Journal of Mathematical Sciences, Vol. 146, No. 2, 2007
ON MULTIPLE INTEGRALS REPRESENTED AS A LINEAR FORM IN 1, ζ(3), ζ(5), . . . , ζ(2k − 1) UDC 511.36
V. Salikhov and A. Frolovichev
Abstract. A theorem on the presentability of a multiple integral as a linear form in 1, ζ(3), ζ(5), . . . , ζ(2k − 1) over Q is proved. This theorem refines the results recently obtained by D. Vasiliev, V. Zudilin, and S. Zlobin.
Consider the integral 2k−1
J2k−1 = [0,1]2k−1
j=1
α −1
xj j
(1 − xj )βj −1 dx1 . . . dx2k−1
(1 − x1 + x1 x2 − x1 x2 x3 + · · · + x1 x2 · · · x2k−2 − x1 x2 · · · x2k−1 )α0
,
(1)
where k ≥ 2 and all αj , βi ∈ N. Theorem 1. Suppose that the parameters of integral (1) satisfy the following conditions: α0 ≤ α1 ; αr + βr ≤ βr+1 + αr+2 ,
(2)
r = 0, . . . , 2k − 3,
(3)
where, for uniformity, we take β0 = 0; max(α2j−1 , α2j ) < min(α2j + β2j , α2j+1 + β2j+1 ),
j = 1, . . . , k − 1;
(4)
if α0 > β1 , then β1 + β3 + · · · + β2k−1 ≥ α0 + 1.
(5)
Then, for some r1 , . . . , rk ∈ Q, we have J2k−1 = r1 + r2 ζ(3) + r3 ζ(5) + · · · + rk ζ(2k − 1).
(6)
Such integrals have been considered by many scientists starting from the article by F. Beukers [2], where k = 2 and all αi , βj are equal natural numbers. D. Vasiliev has proved [4] that, in the same situation for k = 3, the result is analogous to Theorem 1. In 2002, V. Zudilin [7] generalized these results for arbitrary k. In his article, formula (6) was proved in the case where conditions (3) are equalities for r = 1, . . . , 2k − 3. All these results are contained in Theorem 1. The result analogous to Theorem 1 is obtained in [5] under stricter conditions for the parameters of integral (1). Note 1. From (3) and (4) easily follows the inequality αd + β1 + β3 + · · · + β2r−1 ≥ α0 + 1,
r = 2, . . . , k + 1, d ∈ {2r − 1, 2r}.
(4 )
Indeed, if r = 2 and d = 3, then α3 + β1 + β3 ≥ α0 + 1 (see (4) for j = 1 and (3) for r = 0); if r = 2 and d = 4, then α4 + β1 + β3 ≥ α2 + β2 + β1 ≥ β2 + α0 + 1; if r > 2, then (1) α2r−1 + β2r−1 ≥ α2r−2 in view of (4), and (4 ) are satisfied by induction; (2) α2r + β2r−1 ≥ α2r−2 + β2r−2 in view of (3), and (4 ) are satisfied again by induction. Translated from Fundamentalnaya i Prikladnaya Matematika, Vol. 11, No. 6, pp. 143–178, 2005. c 2007 Springer Science+Business Media, Inc. 1072–3374/07/1462–5731
5731
Note 2. Condition (2) is not necessary. At the end of the article we give more general but more awkward conditions than (2)–(5) for parameters αi , βj to provide equality (6). To prove Theorem 1, we use multiple series of a special type. Many results will be proved in a more general form than is necessary for proving Theorem 1. They can be used for calculation of other multiple integrals. For the first time in this situation Vasilyev [4] used twofold series for representation of the integral J5 in form (6). Finally, note that the method used for proving Theorem 1 allowed A. Frolovichev to obtain the corresponding result for the integral J2k : J2k = r0 + r1 ζ(2) + r2 ζ(4) + · · · + rk ζ(2k). 1. Multiple Series Representing Linear Forms in 1, ζ(3), ζ(5), . . . , ζ(2m + 1) over Q Let m ∈ N, Qj (x) =
dj (x + aj,i )νj,i ,
j = 1, . . . , m,
i=1
where dj ∈ N and all aj,i ∈ N , aj,1 , . . . , aj,dj are different, all νj,i ∈ {1, 2}, j = 1, . . . , m, i = 1, . . . , dj , ¯ Qm = {Q1 , . . . , Qm }, qj = deg Qj , j = 1, . . . , m. Let us denote
(7)
Aj = {αj,1 , . . . , αj,dj }, Aj2 = {αj,i ∈ Aj | νj,i = 2}, Aj2 = ∅, Aj1 = Aj \ Aj2 ; ej = min aj,i , fj = max aj,i , Ej = min αj,i , Fj = max αj,i . Aj
Aj
Aj2
Aj2
It is obvious that ej ≤ Ej ≤ Fj ≤ fj . Let P = Pm ∈ Q[x1 , . . . , xm ], qj = degxj Pm , pj = degxj Pm , Rm =
j = 1, . . . , m,
Pm , Q1 (x1 ) · · · Qm (xm )
¯ m−1 = (λ2 , . . . , λm ). λ2 , . . . , λm ∈ Z, λ Denote ¯ m ) = {pj ≤ qj − 2, j = 1, . . . , m}, (I) = I(P, Q ¯ m ) = {fj−1 < Ej + λj , j = 2, . . . , m}, (II) = II(Q ¯ m ) = {Fj−1 < ej + λj , j = 2, . . . , m}. (III) = III(Q ¯ m ) and so on, we denote the corresponding inequalities for By Ij , IIj , and IIIj or, more exactly, Ij (P, Q a fixed j. Consider a multiple series of the form m (−1)m−l Pm (t1 , . . . , tm ) ¯ ¯ , (8) Σm (P, Qm , λm−1 ) = Σm (Rm ) = Σm = Dσ Q1 (t1 ) · · · Qm (tm ) + ρ¯ l=1
m
s1 ,...,sm ∈Z
where an abstract function Dσ is defined by the formula Dσ (f (σ)) = f (0); summation in (8) is carried out over 2m−1 vectors ρ¯m = (ρ1 , ρ2 , . . . , ρm ), ρ1 = 1, ρj ∈ {1, 2}, j = 2, . . . , m; l = l(¯ ρm ) is the number of coordinates of the vector ρ¯m , which are equal to 1; and if j = 1, s1 + σ tj = sj (9) if j = 2, . . . , m, ρj = 1, tj−1 + sj + λj if j = 2, . . . , m, ρj = 2. 5732
¯ m−1 ∈ Zm−1 is admissible for Rm ∈ Q(x1 , . . . , xm ) if, for all ρ¯m , s1 , . . . , sm ∈ Z+ Definition 1. A vector λ and t1 , . . . , tm determined by formulas (9) with σ = 0, the rational function Rm has no poles at the points (t1 , . . . , tm ). ¯ m−1 ∈ Zm−1 is admissible for Q ¯ m if, for all j = 2, . . . , m, r = 2, . . . , j, we have Definition 2. A vector λ λr + · · · + λj + ej ≥ 1.
(10)
¯ m−1 is admissible for Q ¯ m , then it is admissible for Let us show that, if a vector λ Rm =
Pm (x1 , . . . , xm ) . Q1 (x1 ) · · · Qm (xm )
Indeed, t1 = s1 ∈ Z+ and Q1 (t1 ) = 0. If j ≥ 2, then, for ρj = 1, from (9) we obtain tj = sj ∈ Z+ and Qj (tj ) = 0 and, for ρj = 2, we determine r ∈ {2, . . . , j} such that ρr−1 = 1 and ρr = · · · = ρj = 2. Then from (9) we obtain tj = sr−1 + · · · + sj + λr + · · · + λj ≥ λr + · · · + λj , Qj (tj ) = 0 (see (10) and definition ej ). ¯ m ). From III(Q ¯ m ), we have Finally, we show that (10) follows from the set of inequalities III(Q λr + · · · + λj + ej > λr + · · · + λj−1 + Fj−1 ≥ λr + · · · + λj−1 + ej−1 ≥ · · · ≥ λr + er ≥ Fr−1 ≥ 1, and (10) holds. ¯ m ) under some ¯ m ), and III(Q The following proposition about the conservation of properties (7), II(Q ¯ transformations of the system of polynomials Qm will be useful for us. ¯ m−1 ∈ Zm−1 . Then these ¯ m ) are satisfied for Q ¯ m and λ ¯ m ), and III(Q Lemma 1. Suppose that (7), II(Q ¯ ¯ properties are satisfied for Qm−1 and λm−1 in the following four situations: if j = 1, . . . , ν − 2, Qj (x) (1) Qj (x) = Qν−1 (x)Qν (x + i + λν ) if j = ν − 1, if j = ν, . . . , m − 1, Qj+1 (x) if j = 2, . . . , ν − 1, λj λj = λν + i + λi+1 if j = ν, if j = ν + 1, . . . , m − 1, λj+1
(2)
(3)
where ν ∈ {2, . . . , m}, i ∈ Z, i > max(fν−1 − Eν − λν , Fν−1 − eν − λν ); Q2 (x + i + λ2 ) if j = 1, Qj (x) = if j = 2, . . . , m − 1, Qj+1 (x) i + λ2 + λ3 if j = 2, λj = if j = 3, . . . , m − 1, λj+1 where i ∈ Z, i + λ2 + e2 ≥ 1; if j = 1, . . . , ν − 2, Qj (x) Qj (x) = Qν−1 (x)(x + i + λν ) if j = ν − 1, if j = ν, . . . , m − 1, Qj+1 (x) if j = 2, . . . , ν − 1, λj λj = λν + λν+1 if j = ν, if j = ν + 1, . . . , m − 1, λj+1 where ν ∈ {2, . . . , m}, i ∈ Aν ; 5733
(4)
if j = 1, . . . , ν − 2, Qj (x) 2 Qj (x) = (x + i)(x + aν + λν ) if j = ν − 1, if j = ν, . . . , m − 1, (x + αj+1 )2 if j = 2, . . . , ν − 1, λj λj = λν + λν+1 if j = ν, if j = ν + 1, . . . , m − 1, λj+1 where ν ∈ {2, . . . , m}, j ∈ Aν−1 , ak ∈ Ak2 , k = ν, . . . , m.
Proof. All the four cases are considered in almost the same way. In each case, for j ≤ ν − 2 and j > ν ¯ ¯ ¯ ¯ (in the second case for j > 2), IIj (Q m−1 ) and IIIj (Qm−1 ) immediately follow from II(Qm ) and III(Qm ), and, for j ∈ {ν − 1, ν} (in the second case for j = 1), the proof is straightforward. We give the proof for cases (1)–(4) without any comments. / Aν−1,2 , i + λν + Eν > fν−1 , whence (7) (1) We have j + λν + eν > Fν−1 , i.e., i + λν + eν ∈ ¯ follows for Qν−1 (x), because the polynomials Qν−1 (x) and Qν (x + i + λν ) may have common roots only of multiplicity 1 for each polynomial. Besides, it is obvious that eν − 1 = eν − 1, Eν − 1 = Eν − 1, ¯ ¯ = i + λν + Fν , and fν−1 = i + λν + fν . Therefore, IIν−1 (Q Fν−1 m−1 ) and IIIν−1 (Qm−1 ) are satisfied. ¯ Then IIν (Q m−1 ): fν−1 < Eν + λν ⇐⇒ i + λν + fν < Eν+1 + i + λν + λν+1 ,
¯ m ). which is satisfied in view of IIν+1 (Q ¯ IIIν (Qm−1 ): Fν−1 < eν + λν ⇐⇒ i + λν + Fν < eν+1 + i + λν + λν+1 , ¯ m ). Thus, the lemma in case (1) is proved. which is satisfied in view of IIIν+1 (Q ¯ ¯ (2) Here, it is necessary to check only II2 (Q m−1 ) and III2 (Qm−1 ). ¯ II2 (Q m−1 ): i + λ2 + f2 < E2 + λ2 ⇐⇒ i + λ2 + f2 < E3 + i + λ2 + λ3 ¯ m ). which is satisfied in view of II3 (Q ¯ III2 (Qm−1 ): i + λ2 + F2 < e2 + λ2 ⇐⇒ i + λ2 + F2 < e3 + i + λ2 + λ3 , ¯ m ). Finally, i + λ2 + e2 ≥ 1, i.e., the roots of the polynomial Q (x) = which is satisfied in view of III3 (Q 1 Q2 (x + i + λ2 ) are negative, A1 ⊂ N. (3) Since i ∈ Aν , it follows that λν + i ≥ λν + eν > Fν−1 , i.e., the polynomial Qν−1 (x) satisfies (7), ¯ ¯ and, moreover, eν−1 = eν−1 and Eν−1 = Eν−1 . Therefore, IIν−1 (Q m−1 ) and IIIν−1 (Qm−1 ) are satisfied. ¯ IIν (Q m−1 ): we have fν−1 = max(fν−1 , λν + i). Then Eν + λν = Eν+1 + λν+1 + λν > fν + λν ≥ i + λν , Eν + λν > fν + λν ≥ Eν + λν > fν−1 .
. However, in this case, Eν + λν > fν−1 ¯ IIIν (Q ): we have F = F ν−1 ; hence, m−1 ν−1
eν + λν = eν+1 + λν+1 + λν > Fν + λν ≥ Eν + λν > fν−1 ≥ Fν−1 , . eν + λν > Fν−1
The consideration of case (3) is completed. (4) We have aν + λν ≥ Eν + λν > fν−1 ≥ i since aν ∈ Aν2 , i ∈ Aν−1 . Hence, eν−1 = i and = fν−1 = aν + λν . Eν−1 = Fν−1 It is necessary to verify that the following four inequalities hold. 5734
¯ IIν−1 (Q m−1 ):
< Eν−2 + λν−2 ⇐⇒ fν−2 < aν + λν + λν−1 . fν−2
However, aν + λν + λν−1 ≥ Eν + λν + λν−1 > Eν−1 + λν−1 > fν−2 . ¯ IIIν−1 (Q
m−1 ):
Fν−2 < eν−1 + λν−1 ⇐⇒ Fν−2 < i + λν−1 .
However, i + λν−1 ≥ eν−1 + λν−1 > Fν−2 . ¯ IIν (Q m−1 ):
< Eν + λν ⇐⇒ aν + λν < aν+1 + λν + λν+1 . fν−1
However, aν+1 + λν+1 ≥ Eν+1 + λν+1 > fν ≥ Fν ≥ aν . ¯ IIIν (Q m−1 ): ¯ i.e., IIν (Q m−1 ). The lemma is proved.
< eν + λν ⇐⇒ aν + λν < aν+1 + λν + λν+1 , Fν−1
Denote Ωm = {r0 + r1 ζ(3) + r2 ζ(5) + · · · + rm ζ(2m + 1) | ri ∈ Q}.
(11)
Let us formulate and begin the proof (its end is in Sec. 4) of the main result of the article on multiple series. Theorem 1 follows from it (in Sec. 5) quite easily. Theorem 2. Under conditions (7) and (I)–(III), we have ¯ m−1 ) ∈ Ωm . ¯ m, λ Σm (P, Q Proof. We prove the theorem by induction on m. For m = 1 (omitting for brevity the index j = 1), we P (t) into the sum of elementary fractions: get the expansion of Q(t) ci P (t) bi = , + Q(t) (t + i)2 t+i F
f
i=E
i=e
(12)
where all bi , ci ∈ Q, bi = 0 for i ∈ / A1,2 , and ci = 0 for i ∈ / A1 . From (7) for j = 1 we get that deg P ≤ deg Q − 2; hence, f
ci = 0.
(13)
i=e
Similarly, for example, [6], from (8), (9), (12), and (13), we have Σ1 = Dσ
∞ F s=0
i=E
bi ci + (s + i + z)2 s + i + z1 f
i=e
f −1 f −1 i−1 F F 1 1 bj ζ(3) + 2 bi − ck 2 = r1 + r2 ζ(3), = −2 k2 i i=E
i=E
k=1
i=e
k=e
where r1 , r2 ∈ Q, and Theorem 2 is valid for m = 1. In what follows, we assume that m ≥ 2 and for all n = 1, . . . , m − 1 the following assertion is valid: if conditions (7) and (I)–(III) are satisfied, then Σn ∈ Ωn .
(Un ) 5735
To prove the theorem, it is necessary to show that (Um ) holds. First, we prove (Um ) for the most important special case of sums Σm , where P = 1, Qj (x) = (x + aj )2 (here aj ∈ N, dj = 1, qj = 2, ρj = 0, ej = Ej = Fj = fj = aj , conditions (II) and (III) coincide and reduce to the inequalities aj−1 < aj + λj ,
j = 2, . . . , m.
(14)
The statement (Um ) takes the following form. ¯ m−1 ∈ Zm−1 , the inequalities (14) and all Proposition 1. Let m ≥ 2, a ¯m = (a1 , . . . , am ) ∈ Nm , λ statements (U1 ), . . . , (Um−1 ) are satisfied, Sm
¯ m−1 ) = Dσ = Sm (¯ am , λ
m l=1
ρ¯m
s1 ,...,sm ∈Z+
(−1)m−l , (t1 + a1 )2 · · · (tm + am )2
(15)
where l and ρ¯m are defined as in (8), tj are defined by formulas (9). Then ¯ m−1 ) ∈ Ωm ; (1) Sm (¯ am , λ ¯ m−1 ) in form (11), the equality rm = −2 holds. αm , λ (2) in the representation of Sm (¯ To prove Proposition 1, we need some auxiliary results. ¯ m−1 ∈ Zm−1 , Lemma 2. Let a ∈ Z, Rm ∈ Q[x1 , . . . , xm ], λ ¯ m−1 ) = Sm = Sm (a, Rm , λ
m l=1
ρ¯m
(−1)m−l Rm (t1 , . . . , tm ),
(16)
∈Z+
s2 ,...,sm s1 =a
where (t1 , . . . , tm ) are defined by formulas (9) with σ = 0, and Rm have no poles at the points (a, t2 , . . . , tm ) ¯ m−1 ) ∈ Q. for all s2 , . . . , sm ∈ Z+ . Then Sm (a, Rm , λ Proof. Induction on m. For m = 1, we have S1 = R1 (α) ∈ Q. Suppose that the lemma is true for Sm−1 , m ≥ 2. Let us prove it for the sums Sm . Let α ∈ Z. Denote (17) i0 (α) = min(0, α), i1 (α) = max(0, α) − 1. As usual,
i2 i=i1
ai = 0 if i1 > i2 . Let us show that i1 (α)
¯ m−1 ) = ± Sm (a, Rm , λ
∗ ¯ ∗ ), Sm−1 (i, Rm−1 ,λ m−2
(18)
i=i0 (α) ∗ where the plus sign is chosen in the case α > 0 and the minus sign is taken in the case α < 0, Rm−1 = ¯∗ Rm (a, x1 , . . . , xm−1 ), λ = (λ , . . . , λ ). 3 m m−2 Then the lemma follows from (18) even in a stronger form: Sm is a finite linear combination of values of the function Rm at a set of points ¯b ∈ Zm with coefficients ±1.
(1) (2) Consider 2m−2 pairs of vectors ρ¯m , ρ¯m , where
ρ¯(2) ρ¯(1) m = (1, 1, ρ3 , . . . , ρm ), m = (1, 2, ρ3 , . . . , ρm ). (1) (2) Let l = l ρ¯m . Then l ρ¯m = l − 1, l ≥ 2. Let (r − 1) coordinates ρ3 = · · · = ρr+1 = 2, where r ∈ {1, . . . , m − 1} and r = 1 if ρ3 = 1. (1) (1) (1) For vectors ρ¯m , from (9), we have t2 = s2 ; for j ∈ {3, . . . , r + 1}, we have tj = s2 + · · · + sj + (1)
(2)
(2)
λ3 + · · · + λj and tr+2 = sr+2 . For vectors ρ¯m , we similarly have t2 for j ∈ {3, . . . , r + 1}, we have r ≤ m − 2. 5736
(2) tj
= s1 + s2 + λ2 = s2 + a + λ2 ; (2)
(1)
= a + λ2 + s2 + · · · + sj + λ3 + · · · + λj and tr+2 = tr+2 = sr+2 for
By summing over variable s2 in (16), we get ¯ m−1 ) = Sm = Sm (a, Rm , λ
m l=2
ρ¯m
(−1)m−l
s2 ,...,sm ∈Z+
× (Rm (a, s2 , s2 + s3 + λ3 , . . . ) − Rm (a, a + λ2 + s2 , a + λ2 + s2 + s3 + λ3 , . . . )) i1 (α+λ2 )
=±
m
(−1)m−l Rm (α, s2 , s2 + s3 + λ3 , . . . ).
(19)
i=i0 (α+λ2 ) l=2 ρ¯(1) s2 ,...,sm ∈Z+ m s2 =i (1)
ρm−1 ) = l−1. It is obvious that ρ¯m = (1, ρ¯m−1 ), where ρ¯m−1 run over all possible ρ¯m−1 -vectors, l = l(¯ After the change of variables (sj , λj , tj ) = (sj , λj , tj ) for j = 1, . . . , m − 1, from (19) we obtain i1 (α+λ2 )
¯ m−1 ) = ± Sm (a, Rm , λ
m−1
i=i0 (α+λ2 ) l =1 ρ¯m−1 s2 ,...,sm ∈Z+ s1 =i
(−1)m−1−l Rm (a, s1 , t2 , . . . , tm−1 ),
which corresponds to (18), and the lemma is proved. Note 3. From the proof of the lemma we see that the series converges with any admissible definition of partial sums of multiple series (triangular, rectangular, or spherical), because its partial sums stabilize, and the series reduces to a finite sum. 4. From (I) we obtain that the series in (8) and (15) converge absolutely and uniformly for σ ∈ Note − 12 ; 12 (before applying the operator Dσ ), because, for some C > 0, they are dominated by the series C , 1 2 (s1 + 2 ) (s2 + 1)2 · · · (sm + 1)2 + s1 ,...,sm ∈Z
which converges with any admissible definition of partial sums. The same is also valid for the series ∂ obtained from (8) by termwise application of the operator ∂σ . Therefore, we may use any permutation of the series terms and termwise application of an abstract function Dσ . The following transformations of series (8) will be useful: let ν ∈ {1, . . . , m}, α, β ∈ Z. We denote Tν,α,β (Σm ) = Dσ
m−1 l=1
ρ¯m s1 ,...,sm ∈Z+ ρν =1 sν ≥α
(−1)m−l Rm (t1 , . . . , tm ) + Dσ
m−1 l=1
(−1)m−l Rm (t1 , . . . , tm ), (20)
ρ¯m s1 ,...,sm ∈Z+ ρν =2 sν ≥β
where in case ν = 1 the second series is absent since ρ1 = 1. We denote the corresponding operator for ν = 1 by T1,α . Lemma 3. Let α + eν ≥ 1, where ν ∈ {1, . . . , m}, and β > fν−1 − Eν − λν , β > Fν−1 − eν − λ, where ν ∈ {2, . . . , m}. Then, under the conditions of Theorem 2 and the hypotheses (U1 ), . . . , (Um−1 ), we have Tν,α,β (Σm ) − Σm ∈ Ωm−1 ,
ν = 2, . . . , m,
T1,α (Σm ) − Σm ∈ Ωm−1 .
(21) (22)
Proof. 1. Let us begin with the simplest case: we prove (21) for ν = m. Let ρm−1 , 1), ρ¯(1) m = (¯
ρ¯(2) ρm−1 , 2), m = (¯ 5737
where ρ¯m−1 is an arbitrary ρ¯m−1 -vector. In the first case, from (9) we have tm = sm ; in the second case, tm = sm + tm−1 + λm . Summing over variable sm , from (8) and (20), we obtain Tm,α,β (Σm ) − Σm i1 (α)
= − sign(α)
m−1
(−1)m−1−l
i=i0 (α) l =1 ρ¯m−1 s2 ,...,sm−1 ∈Z+ i1 (β)
+ sign(β)
m−1
(−1)m−1−l
Pm (t1 , . . . , tm−1 , i) Q1 (t1 ) · · · Qm−1 (tm−1 )Qm (i)
i=i0 (β) l =1 ρ¯m−1 s1 ,...,sm−1 ∈Z+
×
Pm (t1 , . . . , tm−1 , tm−1 + i + λm ) , Q1 (t1 ) · · · Qm−2 (tm−2 )Qm−1 (tm−1 )Qm (tm−1 + i + λm )
l = l(¯ ρm−1 ).
(23)
For a fixed i, each summand of the first sum has the following form:
1 ¯ ¯ Pm (t1 , . . . , tm−1 , i); Qm−1 ; λm−2 , Σm−1 Qm (i) all conditions (7), (I)–(III) for it are satisfied trivially; Qm (i) = 0, because, for i ≥ 0, this inequality is obvious and, in the case α < 0, i ∈ {α, . . . , −1}, we have i + em ≥ α + em ≥ 1. Therefore, according to (Um−1 ), all summands of the first sum are elements of Ωm−1 . Each summand of the second sum for a fixed i has denominators similar to item (1) of Lemma 1 with ¯ ¯ ¯ ν = m; therefore, (7), II(Q m−1 ), and III(Qm−1 ) are satisfied. It is necessary to check only I(Pm−1 , Qm−1 ), where Pm−1 = Pm (x1 , . . . , xm−1 , xm−1 + i + λm ). We have obvious relations: ρj ≤ ρj , j = 1, . . . , m − 2, ¯ ρm−1 ≤ ρm−1 + ρm ; qj = qj , j = 1, . . . , m − 2, qm−1 = qm−1 + qm , whence I(Pm−1 ,Q m−1 ). By virtue of (Um−1 ), all summands of the second sum in (23) are elements of Ωm−1 , i.e., (21) holds for ν = m. (1) (2) 2. Now we prove (21) for ν ∈ {2, . . . , m−1}. Let us denote ρ¯m = {¯ ρm | ρν = 1}, ρ¯m = {¯ ρm | ρν = 2}, (1) ρ¯m = (¯ ρν−1 , ρ¯∗m−ν+1 ), where ρ¯ν−1 and ρ¯∗m−ν+1 are arbitrary ρ¯ν−1 - and ρ¯m−ν+1 -vectors, respectively; (1)
ρν−1 ), l∗ = l(¯ ρ∗m−ν+1 ), l + l∗ = l = l(¯ ρm ), l ≥ 2. l = l(¯ By analogy with (23), we have
Tν,α,β (Σm ) − Σm = σ1 + σ2 , i1 (α)
σ1 = − sign(α)
i=i0 (α)
σ1,i = Dσ
ν−1
m−ν+1 l∗ =1
ρ¯∗m−ν+1
σ2 = − sign(β)
σ2,i ,
(25)
i=i0 (β)
(−1)ν−1−l Q1 (t1 ) · · · Qν−1 (tν−1 )
sν+1, ...,sm sν =i
σ2,i = Dσ
i1 (β)
l =1 ρ¯ν−1 s1 ,...,sν−1 ∈Z+
×
1 σ1,i , Qν (i)
(24)
(−1)m−ν+1−l ∈Z+
m−1
l=1
(2) ρ¯m
∈Z+
s1 ,...,sm sν =i
∗
Pm (t1 , . . . , tν−1 , sν , tν+1 , . . . , tm ) , Qν+1 (tν+1 ) · · · Qm (tm )
(−1)m−l Pm (t1 , . . . , tm ) . Q1 (t1 ) · · · Qm (tm )
(26)
(27)
¯ m ), Consider σ1,i . Just as in item 1, we have Qν (i) = 0 since α + eν ≥ 1. Similarly to (10), from III(Q we obtain Qj (tj ) = 0, j > ν. Let us sum over variables sν+1 , . . . , sm in (26). By Lemma 2 (see also Note 3), we get that the ∗ ∗ ∗ (t1 , . . . , tν−1 ), Pν−1,i ∈ Q[x1 , . . . , xν−1 ], degxj Pν−1,i ≤ pj for sum in square brackets is a polynomial Pν−1,i 5738
∗ ¯ ν−2 ) hold since (7) and (I)–(III) j = 1, . . . , ν −1. All conditions (7), (I)–(III) for σ1,i = Σν−1 (Pν−1,i , Qν−1 , λ hold for Σm . Therefore, according to (Uν−1 ), we have σ1,i ∈ Ων−1 and, hence, σ1 ∈ Ων−1 (see (25)). (2) Consider (27). We have l(¯ ρm ) = l, tν = tν−1 + i + λν . If ρν+1 = 2, then
tν+1 = tν + sν+1 + λν+1 = tν−1 + sν+1 + i + λν + λν+1 . We make the change of variables (ρj , λj , Qj , sj , tj ) → (ρj , λj , Qj , sj , tj ), where Qj and λj are defined as in item (1) of Lemma 1, if j = 1, . . . , ν − 1, (ρj , sj , tj ) (ρj , sj , tj ) = (28) (ρj+1 , sj+1 , tj+1 ) if j = ν, . . . , m − 1. We denote Pm−1,i = P (x1 , . . . , xν−1 , xν−1 + i + λν , xν , . . . , xν−1 ). The inequalities for β imply analogous ones for i , i.e., the conditions in item (1) of Lemma 1 are satisfied (if i ≥ 0, then everything is clear, because fν−1 − Eν − λν < 0, Fν−1 − eν − λν < 0 follow from IIν and IIIν ; if i < 0, then β < 0, i ≥ β). After the change of variables (28), we get that σ2,i = ¯ ¯ ¯ ¯ −Σm−1 (Pm−1,i , Q m−1 , λm−2 ), and, according to Lemma 1, (7), II(Qm−1 ), and III(Qm−1 ) are satisfied. ¯ We have only to check I(Q m−1 ). Similarly to item 1, we have pj ≤ pj , j = 1, . . . , ν−2, pν−1 ≤ pν−1 +pν , = qν−1 + qν , qj = qj+1 , j = ν, . . . , m − 1. pj ≤ pj+1 , j = ν, . . . , m − 1; qj = qj , j = 1, . . . , ν − 2, qν−1 ¯ Therefore, I(Q m−1 ) follows from I(Qm ). Thus, by virtue of (Um−1 ), we obtain σ2,i ∈ Ωm−1 . From (25), we have σ2 ∈ Ωm−1 and, from (24), we obtain (21) since σ1 ∈ Ων−1 . 3. Let us prove (22). Let ρ¯m = 1, 2, . . . , 2, 1, ρr+2 , . . . , ρm , r ∈ {1, . . . , m}. r−1
Summing over variable s1 , we obtain i1 (α)
T1,α (Σm ) − Σ = − sign(α)
Dσ
m l=1
i=i(α)
ρ¯m
(−1)m−1
s2 ,...,sm ∈Z+ s1 =i
Pm (t1 , . . . , tm ) , Q1 (s1 + σ)Q2 (t2 ) · · · Qm (tm )
(29)
where t1 = s1 + σ = i + σ, tj = s2 + · · · + sj + λ2 + · · · + λj + i + σ, j = 2, . . . , r, tr+1 = sr+1 , tj are defined by formulas (9) with j = r + 2, . . . , m. Since, in this situation, r ∂R ∂R (t1 , . . . , tm ) + (t1 , . . . , tm ) , Dσ (R(t1 , . . . , tm )) = ∂x1 ∂xν σ=0 σ=0 ν=2
where the second summand is absent for r = 1, it follows from (29) that i1 (α)
T1,α (Σm ) − Σm = − sign(α)
i1 (α)
σ1,i − sign(α)
i=i0 (α)
where σ1,i =
m l=1
ρ¯m
m−l
(−1)
s2 ,...,sm ∈Z+ s1 =i
i=i0 (α)
∂ ∂t1
1 σ2,i , Q1 (i)
Pm (t1 , . . . , tm ) Q1 (t1 ) · · · Qm (tm )
(30)
,
(31)
t1 = s1 , tj are defined by formulas (9) for j = 2, . . . , m, σ2,i = Dσ
m−1 l=1
ρ¯m s2 ,...,sm ∈Z+ r>1
(−1)m−l
P (i, t2 , . . . , tm ) , Q2 (t2 ) · · · Qm (tm )
(32)
5739
t2 = s2 + i + λ2 + σ, tj = s2 + · · · + sj + i + λ2 + · · · + λj + σ, j = 3, . . . , r, tr+1 = sr+1 , tj are defined by formulas (9) for j = r + 2, . . . , m. First, consider (31). Taking into consideration the condition α + e1 ≥ 1, we obtain in the standard way that Q1 (i) = 0. For j = 2, . . . , r, similarly to the consideration of (10), we have Qj (tj ) = 0, since tj + ej = s2 + · · · + sj + λ2 + · · · + λj + ei + i ≥ i + e1 . Hence, the rational function in sum (31) satisfies the conditions of Lemma 2, i.e., all σ1,i ∈ Q.
(33)
¯ We make the standard shift of the form (28) in (32): ρ¯m−1 = (ρ1 , ρ3 , . . . , ρm ), Q m−1 = (Q2 (x + i + λ2 ), ¯ Q3 (x), . . . , Qm (x)), λ = (λ + i + λ , λ , . . . , λ ); s = s , j = 1, . . . , m − 1; t1 = s1 + σ, tj defined 3 2 4 m j+1 m−2 j ¯ by formulas (9), where (sj , tj , λj ) → (sj , tj , λj ). From item (2) of Lemma 1, it follows that (7), II(Q m−1 ), ¯ and III(Qm−1 ) are satisfied. From (32), since l(¯ ρm−1 ) = l(ρm ) (the inequality r > 1 implies that ρ2 = 2), we obtain ¯ ¯ m−1 , λ σ2,i = −Σm−1 (Pm (i, x1 + i + λ2 , x2 , . . . , xm−1 ), Q m−2 ). ¯ The conditions I(Pm−1 , Q m−1 ) are checked trivially. Therefore, by (Um−1 ), all σ2,i ∈ Ωm−1 .
(34)
Condition (22) follows from (30), (33), and (34). The lemma is proved. Now we prove Lemma 4 whose fragments are present in the proof of Lemma 3. Lemma 4. Let α, β ∈ Z, and let it satisfy the same conditions as in Lemma 3, ν ∈ {1, . . . , m}. Then ¯ m−1 )) = Σm (R , λ ¯ (35) Tγ,α,β (Σm (Rm , λ m m−1 ), where = Rm (x1 , . . . , xν−1 , xν + α, xν+1 , . . . , xm ), Rm
(36)
¯ λ m−1 = (λ2 , . . . , λν−1 , λν + β − α, λν+1 + α, λν+2 , . . . , λm ),
(37)
where, for ν = 1, Tν,α,β → T1,α , λν + β − α is absent in (37); for ν = m, λν+1 + α is absent in (37). Proof. 1. First, consider the first sum in (20), where ρν = 1. Since sν ≥ α, we have sν = sν + α, sν ∈ Z+ . Let sj = sj , j = 1, . . . , m, j = ν; tj be defined by formulas (9), where (sj , λj ) → (sj , λj ), λj be defined in (37). Then ¯ , λm−1 ) = Dσ Σm = Σm (Rm
m ρ¯m
l=1
(−1)m−l Rm (t1 , . . . , tm ).
(38)
s1 ,...,sm ∈Z+
We prove that the set of terms of series is the same as the set of terms of series (20) for ρ¯m such that ρν = 1. 1.1. Let ρν+1 = 1. Then tν = sν ,
tν+1 = sν+1 ,
tν = tν + α,
tν+1 = tν+1 .
From (37) we have tj = tj , j = 1, . . . , m, j = ν, i.e., (t1 , . . . , tm ) = Rm (t1 , . . . , tν−1 , tν + α, tν+1 , . . . , tm ) = Rm (t1 , . . . , tm ). Rm
1.2. ρν+1 = 2. Here, tν = sν ,
tν+1 = sν+1 + tν + λν+1 = sν+1 + sν − α + λν+1 + α = tν+1 , tν = sν − α = tν − α.
5740
From (37) we again have tj = tj , j = 1, . . . , m, j = ν, and (t1 , . . . , tm ) = Rm (t1 , . . . , tm ) Rm
as in 1.1. 2. Now, consider the second sum in (20), where ρν = 2, and prove the statement which is analogous to item 1 (here, sν = sν + β, sν ∈ Z+ , the rest is as in item 1). 2.1. ρν+1 = 1. Here, tν = tν−1 + sν + λν , = tν−1 + sν + λν = tν−1 + sν − β + λν + β tν = tν + α. Then tν+1 = sν+1 = sν+1 = tν+1 , tν
i.e., as in 1.1, we have
− α = tν − α, i.e., as in 1.1, we have tj = tj , j = ν,
(t1 , . . . , tm ) = Rm (t1 , . . . , tm ). Rm
2.2. ρν+1 = 2. Similarly to item 2.1, e have tν = tν + α. Hence, tν+1 = tν + sν+1 + λν+1 , tν+1 = tν + sν+1 + λν+1 = tν − α + sν+1 + λν+1 + α = tν+1 . Similarly to the previous, we have (t1 , . . . , tm ) = Rm (t1 , . . . , tm ), Rm
and the lemma is proved. 2. Representation of Multiple Integrals in the Form of Multiple Series In this section, we represent integral (1) in the form of series (8). A number of lemmas being proved here will be useful for the calculation of other multiple integrals, in particular, J2k . Moreover, we prove another auxiliary result (Lemma 9), which is necessary for the proof of Proposition 1. The following Gauss’s hypergeometric function will play an essential role: ∞ Γ(s + a)Γ(s + b)Γ(c)z s . F (a, b, c, z) = Γ(a)Γ(b)Γ(s + 1)Γ(s + c) s=0
Lemma 5. Let a, b, c ∈ 1 0
Z+ ,
a + b ≥ c, z ∈ (0; 1). Then
(−1)b+c+1 xa (1 − x)b dx = (1 − xz)c+1 Γ(c + 1)Γ(a + b + 1 − c) ∞
× Dσ
(t − l + 1 + σ) · · · (t + σ)(t − n + 1 + σ) · · · (t + m + σ)(1 − z)t−n+σ , (39)
t=min(n,l)
where n = min(a, c), m = max(a, c) − min(a, c), l = a + b − max(a, c). Proof. By Euler’s formula [1, p. 72, formula (10)], we have 1 0
Γ(a + 1)Γ(b + 1) xa (1 − x)b dx F (c + 1, a + 1, a + b + 2, z). = (1 − xz)c+1 Γ(a + b + 2)
(40)
By [1, p. 82, formula (4)], for n, m, l ∈ Z+ , we have F (c + 1, a + 1, a + b + 2, z) = F (n + 1, n + m + 1, n + m + l + 2, z)
l ln(1 − z) (−1)m+1 (n + m + l + 1)! dn+m m+l d (1 − z) , (41) = l! n! (n + m)! (n + l)! dz n+m dz l z 5741
where n = min(a, c) ∈ Z+ , m = max(a, c) − min(a, c) ∈ Z+ , l = a + b − max(a, c) ∈ Z+ since a + b ≥ c, b ∈ Z+ . For fixed z ∈ (0, 1), we have ∞ 1 ln(1 − z) = Dσ (1 − z)σ , = (1 − z)t . z t=0
Let us denote
(−1)m+1 (n + m + l + 1)! . n! l! (m + l)! (n + m)! From (41) and (42), we have successively ∞ l dn+m m+l d t+σ F (c + 1, a + 1, a + b + 2, z) = λDσ n+m (1 − z) (1 − z) dz dz l
(42)
λ=
= λ(−1)l Dσ
dn+m dz n+m
= λ(−1)m+n+l Dσ
∞
t=0
(t − l + 1 + σ) · · · (t + σ)(1 − z)t+m+σ
t=0 ∞
(t − l + 1 + σ) · · · (t + σ)(t − n + 1 + σ) · · · (t + m + σ)(1 − z)t−n+σ , (43)
t=min(n,l)
where we made use of the fact that the summands of series (43) vanish after the application of the operator Dσ for 0 < t < min(n, l). Furthermore, n! (m + n)! = Γ(a + 1)Γ(c + 1),
l! (m + l)! = Γ(b + 1)Γ(a + b + 1 − c),
(−1)l+n+1 = (−1)b+c+1 .
Therefore, from (42), we have (−1)m+n+l λ =
(−1)b+c+1 Γ(a + b + 2) , Γ(a + 1)Γ(b + 1)Γ(c + 1)Γ(a + b + 1 − c)
and, hence, (39) follows from (40) and (43). Lemma 6. Let µ, ν ∈ Z+ , t ∈ R, t = −1, −2, . . . , z ∈ (0, 1), 1 xν (1 − x)µ (1 − xz)t dx.
Jµ,ν,t (z) = 0
Then (−1)µ Jµ,ν,t (z) = (t + 1) · · · (t + ν + µ + 1) −
∞
∞
(s − t − µ) · · · (s − t − 1)(s + 1) · · · (s + ν)(1 − z)s
s=0
(t + s + µ + 2) · · · (t + s + µ + ν + 1)(s + 1) · · · (s + µ)(1 − z)s+t+µ+1 . (44)
s=0
Proof. Everywhere below we assume that t ∈ R \ Z. If t ∈ Z+ , t = N , then, setting t = N + ε, where ε ∈ (0; 1) in (44), we can make the standard transition ε → 0 + 0. We apply one of 20 Kummer’s relations for Gauss’s hypergeometric function (see [1, p. 115, formula (33)]): F (a, b, c, z) =
5742
Γ(c)Γ(c − a − b) F (a, b, a + b + 1 − c; 1 − z) Γ(c − a)Γ(c − b) Γ(c)Γ(a + b − c) (1 − z)c−a −b F (c − a, c − b, c + 1 − a − b; 1 − z). (45) + Γ(a)Γ(b)
By the Euler’s formula that was applied in Lemma 5, we have Γ(ν + 1)Γ(µ + 1) F (−t, ν + 1, ν + µ + 2; z). Jµ,ν,t (z) = Γ(ν + µ + 2)
(46)
Setting a = −t, b = ν + 1, and c = ν + µ + 2 in (45), we get F (−t, ν + 1, ν + µ + 2; z) ∞
= +
Γ(ν + µ + 2)Γ(t + µ + 1) Γ(−t + s)Γ(ν + s + 1) Γ(−t − µ) (1 − z)s Γ(t + µ + ν + 2)Γ(µ + 1) Γ(−t)Γ(ν + 1)Γ(s + 1) Γ(−t − µ + s) s=0 ∞
Γ(ν + µ + 2)Γ(−t − µ − 1) Γ(−t)Γ(ν + 1)
s=0
Γ(t + µ + ν + 2 + s)Γ(µ + 1 + s) Γ(t + µ + 2) (1 − z)s+t+µ+1 . Γ(t + µ + ν + 2)Γ(µ + 1)Γ(s + 1) Γ(t + µ + 2 + s) (47)
After reduction of some ratios of gamma-function in (47) (the following eight fractions—four for each summand (47)): 1)
Γ(−t + s) = (−t − µ + s) · · · (−t + s − 1) = (s − t − µ) · · · (s − t − 1); Γ(−t − µ + s)
2)
Γ(ν + s + 1) = (s + 1) · · · (s + ν); Γ(s + 1)
3)
1 (−1)µ Γ(−t − µ) = = ; Γ(−t) (−t − 1) · · · (−t − µ) (t + 1) · · · (t + µ)
4)
1 Γ(t + µ + 1) = ; Γ(t + µ + ν + 2) (t + µ + 1) · · · (t + µ + ν + 1)
5)
Γ(t + µ + ν + 2 + s) = (t + s + µ + 2) · · · (t + s + µ + ν + 1); Γ(t + µ + 2 + s)
6)
Γ(µ + 1 + s) = (s + 1) · · · (s + µ); Γ(s + 1)
7)
1 Γ(t + µ + 2) = ; Γ(t + µ + ν + 2) (t + µ + 2) · · · (t + µ + ν + 1)
8)
1 (−1)µ+1 Γ(−t − µ − 1) = = Γ(−t) (−t − µ − 1) · · · (−t − 1) (t + 1) · · · (t + µ + 1)
from 1)–8), (47), and (46) we get (44), and the lemma is proved. For z ∈ (0; 1], we define the function (see (1)) 2k−1
f2k−1 (z) = [0,1]2k−1
j=1
α −1
xj j
(1 − xj )βj −1 dx1 · · · dx2k−1
(1 − x1 + x1 x2 − x1 x2 x3 + · · · + x1 x2 · · · x2k−2 − zx1 x2 · · · x2k−1 )α0
,
(48)
where k ≥ 2. From (1) and (48), we have J2k−1 = f2k−1 (1).
(49)
We denote P10 (x) = (x + α0 − β1 + 1) · · · (x + α0 − 1), Pj1 (x) = (x + 1) · · · (x + α2j−1 − 1),
j = 1, . . . , k; 5743
and, for j = 1, . . . , k − 1, Pj2 (x) = (x + α2j ) · · · (x + α2j + β2j − 1),
Pj3 (x) = (x + α2j−1 ) · · · (x + α2j+1 + β2j+1 − 1),
Gj (x) = (x + 1) · · · (x + β2j−1 − 1); P10 (x) (−1)α0 +β1 +β3 +···+β2k−1 −k Γ(β2 )Γ(β4 ) · · · Γ(β2k−2 ) ; Λ2k−1 = . P12 (x)P13 (x) Γ(α0 )Γ(α1 + β1 − α0 ) Everywhere below, σ ∈ − 12 , 0 ∪ 0; 12 , Dσ∗ (f (σ)) = lim f (σ). Obviously, for functions f (σ) having R1 (x) =
σ→0
a continuous derivative in zero, we have Dσ∗ (f (σ)) = Dσ (f (σ)). Finally, for k ≥ 3, j ∈ {2, . . . , k − 1}, let Rj (xj−1 , xj ) =
Gj (xj − xj−1 − β2j−1 ) . Pj2 (xj )Pj3 (xj )
All this notation is referred to below as the notation of Lemma 7. Lemma 7. Let all conditions (2)–(5) for the parameters αj , βi of the function f2k−1 (z) be satisfied and z ∈ (0; 1]. Then f2k−1 (z) = Λ2k−1 Dσ
k l=1
ρ¯k
s1 ,...,sk
¯ k (t1 , . . . , tk )(1 − z)tk , (−1)k−l R
(50)
∈Z+
Dσ∗ ;
for all z ∈ (0; 1], the series in (50) converges absolutely and where the operator Dσ can be replaced by uniformly with respect to σ ∈ − 12 ; 12 ; l and ρ¯k are defined as in (8) for m = k; and if j = 1, µ0 = min(0, β1 − α0 ), s1 + µ0 + σ tj = sj (51) if j = 2, . . . , k, ρj = 1, tj−1 + sj + β2j−1 if j = 2, . . . , k, ρj = 2, ¯ k (x1 , . . . , xk ) = R1 (x1 )R2 (x1 , x2 ) · · · Rk−1 (xk−2 , xk−1 )Gk (xk − xk−1 − β2k−1 )Pk1 (xk ). R Series (50) converges with any classical definition of partial sums (triangular, rectangular, or spherical ). Proof. Induction on k. For k = 2, by virtue of Lemma 5 with a = α1 − 1, b = β1 − 1, and c = α0 − 1 (by (2), we have c ≤ a , see Lemma 5), we have n = α0 − 1, m = α1 − α0 , and l = α1 + β1 − α1 − 1 = β1 − 1. Making the change of variables z → 1 − x2 + x2 x3 z, from (48) we obtain f3 (z) = [0;1]2
=
α −1 xj j (1
βj −1
− xj )
j∈{2;3}
0
(−1)α0 +β1 −1 Γ(α0 )Γ(α1 + β1 − α0 ) 1
× 0
1
x2s+α2 −α0 +σ (1
Dσ∗
dx2 dx3
(s − β1 + 2 + σ) · · · (s + σ)(s − α0 + 2 + σ) · · · (s + α1 − α0 + σ)
s≥µ0 +α0 −1
β2 −1
− x2 )
x1α1 −1 (1 − x1 )β1 −1 dx1 (1 − x1 (1 − x2 + x2 x3 z))α0
1 dx2
x3α3 −1 (1 − x3 )β3 −1 (1 − x3 z)s−α0 +1+σ dx3 .
(52)
0
In the first integral in (52), we have s + α2 − α0 ≥ 0 for s ≥ µ0 + α0 − 1. Indeed, for β1 < α0 , we have µ0 = β1 − α0 (see (51)), µ0 + α2 − 1 = α2 + β1 − α0 − 1 ≥ 0 due to (3) with r = 0; and, for β1 ≥ α0 , we have µ0 = 0, µ0 + α2 − 1 ≥ 0 since α2 ∈ N. Therefore, this beta-integral converges. We apply Lemma 6 / Z. Calculate the beta-integral by the formula to the second integral in (52), where t = s − α0 + 1 + σ ∈ Γ(β2 ) Γ(β2 ) Γ(β2 )Γ(t + α2 ) = = . Γ(t + α2 + β2 ) (t + α2 ) · · · (t + α2 + β2 − 1) P12 (t) 5744
We make the change of variable s → µ0 + α0 − 1 + s1 , with s1 ∈ Z+ , and let t1 = s1 + µ0 + σ (see (51)). Then t = t1 , and (52) has the following form: P10 (t1 )P11 (t1 ) f3 (z) = Λ3 Dσ∗ P12 (t1 )(t1 + 1) · · · (t1 + α3 + β3 − 1) s1 ∈Z+
× (s2 + 1) · · · (s2 − α3 − 1)(s2 − t1 − β3 + 1) · · · (s2 − t1 + 1)(1 − z)s2 s2 ∈Z+
−
t1 +s2 +β3
(t1 + s2 + β3 + 1) · · · (t1 + s2 + α3 + β3 − 1)(s2 + 1) · · · (s2 + β3 − 1)(1 − z)
.
s2 ∈Z+
After the reduction
1 P11 (t1 ) = (t1 + 1) · · · (t1 + α3 + β3 − 1) P13 (t1 ) for α1 ≤ α3 + β3 (we put off the case α1 > α3 + β3 , see Note 7), in the notation of Lemma 7, we obtain
∗ R1 (t1 )P21 (s2 )G2 (s2 − t1 − β3 )(1 − z)s2 f3 (z) = Λ3 Dσ s1 ,s2 ∈Z+ t1 +s2 +β3 . (53) − R1 (t1 )P21 (s2 + t1 + β3 )G2 (s2 )(1 − z) s1 ,s2 ∈Z+
Let us prove that the series in (53) coincide with the series in (50). Indeed, two cases are possible for the vector ρ¯2 : (1) ρ¯2 = (1, 1). Here, l = 2, (−1)k−l = 1, t2 = s2 , and the first summands in (50) and (53) coincide; (2) ρ¯2 = (1, 2). Here, l = 1, (−1)k−l = −1, t2 = t1 + s2 + β3 and the second summands in (50) and (53) also coincide. For a fixed σ ∈ − 12 , 12 , let us check the convergence of the series in (50) which, for k = 2, have the following form: 2 (−1)l P10 (t1 )G2 (t2 − t1 − β3 )P21 (t2 ) (1 − z)t2 . (54) P (t )P (t ) 12 1 13 1 + ρ¯ l=1
2
s1 ,s2 ∈Z
If ρ¯2 = (1, 1), then t2 = s2 ≥ 0. If ρ¯2 = (1, 2), then we shall prove that 1 t2 ≥ s1 + s2 + . 2 From (51) we have
(55)
1 t2 = s1 + µ0 + σ + s2 + β3 ≥ s1 + s2 + µ0 + β3 − . 2 Similarly to the investigation of the beta-integral in (52), consider two cases: (1) β1 < α0 . Here, µ0 = β1 − α0 , µ0 + β3 − 12 ≥ 12 in view of (5) for k = 2; (2) β1 ≥ α0 . Here, µ0 = 0, µ0 + β3 − 12 ≥ 12 since β3 ∈ N. Inequality (55) is proved. We prove that, for any z0 ∈ (0, 1), (z, σ) ∈ [z0 , 1] × − 12 , 12 , series (54) is dominated by a double numerical series of the form 2 (M1 + 1)β1 +β3 −2 (M2 + 1)α3 +β3 −2 (1 − z0 )M2 , (56) C (M1 + 12 )β2 +α3 +β3 −α1 + l=1 ρ¯2 M1 ,M2 ∈Z
where C > 0,
(M1 , M2 ) =
(s1 , s2 ) (s1 , s1 + s2 )
if ρ2 = 1, if ρ2 = 2. 5745
Let us write the trivial estimates |P10 (t1 )| ≤ C1 (M1 + 1)β1 −1 , |P21 (t2 )| ≤ C3 (M2 + 1)α3 −1 ,
|G2 (t2 − t1 − β3 )| ≤ C2 (M1 + 1)β3 −1 (M2 + 1)β3 −1 ,
1 β2 1 α3 +β3 −α0 |P12 (t1 )| ≥ C4 M1 + , |P13 (t1 )| ≥ C5 M1 + , 2 2
where all Ci > 0, and we made use of the fact that 1 1 ≥ M1 + , 2 2 1 t1 + α1 = M1 + min(α0 , β1 ) + (α1 − α0 ) + σ ≥ M1 + 2 t1 + α2 ≥ M1 + α2 + µ0 −
(see (2)). These estimates prove (56). Since (β2 + α3 + β3 − α1 ) − (β1 + β3 − 2) = β2 + α3 − α1 − β1 + 2 ≥ 2 (see (3) for r = 1), series (56) converges. Evaluations analogous to the aforementioned ones can be applied ∂ n to the series obtained from (54) by termwise application of the operator , n ∈ N. Hence, for any ∂σ 1 1 ∗ z ∈ (0, 1], the series in (54) is an analytic function of σ ∈ − 2 ; 2 , i.e., Dσ can be replaced by Dσ . The lemma is completely proved for k = 2. Let us make a step of induction k → k + 1, where k ≥ 2. From the recurrent equation (suggested by Y. Nesterenko) α −1 xj j (1 − xj )βj −1 f2k−1 (1 − x2k + x2k x2k+1 z) dx2k dx2k+1 f2k+1 (z) = [0,1]2 j∈{2k,2k+1}
and (50), which holds by the induction hypothesis, we obtain f2k+1 (z) = Λ2k+1 Dσ∗
k l=1
1 ×
ρ¯k
xα2k2k −1+tk (1
0
¯ k (t1 , . . . , tk ) (−1)k−l R
s1 ,...,sk ∈Z+ β2k −1
− x2k )
1
α
2k+1 x2k+1
dx2k
−1
(1 − x2k+1 )β2k+1 −1 (1 − x2k+1 z)tk dx2k+1 . (57)
0
First, consider the convergence of the beta-integral in (57). Let us denote (1) ρ¯k = 1, 2, . . . , 2 . k−1
We consider two cases: (1)
(1) ρ¯k = ρ¯k . From (51) we obtain that tk ≥ 0 (if ρk = 1, then tk = sk ≥ 0; and, if ρk = 2, then, for some r ∈ {2, . . . , k−1}, we have ρr = 1, ρr+1 = · · · = ρk = 2, tk = sr +· · ·+sk +β2r+1 +· · ·+β2k−1 ≥ 1). Hence, α2k − 1 + tk ≥ 0 and the beta-integral converges; (1) (2) ρ¯k = ρ¯k . Then tk = s1 + · · · + sk + µ0 + σ + β3 + β5 + · · · + β2k−1 . If β1 ≥ α0 , then µ0 = 0, tk ≥ 0, and the beta-integral converges. If β1 < α0 , then µ0 = β1 − α0 . From (4 ), for r = k, d = 2k, we obtain α2k −1+tk ≥ s1 +· · ·+sk +(β1 +β3 +· · ·+β2k−1 +α2k −α0 −1)− and the beta-integral in (57) converges. 5746
1 1 1 ≥ s1 +· · ·+sk − ≥ − , (58) 2 2 2
Applying Lemma 5 to the second integral in (57) and calculating the beta-integral in (52), in notation of Lemma 7, from (57) we obtain f2k+1 (z) = (−1)β2k+1 −1 Γ(β2k )Λ2k−1 Dσ∗
k l=1
ρ¯k
Γ(β2k ) Pk2 (tk )
just as
(−1)k−l
s1 ,...,sk ∈Z+
1 ¯ k (t1 , . . . , tk ) ×R Pk2 (tk )(tk + 1) · · · (tk + α2k+1 + β2k+1 − 1)
Gk+1 (sk+1 − tk − β2k+1 )Pk+1,1 (sk+1 )(1 − z)sk+1 × sk+1 ∈Z+
−
tk +sk+1 +β2k+1
Gk+1 (sk+1 )Pk+1,1 (tk + sk+1 + β2k+1 )(1 − z)
.
(59)
sk+1 ∈Z+
It should be noted that (−1)β2k+1 −1 Γ(β2k )Λ2k−1 = Λ2k+1 . After the reduction in (59) 1 Pk1 (tk ) = (tk + 1) · · · (tk + α2k+1 + β2k+1 − 1) Pk3 (tk ) ¯ k in (50)), we find that the first summand in (59) corresponds to ρ¯k+1 = (¯ (see the structure of R ρk , 1), ρk , 2), whence (50) is formally satisfied as k → k + 1. and the second one corresponds to ρ¯k+1 = (¯ It remains to consider the convergence of this series. First, consider the behavior of the polynomials Pk2 (x) and Pk3 (x) for x = tk and σ ∈ − 12 , 12 . From (58) we obtain 1 (60) α2k + tk ≥ + s1 + · · · + sk . 2 Analogously, with the help of (4) for r = k and d = 2k − 1, we obtain 1 + s1 + · · · + sk . (61) 2 In particular, Pk2 (tk ) = 0 and Pk3 (tk ) = 0. Similarly to the proof of (55) and (58), with the help of (5) with k → k + 1, we obtain that, for ρ¯k+1 such that ρk+1 = 2, α2k−1 + tk ≥
1 (62) tk+1 ≥ . 2 Now, we should only construct a dominating series similar to (56) for the multiple series (50) as k → k + 1: k+1 (M1 + 1)n1 · · · (Mk+1 + 1)nk+1 (1 − z0 )Mk+1 C , (63) (M1 + 12 )N 1 · · · (M1 + 12 )Nk + l=1 ρ¯ k+1
M1 ,...Mk+1 ∈Z
where we use (60) and (61). In (63), we have C > 0, if j = 1, . . . , k + 1, ρj = 1, sj Mj = sr + · · · + sj if ρr = 1, ρr+1 = · · · = ρj = 2, j = 2, . . . , k + 1, r ∈ 1, . . . , j − 1, Nj = β2j + α2j+1 + β2j−1 − α2j−1 , j = 1, . . . , k, nj = β2j−1 + β2j+1 − 2, j = 1, . . . , k, nk+1 = α2k+1 + β2k+1 − 2. For all j = 1, . . . , k, we have Nj − nj = β2j + α2j+1 − α2j−1 − β2j−1 + 2 ≥ 2 5747
since (3) holds for r = 2j − 1. This proves the convergence of series (63). Similarly to the case k = 2, the operator Dσ∗ may be replaced by Dσ , and (59) coincides with (50) completely as k → k + 1. The lemma is proved. Note 5. While deriving (52) and (57), we introduced the uniform convergence of the series ((39) and (50), respectively) for termwise integration. However, series (39) was considered only for z ∈ (0; 1]. Let us eliminate this inaccuracy. We shall consider only (52), because the situation in (57) is completely analogous. When deriving (52), we consider series (39) at the points z = 1 − x2 + x2 x3 z (z = 0 if x2 = 1, x3 = 0; z = 1 if x2 = 0). Integration on variables x2 and x3 over the rectangle {ε ≤ x2 ≤ 1 − ε, 0 ≤ x3 ≤ 1 − ε}, ε ∈ 0; 12 instead of the square [0; 1]2 , we obtain that z ∈ [ε; 1 − ε2 ], i.e., termwise integration in (52) is correct. The limit transition ε → 0 + 0 is trivial, because series (52) is dominated by series (56) (series (57) is dominated by (63), respectively). Note 6. It should be noted that, in the proof of Lemma 7, conditions (3) are applied only for odd r. Note 7. If, for a certain j ∈ {1, . . . , k − 1}, we have α2j−1 > α2j+1 + β2j+1 , then Γ(x + α2j+1 + β2j+1 ) 1 = . Γ(x + α2j−1 ) (x + α2j+1 + β2j+1 ) · · · (x + α2j−1 − 1) As for the rest, nothing changes in the proof of Lemma 7. Pj3 (x) =
Lemma 8. Let parameters αj , βi of integral (1) satisfy conditions (2)–(5) and k ≥ 2. Then J2k−1 = Λ∗2k−1 Dσ
k−1
l=1 ρ¯k−1 s1 ,...,sk−1 ∈Z+
∗ (−1)k−1−l Rk−1 (t1 , . . . , tk−1 ),
(64)
where
(−1)α0 +β1 +β3 +···+β2k−3 −k+1 Γ(β2 )Γ(β4 ) · · · Γ(β2k−2 )Γ(α2k−1 ) , Γ(α0 )Γ(α1 + β1 − α0 ) in the notation of Lemma 7 Λ∗2k−1 =
∗ = R1 (x1 )R2 (x1 , x2 ) · · · Rk−1 (xk−2 , xk−1 )Gk (xk−1 ). Rk−1
Proof. We apply Lemma 7 and (49). Owing to (62), all terms of series (50) vanish at z = 1 except for those ρk−1 , 1), tk = sk = 0. For z = 1 in (50), we replace ρ¯k by ρ¯k−1 and l = l(¯ ρk−1 ) = l(¯ ρk ) − 1 = with ρ¯k = (¯ l − 1 by (−1)k−l = (−1)k−1−l . Then Pk2 (0) = Γ(α2k−1 ), Gk (−tk−1 − β2k−1 ) = (−1)β2n−1 −1 Gk (tk ), (−1)β2k−1 −1 Γ(α2k−1 )Λ2k−1 = Λ∗2k−1 , and we obtain (64). Note 8. We emphasize equality (64) for k = 2: J3 = Λ∗3 Dσ
P10 (t1 )G2 (t1 ) . P12 (t1 )P13 (t1 ) +
(65)
s1 ∈Z
Considering, in particular, the case α0 = α1 = α2 = α3 = β1 = β2 = β3 = n + 1,
n ∈ Z+ ,
in (65), we obtain P10 (x) = G2 (x) = (x + 1) · · · (x + n),
Λ∗3 = −1,
P12 = P13 = (x + n + 1) · · · (x + 2n + 1).
Conditions (2)–(5) hold trivially, i.e., in this case, we have 2 d
(s + 1) · · · (s + n) J3 = J3 (n) = − ds (s + n + 1) · · · (s + 2n + 1) s∈Z+
∞ (s − 1) · · · (s − n) 2 d =− = r1 + r2 ζ(3), ds s(s + 1) · · · (s + n) s=1
This is the famous result by Beukers [2]. 5748
r1 , r2 ∈ Q.
We distinguish another special case of equality (64), which will be useful in the proof of Proposition 1. Lemma 9. Let
a ¯0m = 1, . . . , 1 ,
¯0 = 1, . . . , 1 λ . m−1
m
Then
m−1
0 0 ¯ ¯m , λ Sm a m−1 = −2ζ(2m + 1).
Proof. Consider integral (1) J2m+1 = J2m+1 (1), where all αj = βi = 1. Then all Pj1 (x) = Gj (x) = 1, Pj2 (x) = Pj3 (x) = x + 1; Λ∗2m+1 = −1; in (51) µ0 = 0, λj = β2j−1 = 1, j = 2, . . . , m. From (61) for ¯ 0 ). In [3] it was proved that J2m+1 (1) = 2ζ(2m + 1). The a0m , λ k = m + 1 we get J2m+1 (1) = −Sm (¯ m−1 lemma is proved. 3. Proof of Proposition 1 In the notation of Lemma 9 and (8), ¯ 0 ) = Σm (R0 , λ ¯0 a0m , λ Sm (¯ m−1 m m−1 ), where 0 = Rm
1 (x1 +
1)2 · · · (xm
+ 1)2
.
We apply the operator T = T1,α1 T2,α2 ,β2 . . . Tm,αm ,βm , where αν = aν − 1 ∈ Z+ , ν = 1, . . . , m,
βν = λν + aν − aν−1 − 1, ν = 2, . . . , m,
0 ,λ ¯ 0 ). By virtue of (14), all βν ∈ Z+ ; therefore, Lemmas 3 and 4 are applicable. Repeatedly to Σm (Rm m−1 applying Lemma 4, we obtain 0 ¯0 ¯ , λm−1 )) = Σm (Rm , λm−1 ), T (Σm (Rm
where, from (36), Rm =
1 (x1 +
a1 )2 · · · (xm
+ am )2
,
for ν = 2, . . . , m, from (37), λν = λ0ν + βν − αν + αν−1 = 1 + (λν + αν − αν−1 − 1) − αν + αν−1 = λν , i.e., we get 0 ¯0 ¯ m−1 ). T (Σm (Rm , λm−1 )) = Sm (¯ am , λ
Applying Lemma 3 together with (35) m times, from Lemma 9 we obtain (m) (m) 0 ¯0 ¯ , λm−1 )) = Σm Rm ,λ Tm,αm ,βm (Σm (Rm m−1 = −2ζ(2m + 1) + wm , m, λ ¯ m ) are defined in Lemma 4 (their explicit form is not used where wm ∈ Ωm−1 by Lemma 3 and (Rm m−1 below). Similarly, (m) (m) (m−1) (m−1) ¯ ¯ ,λ ,λ Tm−1,αm−1 ,βm−1 Σm Rm = Σm Rm = −2ζ(2m + 1) + wm + wm−1 , m−1 m−1
where wm−1 ∈ Ωm−1 by Lemma 3. Finally, 0 ¯0 ¯ m ) = −2ζ(2m + 1) + wm + wm−1 + · · · + w1 , T (Σm (Rm , λm−1 )) = Sm (¯ am , λ
where all wi ∈ Ωm−1 , and both assertions of Proposition 1 are proved. 5749
4. Completion of the Proof of Theorem 2 Let us introduce an equivalence relation on R in the following way: a ∼ b if (a − b) ∈ Ωm−1 . As was stated in Sec. 1, it is necessary to make the step of induction {(U1 ), . . . , (Um−1 )} → (Um ). Let us prove that (see (8)) ¯ m−1 ), r(¯ am )Sm (¯ am , λ (66) Σm ∼ a ¯m
where all r(¯ am ) ∈ Q, a ¯m = (a1 , . . . , am ), aj ∈ Aj2 , j = 1, . . . , m. Conditions (14) can be directly verified. Then (Um ) follows from Proposition 1, and Theorem 2 holds by induction. The proof of (66) will be divided into three stages. 1. Consider the partial-fraction expansion with respect to variable xm for fixed x1 , . . . , xm−1 fm Fm P (x1 , . . . , xm ) Bi Ci = , + Qm (xm ) (xm + i)2 xm + i
(67)
i=em
i=Em
¯ m ), where all Bi , Ci ∈ Q[x1 , . . . , xm−1 ], Bi = 0 for i ∈ / Am,2 , and Ci = 0 for i ∈ / Am . By virtue of Im (P, Q we have fm Ci = 0. (68) i=em
Moreover, for all Bi and Ci , we have degxj Bi ≤ pj , degxj Ci ≤ pj , because
j = 1, . . . , m − 1,
(69)
P (x1 , . . . , xm )(xm + i)2 , for i ∈ Am,2 , Bi = Qm (xm ) xm =−i P (x1 , . . . , xm )(xm + i)2 , for i ∈ Am,1 , Ci = Qm (xm ) xm =−i
P (x1 , . . . , xm )(xm + i)2 ∂ , for i ∈ Am,2 . Ci = ∂xm Qm (xm ) xm =−i
The following proposition analogous to [4, Lemma 7] will be useful below. Lemma 10. Let e, f ∈ N, e ≤ f , Ce , . . . , Cf ∈ R, (1)
f M s=0 i=e
(2)
f ∞ s=0 i=e
f i=e
Ci = 0, ∆ ∈ R, ∆ + e = 0, −1, −2, . . . . Then
f −1 f f −1 i 1 1 Ci = − , Cν Cν s+∆+i ∆ + i M + ∆+i ν=e r=e i=e
Ci = s+∆+i
i=e+1
f −1 i i=e
M ∈ Z+ ,
Cν
ν=e
1 . ∆+i
Proof. Let us prove (1). Then (2) follows from (1) as M → ∞. We have
M f +f f f M i−1 Ci 1 1 1 = − − Ci s+∆+i ν + ∆ ν=e ν + ∆ M +ν+∆ ν=e s=0 i=e
=−
i=e
f i=e+1
5750
Ci
i−1 ν=e
ν=i+1
f f −1 1 1 − Ci ν+∆ M +ν+∆ i=e
ν=i+1
=−
f −1
j=e
=−
f
Ci
i=j+1
f −1 j j=e
f j−1 1 1 − Ci ∆+j M +∆+j j=e+1
Ci
i=e
i=e
f j−1 1 1 − . Ci ∆+j M +∆+j j=e+1
i=e
Thus, (1) and, hence, the lemma are proved. Substituting (67) into (8) we obtain ¯ m−1 ) = Σ(1) + Σ(2) , ¯ m, λ Σm (Pm , Q m m
(70)
where Σ(1) m
Fm
=
Dσ
i=Em i∈Am2
Σ(2) m
= Dσ
m ρ¯m
l=1
m l=1
ρ¯m
s1 ,...,sm ∈Z+
s1 ,...,sm ∈Z+
(−1)m−l Bi (t1 , . . . , tm−1 ) , Q1 (t1 ) · · · Qm−1 (tm−1 )(tm + i)2
fm (−1)m−l Ci (t1 , tm−1 ) , Q1 (t1 ) · · · Qm−1 (tm−1 ) sm + ∆ + i
(72)
i=em
0 ∆= tm−1 + λm
and
(71)
if ρm = 1, if ρm = 2.
In series (72) which converges absolutely (see (1) of Lemma 10) we can, as before, permute the terms of the series. ρm−1 , ρm ). Then Let us sum over sm ∈ Z+ in (72) with the help of Lemma 10 (see (68)). Let ρ¯m = (¯ ρm−1 ) = l − 1 if ρm = 1 and l ≥ 2. In the second case, we make l(¯ ρm−1 ) = l if ρm = 2 and l ≤ m − 1; l(¯ the standard change l − 1 → l. Thus, we have Σ(2) m
= Dσ
−
m−1
l=1 ρ¯m−1 s1 ,...,sm−1 ∈Z+ f m −1 i=em
Dσ
m−1
f i m −1 (−1)m−1−l 1 Cν (t1 , . . . , tm−1 ) Q1 (t1 ) · · · Qm−1 (tm−1 ) i ν=e
l=1 ρ¯m−1 s1 ,...,sm−1 ∈Z+
i=em
i 1 (−1)m−1−l . (73) Cν (t1 , . . . , tm−1 ) Q1 (t1 ) · · · Qm−1 (tm−1 ) ν=e tm−1 + λm + i m
For i ∈ {em , . . . , fm−1 }, we denote Qm−1 (x) = Qm−1 (x)(x + λm + i). We are in situation (3) of Lemma 1 with ν = m. Therefore, (7), (II), and (III) are satisfied. The same is also valid for the first sum in (73). The conditions (I) are satisfied according to (69). By (Um−1 ), we obtain that (74) Σm ∼ Σ(1) m . 2. Step of induction. Suppose that, for some d ∈ {2, . . . , m − 1}, the following formula is proved: Σm ∼
a ¯∗m−d
Dσ
m l=1
ρ¯m
s1 ,...,sm ∈Z+
(−1)m−l Ba¯∗m−d (t1 , . . . , td ) , m Q1 (t1 ) · · · Qd (td ) (tj + aj )2
(Vd )
j=d+1
where a ¯∗m−d = (ad+1 , . . . , am ), aj ∈ Aj2 , j = d + 1, . . . , m; all Ba¯∗m−d ∈ Q[x1 , . . . , xd ], degxj Ba¯∗m−d ≤ pj , j = 1, . . . , d. In particular, by virtue of (71), relation (71) has the form (Vm−1 ) (the base of induction). Let us prove that we can proceed from (Vd ) to (Vd−1 ). 5751
For fixed a ¯∗m−d , x1 , . . . , xd−1 , we have the partial-fraction expansion fd Fd Bi Ci B(x1 , . . . , xd ) = , + 2 Qd (xd ) (xd + i) xd + i
(75)
i=ed i∈Ad
i=Ed i∈Ad2
where the index a ¯∗m−d is omitted for brevity and, similarly to (67)–(69), Bi , Ci ∈ Q[x1 , . . . , xd−1 ], fd
Ci = 0,
(76)
i=ed
degxj Bi ≤ pj , degxj Ci ≤ pj , Substituting (75) into (Vd ), we obtain
j = 1, . . . , d − 1.
(77)
(2) Σm ∼ Σ(1) m + Σm ,
(78)
where Fd
i=Ed i∈Ad2
a ¯∗m−d
Σ(1) m =
Σ(2) m
=
Dσ
m l=1
ρ¯m
s1 ,...,sm
(−1)m−l Bi,¯a∗m−d (t1 , . . . , td−1 )
∈Z+
d−1 j=1
Dσ
a ¯∗m−d
m l=1
ρ¯m
s1 ,...,sm ∈Z+
Qj (tj )
m
(tj + aj
,
j=d
fd Ci,¯a∗m−d (t1 , . . . , td−1 ) (−1)m−l , m d−1 2 i=e (∆ + s + i) (t + a ) d j j d Qj (tj ) i∈A d
j=1
0 ∆= td−1 + λd
(79)
)2
(80)
j=d+1
if ρd = 1, if ρd = 2.
As above, series (80) converges absolutely. We prove that
(81) Σ(2) m ∈ Ωm−1 . Then the required transition from (Vd ) to (Vd−1 ) follows from (78) and (79). For fixed a ¯∗m−d , consider four groups of series in (80) which have fixed coordinates ρd and ρd+1 (below, (1)
(4)
¯∗m−d . We sum over variable sd in each group of the ρ¯m , . . . , ρ¯m ). Again, for brevity, we omit the index a series. For brevity, we choose only multipliers depending on the variable sd . Let us denote fd−1 i i−1 i 1 (2) (3) (1) C = Cν , Ci = Cν , Ci = Cν . (82) i ν=e ν=e ν=e i=ed
I. ρd = ρd+1 = 1
(1) (¯ ρm ).
d
d
d
Then td = sd , td+1 = sd+1 , ∆ = 0. By Lemma 10 and (76), we have fd ∞ sd =0 i=ed
Ci = C (1) . sd + i
(83)
(2)
ρm ). Let ρd+1 = · · · = ρd+r = 2, ρd+r+1 = 1, where r ∈ {1, . . . , m − d}. We II. ρd = 1, ρd+1 = 2 (¯ denote M = sd + sd+1 , M ∈ Z+ , sd ∈ {0, . . . , M }. We have td = sd ,
td+1 = M + λd+1 ,
tj = tj−1 + sj + λj , j = d + 2, . . . , d + r,
td+r+1 = sd+r+1 .
By Lemma 10 for ∆=0, taking into account (82), (83), and (76), we have fd M sd =0 i=ed
5752
fd Ci 1 (2) (1) =C − . Ci sd + i M +i i=ed +1
(84)
(3)
III. ρd = 2, ρd+1 = 1 (¯ ρm ). Here, td = td−1 + sd + λd , td+1 = sd+1 , ∆ = td−1 + λd . We have fd ∞ sd =0 i=ed
f (3) d −1 Ci Ci = . sd + ∆ + i td−1 + λd + i
(85)
i=ed
(4)
ρm ). For some r ∈ {1, . . . , m − d}, we have ρd = · · · = ρd+r = 2, ρd+r+1 = 1. IV. ρd = ρd+1 = 2 (¯ Similarly to case II, we set M = sd + sd+1 , M ∈ Z+ , sd ∈ {0, . . . , M },
td+j
td = sd + ∆, where ∆ = td−1 + λd , td+1 = sd+1 + td + λd+1 = M + td−1 + λd + λd+1 , = td+j−1 + sd+j + λd+j , j = 2, . . . , r, td+r+1 = sd+r+1 ,
fd M sd =0 i=ed
f fd (3) (2) d −1 Ci Ci Ci = − sd + ∆ + i td−1 + λd + i M + td−1 + λd + i
(86)
i=ed +1
i=ed
(as usual, see Lemma 10, (82), and (76)). (2) After summing over sd in (80), we choose three summands according to the presence of C (1) , Ci , (3) and Ci in them, respectively: Σ(2) m = σ1 + σ2 + σ3 ,
(87)
where (83) and the first summand from (84) are taken in σ1 , the second summands from (84) and (86) are taken in σ2 , and (85) and the first summand from (86) are taken in σ3 . (a) Calculation of σ1 . Let Qj (x) = (x + aj )2 for j = d + 1, . . . , m; if j = 1, . . . , d − 1, (ρj , λj , Qj , sj , tj ) (ρj , λj , Qj , sj , tj ) = (88) (ρj+1 , λj+1 , Qj+1 , sj+1 , tj+1 ) if j = d, . . . , m − 1, except for j = 1 and j = d in the definition of λj (λ1 is absent, λd = λd + λd+1 ), ρ¯m−1 = (ρ1 , . . . , ρm−1 ), = C (1) (x1 , . . . , xd−1 ), Pm−1
¯ Q m−1 = (Q1 , . . . , Qm−1 ),
Rm−1 =
¯ λ m−2 = (λ2 , . . . , λm−1 ),
Pm−1 , Q1 (x1 ) · · · Qm−1 (xm−1 )
¯ Σm−1 = Σm−1 (Rm−1 ,λ m−2 ).
(1) (2) ρm−1 ) = l ρ¯m − 1, l ρ¯m = l + 1, We have l = l(¯ σ 1 = Dσ
m−1
l =1 ρ¯m−1 ρ¯d =1
s1 ,...,sm−1 ∈Z+
(−1)m−1−l Rm−1 (t1 , . . . , tm−1 )
−
m−1−l
(−1)
Rm−1 (t1 , . . . , tm−1 )
,
s1 ,...,sm−1 ∈Z+ sd ≥λd+1
where two changes are made in the second sum as compared to notation (88), namely, sd = M + λd+1 ≥ λd+1 and ρd = 1 instead of ρd+1 = 2 (then (88) implies that ρd = 2). From (20) we obtain that σ1 = Σm−1 − Td,λd+1 ,0 (Σm−1 ). For α = λd+1 , the condition of Lemma 3 is satisfied for ν = d: α+ed ≥ 1, because ed = ad+1 ≥ Ed+1 in view of ad+1 ∈ Ad+1,2 , α+ed = λd+1 +ad+1 ≥ λd+1 + Ed+1 > fd ≥ 1. Therefore, by Lemma 3, σ1 ∈ Ωm−2 (conditions (7), (I), and (III) can be easily verified). 5753
(b) Calculation of σ2 = −
fd i=ej +1
σ2,i .
Let us keep notation (88) with the following changes for j = d: sd = M + λd+1 ≥ λd+1 , Qd = (2) (4) ρm−1 ), (x + i − λd+1 )(x + ad+1 )2 , ρd = ρd , λd = λd , td = td , where sd → sd , l = l ρ¯ m = l ρ¯m = l(¯ because the coordinate ρd+1 = 2 is withdrawn from ρ¯m . Therefore, (−1)m−l = −(−1)m−1−l , σ2,i = −Dσ
m−1
l =1 ρ¯m−1 s1 ,...,sm−1 ∈Z+ sd ≥λd+1
(3)
(−1)m−1−l Ci (t1 , . . . , td−1 ) Q1 (t1 ) · · · Qm−1 (tm−1 )
(3) ¯ ¯ m−1 , λ = −Td,λd+1 ,λd+1 Σm−1 Ci , Q m−2
(see (8) and (20)). Let us verify that the conditions of Lemma 4, where ν = d, α = β = λd+1 , are satisfied. Since ad+1 + λd+1 ≥ Ed+1 + λd+1 > fd ≥ i, the parameters of the polynomial Qd are ed = i − λd+1 , Ed = Fd = fd = ad+1 . We have α + ed = λd+1 + i − λd+1 ≥ ed + 1 > 1; therefore, β > fd−1 − Ed − λd ⇐⇒ λd+1 + ad+1 + λd > fd−1 , which is obvious; β > Fd−1 − ed − λd ⇐⇒ λd+1 > Fd−1 − (i − λd+1 ) − λd ⇐⇒ i + λd > Fd−1 . However, i + λd > ed + λd > Fd−1 . By Lemma 4, (3) ¯ ¯ σ2,i = −Σm−1 (C˜i , Q m−1 , λm−2 ),
(89)
(3) (3) (3) where, in accordance with (35)–(37), we have C˜i = Ci since Ci ∈ Q{x1 , . . . , xd−1 },
Qd (x) = Qd (x + λd+1 ) = (x + i)(x + ad+1 + λd+1 )2 ,
Qj (x) = Qj (x), j = d,
¯ λ m−2 = (λ2 , . . . , λd , λd+1 + λd+1 , λd+2 , . . . , λm−1 ) = (λ2 , . . . , λd , λd+1 + λd+2 , λd+2 , . . . , λm−1 ).
¯ ), and We are in conditions (4) of Lemma 1, where ν = d + 1. Therefore (see (89)), conditions (7), II(Q m−1 ¯ ¯ ) are satisfied for σ2,i . The fulfillment of the conditions I C (3) , Q III(Q follows from (77) and (82). m−1 m−1 i By (Um−1 ), all σ2,i ∈ Ωm−1 ; hence, σ2 ∈ Ωm−1 . f d −1 (c) Calculation of σ3 = σ2,i . i=ed
Again, we keep notation (88) with the following changes: Qd−1 (x) = Qd−1 (x)(x + λd + i), λd = (3) (4) ρm−1 ) = l ρ¯m = l ρ¯m , λd + λd+1 , and, for the first summand in (86), M = sd ∈ Z+ . We have l = l(¯ because ρ¯m−1 is obtained from ρ¯m by eliminating ρd = 2, (−1)m−l = −(−1)m−1−l , σ3,i = −Dσ
m−1
l =1 ρ¯m−1 s1 ,...,sm−1 ∈Z+
(3)
(−1)m−1−l Ci (t1 , . . . , td−1 ) . Q1 (t1 ) · · · Qm−1 (tm−1 )
¯ ¯ We are in conditions (3) of Lemma 1; therefore (7), II(Q m−1 ), and III(Qm−1 ) are satisfied. The conditions (3) ¯ I Ci , Q m−1 are satisfied according to (77) and (82). By (Um−1 ), all σ3,i ∈ Ωm−1 , i.e., σ3 ∈ Ωm−1 . Membership (81) follows from (87) and (a), (b), and (c). Thus, the inductive transition from (Vd ) to (Vd−1 ) is completed. Thus, (V1 ) is proved by induction, and we have only to make the last step from (V1 ) to (66). 5754
3. Proof of (66). For fixed a ¯∗m−1 , we have f1 F1 Bi Ci B(x1 ) = , + Q1 (x1 ) (x1 + i)2 x1 + i
(90)
i=e1 i∈A1
i=E1 i∈A12
Bi , Ci ∈ Q, f1
Ci = 0,
(91)
i=e1
and, as usual, the index a ¯∗m−1 is omitted. Substituting (90) into (V1 ), we get ¯ m−1 ) + Σ(3) , Ba¯m Sm (¯ am , λ Σm ∼ m
(92)
a ¯m
where a ¯m is defined as in (66), Σ(3) m =
σa∗m−1 = Dσ
m l=1
ρ¯m
a ¯∗m−1
σa¯∗m−1 ,
s1 ,...,sm
f1 (−1)m−l Ci,a∗m−1 . m t +i (tj + aj )2 i=e1 1
∈Z+
(93)
j=2
We consider σa¯∗m−1 for a fixed
a ¯∗m−1 .
We omit the index a ¯∗m−1 and prove that σ ∈ Ωm−1 .
(94)
Then (66) follows from (92) and (93). This completes the proof of Theorem 2. Consider two types of ρ¯m just as in item 2. (1) I. ρ2 = 1 (¯ ρm ). Here, t1 = s1 + σ and t2 = s2 . Denote f i 1 −1 1 . Cν γ(σ) = i + σ ν=e i=e1
(95)
1
By virtue of Lemma 10, taking into account (91), we obtain f1 ∞ s1 =0 i=e1
Ci = γ(σ). s1 + σ + i
(96)
(2)
ρm ). For some r ∈ {2, . . . , m}, we have ρ2 = · · · = ρr = 2, ρr+1 = 1, t1 = s1 + σ, II. ρ2 = 2 (¯ t2 = s1 + s2 + λ2 + σ = M + λ2 + σ, where M = s1 + s2 ∈ Z+ , s1 ∈ {0, . . . , M }, tj = M + λ2 + σ + s3 + . . . + sj + λ3 + . . . + λj , j = 3, . . . , r, tr+1 = sr+1 in the case r < m. Again by Lemma 10, taking into account (91), (82), and (95), we have f1 M s1 =0 i=e1
f1 (2) Ci Ci = γ(σ) − . s1 + σ + i M +i+σ
(97)
i=e1 +1
Similarly to item 2, let σ = σ1 + σ2 ,
(98)
where (96) and the first summand from (97) are taken in σ1 and the second summand from (97) is taken in σ2 . 5755
(a) Calculation of σ1 . (2) We again apply (88) for d = 1 with the following changes: ρ1 = ρ1 = 1; for ρ¯m (see the first summand (1) in (97)) s1 = M + λ2 ≥ λ2 and, for ρ¯m (see (96)), t1 = s1 . (2) (1) Then l = l(¯ ρm−1 ) = l ρ¯m = l ρ¯m − 1. Thus, m−1 (−1)m−1−l σ1 = Dσ γ1 (σ) (t1 + a1 )2 · · · (tm−1 + am−1 )2 + l =1 ρ¯m−1
−
s1 ,...,sm−1 ∈Z t1 =s1
s1 ,...,sm−1 ∈Z+ t1 =s1 ≥λ2
(−1)m−1−l (t1 + a1 )2 · · · (tm−1 + am−1 )2 i1 (λ2 )
= γ (0) sign(λ2 )
m−1
n=i0 (λ2 ) l =1 ρ¯m−1 s1 ,...,sm−1 ∈Z+ t1 =s =n
− γ(0)Dσ
m−1
(−1)m−1−l (t1 + a1 )2 · · · (tm−1 + am−1 )2
l =1 ρ¯m−1 s1 ,...,sm−1 ∈Z+ s1 ≥λ2 t1 =s1 +σ
(−1)m−1−l . (t1 + a1 )2 · · · (tm−1 + am−1 )2
(99)
From (95) we have γ(0), γ (0) ∈ Q. The first series in (99) is a rational number by Lemma 10. By Lemma 4, the second series has the following form: ¯ ¯ ), am−1 , λ am−1 , λ T1,λ2 (Σm−1 (¯ m−2 )) = Σm−1 (¯ m−2 where a1 = a1 + λ2 , λ2 = λ2 + λ2 , the other aj and λj coincide with aj and λj . The condition α + e1 ≥ 1 ¯ ) am−1 , λ of Lemma 4 is satisfied since λ2 +a1 = λ2 +a2 ≥ λ2 +E2 > f1 ≥ 1. Conditions (14) for Σm−1 (¯ m−2 are verified trivially. Therefore, by Proposition 1, the second series in (99) is an element of Ωm−1 , i.e., σ1 ∈ Ωm−1 finally. (b) Calculation of σ2 . In the notation of item (a), from (93) and (97), we obtain σ2 =
f1 i=e1 +1
σ2,i = Dσ
m−1
(2)
Ci σ2,i ,
l =1 ρ¯m−1 s2 ,...,sm−1 ∈Z+ s1 =M +λ2 ≥λ2 t1 =s1 +σ
(−1)m−1−l . m−1 2 (t1 + i − λ2 ) (tj + aj ) j=1
We again lift the restriction s1 ≥ λ2 in accordance with Lemma 4 with the help of the change of variables (x1 + i − λ2 )(x1 + a1 )2 → (x1 + i)(x1 + a1 + λ2 ),
λ2 → λ2 + λ2 .
The condition α + e1 ≥ 1 in this lemma takes the form i ≥ 1, because a1 + λ2 ≥ E2 + λ2 > f1 ≥ i ⇒ i − λ2 < a1 , e1 = i − λ2 . We are in conditions (4) of Lemma 1 for ν = 2, i.e., conditions (7), (I)–(III) are satisfied; σ2,i ∈ Ωm−1 by (Um−1 ). However, in this case, σ2 ∈ Ωm−1 . Condition (94) follows from (a) and (b) and (98). The proof of Theorem 2 is completed. 5756
5. Proof of Theorem 1 ∗ According to Lemma 8 it is necessary to verify conditions (7) and (I)–(III) for the function Rk−1 in (64) for m = k − 1 and to reduce (51) to the standard form (9). Then Theorem 1 follows from Theorem 2. According to Lemma 4, from (64) and (51), in the notation of Lemma 7, for α = µ0 and ν = 1, we have ¯ k−2 ), ¯ k−1 , λ J2k−1 = Λ∗ Σk−1 (P, Q 2k−1
where Q1 (x) = P12 (x + µ0 )P13 (x + µ0 ),
Qj (x) = Pj2 (x)Pj3 (x), j = 2, . . . , k − 1,
P = P (x1 , . . . , xk−1 ) = P10 (x1 + µ0 )G2 (x2 − x1 − µ0 − β3 )
k−1
Gj (xj − xj−1 − β2j−1 )Gk (xk−1 ),
(100) (101)
j=3
¯ k−2 = (β3 + µ0 , β5 , . . . , β2k−3 ) if k ≥ 3. λ
(102)
1. The condition α + e1 ≥ 1 of Lemma 4 takes the form µ0 + min(α1 , α2 ) ≥ 1. Actually, it was verified in the proof of Lemma 7 while considering the beta-integral in (52). For β1 ≥ α0 , everything is obvious (µ0 = 0). For β1 < α0 and α1 < α2 , we have µ0 + min(α1 , α2 ) = β1 − α0 + α1 ≥ β1 ≥ 1 in view of (2). Finally, in the last case, for β1 < α0 and α2 < α1 , we have µ0 + min(α1 , α2 ) = β1 − α0 + α2 ≥ 1 due to (3) for r = 0 (just this very case is considered in Lemma 7). 2. Condition (7) follows from (100) directly as doubts about A1 ⊂ N are dispelled in item 1; see also the structure of polynomials Pj2 (x) and Pj3 (x) in the notation of Lemma 7. ¯ k−1 ). From (101) it follows that degx P = pj = β2j−1 + β2j+1 − 2 for 3. Now, we verify I(P, Q j j = 1, . . . , k − 1 and, from (100), we obtain qj = β2j + α2j+1 + β2j+1 − α2j−1 for j = 1, . . . , k − 1. Therefore, pj − qj + 2 = β2j−1 + α2j−1 − β2j − α2j+1 ≤ 0 due to (3) for r = 2j − 1, j = 1, . . . , k − 1; i.e., ¯ k−1 ) are satisfied. the conditions I(P, Q ¯ k−1 ). Assume that, for j = 1, . . . , k − 1, ¯ k−1 ) and III(Q 4. We verify II(Q ej = min(α2j−1 , α2j ), Fj = min(α2j + β2j − 1, α2j+1 + β2j+1 − 1),
Ej = max(α2j−1 , α2j ),
(103)
fj = max(α2j + β2j − 1, α2j+1 + β2j+1 − 1).
From (100) it follows that numbers (103) are parameters of polynomials Qj (x) for j = 2, . . . , k − 1, and we must add µ0 to them for j = 1. First, we consider the inequalities IIj and IIIj for j = 3, . . . , k − 1: max(α2j−2 + β2j−2 − 1, α2j−1 + β2j−1 − 1) < max(α2j−1 , α2j ) + β2j−1 ,
(104)
min(α2j−2 + β2j−2 − 1, α2j−1 + β2j−1 − 1) < min(α2j−1 , α2j ) + β2j−1 .
(105)
System (104), (105) has the following form for a, b, c ∈ N: max(a, b) < max(b, c),
(106)
min(a, b) < min(b, c).
(107)
If a > b, then (106) is equivalent to a ≤ c, i.e., (107) is satisfied. If a ≤ b, then (106) is satisfied, and (107) is also equivalent to a ≤ c, i.e., in the notation of (104) and (105), this system is equivalent to the inequality α2j−2 + β2j−2 ≤ α2j + β2j−1 , or inequality (3) for r = 2j − 2. Thus, IIj and IIIj are satisfied. The inequalities II2 and III2 imply a system similar to (104) and (105) for j = 2, because, as was stated above, µ0 must be added to the left-hand sides of (104) and (105) (F1 → F1 + µ0 , f1 → f1 + µ0 ). The same procedure must also be executed for the right-hand sides, because λ2 = β3 + µ0 (see (102)). ¯ k−1 ). ¯ k−1 ) and III(Q Thus, set (3) of inequalities for even r = 2, 4, . . . , 2k − 4 implies II(Q 5. From the symmetry (c ↔ a) of Gauss’s function in (40) (see also the application of Lemma 5 in Lemma 7 and its influence on parameters of polynomials (100) and (101)), we obtain that the change of 5757
parameters (α0 , α1 , β1 ) → (α1 , α0 , α1 + β1 − α0 ) is possible in integral (1), under which integral (1) is multiplied by the coefficient Γ(α0 )Γ(α1 + β1 − α0 ) Γ(α1 )Γ(α0 + β1 − α1 ) (see (39)). Thus, in the case α0 > α1 , we can form the list of conditions, analogous to (3)–(5), ensuring (6). REFERENCES 1. H. Bateman and A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, New York (1953). 2. F. Beukers, “A note on the irrationality of ζ(2) and ζ(3),” Bull. London Math. Soc., 11, 268–272 (1979). 3. D. V. Vasilyev, “Some formulas for the Riemann zeta function at integers,” Vestn. Mosk. Univ. Ser. 1 Mat. Mekh., No. 1, 81–84 (1996). 4. D. V. Vasilyev, On Small Linear Forms for the Values of the Riemann Zeta-Function at Odd Integers, Preprint No. 1 (558), Nat. Acad. Sci. Belarus, Institute Math., Minsk (2001). 5. S. A. Zlobin, “Expanding multiple integrals into linear forms,” Mat. Zametki, 77, No. 5, 683–706 (2005). 6. W. Zudilin, “Arithmetics of linear forms involving odd zeta values,” J. Th´eor. Nombres Bordeaux, 16, No. 1, 251–291 (2004). 7. W. Zudilin, “Well-poised hypergeometric transformations of Euler-type multiple-integrals,” J. London Math. Soc. (2), 70, 215–230 (2004). V. Salikhov and A. Frolovichev Bryansk State Technical University E-mail:
[email protected]
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