Article Families of Integrals of Polylogarithmic Functions †
Anthony Sofo Victoria University, P. O. Box 14428, Melbourne City, Victoria 8001, Australia; [email protected] † Lovingly dedicated to Mia Eva Manca.
Received: 30 December 2018; Accepted: 26 January 2019; Published: 3 February 2019
Abstract: We give an overview of the representation and many connections between integrals of products of polylogarithmic functions and Euler sums. We shall consider polylogarithmic functions with linear, quadratic, and trigonometric arguments, thereby producing new results and further reinforcing the well-known connection between Euler sums and polylogarithmic functions. Many examples of integrals of products of polylogarithmic functions in terms of Riemann zeta values and Dirichlet values will be given. Suggestions for further research are also suggested, including a study of polylogarithmic functions with inverse trigonometric functions.
Keywords: Euler sums; zeta functions; Dirichlet functions; series representation; harmonic number
MSC: 11M06; 11M32; 33B15
1. Introduction and Preliminaries It is well known that integrals of products of polylogarithmic functions can be associated with Euler sums, see Reference [1]. In this paper, we investigate the representations of integrals of the type
1 Z m x Liq (±x) Lit (±x) dx, 0
1 Z 1 Li (±x) Li x2 dx, x q t 0 for m ≥ −2, and for integers q and t. For m = −2, −1, 0, we give explicit representations of the integrals 1 R m in terms of Euler sums. For the case x Lit (−x) Liq (−x) dx, for m ≥ 0, we show that the integral 0 satisfies a certain recurrence relation. We also mention two specific integrals with a trigonometric argument in the polylogarithm of the form
1 Z sin πx Lip (sin πx) dx. 0
Some examples are highlighted, almost none of which are amenable to a computer mathematical package. This work extends the results given in Reference [1], where the author examined integrals with positive arguments of the polylogarithm. Devoto and Duke [2] also list many identities of lower order polylogarithmic integrals and their relations to Euler sums. Some other important sources of information on polylogarithm functions are the works of References [3] and [4]. In References [5] and [6], the authors explore the algorithmic and analytic properties of generalized harmonic Euler sums systematically, in order to compute the massive Feynman integrals which arise in quantum field
Mathematics 2019, 7, 143; doi:10.3390/math7020143 www.mdpi.com/journal/mathematics Mathematics 2019, 7, 143 2 of 20 theories and in certain combinatorial problems. Identities involving harmonic sums can arise from their quasi-shuffle algebra or from other properties, such as relations to the Mellin transform
1 Z M[ f (z)](N) = dz zN f (z) , 0 where the basic functions f (z) typically involve polylogarithms and harmonic sums of lower weight. Applying the latter type of relations, the author in Reference [7] expresses all harmonic sums of the above type with weight w = 6, in terms of Mellin transforms and combinations of functions and constants of lower weight. In another interesting and related paper [8], the authors prove several identities containing infinite sums of values of the Roger’s dilogarithm function. defined on x ∈ [0, 1], by Li (x) + 1 ln x ln (1 − x) ; 0 < x < 1 2 2 LR (x) = 0; x = 0 . ζ (2) ; x = 1 The Lerch transcendent, ∞ zm = Φ (z, t, a) ∑ t m =0 (m + a) is defined for |z| < 1 and < (a) > 0 and satisfies the recurrence
Φ (z, t, a) = z Φ (z, t, a + 1) + a−t.
The Lerch transcendent generalizes the Hurwitz zeta function at z = 1,
∞ 1 = Φ (1, t, a) ∑ t m =0 (m + a) and the polylogarithm, or de-Jonquière’s function, when a = 1,
∞ zm ( ) = ∈ | | < < ( ) > | | = Lit z : ∑ t , t C when z 1; t 1 when z 1. m =1 m
Let n 1 Z 1 1 − tn ∞ n H = = dt = γ + ψ (n + 1) = , H := 0 n ∑ − ∑ ( + ) 0 r=1 r 0 1 t j=1 j j n be the nth harmonic number, where γ denotes the Euler–Mascheroni constant,
n m−1 1 n (m) 1 (−1) Z 1 − t H = = lnm−1 (t) dt (1) n ∑ m ( − ) − r=1 r m 1 ! 0 1 t is the mth order harmonic number and ψ(z) is the digamma (or psi) function defined by
d Γ0(z) 1 ψ(z) := {log Γ(z)} = and ψ(1 + z) = ψ(z) + , dz Γ(z) z moreover, ∞ 1 1 ψ(z) = −γ + − . ∑ + + n=0 n 1 n z Mathematics 2019, 7, 143 3 of 20
More generally, a non-linear Euler sum may be expressed as
n t qj r ! mk (±1) (αj) (βk) ∑ p ∏ Hn ∏ Jn n≥1 n j=1 k=1 where p ≥ 2, t, r, qj, αj, mk, βk are positive integers and
!q + !m q n 1 m n (−1)j 1 (α) = (β) = Hn ∑ α , Jn ∑ β . j=1 j j=1 j
If, for a positive integer, t r λ = ∑ αjqj + ∑ βjmj + p, j=1 j=1 then we call it a λ-order Euler sum. The polygamma function
dk ∞ 1 ψ(k)(z) = {ψ(z)} = (− )k+1 k k 1 ! ∑ k+1 dz r=0 (r + z) and satisfies the recurrence relation
(−1)k k! ψ(k)(z + 1) = ψ(k)(z) + . zk+1 The connection of the polygamma function with harmonic numbers is
α (α+ ) (−1) H 1 = ζ (α + 1) + ψ(α) (z + 1) , z 6= {−1, −2, −3, ...} . (2) z α! and the multiplication formula is
− 1 p 1 j ψ(k)(pz) = δ p + ψ(k)(z + ) m,0 ln k+1 ∑ (3) p j=0 p for p is a positive integer and δp,k is the Kronecker delta. We define the alternating zeta function (or Dirichlet eta function) η (z) as
+ ∞ (−1)n 1 ( ) = = − 1−z ( ) η z : ∑ z 1 2 ζ z (4) n =1 n where η (1) = ln 2. If we put + (p) ∞ (−1)n 1 H = n S (p, q) : ∑ q , n =1 n in the case where p and q are both positive integers and p + q is an odd integer, Flajolet and Salvy [9] gave the identity ! p p p + i − 1 2S (p, q) = 1 − (−1) ζ (p) η (q) + 2 (−1) ∑ ζ (p + i) η (2k) i+2k=q p − 1
! (5) q + j − 1 j +η (p + q) − 2 ∑ (−1) η (q + j) η (2k) , j+2k=p q − 1 Mathematics 2019, 7, 143 4 of 20
1 1 where η (0) = 2 , η (1) = ln 2, ζ (1) = 0, and ζ (0) = − 2 in accordance with the analytic continuation of the Riemann zeta function. We also know, from the work of Reference [10] that for odd weight w = (p + q) we have ! (p) p ∞ Hn p [ 2 ] p + q − 2j − 1 BW (p, q) = ∑ q = (−1) ∑ ζ (p + q − 2j) ζ (2j) n =1 n j =1 p − 1 p ! p+1 p [ ] p + q − 2j − 1 + 1 1 + (−1) ζ (p) ζ (q) + (−1) ∑ 2 ζ (p + q − 2j) ζ (2j) (6) 2 j =1 q − 1 ! !! ( + ) + p + q − 1 + p + q − 1 + ζ p q 1 + (−1)p 1 + (−1)p 1 , 2 p q where [z] is the integer part of z. It appears that some isolated cases of BW (p, q) , for even weight (p + q) , can be expressed in zeta terms, but in general, almost certainly, for even weight (p + q) , no general closed form expression exits for BW (p, q) . Two examples with even weight are
( ) ( ) ∞ H 2 1 ∞ H 4 37 n = 2 ( ) − ( ) n = ( ) − 2 ( ) ∑ 4 ζ 3 ζ 6 , ∑ 2 ζ 6 ζ 3 . n =1 n 3 n =1 n 12 There are also the two general forms of identities
( + ) ∞ H 2p 1 n = 2 ( + ) + ( + ) 2 ∑ 2p+1 ζ 2p 1 ζ 4p 2 , n =1 n and ( ) 2 ! ∞ H 2p (4p)!B n = − 2p ( ) 2 ∑ 2p 1 2 ζ 4p n =1 n 2 ((2p)!) B4p for p ∈ N and where Bp are the signed Bernoulli numbers of the first kind. At even weight w = 8, neither BW (2, 6) nor BW (6, 2) currently have a reduced expression in terms of zeta values and their products. Using (1) we can express
( ) ∞ H 2 5 Z 1 ln x Li (x) BW (2, 6) = n = ζ (8) − 6 dx ∑ 6 − n =1 n 3 0 1 x and ( ) ∞ H 6 5 1 Z 1 ln5 x Li (x) BW (6, 2) = n = ζ (8) + 2 dx. ∑ 2 − n =1 n 3 5! 0 1 x The work in this paper extends the results of Reference [1] and later Reference [11], in which they gave identities of products of polylogarithmic functions with positive argument in terms of zeta functions. Other works, including References [12–23], cite many identities of polylogarithmic integrals and Euler sums. The following lemma was obtained by Freitas, [1]:
Lemma 1. For q and t positive integers
1 (+,+) R Lit(x) Liq(x) I (q, t) = x dx 0 (7) q−1 j+1 q+1 = ∑j =1 (−1) ζ (t + j) ζ (q − j + 1) + (−1) EU (t + q) where EU (m) is Euler’s identity given in the next lemma. Mathematics 2019, 7, 143 5 of 20
Lemma 2. The following identities hold: for m ∈ N. The classical Euler identity, as given by Euler, states
∞ Hn m EU (m) = ∑n =1 nm = ( 2 + 1)ζ (m + 1) (8) 1 m−2 − 2 ∑j=1 ζ (m − j) ζ (j + 1) .
The following identities are new. For p a positive even integer,
∞ Hn p HE (p) = ∑ p = (ζ (p + 1) + η (p + 1)) − (ζ (p) + η (p)) ln 2 n =1 (2n + 1) 2 (9) p −1 1 2 − ∑ (ζ (p + 1 − 2j) + η (p + 1 − 2j))(ζ (2j) + η (2j)) . 2 j=1
For p a positive odd integer,
∞ Hn p HO (p) = ∑ p = (ζ (p + 1) + η (p + 1)) − (ζ (p) + η (p)) ln 2 n =1 (2n + 1) 4
p−1 1 1 + (−1) 2 p + 1 p + 1 − ζ + η (10) 4 2 2 2
1 b − ∑ (ζ (p − 2j) + η (p − 2j))(ζ (2j + 1) + η (2j + 1)) 2 j=1
p−1 ! h p−1 i 1+(−1) 2 where η (z) is the Dirichlet eta function, b = 4 − 2 , and [z] is the greatest integer less than z. For p and t positive integers we have
n+1 ∞ (−1) F (p, t) = ∑ t n =1 np(n+1) ! p − p + t − r − 1 = (−1)p r η (r) ∑r =1 − p r (11) ! p+1 p + t − s − 1 + ∑t (−1) (1 − η (s)) , s =1 t − s
! ∞ 1 p+1 p + t − 1 G (p, t) = ∑ t = (−1) n =1 np(n+1) p
! p p−r p + t − r − 1 + ∑ (−1) ζ (r) (12) r =2 p − r
! p p + t − s − 1 + ∑t (−1) ζ (s) , s =1 t − s Mathematics 2019, 7, 143 6 of 20
! n+1 ∞ (−1) Hn t+1 q + t − 2 1 2 HA (t, q) = ∑ q = (−1) ζ (2) − ln 2 n =1 nt(n+1) t − 1 2 ! t−r q + t − r − 1 + ∑t (−1) S (1, r) (13) r =2 t − r ! q t q + t − s − 1 + ∑ (−1) (η (s + 1) − S (1, s)) , s =2 q − s and ! ∞ Hn p+1 p + t − 2 HG (p, t) = ∑ t = (−1) ζ (2) n =1 np(n+1) p − 1 ! p p−r p + t − r − 1 + ∑ (−1) EU (r) (14) r =2 p − r ! p p + t − s − 1 + ∑t (−1) (EU (s) − ζ (s + 1)) . s =2 t − s
Proof. The identity (8) is the Euler relation and by manipulation we arrive at (9) and (10). The results (9) and (10) are closely related to those given by Nakamura and Tasaka [24]. For the proof of (11) we notice that
p ! 1 − p + t − r − 1 1 = (−1)p r p t ∑ − r n (n + 1) r =1 p r n ! t p + t − s − 1 1 + (−1)p ∑ − s s =1 t s (n + 1) therefore, summing over the integers n,
∞ n+1 p ! (−1) − p + t − r − 1 F (p, t) = = (−1)p r η (r) ∑ p t ∑ − n =1 n (n + 1) r =1 p r ! t p + t − s − 1 + (−1)p (1 − η (s)) ∑ − s =1 t s and hence (11) follows. Consider
p+1 ! p ! 1 (−1) p + t − 2 − p + t − r − 1 1 = + (−1)p r p t ( + ) ∑ r n (n + 1) n n 1 p − 1 r =2 p − r n
t ! p p + t − s − 1 1 + ∑ (−1) s , s =2 t − s (n + 1) and summing over the integers n produces the result (12). The proof of (13) and (14) follows by ∞ Hn summing ∑ t in partial fraction form. Here is an example from (10): n =1 np(n+1)
∞ H 9207 961 889 511 n = ( ) − 2 ( ) − ( ) ( ) − ( ) ∑ 9 ζ 10 ζ 5 ζ 7 ζ 3 ζ 9 ln 2 n =1 (2n + 1) 2048 1024 512 256 Mathematics 2019, 7, 143 7 of 20 and from (9):
∞ H 381 93 105 63 n = ( ) − ( ) ( ) − ( ) ( ) − ( ) ∑ 6 ζ 7 ζ 5 ζ 2 ζ 4 ζ 3 ζ 6 ln 2. n =1 (2n + 1) 64 64 64 32
2. Summation Identity We now prove the following theorems:
Theorem 1. For positive integers q and t, the integral of the product of two polylogarithmic functions with negative arguments
1 0 (−,−) R R I0 (q, t) = Lit (−x) Liq (−x) dx = Lit (x) Liq (x) dx 0 −1 q−1 j+1 (15) = ∑j =1 (−1) η (q − j + 1) F (t, j) q q + (−1) (F (t, q + 1) − (F (t, q) − G (t, q)) ln 2) + (−1) Wn (q, t) where the sum ! ∞ 1 1 1 W (q t) = H − + n , ∑ n t q t q t q (16) n =1 (2n) (2n + 1) n (n + 1) (2n + 1) (2n + 2) is obtained from (8), (9), (10), and the terms F (·, ·) , G (·, ·) are obtained from (11) and (12), respectively.
Proof. By the definition of the polylogarithmic function, we have
1 ∞ ∞ n+r − − Z (−1) ( , ) ( ) = (− ) (− ) = I0 q, t Lit x Liq x dx ∑ ∑ t q = = n r (n + r + 1) 0 n 1 r 1
+ q + ! ∞ ∞ (−1)n r (−1)q (−1)j 1 = + ∑ ∑ t q ∑ j q−j+1 n =1 r =1 n (n + r + 1)(n + 1) j =1 (n + 1) r
+ + q + ! ∞ (−1)n r (−1)q 1 1 1 (−1)j 1 η (q − j + 1) = q H n+1 − H n + ∑ t 2 2 ∑ j n =1 n (n + 1) 2 2 j =1 (n + 1)
q−1 ∞ n+1 ∞ n+1 + (−1) (−1) = (−1)j 1 η (q − j + 1) + (−1)q ∑ ∑ t j ∑ t q+1 j =1 n =1 n (n + 1) n =1 n (n + 1) ∞ n+1 q (−1) 1 1 + (−1) q H n+1 − H n − ln 2 . ∑ t 2 2 n =1 n (n + 1) 2 2
Now we utilize the double argument identity (3) together with (11), we obtain
q−1 (−,−) j+1 q I0 (q, t) = ∑ (−1) η (q − j + 1) F (t, j) + (−1) F (t, q + 1) j =1
∞ n+1 q (−1) + (−1) Hn − H n − 2 ln 2 , ∑ t q 2 n =1 n (n + 1) Mathematics 2019, 7, 143 8 of 20 and we can use the alternating harmonic number sum identity (5) to simplify the last sum. However, we shall simplify further as follows:
q−1 (−,−) j+1 q I0 (q, t) = ∑ (−1) η (q − j + 1) F (t, j) + (−1) F (t, q + 1) j =1
∞ n+1 q (−1) n+1 n + (−1) q (−1) H n − Hn − 1 + (−1) ln 2 ∑ t [ 2 ] n =1 n (n + 1) where [z] is the integer part of z. Now,
q−1 (−,−) j+1 q I0 (q, t) = ∑ (−1) η (q − j + 1) F (t, j) + (−1) F (t, q + 1) j =1
q q − (−1) (F (t, q) − G (t, q)) ln 2 + (−1) Wn (q, t) where ! ∞ 1 1 1 W (q t) = H − + n , ∑ n t q t q t q n =1 (2n) (2n + 1) n (n + 1) (2n + 1) (2n + 2) and the infinite positive harmonic number sums are easily obtainable from (8), (9), (10), hence the identity (15) is achieved.
The next theorem investigates the integral of the product of polylogarithmic functions divided by a linear function.
Theorem 2. Let (t, q) be positive integers, then for t + q an odd integer
1 0 Z Lit (−x) Liq (−x) Z Lit (x) Liq (x) I(−,−) (q, t) = dx = − dx x x 0 −1 q−1 (17) + + = ∑ (−1)j 1 η (t + j) η (q − j + 1) + (−1)q 1 (ζ (t + q) + η (t + q)) ln 2 j =1 + + + (−1)q 1 2−t−q − 1 EU (q + t) + (−1)q 1 HO (q + t) .
For t + q an even integer
(−,−) q−1 j+1 I (q, t) = ∑j =1 (−1) η (t + j) η (q − j + 1) + + + (−1)q 1 (ζ (t + q) + η (t + q)) ln 2 + (−1)q 1 2−t−q − 1 EU (q + t) (18) + + (−1)q 1 HE (q + t) .
Proof. Consider
1 Z Lit (−x) Liq (−x) I(−,−) (q, t) = dx x 0 1 (−1)n Z = n−1 (− ) ∑ t x Liq x dx, ≥ n n 1 0 Mathematics 2019, 7, 143 9 of 20 and successively integrating by parts leads to
q− + + 1 (−1)n 1 (−1)n q 1 Z (−,−) ( ) = ( − + ) + n−1 (− ) I q, t ∑ t+j ∑ η q j 1 ∑ t+q−1 x Li1 x dx. ≥ n = ≥ n n 1 j 1 n 1 0
Evaluating the inner integral,
1 1 Z Z n−1 n−1 x Li1 (−x) dx = − x ln (1 + x) dx 0 0
1 1 1 = H n − H n−1 − ln 2 , n 2 2 2 2 so that
− (−1)n q 1 (−,−) ( ) = (− )j ( − + ) I q, t ∑ t+j ∑ 1 η q j 1 n≥1 n j =1 + + (−1)n q 1 1 1 + H n − H n−1 − ln 2 ∑ t+q 2 2 n≥1 n 2 2
q−1 + = ∑ (−1)j 1 η (q − j + 1) η (t + j) j =1 + + (−1)n q 1 1 1 + H n − H n−1 − ln 2 . ∑ t+q 2 2 n≥1 n 2 2
If we now utilize the multiplication formula (3), we can write
q−1 n+1 (−,−) j+1 q+1 (−1) I (q, t) = (−1) η (q − j + 1) η (t + j) + (−1) Hn − H n . ∑ ∑ t+q 2 j =1 n≥1 n
Now consider the harmonic number sum n + + 1 − (−1) ln 2 (−1)n 1 (−1)n 1 Hn − H n = ∑ t+q 2 ∑ t+q n≥1 n n≥1 n n+1 + (−1) H n − Hn [ 2 ]
+ ! (−1)n 1 n (−1)j = − − n + − n+1 ∑ t+q 1 ( 1) ln 2 ( 1) ∑ n≥1 n j=1 j + (−1)n 1 1 H H = − − n + − n + n ∑ t+q 1 ( 1) ln 2 ∑ t+q 1 t+q ∑ t+q n≥1 n n≥1 2 n n≥1 (2n + 1) 1 H H = + + + + − n + n (ζ (t q) η (t q)) ln 2 ∑ t+q 1 t+q ∑ t+q n≥1 2 n n≥1 (2n + 1) Mathematics 2019, 7, 143 10 of 20 where [z] is the integer part of z. Hence,
q−1 + + I(−,−) (q, t) = ∑ (−1)j 1 η (q − j + 1) η (t + j) + (−1)q 1 (ζ (t + q) + η (t + q)) ln 2 j =1 1 H H + − q+1 − n + − q+1 n ( 1) ∑ t+q 1 t+q ( 1) ∑ t+q n≥1 2 n n≥1 (2n + 1)
q−1 + + = ∑ (−1)j 1 η (q − j + 1) η (t + j) + (−1)q 1 (ζ (t + q) + η (t + q)) ln 2 j =1 + 1 + (−1)q 1 − 1 EU (q + t) 2t+q HO (q + t) , for t + q odd + + (−1)q 1 , HE (q + t) , for t + q even hence (17) and (18) follow.
Remark 1. It is interesting to note that for m ∈ R,
1 m m 1 Z Lit (−x ) Liq (−x ) 1 Z Lit (−x) Liq (−x) dx = dx. x m x 0 0
The next theorem investigates the integral of the product of polylogarithmic functions divided by a quadratic factor.
Theorem 3. For positive integers q and t, the integral of the product of two polylogarithmic functions with negative arguments
1 − − 0 (−,−) ( ) = R Lit( x) Liq( x) = R Lit(x) Liq(x) I2 q, t x2 dx x2 dx 0 −1 q−1 j (19) = η (q + 1) + ∑j =1 (−1) η (q − j + 1) F (j, t) q q + (−1) (F (q, t) + G (q, t)) ln 2 + (−1) Wn (t, q) where the sum ! ∞ 1 1 1 Wn (t, q) = Hn − + (20) ∑ q t q t q t n =1 (2n) (2n + 1) n (n + 1) (2n + 1) (2n + 2) is obtained from (8), (9), (10), and the terms F (·, ·) , G (·, ·) are obtained from (11) and (12), respectively.
Proof. Follows the same process as in Theorem2.
The following recurrence relation holds for the reduction of the integral of the product of polylogarithmic functions multiplied by the power of its argument. Mathematics 2019, 7, 143 11 of 20
Lemma 3. For (q, t) ∈ N and m ≥ 0, let
1 0 Z Z m m m J (m, q, t) = x Lit (−x) Liq (−x) dx = (−1) x Lit (x) Liq (x) dx 0 −1 then (m + 1) J (m, q, t) = η (q) η (t) − J (m, q, t − 1) − J (m, q − 1, t) .
For q = 1,
(m + 1) J (m, 1, t) = η (t) + mJ (m − 1, 1, t) + J (m − 1, 1, t − 1) − J (m, 1, t − 1)
−mK (m, t) − K (m, t − 1) where 1 Z m K (m, t) = x Lit (−x) dx. 0
Proof. The proof of the lemma follows in a straightforward manner after integration by parts.
Theorem 4. For positive integers q and t, the integral
1 0 (+,−) Z Z I0 (q, t) = Liq (x) Lit (−x) dx = Liq (−x) Lit (x) dx 0 −1 q−1 = ∑ (−1)j ζ (q − j + 1) F (t, j) j =1
+ (−1)q F (t, q + 1) + (−1)q HA (t, q) .
Let (q, t) be positive integers, then for t + q an even integer,
1 0 Z Liq (x) Lit (−x) Z Liq (−x) Lit (x) I(+,−) (q, t) = dx = − dx x x 0 −1 (21) q−1 = ∑ (−1)j η (t + j) ζ (q − j + 1) + (−1)q S (1, q + t) , j =1
1 0 Z Liq (−x) Lit (x) Z Liq (x) Lit (−x) I(−,+) (q, t) = dx = − dx x x 0 −1 (22) t−1 = ∑ (−1)j η (q + j) ζ (t − j + 1) + (−1)tS (1, q + t) j =1 Mathematics 2019, 7, 143 12 of 20
For positive integers q and t,
1 0 (+ −) Z Liq (x) Lit (−x) Z Liq (−x) Lit (x) I , (q, t) = dx = dx 2 x2 x2 0 −1 q−1 + = −ζ (q + 1) + ∑ (−1)j 1 ζ (q − j + 1) F (j, t) j =1
+ + (−1)q 1 HA (q, t) ,
where the terms F (·, ·) , HA (·, ·) and S (·, ·) are obtained from (11)(13) and (5), respectively.
Proof. The proof is similar to the one outlined in the previous theorems.
Theorem 5. For positive integers q and t, the integral of the product of two polylogarithmic functions,
1 2 (+,+) R Liq(x) Lit(x ) U (q, t) = x dx 0 t−1 q−1 j+1 q+1 (23) = 2 ∑j =1 (−1) ζ (t + j) ζ (q − j + 1) + (−1) EU (t + q) t−1 q−1 j q +2 ∑j =1 (−1) η (t + j) ζ (q − j + 1) + (−1) S (1, q + t) where the terms EU (·, ·) , S (·, ·) are obtained from (8) and (5), respectively.
Proof. By the definition of the polylogarithmic function, we have
1 2 ∞ 1 Z Liq (x) Lit x 1 Z (+,+) ( ) = = 2n−1 ( ) U q, t dx ∑ t x Liq x dx. x = n 0 n 1 0
Successively integrating by parts leads to
q−1 j+1 U(+,+) (q, t) = ∑ (−1) ζ (q − j + 1) ∑ 1 j =1 n≥1 nt(2n)j q+1 1 (−1) R 2n−1 + ∑ q−1 x Li1 (x) dx, n≥1 nt(2n) 0 (24) q+1 q−1 j+1 1 (−1) H2n = ∑ (−1) ζ (q − j + 1) ∑ + ∑ q j =1 n≥1 nt(2n)j n≥1 nt(2n) j+1 q+1 q−1 (−1) (−1) H2n = ζ (q − j + 1) ζ (t + j) + q . ∑j =1 2j 2 ∑n≥1 nt+q
Alternatively, we have the relation
2 t−1 Lit x = 2 (Lit (x) + Lit (−x)) , (25) from which we obtain
1 2 1 R Liq(x) Lit(x ) t−1 R Liq(x)(Lit(x)+Lit(−x)) x dx = 2 x dx 0 0 t−1 (+,+) (+,−) = 2 I (q, t) + I (q, t) (26) t−1 q−1 j+1 q+1 = 2 ∑j =1 (−1) ζ (t + j) ζ (q − j + 1) + (−1) EU (t + q) t−1 q−1 j q +2 ∑j =1 (−1) η (t + j) ζ (q − j + 1) + (−1) S (1, q + t) , Mathematics 2019, 7, 143 13 of 20 hence the identity (23) is achieved. An alternative expansion of the integral
1 2 Z Liq (x) Lit x 1 1 (+,+) ( ) = = U q, t dx ∑ q ∑ t . x ≥ n ≥ r (n + 2r) 0 n 1 r 1
By partial fraction expansion,
H n (+,+) 1 1 t−1 t−1 2 U (q t) = q = (− ) , ∑n≥1 n ∑r≥1 rt(n+2r) 1 2 ∑n≥1 nt+q (27) t−1 j+1 j−1 + ∑j =1 (−1) 2 ζ (q + j) ζ (t − j + 1) .
Remark 2. We note that from (23) and (24), we are able to isolate a new Euler identity of weight q + t + 1, in the form H W (a k) = an , ∑ k . n≥1 n For a = 2, k = q + t W ( q + t) = H2n = (− )q+1 q+t−1 I(+,+) (q t) + I(+,−) (q t) 2, ∑n≥1 nt+q 1 2 , , (28) q−1 j+q q−j − ∑j =1 (−1) 2 ζ (q − j + 1) ζ (t + j) .
1 Similarly, from (26) and (27), we obtain, for a = 2 , k = q + t, another new Euler identity
H n W 1 q + t = 2 = (− )t+1 I(+,+) (q t) + I(+,−) (q t) 2 , ∑n≥1 nt+q 1 , , (29) t t−1 j+1 j−1 + (−1) ∑j =1 (−1) 2 ζ (q + j) ζ (t − j + 1) .
The difference of (28) and (29) delivers the delightful identity
q q−1 j 1 t t−1 (−1) (−1) (−1) 2 H n − H n = ζ (q − j + 1) ζ (t + j) ∑ t+q 2 q 2 ∑ j n≥1 n 2 j =1 2
t−1 + + ∑ (−1)j 1 2j−1ζ (q + j) ζ (t − j + 1) . j =1
The next theorem investigates the integral of the product of polylogarithmic functions with one positive and one negative argument.
Theorem 6. Let (q, t) be positive integers, then for t + q an even integer,
1 2 0 2 (−,+) R Liq(−x) Lit(x ) R Liq(x) Lit(x ) U (q, t) = x dx = x dx 0 −1 t−1 t−1 j = 2 ∑j =1 (−1) η (q + j) ζ (t − j + 1) + S (1, q + t) t−1 q−1 j+1 q+1 (30) +2 ∑j =1 (−1) η (t + j) η (q − j + 1) + (−1) (ζ (t + q) + η (t + q)) ln 2 HO (q + t) , for q + t odd t−1 q+1 −t−q q+1 +2 (−1) 2 − 1 EU (t + q) + (−1) . HE (q + t) , for q + t even Mathematics 2019, 7, 143 14 of 20
Proof. From (25), we have
1 2 1 (−,+) R Liq(−x) Lit(x ) t−1 R Liq(−x)(Lit(x)+Lit(−x)) U (q, t) = x dx = 2 x dx (31) 0 0 = 2t−1 I(−,+) (q, t) + I(−,−) (q, t) , replacing I(−,+) (q, t), and I(−,−) (q, t), with (22) and (17), respectively, leads to (17), as required. We can also consider
1 2 Z Liq (−x) Lit x U(−,+) (q, t) = dx x 0 1 (−1)n Z = n−1 2 ∑ q x Lit x dx, ≥ n n 1 0 and successively integrating by parts, as in Theorem1, leads to
n+1 t−1 (−1) H n (−,+) = − j − + + + − t 2 U (q, t) ∑ ( 1) ζ (t j 1) η (q j) ( 1) ∑ q+t . (32) j =1 n≥1 n
Another expansion of
1 1 2 n n+2r−1 Z Liq (−x) Lit x Z (−1) x (−,+) ( ) = = U q, t dx ∑ ∑ q t dx x ≥ ≥ n r 0 0 r 1 n 1 (−1)n = . ∑ ∑ q t ( + ) r≥1 n≥1 n r n 2r
Partial fraction expansion, and summing
q−1 U(−,+) (q, t) = (−1)q ∑ (−1)j 2j−qη (1 + j) ζ (q + t − j) j =1 + (−1)q 1 H H + n − 2n q ∑ q+t q+t , 2 n≥1 n n
q− + 1 (−1)q 1 = − q − j j−q + + − + + + + ( 1) ∑ ( 1) 2 η (1 j) ζ (q t j) q (EU (q t) W (2, q t)) . j =1 2 hence (17) follows. We note that from (18) and (32), we obtain another new Euler sum, namely
n+1 (−1) H n 2 = (−1)t I(−,+) (q, t) + I(−,−) (q, t) ∑n≥1 nq+t (33) t t−1 j+1 j−t + (−1) ∑j =1 (−1) 2 ζ (t − j + 1) η (q + j) .
Example 1. Some examples follow.
1 2 Z Li3 (x) Li5 x 521 1 U(+,+) (3, 5) = dx = ζ (9) − 8ζ (2) ζ (7) − 2ζ (4) ζ (5) − ζ (6) ζ (3) . x 32 2 0 Mathematics 2019, 7, 143 15 of 20
1 2 Z Li4 (−x) Li4 x 501 11 U(−,+) (4, 4) = dx = ζ (9) − 4ζ (2) ζ (7) − ζ (4) ζ (5) . x 64 8 0
H 2059 ( + ) = 2n = ( ) − ( ) ( ) W 2, 6 4 ∑ 10 ζ 11 256ζ 9 ζ 2 n≥1 n 4
−4ζ (8) ζ (3) − 64ζ (7) ζ (4) − 16ζ (6) ζ (5) .
H ( + ) = 2n = ( ) − ( ) W 2, 2 1 ∑ 3 5ζ 4 4L 3 n≥1 n where 11 7 1 1 1 L (3) = ζ (4) − ζ (3) ln 2 + ζ (2) ln2 2 − ln4 2 − 2Li . 4 4 2 12 4 2 1 3 1 17 2 517 2 H n + H n = ζ (3) ζ (7) + ζ (5) − ζ (10) . ∑ 9 2 5 2 n≥1 n 2 4 32
1 H n 521 1 + = 2 = ( ) − ( ) ( ) W , 5 3 ∑ 8 ζ 9 ζ 2 ζ 7 2 n≥1 n 512 64 1 1 − ζ (4) ζ (5) − ζ (3) ζ (6) . 16 4
n+1 (−1) H n 49 31 635 2 = ( ) ( ) + ( ) ( ) − ( ) ∑ 6 ζ 4 ζ 3 ζ 5 ζ 2 ζ 7 . n≥1 n 32 32 128
The next two results deal with Polylog integrals involving the inverse of the argument.
Theorem 7. For positive integers q and t, the integral of the product of two polylogarithmic functions,
1 1 2 R Liq(− x ) Lit(x ) I3 (q, t) = x dx 0 1 1 (34) = − ζ (q + t + 1) + q (EU (q + t) − W (2, q + t)) 2q+1 2 q−1 − 1 ζ (q + t) η (q − j + 1) ∑j =1 2j where the terms EU (·) , W (·, ·) are obtained from (8) and (28), respectively. For positive integers q and t, the integral of the product of two polylogarithmic functions,
1 1 Liq Lit(x) R x2 I4 (q, t) = x dx 0 q q−1 1 (35) = −2 ζ (q + t + 1) + 2 W 2 , q + t q−1 j−1 q−1 + ∑j =1 2 ζ (j + t) ζ (q − j + 1) − 2 iπζ (q + t) √ 1 where W 2 , q + t is obtained from (29), and i = −1.
Proof. Following the same process as in Theorem2, we have 1 − 1 2 1 Z Liq x Lit x 1 Z 1 ( ) = = 2n−1 − I3 q, t dx ∑ t x Liq dx. x ≥ n x 0 n 1 0 Mathematics 2019, 7, 143 16 of 20
Integrating by parts, as in Theorem2, we have
1 q−1 Z 1 1 2n−1 1 I3 (q, t) = − η (q − j + 1) ζ (t + j) + x Li1 − dx ∑ j ∑ t q−1 = 2 ≥ n (2n) x j 1 n 1 0 − q 1 1 1 1 1 1 1 = − η (q − j + 1) ζ (t + j) − ζ (t + q) + Hn − H 1 − ln 2 , ∑ j q q ∑ t+q n− 2 j =1 2 2 2 n≥1 n 2 2 from the multiplication Formula (3), we can simplify so that
− q 1 1 1 1 1 ( ) = − ( − + ) ( + ) − ( + ) − ( − ) I3 q, t ∑ j η q j 1 ζ t j q ζ t q q ∑ t+q H2n Hn j =1 2 2 2 n≥1 n and the proof of the first part of the theorem is finalized. The second result is proved as in Theorem2.
Example 2. 1 − 1 2 Z Li4 x Li2 x 119 7 I (4, 2) = dx = ζ (7) − ζ (2) ζ (5) − ζ (4) ζ (3) . 3 x 64 16 0 1 Li 1 Li (x) Z 3 x2 5 31 503 I (3, 5) = dx = ζ (2) ζ (7) − ζ (9) 4 x 16 128 0 1 − ζ (4) ζ (5) − iπζ (8) . 4
3. Some Extensions Some other valuable research on the representation of integrals of polylogarithmic functions with a trigonometric argument has been carried out, and we highlight the following results: Espinosa and Moll [25,26] give for p ∈ N0
h p+1 i 1 ! Z 2 p + 1 (2j)! (p + ) xp ( ( x)) dx = − + (− )j ( j + ) 1 ln sin π ln 2 ∑ 1 2j ζ 2 1 = 2j (2π) 0 j 1 where [m] is the integer part of m. For a non-trigonometric argument, the work of Espinosa and Moll may be extended in the following way: For (p, q) ∈ N, let
1 q Z ln x Lip (x) M (p, q) = dx. 1 − x x 0
Rewriting ! 1 n + q − 1 n q = ∑ x , (1 − x) n ≥0 n then after integration and simplification ! (−1)q n + q − 1 1 M (p, q) = ζ (p + q + 1) + , ∑ ∑ p q+1 q! j ≥1 n ≥1 n j (j + n) Mathematics 2019, 7, 143 17 of 20 which results in the product of zeta functions. For the case of q = 2, we may write
1 2 Z ln x Lip (x) M (p, 2) = dx = 2ζ (p + 3) − 2ζ (p + 3) ζ (p − 1) 1 − x x 0
+2ζ (p)(ζ (2) + ζ (3)) − 2BW (2, p) − 2BW (3, p) + 2BW (3, p − 1) , where BW (·, ·) is given by (6). For p = 4, we find
1 Z ln x 2 Li (x) 31 M (4, 2) = 4 dx = ζ (6) − 3ζ2 (3) 1 − x x 16 0
+20ζ (2) ζ (5) + 2ζ (3) ζ (4) − 34ζ (7) .
Choi [27,28] has also produced many interesting examples of log-sine and log-cosine integrals. Mezo [29] continues the study and considers the class of integrals
1 1 Z Z Ip = Lip (sin πx) dx and Jp = Lip (cos πx) dx, 0 0 such that ! 1 2j 1 = Jp p ∑ j p 2 j ≥1 j 4 j and 2 (2j − 2)!! 1 I = J + , p p ∑ ( − ) p π j ≥1 2j 1 !! (2j − 1) where 1, for m = −1, 0 m!! = m (m − 2) . . . 6.4.2, for m > 0 even . m (m − 2) . . . 5.3.1, for m > 0 odd Mezo [29] gives the following examples:
1 Z π π3 Li (sec πx) dx = i ln2 2 + 3 4 48 0 and 1 1 Z Z 4 Li sin2 πx dx = Li cos2 πx dx = 2ζ (3) − 2ζ (2) ln 2 + ln3 2. (36) 3 3 0 0 It is possible to generalize (36), so that for m ∈ N
1 ! Z 1 2mn 2m = Lip sin πx dx ∑ 2mn p , ≥ 2 n mn 0 n 1 and 1 Z 9 2 Li sin2 πx dx = ζ (4) + 2ζ (2) ln2 2 − 4ζ (3) ln 2 − ln4 2. 4 4 3 0 Mathematics 2019, 7, 143 18 of 20
There are a number of extensions that may be very interesting to investigate further in regard to the representation of integrals of polylogarithmic functions. It would be beneficial to study the integrals
1 1 Z Z arcsin x a (sin πx)a Lib (sin πx) dx, Lib x2 dx p x p 0 0 where, for instance
1 1 4 Z πx Z π sin πx Li tan4 dx = 4π sin 4πy Li tan4 πy dy 4 4 4 0 0 = 6ζ (4) + 32ζ (2) ln 2 − 32ζ (3) ,
1 Z π 1 π sin πx Li (sin πx) dx = − 1 + ζ (2) + (2 + ln 2) ln 2 4 2 8 4 0 1 2n n! + ∑ 3 , 4π n≥0 (n + 1) (2n + 3)!!
1 ! Z arcsin x π π 1 2n Li x2 dx = ζ (p + 1) − , p ∑ 2n p+1 x 4 4 ≥ 2 n n 0 n 1 and
1 1 Z arcsin x 9 1 1 Li x2 dx = ζ (4) ln 2 + ζ (2) ln3 2 − ln5 2 π x 4 8 3 15 0 1 5 + ζ (3) ζ (2) − 2 ln2 2 − ζ (5) . 2 4 We may also study 1 Z a b Lip (sin πx) Liq (cos πx) dx, 0 and the moments 1 Z a b x Lip (sin πx) dx, 0
1 Z a b x Lip (arcsin x) dx, 0 where (a, b, p, q) ∈ N. In this paper we have highlighted various new identities of general Euler type sums, which represent the analytical solution of integrals with polylogarithm functions containing linear, quadratic, and trigonometric type arguments. In a series of papers [7,30,31], the authors explore linear combinations of associated harmonic polylogarithms and nested harmonic numbers. The multiple zeta value data mine, computed by Blumlein et al. [32], is an invaluable tool for the evaluation of harmonic numbers, in which values with weights of twelve, for alternating sums, and weights above twenty for non-alternating sums are presented. Further areas of fruitful research are those related to trigonometric, inverse trigonometric, and linear arguments for products of integrals of polylogarithmic functions. Mathematics 2019, 7, 143 19 of 20
Funding: This research received no external funding. Conflicts of Interest: The author declares no conflict of interest.
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