Kyushu J. Math. 65 (2011), 15–24 doi:10.2206/kyushujm.65.15

POLYLOGARITHMS AND POLY-BERNOULLI POLYNOMIALS

Abdelmejid BAYAD and Yoshinori HAMAHATA (Received 18 February 2010)

Abstract. In this paper we investigate special generalized Bernoulli polynomials that generalize classical Bernoulli polynomials and numbers. We call them poly-Bernoulli polynomials. We prove a collection of extremely important and fundamental identities satisfied by our poly-Bernoulli polynomials and numbers. These properties are of arithmetical nature.

1. Introduction and main results

The classical polylogarithmic functions Lis (x) are ∞ xk Li (x) = . (1.1) s ks k=1 If s is a negative integer, say s =−r, then the polylogarithmic function converges for |x| < 1 and equals r  r  r−j j=0 j x Li−r (x) = , (1.2) (1 − x)r+1  r   r  where the j are the Eulerian numbers. The Eulerian number j is the number of permutations of {1,...,r} with j permutation ascents. One has

+ r j 1 r + 1 = (−1) (j − + 1)r . (1.3) j =0

We record Li−r (x) for some r: x x Li0(x) = , Li−1(x) = , 1 − x (1 − x)2 x2 + x x3 + 4x2 + x Li−2(x) = , Li−3(x) = , (1 − x)3 (1 − x)4 x4 + 11x3 + 11x2 + x x5 + 26x4 + 66x3 + 26x2 + x Li− (x) = , Li− (x) = . 4 (1 − x)5 5 (1 − x)6

2000 Mathematics Subject Classification: Primary 11B73, 11A07. Keywords: polylogarithms; zeta functions; poly-Bernoulli numbers; poly-Bernoulli polynomials.

c 2011 Faculty of Mathematics, Kyushu University 16 A. Bayad and Y. Hamahata

From (1.2) we immediately deduce that, when k is a negative integer, Lik(x) is a rational function whose denominator is (1 − x)|k|+1. Precisely, for an integer k ∈ Z, the formal power series Lik(x) is the kth polylogarithm if k ≥ 1, and a rational function if k ≤ 0. When k = 1,

Li1(x) =−log(1 − x). (1.4)

Using Lik(x), one can introduce (in the natural way) poly-Bernoulli numbers and polynomials. (k) For every integer k, we define a sequence of polynomials (Bn (x))n∈N, which we call poly-Bernoulli polynomials,by −t n Lik(1 − e ) t ext = B(k)(x) . (1.5) 1 − e−t n n! n≥0

(k) xt −xt We remark that Bn (x) are defined in [3] by replacing e by e in the left-hand side of (1.5). (k) (k) The numbers Bn := Bn (0) are the poly-Bernoulli numbers. These numbers are introduced by Kaneko [6], and then investigated in [1, 2, 4, 5, 7–9]. On the other hand, from (1.4) and (1.5), it is easy to see that, for any n ≥ 0, − n (1) − = ( 1) Bn ( x) Bn(x) (1.6) where Bn(x) are the classical Bernoulli polynomials given by their generating function ∞ text tn = Bn(x) . (1.7) et − 1 n! n=0 In this paper, we establish a series of properties, satisfied by our poly-Bernoulli polynomials and numbers, which are important and fundamental. These properties are of arithmetical nature.

THEOREM 1.1. (Explicit formula) For k ∈ Z,n≥ 0, we have n 1 m m B(k)(x) = (−1)j (x − j)n. (1.8) n (m + 1)k j m=0 j=0

THEOREM 1.2. (Recurrence formula 1) For all k ≥ 1, n ≥ 0, we have n m − l (k) m n (k−1) ( 1) m B (x) = (−1) B − Bl(x). (1.9) n m n m n − l + 1 l m=0 l=0 THEOREM 1.3. (Recurrence formula 2) For k ∈ Z,n≥ 2, we have − 1 n 1 n n B(k)(x) = B(k−1)(x) + xB(k)(x) − − x B(k)(x) n n + 1 n 0 m − 1 m m m=1 and (k) = (k) = 1 (k−1) + (k) B0 (x) 1,B1 (x) 2 (B1 (x) xB0 (x)). THEOREM 1.4. (Appell sequence) For k ∈ Z,n≥ 0, we have d B(k) (x) = (n + 1)B(k)(x). (1.10) dx n+1 n Polylogarithms and poly-Bernoulli polynomials 17

THEOREM 1.5. (Addition formula) For k ∈ Z,n≥ 0, we have n n − B(k)(x + y) = B(k)(x)yn m. (1.11) n m m m=0 For m, n ≥ 0, set m m C(−m)(x, y) = B(−k)(x)ym−k. n k n k=0 Then we have the following result. (−m) THEOREM 1.6. (Symmetric formula) For m, n and Cn (x, y) as above, we have ∞ ∞ ∞ ∞ tn um tn um C(−m)(x, y) = C(−n)(y, x) n n! m! m n! m! n=0 m=0 n=0 m=0 ext+yuet+u = . et + eu − et+u As a corollary, we have the following result.

THEOREM 1.7. (Duality) We have (−m) = (−n) Cn (x, y) Cm (y, x). (1.12) This is a dual relation for addition formula. In particular, for x = y = 0, we obtain (−m) = (−n) Bn Bm , (1.13) the duality property of poly-Bernoulli numbers.

THEOREM 1.8. (Inversion formula) For m, n ≥ 0, we have m − m − − B( m)(x) = C( k)(x, y)(−y)m k. (1.14) n k n k=0

THEOREM 1.9. (Closed formula) For m, n ≥ 0, we have ∞ n m − − n a − m b C( m)(x, y) = (j!)2 (x + 1)n a (y + 1)m b , (1.15) n a j b j j=0 a=0 b=0 where n (−1)m m m = (−1)l ln m m! l l=0 are the so-called Stirling numbers of the second kind. For k ∈ Z we set ∞ ( − e−t ) := 1 Lik 1 −xt s−1 Zk(s, x) −t e t dt, (s) 0 1 − e the Laplace–Mellin integral. It is defined for Re(s) > 0andx>0ifk ≥ 1, and for Re(s) > 0 and x>|k|+1ifk ≤ 0. 18 A. Bayad and Y. Hamahata

THEOREM 1.10. (Interpolation formula) The function s → Zk(s, x) has analytic continua- tion to an entire function on the whole complex s-plane and − = − n (k) − Zk( n, x) ( 1) Bn ( x) (1.16) holds for n ≥ 0 and x>0. Besides, this zeta function can be rewritten as follows. For k ∈ Z, we have m 1 j m 1 Zk(s, x) = (−1) . (1.17) (m + 1)k j (x + j)s m≥0 j=0 In the case k ≤ 0, we have a more precise expression for this zeta function. For k ≤ 0, |k| | | |k|−j |k|−j − m k m ( 1) Zk(s, x) = . (1.18) j (x + m −|k|−1)s j=0 m=0 It should be noted that the first result of the above theorem has been independently obtained by Coppo and Candelpergher [3] and the present authors. Remark 1.11. We can make the following two remarks. (1) In particular, for k = 1wehave ∞ − −t 1 Lik(1 e ) −xt s−1 Zk(s, x) = e t dt (s) 1 − e−t 0 1 ∞ t = e−xtts−1 dt (s) 1 − e−t 0 (s + ) ∞ ts = 1 1 −xt −t e dt (s) (s + 1) 0 1 − e = sζ(s + 1,x), (1.19) where ζ(s, x)is the Hurwitz zeta function. Then, for k = 1, s =−n, we get from (1.6), (1.10) and (1.19) that − = − n (1) − = Z1( n, x) ( 1) Bn ( x) Bn(x) and Z1(−n, x) =−nζ(1 − n, x). Hence, we derive the famous classical interpolation relation:

Bn(x) =−nζ(1 − n, x).

(2) For k ≤ 0, from the formula (1.18) we derive another new explicit formula:

| | | |− k |k| k j |k|−j B(k)(x) = (−1)m(x + m −|k|−1)n. (1.20) n j m j=0 m=0

2. Proofs of main results

In this section we will give proofs of the theorems mentioned in the previous section. Polylogarithms and poly-Bernoulli polynomials 19

Proof of Theorem 1.1. Since ∞ ( − e−t ) ( − e−t )m−1 Lik 1 = 1 1 − e−t mk m=1 ∞ ( − e−t )m = 1 (m + 1)k m=0 ∞ m 1 m − = (−1)j e jt, (m + 1)k j m=0 j=0 we have −t ∞ m Lik(1 − e ) 1 m − ext = (−1)j e(x j)t 1 − e−t (m + 1)k j m=0 j=0 ∞ ∞ 1 m m tn = (−1)j (x − j)n (m + 1)k j n! m=0 j=0 n=0 ∞ n 1 m m tn = (−1)j (x − j)n . (m + 1)k j n! n=0 m=0 j=0 This completes the proof. 2 Proof of Theorem 1.2. Since − − ( − e t ) et t − ( − e s) Lik 1 = −s Lik 1 1 −t t e −s ds, 1 − e e − 1 0 1 − e we have ∞ tn B(k)(x) n n! n=0 ∞ ∞ ∞ tn−1 t (−s)n sn = B (x) B(k−1) ds n n! n! n n! n=0 0 n=0 n=0 ∞ n−1 t ∞ n n t − n − s = B (x) (−1)n m B(k 1) ds n n! m m n! n=0 0 n=0 m=0 ∞ n−1 ∞ n n+1 t n−m n (k−1) t = Bn(x) (−1) B n! m m (n + 1)! n=0 n=0 m=0 ∞ n l n l−m l (k−1) t = Bn−l (x) (−1) B m m (l + 1)!(n − l)! n=0 l=0 m=0 ∞ n l n Bn−l (x) n − l − t = (−1)l m B(k 1) l + l m m n! n= l= 1 m= 0 0 0 n l n n − m by = l m m n − l 20 A. Bayad and Y. Hamahata ∞ n n − l−m − n (k−1) n ( ) n m t = B Bn−l (x) m m l + 1 n − l n! n=0 m=0 l=m  put l = n − l ∞ n n−m − n−l−m − n (k−1) n ( 1) n m t = B Bl (x) m m n − l + 1 l n! n=0 m=0 l=0  put m = n − m  ∞ n m m+l  n (k−1) n (−1) m t = B  Bl (x) . n−m m n − l + 1 l n! n=0 m=0 l=0 The theorem therefore follows. 2 −t Proof of Theorem 1.3. Put z = 1 − e .Thent = Li1(z). From (1.5), we have ∞ n Li1(z) Li (z)exLi1(z) = z B(k)(x) . (2.1) k n n! n=0 Note that d (1/z)Lik−1(z) (k > 1), Lik(z) = dz 1/(1 − z) (k = 1). Differentiating (2.1) with respect to z,wehave

1 xLi (z) x xLi (z) Lik− (z)e 1 + Lik(z) e 1 z 1 1 − z ∞ ∞ Li (z)n z Li (z)n = B(k)(x) 1 + B(k) (x) 1 . n n! 1 − z n+1 n! n=0 n=0 Using z/(1 − z) = et − 1, we obtain ∞ n ∞ n ∞ n ∞ n − t t t t B(k 1)(x) + x(et − 1) B(k)(x) = B(k)(x) + (et − 1) B(k) (x) . n n! n n! n n! n+1 n! n=0 n=0 n=0 n=0 This yields ∞ ∞ ∞ tn tn n n tn B(k−1)(x) − B(k)(x) + xB(k)(x) n n! n n! m m n! n=0 n=0 n=0 m=0 ∞ ∞ ∞ tn tn n n = B(k)(x) − B(k) (x) + B(k) (x). n n! n+1 n! m m+1 n=0 n=0 n=0 m=0 Comparing both sides of this identity gives n n n n B(k−1)(x) − xB(k)(x) + xB(k)(x) = B(k)(x) − B(k) + B(k) (x). n n m m n n+1 m m+1 m=0 m=0 From this, we obtain n−1 n−1 − n n (n + 1)B(k)(x) = B(k 1)(x) + x B(k)(x) − B(k)(x), n n m m m − 1 m m=0 m=1 which leads to the result. 2 Polylogarithms and poly-Bernoulli polynomials 21

Proof of Theorem 1.4. Differentiation of both sides of (1.5) with respect to x gives ∞ tLi (1 − e−t ) d tn k ext = B(k)(x) . 1 − e−t dx n n! n=1 From this, we have −t ∞ n Lik(1 − e ) d t ext = (n + 1) B(k)(x) . 1 − e−t dx n n! n=0 This completes the proof. 2 Proof of Theorem 1.5. From the definition in (1.5), we have ∞ tn Li (1 − e−t ) B(k)(x + y) = k e(x+y)t n n! 1 − e−t n=0 Li (1 − e−t ) = k exteyt 1 − e−t ∞ ∞ tm yltl = B(k)(x) m m! l! m=0 l=0 ∞ n n tn = B(k)(x)yn−m . m m n! n=0 m=0 The proof follows from this. 2 (−m) Proof of Theorem 1.6. Using the definition of Cn (x, y), the left-hand side can be rewritten as ∞ ∞ m n m − − t u LHS = B( k)(x)ym k n n! k!(m − k)! n=0 m=0 k=0 put l = m − k ∞ ∞ ∞ tn uk ul = B(−k)(x)yl n n! k! l! n=0 k=0 l=0 ∞ ∞ tn uk = eyu B(−k)(x) n n! k! n=0 k=0 (−k) by definition of Bn (x) ∞ ∞ tn uk = eyu ext B(−k) n n! k! k=0 n=0 ∞ ∞ tn uk = ext+yu B(−k) . n n! k! k=0 n=0 Here, using the expression ∞ ∞ tn uk et+u B(−k) = n n! k! et + eu − et+u k=0 n=0 in Kaneko [6], the proof is completed. 2 22 A. Bayad and Y. Hamahata

(−m) Proof of Theorem 1.8. To prove the theorem, we calculate the generating function Bn . ∞ ∞ m m tn um C(−k)(x, y)(−y)m−k k n n! m! n=0 m=0 k=0 put j = m − k ∞ ∞ ∞ (−yu)j tn uk = C(−k)(x, y) n j! n! k! n=0 j=0 k=0 ∞ ∞ tn uk = e−yu C(−k)(x, y) n n! k! n=0 k=0 by Theorem 1.6 ext+t+u = et + eu − et+u ∞ ∞ tn um = B(−m)(x) . n n! m! n=0 m=0 This completes the proof. 2 Proof of Theorem 1.9. Using Theorem 1.6, we have ∞ ∞ tn um C(−m)(x, y) n n! m! n=0 m=0 ext+yu+t+u = et + eu − et+u e(x+1)t+(y+1)u = 1 − (et − 1)(eu − 1) ∞ + + + = e(x 1)t (y 1)u (et − 1)j (eu − 1)j j=0 ∞ + + = e(x 1)t(et − 1)j e(y 1)u(eu − 1)j j=0 ∞ n un (eu − 1)k by = k n! k n=0 ∞ ∞ ∞ ∞ ∞ (x + 1)ntn m tm (y + 1)nun m um = j! j! n! j m! n! j m! j=0 n=0 m=0 n=0 m=0 ∞ ∞ l l ∞ p p − l m t − p r u = j! (x + 1)l m j! (y + 1)p r m j l! r j p! j=0 l=0 m=0 p=0 r=0 ∞ ∞ l p ∞ l p t u − l m − p r = (j!)2 (x + 1)l m (y + 1)p r , l! p! m j r j l=0 p=0 j=0 m=0 r=0 which proves the theorem. 2 Polylogarithms and poly-Bernoulli polynomials 23

Proof of Theorem 1.10. Since the proof of the first part of the theorem is the same as that of Theorem 2 of Coppo and Candelpergher [3], we omit it. We now prove (1.17) of Theorem 1.10. We have ∞ 1 ∞ (1 − e−t )m−1 Z (s, x) = e−xtts−1 dt. k (s) nk 0 n=1 For k ≥ 1, it is easy to see that, for all N ≥ 1andt>0, ∞ ∞ ( − e−t )n−1 ( − e−t )n−1 1 ≤ 1 k n=N n n=N n ∞ ( − e−t )n−1 ≤ 1 n n=1 t = , 1 − e−t which ensures the convergence of the integral absolutely. Then we can invert and . Hence ∞ ∞ 1 1 − − − − Z (s, x) = (1 − e t )n 1e xtts 1 dt k (s) nk n=1 0 ∞ ∞ n 1 1 n − + = (−1)j e (x j)tts dt (s) (n + 1)k j n=0 0 j=0 ∞ n ∞ 1 n 1 − + − = (−1)j e (x j)tts 1 dt. (n + 1)k j (s) n=0 j=0 0 We thus obtain (1.17). We prove the case k ≤ 0. Since |k| | | 1 k |k|−j Lik(x) = x , ( − x)|k|+1 j 1 j=0 we have |k| ∞ 1 |k| − | |− | |+ − − Z (s, x) = (1 − e t ) k j e( k 1)te xtts 1 dt k (s) j j=0 0 |k| |k|−j ∞ 1 |k| |k|−j − + −| |− − = (−1)m e (x m k 1)tts 1 dt. (s) j m j=0 m=0 0 This yields (1.18) for all k ≤ 0, Re(x) > |k|+1. 2

Acknowledgement. We would like to thank the referee for informing us of the existence of [3]. The second named author is supported by a Grant-in-Aid for Scientific Research (No. 20540026), from the Japan Society for the Promotion of Science. 24 A. Bayad and Y. Hamahata

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Abdelmejid Bayad Departement´ de Mathematiques´ Universited’´ Evry´ Val d’Essone Boulevard F. Mitterrand 91025 Evry´ Cedex France (E-mail: [email protected])

Yoshinori Hamahata Graduate School of Mathematical Sciences The University of Tokyo 3-8-1 Komaba, Meguro, Tokyo 153-8914 Japan (E-mail: [email protected])