ( z ) be a solution of the Pad´e problem P r,n . By Lemma 1,there exist some polynomials ˇ A ρ,r,n ( z ), ˇ B ρ,r,n ( z ) and ˇ C r,n ( z ) of degree ≤ n + 1 such that b S r,n ( z ) := z n +1 n ! S ( n +1) r,n ( z ) = r X ρ =0 (cid:20) ˇ A ρ,r,n ( z )(1 − z ) n +1 Li { ℓs } ρ ℓ { } ρ +2 (cid:18) z (cid:19) + ˇ B ρ,r,n ( z )(1 − z ) n +1 Li { sℓ } ρ { } ρ +1 (cid:18) z (cid:19)(cid:21) + ˇ C r,n ( z )(1 − z ) n +1 = O (cid:18) z ( r +1)( n +1) (cid:19) . (5.1) As in § P r,n ( z ) defined by P r,n ( z ) = r X ρ =0 (cid:20) A ρ,r,n ( z ) ω { ℓs } ρ ℓ { } ρ +1 ( z ) + B ρ,r,n ( z ) ω { ℓs } ρ ℓ { } ρ +2 ( z ) + C ρ,r,n ( z ) ω { sℓ } ρ { } ρ +1 ( z ) (cid:21) . Since it is obtained from S r,n by replacing each polylogarithm by its weight, it obeys thesame derivation rules (see the remark before Lemma 9). This implies that b P r,n ( z ) := z n +1 n ! P ( n +1) r,n ( z ) = r X ρ =0 (cid:20) ˇ A ρ,r,n ( z )(1 − z ) n +1 ω { ℓs } ρ ℓ { } ρ +2 ( z ) + ˇ B ρ,r,n ( z )(1 − z ) n +1 ω { sℓ } ρ { } ρ +1 ( z ) (cid:21) = r X ρ =0 (cid:20) ˇ A ρ,r,n ( z )(1 − z ) n +1 Li { sℓ } ρ { } ρ +1 ( z ) + ˇ B ρ,r,n ( z )(1 − z ) n +1 Li { ℓs } ρ − ℓ { } ρ ( z ) (cid:21) (5.2)with the same polynomials ˇ A ρ,r,n ( z ) and ˇ B ρ,r,n ( z ); here we have used Eqs. (4.3) and (4.4)in Lemma 9 to compute the weights.Now, by Lemmas 2, 10 and the Pad´e conditions at z = 1 in P r,n for U j,r,n and V j,r,n , thefunction b P r,n ( z ) is necessarily of the form b P r,n ( z ) = r X j =0 (cid:20) h j +1 ( z ) Li { sℓ } j { } j +1 ( z ) + h j ( z ) Li { ℓs } j − ℓ { } j ( z ) (cid:21) (5.3)for some functions h j holomorphic at z = 1. Now we have obtained two expressions for b P r,n ( z ), namely Eqns. (5.2) and (5.3). Using Lemma 7 they have to coincide, that is ˇ A ρ,r,n ( z )(1 − z ) n +1 = h ρ +1 ( z ) and ˇ B ρ,r,n ( z )(1 − z ) n +1 = h ρ ( z ) for any ρ = 0 , . . . , r . Therefore (1 − z ) n +1 dividesˇ A ρ,r,n ( z ) and ˇ B ρ,r,n ( z ).We now claim that (1 − z ) n +1 also divides ˇ C r,n ( z ). To prove this, we use the integralrepresentation for S r,n ( z ) given by Lemma 11. Differentiating n +1 times under the integral,we obtain b S r,n ( z ) = ( n + 1)( − z ) n +1 Z P r,n ( x )( z − x ) n +2 d x. Again by Lemma 10 and the Pad´e conditions at z = 1 in P r,n for U r,n,j and V r,n,j , wededuce that P r,n ( x ) = O (cid:0) (1 − x ) n +1 (1 + | log(1 − x ) | r +1 ) (cid:1) as x → x <
1. Therefore the singularity of b S r,n ( z ) at z = 1 is at most a power oflogarithm. The expression (5.1) for b S r,n ( z ), together with the above deductions made forˇ A ρ,r,n ( z ) and ˇ B ρ,r,n ( z ), implies the claim. We can summarize the above results as follows: there exist polynomials b A ρ,r,n ( z ), b B ρ,r,n ( z )( ρ ∈ { , . . . , r } ) and b C r,n ( z ), all of degree at most n , such that b S r,n ( z ) = r X ρ =0 (cid:20) b A ρ,r,n ( z ) Li { ℓs } ρ ℓ { } ρ +2 (cid:18) z (cid:19) + b B ρ,r,n ( z ) Li { sℓ } ρ { } ρ +1 (cid:18) z (cid:19)(cid:21) + b C r,n ( z ) = O (cid:18) z ( r +1)( n +1) (cid:19) . (5.4)5.2. Second reduction.
We want to find further Pad´e conditions involving the polyno-mials b A ρ,r,n ( z ), b B ρ,r,n ( z ) ( ρ ∈ { , . . . , r } ) and b C r,n ( z ). For this, we form the functions Q j,r,n := r X ρ = j (cid:20) − A ρ,r,n ( z ) Li { sℓ } ρ − j { } ρ − j ( z ) + B ρ,r,n ( z ) Li { sℓ } ρ − j { } ρ − j +1 ( z ) + C ρ,r,n ( z ) Li { ℓs } ρ − j − ℓ { } ρ − j ( z ) (cid:21) where j = 0 , . . . , r , and A ρ,r,n ( z ), B ρ,r,n ( z ), C ρ,r,n ( z ) are the polynomials in our initial Pad´eproblem P r,n . Each Q j,r,n ( z ) is holomorphic at z = 0 and the rules of differentiation ofmultiple polylogarithms (see § b Q j,r,n ( z ) := z n +1 n ! Q ( n +1) j,r,n ( z )= r X ρ = j (cid:20) b A ρ,r,n ( z ) Li { sℓ } ρ − j { } ρ − j +1 ( z ) + b B ρ,r,n ( z ) Li { ℓs } ρ − j − ℓ { } ρ − j ( z ) (cid:21) = O ( z n +1 )for all j = 0 , . . . r . The main point here is that the polynomials b A ρ,r,n ( z ) and b B ρ,r,n ( z ) arethe same as in Eq. (5.4).5.3. The intermediate Pad´e problem Q r,n . The previous two sections show that anysolution S r,n ( z ) to the problem P r,n yields (by differentiating n + 1 times and multiplyingby z n +1 /n !) a solution to the following problem: given non-negative integers r and n , findpolynomials b A ρ,r,n ( z ), b B ρ,r,n ( z ) (for 0 ≤ ρ ≤ r ) and b C r,n ( z ), of degrees ≤ n , such that thefollowing holds: b S r,n ( z ) := r X ρ =0 (cid:20) b A ρ,r,n ( z ) Li { ℓs } ρ ℓ { } ρ +2 (cid:18) z (cid:19) + b B ρ,r,n ( z ) Li { sℓ } ρ { } ρ +1 (cid:18) z (cid:19)(cid:21) + b C r,n ( z ) = O (cid:18) z ( r +1)( n +1) (cid:19) , b Q j,r,n ( z ) := r X ρ = j (cid:20) b A ρ,r,n ( z ) Li { sℓ } ρ − j { } ρ − j +1 ( z ) + b B ρ,r,n ( z ) Li { ℓs } ρ − j − ℓ { } ρ − j ( z ) (cid:21) = O ( z n +1 ) , j = 0 , . . . , r. We shall denote this Pad´e approximation problem by Q r,n . It amounts to solving a linearsystem of (3 r + 4)( n + 1) − r + 4)( n + 1) unknowns (the coefficients of the polynomials). Hence it has a least one non trivial solution and our next task is to provethat is has exactly one solution up to a multiplicative constant.To do so, we will identify the problem with one already solved by Sorokin [14]. We firstobserve the effect of changing z to 1 − z in the Pad´e problem Q r,n . Lemma 12.
For any z ∈ C \ [0 , , we have Li { sℓ } ρ { } ρ +1 (cid:18) z (cid:19) = ( − ρ +1 Li { s } ρ { } ρ (cid:18) − z (cid:19) , Li { ℓs } ρ ℓ { } ρ +2 (cid:18) z (cid:19) = ( − ρ +1 Li { s } ρ { } ρ +1 (cid:18) − z (cid:19) . Proof.
We prove these identities by induction on ρ . They hold trivially for ρ = 0 and bydifferentiation of both sides at level ρ , we get the identity at level ρ −
1. We deduce that theidentity at level ρ holds, up to some additive constant. This constant must be 0 becauseboth sides vanish at z = ∞ . (cid:3) Therefore, when we change z to 1 − z , the Pad´e problem Q r,n becomes b S r,n (1 − z ) := r X ρ =0 ( − ρ +1 (cid:20) b A ρ,r,n (1 − z ) Li { s } ρ { } ρ +1 (cid:18) z (cid:19) + b B ρ,r,n (1 − z ) Li { s } ρ { } ρ (cid:18) z (cid:19)(cid:21) + b C r,n (1 − z ) = O (cid:18) − z ) ( r +1)( n +1) (cid:19) = O (cid:18) z ( r +1)( n +1) (cid:19)b Q j,r,n (1 − z ) := r X ρ = j (cid:20) b A ρ,r,n (1 − z ) Li { sℓ } ρ − j { } ρ − j +1 (1 − z )+ b B ρ,r,n (1 − z ) Li { ℓs } ρ − j − ℓ { } ρ − j (1 − z ) (cid:21) = O ((1 − z ) n +1 ) , j = 0 , . . . , r. Let us define e A ρ,r,n ( z ) = ( − ρ +1 b A ρ,r,n (1 − z ) , e B ρ,r,n ( z ) = ( − ρ +1 b B ρ,r,n (1 − z ) , e C r,n ( z ) = b C r,n (1 − z ) , e S r,n ( z ) = b S r,n (1 − z ) , e Q j,r,n ( z ) = − b Q j,r,n (1 − z ) . With these notations, the Pad´e problem Q r,n now reads e S r,n ( z ) := r X ρ =0 (cid:20) e A ρ,r,n ( z ) Li { s } ρ { } ρ +1 (cid:18) z (cid:19) + e B ρ,r,n ( z ) Li { s } ρ { } ρ (cid:18) z (cid:19)(cid:21) + e C r,n ( z ) = O (cid:18) z ( r +1)( n +1) (cid:19)e Q j,r,n ( z ) := r X ρ = j ( − ρ (cid:20) e A ρ,r,n ( z ) Li { sℓ } ρ − j { } ρ − j +1 (1 − z ) + e B ρ,r,n ( z ) Li { ℓs } ρ − j − ℓ { } ρ − j (1 − z ) (cid:21) = O ((1 − z ) n +1 ) , j = 0 , . . . , r. In spite of different notations, we recognize here Sorokin’s problem [14] for π of weight2 r + 2, which we denote by R r,n from now on. Sorokin proved that this problem has aunique solution up to proportionality. Therefore the same property holds for Q r,n , and also for P r,n . This concludes the proof of Theorem 1, except for the integral representation(1.3) of S r,n ( z ) that we shall prove now.5.4. Hypergeometric integrals for ˜ S r,n ( z ) and S r,n ( z ) . Sorokin has found an explicitintegral formula for the solution e S r,n ( z ) of his Pad´e problem R r,n stated in § e S r,n ( z ) = ( − ( r +1) n Z [0 , r +2 r +1 Y j =1 x nj (1 − x j ) n y nj (1 − y j ) n (cid:0) zx y ··· x j − y j − − x j y j (cid:1) n +1 d x j d y j . (5.5)In this and the next sections we shall deduce from it the integral expression (1.3) of S r,n ( z ),using the relation z n +1 n ! S ( n +1) r,n ( z ) = e S r,n (1 − z ) (5.6)and the integral operator defined in § R r,n re-cursively and showed that, for any integer r ≥ z ∈ C \ [0 , S r − ,n ( z ) = 1 n ! z n +1 (1 − z ) n +1 (cid:0) z n +1 ˜ S ( n +1) r,n ( z ) (cid:1) ( n +1) (5.7)and ˜ S ,n ( z ) = Z Z x n (1 − x ) n y n (1 − y ) n ( z − xy ) n +1 d x d y. It is not hard to see that, with the notation of § z ∈ C \ [0 , S ,n ( z ) = H n +1 n +1 , (cid:18)Z x n (1 − x ) n ( z − x ) n +1 (cid:19) = H n +1 n +1 , H n +1 n +1 ,n +1 ( ) , (5.8)where is the constant function equal to 1 on C \ [0 , § § f ∂ ( z ) := f (1 − z ) and we denote by H k = H ◦ H ◦ · · · ◦ H the composition of an integraloperator H with itself k times. Proposition 2.
For any z ∈ C \ [0 , and any integer r ≥ , we have ˜ S r,n ( z ) = ( H n +1 n +1 , H n +1 n +1 ,n +1 ) r +1 ( )( z ) (5.9) and ˜ S ∂r,n ( z ) = ( H n +10 ,n +1 H n +1 n +1 ,n +1 ) r +1 ( )( z ) . (5.10)Eq. (5.9) follows immediately from Eq. (5.8) and the relation˜ S r,n = H n +1 n +1 , H n +1 n +1 ,n +1 ( ˜ S r − ,n ) , which is just a translation of Eq. (5.7) (using Lemma 3). Then Eq. (5.10) follows from (5.9)by means of Lemma 6. Now Eq. (5.6) reads z n +1 n ! S ( n +1) r,n ( z ) = ˜ S ∂r,n ( z ) (5.11)and lim z →∞ S r,n ( z ) = 0 for any r ≥
0, so that Lemma 3 yields S r,n ( z ) = H n +1 n +1 , ( e S ∂r,n )( z ) . Hence, by (5.10) in Proposition 2, we obtain the following result (using also Lemma 5 totake limits as z → Proposition 3.
For any z ∈ C \ [0 , and any integer r ≥ , we have S r,n ( z ) = H n +1 n +1 , ( H n +10 ,n +1 H n +1 n +1 ,n +1 ) r +1 ( )( z ) . (5.12) Moreover, both sides of (5.12) are defined and equal for z = 1 . Explicit multiple integrals.
The integral expression for S r,n ( z ) given in Theorem 1is simply the explicit “expansion” of the formula (5.12) given in Proposition 3 above. Letus provide details on this expansion.For any function F analytic on C \ [0 ,
1] and at infinity, Eq. (3.1) in § H n +1 n +1 ,n +1 ( F )( z ) = ( − n +1 Z u n (1 − u ) n ( u − z ) n +1 F (cid:16) zu (cid:17) d u. This function H n +1 n +1 ,n +1 ( F )( z ) is analytic on C \ [0 ,
1] and at infinity, and vanishes to anorder ≥ n + 1 at ∞ (using Lemma 3). The same property can be proved in the same wayfor the following function: H n +10 ,n +1 H n +1 n +1 ,n +1 ( F )( z ) = z n +1 Z v − (1 − v ) n ( v − z ) n +1 Z u n (1 − u ) n ( u − z/v ) n +1 F (cid:16) zuv (cid:17) d u d v = z n +1 Z Z v n (1 − v ) n u n (1 − u ) n ( v − z ) n +1 ( uv − z ) n +1 F (cid:16) zuv (cid:17) d u d v. By induction on r ≥ S ∂r,n ( z ) = ( H n +10 ,n +1 H n +1 n +1 ,n +1 ) r +1 ( )( z ) = z ( r +1)( n +1) × Z [0 , r +1) r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) ( z − u v · · · u j − v j − u j ) n +1 ( z − u v · · · u j v j ) n +1 (cid:1) d u d v . Therefore the equality H n +1 n +1 , ( H n +10 ,n +1 H n +1 n +1 ,n +1 ) r +1 ( )( z ) = ( − n +1 Z u − (1 − u ) n ˜ S ∂r,n ( z/u )d u yields, using Proposition 3: S r,n ( z ) = ( − n +1 z ( r +1)( n +1) × Z [0 , r +3 u ( r +1)( n +1) − (1 − u ) n r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) ( z − u u v · · · u j − v j − u j ) n +1 ( z − u u v · · · u j v j ) n +1 (cid:1) d u d v . This completes the proof of Theorem 1.6.
Beyond Vasilyev’s conjecture: irrationality of odd zeta values
A natural problem is to find a proof that the numbers ζ (2 r + 1), r ≥
0, span an infinite-dimensional Q -vector space [2, 11] that would be analogous to Sorokin’s proof that π istranscendental [14] (since Sorokin’s result is equivalent to the fact that the numbers ζ (2 r ), r ≥
0, span an infinite-dimensional Q -vector space). In particular, such a proof wouldinvolve a Pad´e approximation problem with multiple polylogarithms.Let σ be an integer such that 1 ≤ σ ≤ r + 2. To achieve this goal, it is enough to relatethe very-well-poised hypergeometric series ∞ X k =1 ( k + n k − σn ) σn ( k + n + 1) σn ( k ) r +4 n +1 , (6.1)which can be used to prove the above mentioned result (see for instance [7]), to such aPad´e approximation problem. An analogous work has been done in [8], where this series isrelated to a Pad´e approximation problem involving only classical polylogarithms, namelyof depth 1.We shall prove now that for σ = 1 the hypergeometric series (6.1) is equal (up to a sign)to S r,n (1), thereby providing in this case the relation we are looking for. For any σ we shallprove that this series is the value at z = 1 of a function S r,n,σ ( z ) which generalizes S r,n ( z );what is missing is a Pad´e approximation problem of which S r,n,σ ( z ) would be a solution.We believe that a suitable generalisation of the problem P r,n solved in Theorem 1 couldhave this property.With this aim in view, we consider the function S r,n,σ ( z ) defined by z n +1 n ! S ( σn +1) r,n,σ ( z ) = ˜ S ∂r,n ( z )and lim z →∞ S r,n,σ ( z ) = 0; in this way we have S r,n, ( z ) = S r,n ( z ) (see Eq. (5.11)). We have S r,n,σ ( z ) = H σn +1 n +1 , ( e S ∂r,n )( z ) . The equality H σn +1 n +1 , ( H n +10 ,n +1 H n +1 n +1 ,n +1 ) r +1 ( )( z ) = ( − σn +1 z ( σ − n Z u (1 − σ ) n − (1 − u ) σn ˜ S ∂r,n ( z/u )d u yields, using Proposition 3: S r,n,σ ( z ) = ( − σn +1 z ( r + σ ) n + r +1 × Z [0 , r +3 u ( r − σ +2) n + r (1 − u ) σn r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) ( z − u u v · · · u j − v j − u j ) n +1 ( z − u u v · · · u j v j ) n +1 (cid:1) d u d v . This function has the following value at z = 1: S r,n,σ (1) =( − σn +1 Z [0 , r +3 u ( r − σ +2) n + r (1 − u ) σn r +1 Y j =1 (cid:0) ( u j v j ) ( r − j +2)( n +1) − (1 − u j ) n (1 − v j ) n (cid:1) r +1 Y j =1 (cid:0) (1 − u u v · · · u j − v j − u j ) n +1 (1 − u u v · · · u j v j ) n +1 (cid:1) d u d v . Using Proposition 17 of [6] (which amounts to a change of variables) one obtains S r,n,σ (1) = ( − σn +1 Z [0 , a − Q a − j =1 x σnj (1 − x j ) n (1 − x x · · · x a − ) σn +1 Q ≤ j ≤ a − j even (1 − x x · · · x j ) n +1 d x with a = 2 r + 4. Then using Zlobin’s result [16] or another change of variables (namelyTh´eor`eme 10 of [6]), one obtains the Vasilyev-type integral S r,n,σ (1) = ( − σn +1 Z [0 , a − Q a − j =1 x σnj (1 − x j ) n Q a − ( x , · · · , x a − ) σn +1 d x . Now Theorem 5 of [18] yields S r,n,σ (1) = ( − σn +1 ∞ X k =1 ( k + n k − σn ) σn ( k + n + 1) σn ( k ) an +1 . Up to a sign, this is exactly the very-well poised hypergeometric series (6.1).
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