Families of Log Legendre Chi Function Integrals

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Appl. Anal. Discrete Math. x (xxxx), xxx–xxx. doi:10.2298/AADMxxxxxxxx

FAMILIES OF LOG LEGENDRE CHI FUNCTION INTEGRALS

Anthony Sofo

In this paper we investigate the representation of integrals involving the product of the Legendre Chi function, polylogarithm function and log function. We will show that in many cases these integrals take an explicit form involving the Riemann zeta function, the Dirichlet Eta function, Dirichlet lambda function and many other special functions. Some examples illustrating the theorems will be detailed.

1. INTRODUCTION PRELIMINARIES AND NOTATION In this paper we investigate the representations of integrals of the type

1 Z a b m (1) x χp (x) Liq(δx ) ln (x) dx, 0 in terms of special functions such as zeta functions, Dirichlet eta functions, polylogarithmic functions, beta functions and others. We note that χp (x) is the Legendre- b Chi function (LCF), Liq(δx ) is the polylogarithmic function, a ∈ R, a ≥ −2, b ∈ R+, p ∈ N, q ∈ N, δ ∈ [−1, 1] \{0} and m ∈ N for the set of complex numbers C, natural numbers N, the set of real numbers R and the set of positive real numbers, R+. The following notation and results will be useful in the subsequent λ sections of this paper. A generalized binomial coeﬃcient µ (λ, µ ∈ C) is deﬁned, in terms of the familiar (Euler’s) gamma function, by λ Γ(λ + 1) := (λ, µ ∈ ), µ Γ(µ + 1) Γ (λ − µ + 1) C

2020 Mathematics Subject Classiﬁcation. 11M06, 33B15, 65B10 Keywords and Phrases. Legendre Chi Function, Polylogarithm Function, Euler Sum, Dirichlet Lambda Function, Dirichlet Beta Function

1 2 Anthony Sofo which, in the special case when µ = n (n ∈ N0 := N ∪ {0}), yields λ λ λ (λ − 1) ··· (λ − n + 1) (−1)n (−λ) := 1 and := = n (n ∈ ), 0 n n! n! N where (λ)ν (λ, ν ∈ C) is the Pochhammer symbol deﬁned, also in terms of the Gamma function, by

 1 (ν = 0; λ ∈ \{0}) Γ(λ + ν)  C (λ) := = ν Γ(λ)  λ(λ + 1) ··· (λ + n − 1) (ν = n ∈ N; λ ∈ C), it being understood conventionally that (0)0 := 1 and assumed tacitly that the Γ- (p) quotient exists. The generalized p−order harmonic numbers, Hn (α, β) are deﬁned as the partial sums of the modiﬁed Hurwitz zeta function

X 1 ζ (p, α, β) = . (αn + β)p n≥0

The classical Hurwitz zeta function X 1 ζ (p, a) = (n + a)p n≥0 for Re (p) > 1 and by analytic continuation to other values of p 6= 1, where any term of the form (n + a) = 0 is excluded. Therefore

n X 1 H(p) (α, β) = n (αj + β)p j=1

(p) (p) and the ”ordinary” p−order harmonic numbers Hn = Hn (1, 0) . Many functions can be expanded through the generalized p−order harmonic numbers, such as the Dirichlet Beta cases X n! X (n − 1)! β (1) = = 2n+1 3 2n+1 1 n≥0 2 n n≥1 2 n and X n! (1 + Hn (2, 1)) X (n − 1)!hn β (2) = = , 2n+1 3 2n+1 1 n≥0 2 n n≥1 2 n 1 here β (2) is Catalan’s constant and hn = H2n − 2 Hn. Two special cases of the Legendre-Chi function are

n X (−1) n!x2n+1 χ1 (x) = 2 n+1 3 n≥0 (1 − x ) 2 n Familes of Legendre Chi Function Integrals 3 and n 2n+1 ! X n! (−1) x Hn (2, 1) χ2 (x) = 1 + . 3 2 n+1 n≥0 2 n (1 − x ) The Catalan constant ∞ n+1 X (−1) G = β (2) = 2 ≈ 0.91597 n=1 (2n − 1) is a special case of the Dirichlet beta function

∞ n+1 X (−1) β (z) = , for Re (z) > 0(2) (2n − 1)z n=1 1 1 3 = ψ(z−1) − ψ(z−1) , (−2)2z (z − 1)! 4 4 with functional equation 2 z πz β (1 − z) = sin Γ(z) β (z) π 2 extending the Dirichlet Beta function to the left hand side of the complex plane Re(z) ≤ 0. The Lerch transcendent,

∞ X zm Φ(z, t, a) = t m =0 (m + a) is deﬁned for |z| < 1 and Re (a) > 0 and satisﬁes the recurrence:

Φ(z, t, a) = z Φ(z, t, a + 1) + a−t.

It is known that the Lerch transcendent extends by analytic continuation to a function Φ (z, t, a) which is deﬁned for all complex t, z ∈ C − [1, ∞) and a > 0, which can be represented, [15], by the integral formula

∞ 1 1 Z xt−1e−(t−1)x 1 Z xa−1 ln 1 Φ(z, t, a) = dx = x dx Γ(t) ex − z Γ(t) 1 − xz 0 0 for Re (t) > 0. The Lerch transcendent generalizes the Hurwitz zeta function at z = 1, ∞ X 1 Φ (1, t, a) = ζ (t, a) = t m =0 (m + a) and the polylogarithm, or de-Jonqui`ere’sfunction, when a = 1,

∞ X zm Li (z) := , t ∈ when |z| < 1; Re (t) > 1 when |z| = 1. t mt C m =1 4 Anthony Sofo

The polylogarithm of negative integer order arises in the sums of the form

n−1 X 1 X n jnzj = Li (z) = zn−i −n n+1 i j≥1 (1 − z) i =0

n n + 1 where the Eulerian number = Pi+1 (−1)j (i − j + 1)n. The i j =0 j Legendre-Chi function is a special case of the Lerch transcendent

∞ 1 X x2n+1 χ (x) = 2−p x Φ x2, p, = p 2 (2n + 1)p n =0 and is related to the polylogarithm by 1 χ (x) = (Li (x) − Li (−x)) = Li (x) − 2−pLi (x2). p 2 p p p p There are many special values of the LCF, from Lewin [19]

x 1 Z 1 + t dt χ (x) = ln 2 2 1 − t t 0 and 1 − x 3 1 1 + x χ (x) + χ (x) = ζ (2) + ln x ln , 2 1 + x 2 4 2 1 − x hence, √ 1 3 χ 5 − 2 = ζ (2) − ln2 (φ) 2 4 4 √ 1 where the golden ratio φ = 2 1 + 5 . The Dirichlet lambda function, λ (p) is

(3) 2λ (p) = 2χp (1) = ζ (p) + η (p) where ∞ n+1 X (−1) η (p) = = 1 − 21−p ζ (p) np n =1 is the alternating zeta function and η (1) = ln 2. In the case of the summation of harmonic numbers, we know that the famous Euler identity states, for m ∈ N ≥ 2,

∞ m−2 X Hn 1 1 X (4) EU (m) = = (m + 2)ζ (m + 1) − ζ (m − j) ζ (j + 1) , nm 2 2 n =1 j=1 for odd powers of the denominator, Georghiou and Philippou [14] established the identity: ∞ 2m X Hn 1 X = (−1)r ζ(r) ζ(2m + 2 − r), m ≥ 1. n2m+1 2 n=1 r=2 Familes of Legendre Chi Function Integrals 5

We know that for n ≥ 1, ψ(n + 1) − ψ(1) = Hn with ψ(1) = −γ, where γ is the Euler Mascheroni constant and ψ(n) is the digamma function. For real values of x, ψ(x) is the digamma (or psi) function deﬁned by

d Γ0(x) ψ(x) := {log Γ(x)} = , dx Γ(x)

∞ ∞ 1 X 1 1 X 1 1 ψ(x) = −γ − + − = −γ + − , x n x + n n x + n − 1 n=1 n=1 leading to the telescoping sum:

∞ X 1 1 1 ψ(1 + x) − ψ(x) = − = . x + n − 1 x + n x n=1

The polygamma function

k ∞ d k+1 X 1 ψ(k)(z) = {ψ(z)} = (−1) k! dzk k+1 r=0 (r + z) and has the recurrence

(−1)k k! ψ(k)(z + 1) = ψ(k)(z) + . zk+1

The connection of the polygamma function with harmonic numbers is,

(−1)m (5) H(m+1) = ζ (m + 1) + ψ(m) (z + 1) , z 6= {−1, −2, −3, ...} . z m! 1 (−1)m Z (1 − tz) = lnm t dt m! 1 − t 0 and the multiplication formula is

p−1 1 X j (6) ψ(k)(pz) = δ ln p + ψ(k)(z + ) p,0 pk+1 p j=0

+ (p) for p a positive integer and δp,k is the Kronecker delta. Let a ∈ R and Hn denote (p) P∞ Han harmonic numbers of order p, and put BW (a, p.q) = n =1 nq . We have, from 6 Anthony Sofo the work of [4] that for odd weight (p + q)

∞ (p) X Hn BW (1, p.q) = nq n =1

p [ 2 ] p X p + q − 2j − 1 = (−1) ζ (p + q − 2j) ζ (2j) p − 1 j =1 1 + 1 + (−1)p+1 ζ (p) ζ (q) 2

p [ 2 ] p X p + q − 2j − 1 + (−1) ζ (p + q − 2j) ζ (2j)(7) q − 1 j =1

ζ (p + q) p + q − 1 p + q − 1 + 1 + (−1)p+1 + (−1)p+1 , 2 p q where [z] is the integer part of z. Let us denote

∞ n+1 (p) X (−1) Han S (a, p, q) := , nq n =1 then in the case where p and q are both positive integers and p+q is an odd integer, Flajolet and Salvy [12] gave the identity:

p p X p + i − 1 2S (1, p, q) = (1 − (−1) ) ζ (p) η (q) + 2 (−1) ζ (p + i) η (2k) p − 1 i+2k=q X q + j − 1 j (8) + η (p + q) − 2 (−1) η (q + j) η (2k) , q − 1 j+2k=p

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. In particular

q −1 X2 (9) 2S (1, 1, q) = (1 + q) η (1 + q) − ζ (1 + q) − 2 η (2j) ζ (1 + q − 2j) . j=1

It is interesting to note that recently [1], established that for p ∈ N\{1}

p 1 1 X (10) S (1, p, 1) = pζ (p + 1) − η (j) η (p − j + 1) . 2 2 j=1 Familes of Legendre Chi Function Integrals 7

From [26] we have the identities

∞ (p) 1 X H n BW , p.q = 2 2 nq n =1   p BW (1, r, p + q − r) q X p − 1 + q − r = (−1) 2p−1 p − r   r=1 −S (1, r, p + q − r) p q+1 X 1 p − 1 + q − r (11) + (−1) ζ (r) ζ (p + q − r) 2q−r p − r r=2 q−1 k q+1 X (−1) p − 1 + q − k + (−1) ζ (k) ζ (p + q − k) , 2q−k q − k k=2

1 1 (12) S , p.q = BW , p.q − 21−qBW (1, p.q) 2 2 and

∞ H(p) X n− 1 1 1 (13) 2 2 = BW , p.q − S , p.q . nq 2 2 n =1 The next lemma relates the sum of the double and unitary argument of harmonic numbers in closed form and will be useful in the following section. Lemma 1. Let m ∈ N ≥ 2, p ∈ N then

∞ (p) X H HE (p, m) = 2n (2n − 1)m n =1 p + m − 2 = (−1)m+1 ln 2 p − 1 1 + (BW (1, p, m) + S (1, p, m)) 2 p m X (−1) p + m − 1 − r + ζ (r) 2r p − r r=2 m m−k X (−1) p + m − 1 − k + λ (k) . 2 p − 1 k=2 For the unitary argument

∞ (p) X Hn = 2p−1 (BW (1, p, m) + S (1, p, m)) − 2p−1η (p) λ (m) (2n + 1)m n =1 1 1 1 − BW , p, m + S , p, m 2 2 2 8 Anthony Sofo where BW (p, m) is the Borwein identity (7), S (p, m) is evaluated from (8) and λ (j) is deﬁned by (3).

Proof.

∞ (p) ∞ (p) ∞ n+1 (p) ! X H 1 X Hn X (−1) Hn HE (p, m) = 2n = − (2n − 1)m 2 (n − 1)m (n − 1)m n =1 n =2 n =2  ∞ 1 (p) ∞ n+1 1 (p)  (−1) p + Hn 1 X (n+1)p + Hn X (n+1) = + 2  nm nm  n =1 n =1 ∞ 1 X 1 = (BW (p, m) + S (p, m)) + . 2 (2n)p (2n − 1)m n =1 Expanding in partial fraction form gives us ∞ m+1 1 X (−1) p + m − 2 HE (p, m) = (BW (p, m) + S (p, m)) + 2 2n (2n − 1) p − 1 n =1

∞ p m m m−k ! X X (−1) p + m − 1 − r X (−1) p + m − 1 − k + 2rnr p − r k p − 1 n =1 r=2 k=2 (2n − 1) and the result follows. For the unitary argument the proof follows the same pattern as above.

The following partial fraction decomposition holds. Lemma 2. For y ∈ R, m, p ∈ N\{0} we have 1 (−1)p+1 m + p − 1 1 = (2n − 1)p (2n + y)m+1 (1 + y)m+p−1 m (2n − 1) (2n + y) m + p − r k m + p − k m+1 p (−1) p X m + 1 − r p X m + (−1) m+p+1−r r + (−1) m+p+1−k k . r=2 (1 + y) (2n + y) k=2 (1 + y) (2n − 1) Proof. Follows simply by expansion.

The following lemma will also be useful in the evaluation of integrals of the type (1). Lemma 3. Let m ∈ N ≥ 2. then ∞ X Hn m+1 HM (m) = = mλ (m + 1) + 2 (−1) − λ (m) ln 2 (2n − 1)m n =1 m m−2 X m+j 2 X (14) + (−1) λ (j) − (m − k − 1) λ (k + 1) λ (m − k) (m − 1) j=2 k=1 Familes of Legendre Chi Function Integrals 9 where λ (m) is given by (3).

Proof. We have, from [24], theorem 1, for x a real number, x 6= −1, −2, −3, ··

∞ m X Hn 1 (−1) (m − 1)! = (ψ(x) + γ) ψ(m−1)(x) − ψ(m)(x)(15) (n + x)m 2 n =1 m−2 X m − 2 + ψ(m)(x)ψ(m−k−1)(x). k k=1

1 1 m (m) 1 Choosing x = 2 , we have ψ( 2 ) + γ = −2 ln 2 and from (5), (−1) ψ ( 2 ) = −2mm!λ (m + 1), therefore substituting into the above equation (16) and simplifying leads to

∞ X Hn = mλ (m + 1) − 2λ (m) ln 2 (2n + 1)m (16) n =1 m−2 2 P − (m−1) (m − k − 1) λ (k + 1) λ (m − k) . k=1 Identity (16) corrects a minor error in the paper [20], now reordering the counter in (16) we obtain the identity (14).

Lemma 4. We deﬁne

nπ n+1 ! X sin 4 X (−1) 2 2 1 W (3) := n = 2n 3 + 3 + 3 2 2 n3 2 n≥1 n≥1 (4n − 3) (4n − 2) (4n − 1)  8 + 8 + 4  1 (8n−7)3 (8n−6)3 (8n−5)3 X   = 4n 2  2 2 1  n≥1 − − − (8n−3)3 (8n−2)3 (8n−1)3 1 1 1 1 1 1 3 = 2Φ − , 3, + 2Φ − , 3, + Φ − , 3, . 256 4 4 4 2 4 4 where Φ (·, ·, ·) is the Lerch transcendent.

Proof. Consider the polylogarithms

inπ nπ nπ 1 ± i X exp ± 4 X cos 4 ± i sin 4 Li3 = n = n , 2 2 2 n3 2 2 n3 n≥1 n≥1 adding produces

nπ 1 + i 1 − i X cos 4 Li3 + Li3 = 2 n 2 2 2 2 n3 n≥1 35 1 5 = ζ (3) + ln3 2 − ζ (2) ln 2, 32 24 16 10 Anthony Sofo by a result from [16]. By the circular nature of the trigonometric function, the BBP type sum for

nπ n+1 ! X cos 4 X (−1) 2 1 1 n = 2n 3 − 3 − 3 2 2 n3 2 n≥1 n≥1 (4n − 3) (4n − 1) (4n) 1 1 1 1 3 1 = 2Φ − , 3, − Φ − , 3, − Φ − , 3, 1 256 4 4 4 4 4 ! X 1 8 1 1 = + + 24n 3 3 3 n≥1 (8n − 7) (8n − 1) (8n) ! X 1 4 4 2 − + + . 24n 3 3 3 n≥1 (8n − 5) (8n − 4) (8n − 3)

By a similar argument we have

nπ 1 + i 1 − i X sin 4 Li3 − Li3 = 2i n 2 2 2 2 n3 n≥1 and the nπ 1 + i X sin 4 Im Li3 = n 2 2 2 n3 n≥1 n+1 ! X (−1) 2 2 1 = + + 22n 3 3 3 n≥1 (4n − 3) (4n − 2) (4n − 1)  8 + 8 + 4  1 (8n−7)3 (8n−6)3 (8n−5)3 X   = 4n 2  2 2 1  n≥1 − − − (8n−3)3 (8n−2)3 (8n−1)3 1 1 1 1 1 1 3 = 2Φ − , 3, + 2Φ − , 3, + Φ − , 3, . 256 4 4 4 2 4 4

Since the Legendre-Chi function can be expressed as the diﬀerence of two polylogarithmic functions then we expect that integrals of the type (1) may be represented as Euler sums and therefore in terms of special functions such as the Riemann zeta function. A search of the current literature has not found many examples for the representation of the integral (1) and certainly not a systematic study of (1). Many papers, [11], [13], [27], [28] examined some polylogarithmic integrals in terms of Euler sums. Some other important sources of information on Legendre-Chi functions are the works of [2], [6], [8], [9], [10] and the excellent books [19], [29] and [30]. Other useful references related to the representation of Euler sums in terms of special functions include [1], [3], [21], [22], [23], [24], [25]. Familes of Legendre Chi Function Integrals 11

Some examples are highlighted, most of which are not amenable to a computer mathematical package.

2. MAIN RESULTS

Theorem 1. Let a ∈ R ≥ −2, b ∈ R+, δ = 1, (m, p, q) ∈ N ∪ {0} , the integral of the product of the Legendre-Chi, polylogarithm and log functions,

1 Z a b m P (a, b, m, p, q) = x χp (x) Liq x ln (x) dx 0 (j+1) m q−j−1 m H 2n+a X (−b) (−1) m!(q)m−j X = b (m − j)! p q+m−j j =0 n≥1 (2n − 1) (2n + a) m q−j−1 m X (−b) (−1) m!(q)m−j X ζ (j + 1) − (m − j)! p q+m−j j =1 n≥1 (2n − 1) (2n + a) q X q−r m + (−b) (−1) (q + 1 − r)m ζ (r) r=2 X 1 × p q+m+1−r n≥1 (2n − 1) (2n + a)

(m+1) where H 2n+a are harmonic numbers of order (m + 1) , ζ (·) is the Riemann zeta b function, χp (x) is the Legendre-Chi function, Liq (·) is the polylogarithm function and (q)j is the Pochhammer symbol.

Proof. From the Taylor series expansion of the LCF function and the polylogarithm functions we have

1 1 Z X 1 X 1 Z xa χ (x) Li xb dx = x2n+bj+a−1 dx p q (2n − 1)p jq 0 n≥1 j≥1 0 X X 1 = . (2n − 1)p jq (2n + bj + a) n≥1 j≥1 12 Anthony Sofo

By partial fraction expansion, from Lemma 2

q (−b)q−r 1  P  Z 1 r=2 jr (2n+a)q+1−r a b X X   x χp (x) Liq x dx = p   (2n − 1) (−b)q−1 0 n≥1 j≥1 − j(2n+a)q−1(2n+bj+a) q−1 X (−b) H 2n+a = b (2n − 1)p (2n + a)q n≥1 q q−r X X (−b) ζ (r) + p q+1−r . r=2 n≥1 (2n − 1) (2n + a) Diﬀerentiating 0m0 times under the integral sign, with respect to the real parameter a, we obtain

H 2n+a q−1 m m + q − 1 X P (a, b, m, p, q) = (−b) (−1) m! b q − 1 p q+m n≥1 (2n − 1) (2n + a) m X q−j−1 m+j m m + q − 1 − j + (−b) (−1) (m − j)! j q − 1 j =1 (m) 2n+b+a X ψ × b j p q+m+1−j n≥1 b (2n − 1) (2n + a) q X q−r m m + q − r X 1 + (−b) (−1) m! ζ (r) , q − r p q+m+1−r r=2 n≥1 (2n − 1) (2n + a) simplifying by the use of the Polygamma identity (5) and applying the Pochhammer symbol, we obtain

1 Z a b m P (a, b, m, p, q) = x χp (x) Liq x ln (x) dx 0 (m+1) m q−j−1 m H 2n+a X (−b) (−1) m!(q)m−j X = b (m − j)! p q+m−j j =0 n≥1 (2n − 1) (2n + a) m q−j−1 m X (−b) (−1) m!(q)m−j X ζ (j + 1) − (m − j)! p q+m−j j =1 n≥1 (2n − 1) (2n + a) q X q−r m X 1 + (−b) (−1) (q + 1 − r)m ζ (r) p q+m+1−r r=2 n≥1 (2n − 1) (2n + a) and Theorem 1 follows.

There are some interesting special cases of Theorem 1 and we present these in the next corollaries before listing some illustrative examples. Familes of Legendre Chi Function Integrals 13

Corollary 1. Let the conditions of Theorem 1 hold then, for p = 0

1 Z a b m P (a, b, m, 0, q) = x χ0 (x) Liq x ln (x) dx 0  (m+1)  m H (−1) m! X bn+a = ζ (m + 1) ζ (q) − 2 , 2m+1  nq  n≥1

x where χ0 (x) = 1−x2. . Proof. Applying the same technique as in Theorem 1, we have

1 Z a b m P (a, b, m, 0, q) = x χ0 (x) Liq x ln (x) dx 0 m m X X (−1) m! X 1 X (−1) m! = = q m+1 jq m+1 n≥1 j≥1 j (2n + bj + a) j≥1 n≥1 (2n + bj + a) m X 1 bj + a = (−1) ψ(m) + 1 jq 2 j≥1  (m+1)  m H (−1) m! X bj+a = ζ (m + 1) ζ (q) − 2 , 2m+1  jq  j≥1 replacing the counter 0j0 for 0n0 the proof is completed.

Corollary 2. Let the conditions of Theorem 1 hold then, for q = 0

1 Z a b m P (a, b, m, p, 0) = x χp (x) Li0 x ln (x) dx 0

 (m+1)  m+1 H (−1) m! X 2n+a = b − ζ (m + 1) λ (p) , bm+1  (2n − 1)p  n≥1

b xb where Li0 x = 1−xb. .

Proof. Following the same steps as in corollary 2, we have

(m) 2n+a 1 X ψ + 1 P (a, b, m, p, 0) = b , bm+1 (2n − 1)p n≥1 14 Anthony Sofo replacing the polygamma function with harmonic numbers  (m+1)  m+1 H − ζ (m + 1) (−1) m! X 2n+a P (a, b, m, p, 0) = b bm+1  (2n − 1)p  n≥1 and proof is complete.

Some illustrative examples of Theorem 1 and its corollaries follow. To evaluate the resultant Euler sums for these examples we need a mixture of identities, including (4), (5), (6), (7), (8), (9), (10), (11), (12), (13), Lemma 1, Lemma 2, Lemma 3 and Lemma 4. Examples. 1. Let a = −1, b = 1, then 1 Z −1 m P (−1, 1, m, p, q) = x χp (x) Liq (x) ln (x) dx 0 m q+m−j−1 (m+1) X (−1) m!(q)m−j X H = 2n−1 (m − j)! m+p+q−j j =0 n≥1 (2n − 1) m q+m−j−1 X (−1) m!(q)m−j − ζ (j + 1) λ (m + p + q − j) 2 (m − j)! j =1 q X m+q−r + (−1) (q + 1 − r)m ζ (r) λ (m + p + q + 1 − r) . r=2

H(m) P 2n−1 The Euler sums n≥1 (2n−1)p can be evaluated from Lemma 1. If we let m = p = q = 2, 1 Z −1 2 P (−1, 1, 2, 2, 2) = x χ2 (x) Li2 (x) ln (x) dx 0 67 889 = ζ (5) ζ (2) − ζ (7) . 8 64 1 2. For a = 0, b = 2 , m = p = q = 1 1 1 Z 1 P 0, , 1, 1, 1 = χ (x) Li x 2 ln (x) dx 2 1 1 0 (2) X H4n X ζ (2) − H = + 4n . 2 n (2n − 1) n≥1 (2n) (2n − 1) n≥1 Here we require the Euler sums

X H4n 67 X H4n π 5 1 = ζ (3) − 2πG, = + ζ (2) − ln2 2 n2 8 n (2n − 1) 2 4 2 n≥1 n≥1 Familes of Legendre Chi Function Integrals 15

(2) X H 1 π 13 1 1 4n = 2ζ (3) + G − πG + − ζ (2) − ζ (2) ln 2 − ln 2, n (2n − 1) 2 4 16 8 4 n≥1 which may be evaluated by the techniques described in [5], [7], [18], [26], [31], [32] therefore

1 1 61 3π 9 1 1 P 0, , 1, 1, 1 = πG − 2G − ζ (3) − + ζ (2) + ζ (2) ln 2 + ln 2 + ln2 2. 2 2 32 4 4 2 4

3. For a = 0, b = 2, m = p = 1, q = 5

1 Z 2 P (0, 2, 1, 1, 5) = χ1 (x) Li5 x ln (x) dx 0 (2) 5 X Hn 1 X ζ (2) − Hn = − + 4 n6 (2n − 1) 4 n5 (2n − 1) n≥1 n≥1 5 X 5−r 6 − r X 1 − (−2) ζ (r) . 5 − r 7−r r=2 n≥1 (2n − 1) (2n)

The Euler sums may be evaluated from Lemma 1, therefore

5 23 3 P (0, 2, 1, 1, 5) = ζ (7) − ζ (6) + ζ2 (3) − ζ (2) ζ (5) − ζ (5) ln 2 − 12ζ (3) ln 2 2 12 4 1 21 + 50ζ (3) + 4ζ (4) ln 2 − 4ζ (2) ζ (3) − ζ (4) ζ (3) + ζ (5) 2 2 + 32ζ (2) ln 2 − 9ζ (4) + 8ζ (2) + 80 ln2 2 − 12 ln 2.

1 1 4. For a = − 2 , b = 2 , m = p = q = 1

1 Z − 1 1 x 2 χ1 (x) Li5 x 2 ln (x) dx 0 (2) X ζ (2) − H X H4n−1 = 4 4n−1 − 4 . (2n − 1) (4n − 1) 2 n≥1 n≥1 (2n − 1) (4n − 1)

Here we require various Euler sums and we just highlight the result

(2) X H π 11 1 π3 7 4n = − ζ (2) + G − πG − − ln 2 (2n − 1) (4n − 1) 2 16 4 48 4 n≥1 7 1 1 + ζ (2) ln 2 + G ln 2 + 2W (3) − π ln 2 16 2 16 16 Anthony Sofo from which 1 Z 3 − 1 1 15 π 15 x 2 χ (x) Li x 2 ln (x) dx = ζ (2) − 8G + πG + − ζ (2) ln 2 1 5 4 32 4 0 21 1 3 + 4 ln 2 + ζ (3) + π ln2 2 − ln2 2 4 8 2 1 + π ln 2 − 4W (3) − 2π 2 where W (3) is deﬁned in Lemma 4. 5. For a = 2, b = 2, m = 4, p = 0, q = 5

1  (5)  Z 3 X H x2 χ (x) Li x2 ln4 (x) dx = ζ2 (5) − n+1 0 5 4  n5  0 n≥1 and evaluating the Euler sum gives 3 3 15 105 189 P (2, 2, 4, 0, 5) = ζ2 (5) − ζ (10) + ζ (4) + ζ (2) − . 8 8 2 2 2 6. For a = −1, b = 2, m = p = 3, q = 0

1  (4)  Z H 1 −1 2 3 3 X n− 2 1 x χ3 (x) Li0 x ln (x) dx = −  − ζ (3) λ (4) , 8 (2n − 1)3 2 0 n≥1 and using Lemma 1

1 Z 135 45 381 x−1 χ (x) Li x2 ln3 (x) dx = ζ (4) ζ (3) − ζ (5) ζ (2) − ζ (7) . 3 0 32 32 128 0 1 7. For the degenerate case a = 2, b = 2, m ∈ N, δ = t for 1 ≤ t ≤ −1, p = q = 0, then by expansion

1 1 Z 1 Z x5 x2 χ (x) Li x2 lnm (x) dx = lnm (x) dx 0 0 t (x2 − 1) (x2 − t) 1 0 n 0 n m X X 1 Z X X (−1) m! = x2n+3 lnm (x) dx = tj 2m+1tj (n + 2)m+1 n≥1 j=1 0 n≥1 j=1 m+1 1 (−1) m! X 1 − n = t 2m+1 (1 − t) m+1 n≥1 (n + 2) (−1)m+1 m! 1 1 1 = −1 − + ζ (m + 1) − Φ , m + 1, 3 2m+1 (1 − t) 2m+1 t t (−1)m+1 m! 1 = t − 1 + ζ (m + 1) − t2Li . 2m+1 (1 − t) m+1 t Familes of Legendre Chi Function Integrals 17

Next we investigate the case of the integral in question containing a polylogarithmic function with negative argument. Theorem 2. Let a ∈ R ≥ −2, b ∈ R+, δ = −1, (m, p, q) ∈ N ∪ {0} , then

1 Z a b m Q (a, b, m, p, q) = x χp (x) Liq −x ln (x) dx 0 (j+1) (j+1) m j+1 2H 2n+a − 2 H 2n+a X q−1 m X = (−b) (−1) (m + 1 − j) (q) 2b b j m−j j j+1 p q+m−j j =0 n≥1 b 2 (2n − 1) (2n + a) m X q−1 m X 1 + (−b) (−1) (m + 1 − j)j (q)m−j η (j) p q+m−j j =1 n≥1 (2n − 1) (2n + a) q X q−r m X 1 − (−b) (−1) (q + 1 − r)m η (r) p q+m+1−r r=2 n≥1 (2n − 1) (2n + a) where η (·) is the alternating zeta function.

Proof. From the Taylor series expansion of the LCF function and the polylogarithm functions we have

1 1 j Z X 1 X (−1) Z xa χ (x) Li −xb d = x2n+bj+a−1 dx p q (2n − 1)p jq 0 n≥1 j≥1 0 j X X (−1) = . (2n − 1)p jq (2n + bj + a) n≥1 j≥1

By partial fraction expansion, from Lemma 2

1  q−r  Z Pq (−b) a b X 1 X j r=2 jr (2n+a)q+1−r x χp (x) Liq −x dx = p (−1)  (−b)q−1  (2n − 1) + q−1 0 n≥1 j≥1 j(2n+a) (2n+bj+a) q−1 2n+a+b q q−r X (−1) Φ −1, 1, − ln 2 X X (−b) η (r) = b − (2n − 1)p (2n + a)q p q+1−r n≥1 r=2 n≥1 (2n − 1) (2n + a) q−1 1 2n+a+b 1 2n+a+b 1 X (−1) ψ − ψ + − ln 2 = 2 b 2 b 2 (2n − 1)p (2n + a)q n≥1 q q−r X X (−b) η (r) − p q+1−r . r=2 n≥1 (2n − 1) (2n + a)

Diﬀerentiating 0m0 times under the integral sign, with respect to the real parameter 18 Anthony Sofo a, we obtain

Q (a, b, m, p, q) (j+1) (j+1) m j+1 2H 2n+a − 2 H 2n+a X q−1 m X = (−b) (−1) (m + 1 − j) (q) 2b b j m−j j j+1 p q+m−j j =0 n≥1 b 2 (2n − 1) (2n + a) m X q−1 m X 1 + (−b) (−1) (m + 1 − j)j (q)m−j η (j) p q+m−j j =1 n≥1 (2n − 1) (2n + a) q X q−r m X 1 − (−b) (−1) (q + 1 − r)m η (r) p q+m+1−r , r=2 n≥1 (2n − 1) (2n + a) where Φ (·, ·, ·) is the Lerch transcendent and (q)m is the Pochhammer symbol, the proof of the theorem is ﬁnished.

There are some interesting special cases of Theorem 2 and we present these in the next corollaries before listing some illustrative examples. Corollary 3. Let a = −1, b = 1, (m.p.q) ∈ N ∪ {0} , then

1 Z −1 m Q (−1, 1, m, p, q) = x χp (x) Liq (−x) ln (x) dx 0 m 2H(j+1) − 2j+1H(j+1) X q+m−1 X n− 1 2n−1 = (−1) (m + 1 − j) (q) 2 j m−j j+1 m+p+q−j j =0 n≥1 2 (2n − 1) m 1 X m+q−1 + (−1) (m + 1 − j) (q) η (j + 1) λ (m + p + q − j) 2 j m−j j =1 q 1 X m+q−r − (−1) (q + 1 − r) η (r) λ (m + p + q + 1 − r) 2 m r=2

H(j+1) n− 1 P 2 where η (·) is the Dirichlet eta function and the Euler sums n≥1 (2n−1)m can be evaluated by (13). Corollary 4. Let the conditions of Theorem 2 hold and put p = 0, then

1 Z a b m Q (a, b, m, 0, q) = x χ0 (x) Liq −x ln (x) dx 0  (m+1)  m H (−1) m! X bn+a = 2 − ζ (m + 1) η (q) . 2m+1  nq  n≥1 Familes of Legendre Chi Function Integrals 19

Proof. The proof follows the same pattern as in Corollary 1.

Corollary 5. Let the conditions of Theorem 1 hold then, for q = 0

1 Z a b m Q (a, b, m, p, 0) = x χp (x) Li0 −x ln (x) dx 0

 (m+1) (m+1)  m 2m+1H − 2H (−1) m! X 2n+a 2n+a = b 2b − 2mη (m + 1) λ (p) , m+1  (2n − 1)p  (2b) n≥1

b xb where Li0 −x = − 1+xb. . Proof. The proof follows the same pattern as in Corollary 2.

Some illustrative examples of Theorem 2 and its corollaries follow. To evaluate the resultant Euler sums for these examples we need a mixture of identities, including (4), (5), (6), (7), (8), (9), (10), (11), (12), (13), Lemma 1, Lemma 2, Lemma 3 and Lemma 4. Examples. 1. Let a = −1, b = 1, m = 1, p = 3, q = 2 then

1 Z −1 Q (−1, 1, 1, 3, 2) = x χ3 (x) Li2 (−x) ln (x) dx 0

H(2) − 2H(2) X Hn− 1 − H2n−1 1 X n− 1 2n−1 = 2 2 + 2 + η (2) λ (5) . 6 2 5 n≥1 (2n − 1) n≥1 (2n − 1) The Euler sums can be evaluated by the results in Lemma 1, so that ﬁnally

1 Z 381 13 x−1 χ (x) Li (−x) ln (x) dx = ζ (7) − ζ (5) ζ (2) . 3 2 128 8 0 2. Let a = 3, b = 1, m = p = q = 1 then

1 Z 3 Q (3, 1, 1, 1, 1) = x χ1 (x) Li1 (−x) ln (x) dx 0 4H(2) − 2H(2) − 4η (2) X H2n+3 − Hn+ 3 X 2n+3 n+ 3 = 2 2 + 2 2 2 4 (2n − 1) (2n + 3) n≥1 (2n − 1) (2n + 3) n≥1 and evaluating the Euler sums gives 43 13 829 7 Q (3, 1, 1, 1, 1) = ζ (2) + ln 2 − − ζ (3) . 192 12 864 64 20 Anthony Sofo

3. Let a = 0, b = m = p = 1, q = 3 then

1 Z Q (0, 1, 1, 1, 3) = χ1 (x) Li3 (−x) ln (x) dx 0 1 (2) (2) 3 3 X H2n − Hn + η (3) 1 X 2H − Hn − η (2) = 4 + 2n 8 16 n4 (2n − 1) 16 n3 (2n − 1) n≥1 n≥1 and evaluating the Euler sums gives 5 7 1 7 1 1 Q (0, 1, 1, 1, 3)= ζ (5) − ζ (2) ζ (3) − ζ (3) ln 2 + ζ (3) − Li ( ) − ln2 2 8 16 8 16 4 2 24 1 3 5 45 7 + ζ (2) ln2 2 + ln2 2 − ζ (2) ln 2 − 4 ln 2 + ζ (4) + ζ (2) . 4 2 2 32 4

1 1 4. Let a = − 2 , b = 2 , m = p = q = 1, then

1 Z − 1 1 Q (0, 1, 1, 1, 3) = x 2 χ1 (x) Li1 −x 2 ln (x) dx 0 2H(2) − H(2) − 2η (2) X H4n−1 − H2n− 1 X 2n− 1 4n−1 = 4 2 + 2 . 2 (2n − 1) (4n − 1) n≥1 (2n − 1) (4n − 1) n≥1

From the multiplication identity (6) we can write

1 X 2 ln 2 − 4n − H4n + H2n Q (0, 1, 1, 1, 3) = 4 2 n≥1 (2n − 1) (4n − 1) 7H(2) + 1 − 2H(2) − 4ζ (2) − 2η (2) X 4n (4n)2 2n + . (2n − 1) (4n − 1) n≥1

Evaluating the various Euler sums and highlighting

3 X H4n 1 1 π 3 7π 3 = W (3) − G + + ζ (2) − − ln 2 2 2 2 8 8 256 4 n≥1 (4n − 1) 1 1 21 + G ln 2 − π ln2 2 + ζ (3) , 4 64 32 we obtain 19 11π3 Q (0, 1, 1, 1, 3) = 12W (3) − 4G − 2π + πG + 8G ln 2 + ζ (2) − 4 32 1 21 5 3 7 + 4 ln 2 + π ln 2 − ζ (2) ln 2 − ln2 2 − π ln2 2 + ζ (3) . 2 4 2 8 4 Familes of Legendre Chi Function Integrals 21

5. Let a = 2, b = 1, m = q = 2, p = 1, then 1 Z 2 2 X H2n+2 − Hn+1 Q (2, 1, 2, 1, 2) = x χ1 (x) Li2 (−x) ln (x) dx = 6 (2n − 1) (2n + 2)4 0 n≥1 (2) (2) (3) (3) X 2H − H − 3η (2) 1 X 4H − H − 4η (3) + 2 2n+2 n+1 + 2n+2 n+1 . 3 2 2 n≥1 (2n − 1) (2n + 2) n≥1 (2n − 1) (2n + 2) After a shift in index and some simpliﬁcation, we require the evaluation of a number of Euler sums including,

X H2n 1 8 1 5 = ζ (2) + ln 2 − ln2 2 − , n (2n − 3) 3 9 3 9 n≥1 (2) X H 2 3 17 20 13 2n = ζ (3) − ζ (2) ln 2 − ζ (2) + ln 2 − , n (2n − 3) 3 4 36 27 54 n≥1 (3) X H 19 29 1 67 56 35 2n = ζ (4) − ζ (3) − ζ (3) ln 2 − ζ (2) + ln 2 − . n (2n − 3) 24 72 2 216 81 324 n≥1 To show the process required for the above three identities, we now give a detailed description of another required Euler identity, namely (17) (3) X Hn 2 7 8 49 448 16 8 = ζ (3) ln 2 − ζ (4) − ζ (3) − ζ (2) + ln 2 − + L (3) , n (2n − 3) 3 16 9 27 81 81 3 n≥1 where n+1 X (−1) Hn 11 7 1 1 1 L (3) = = ζ (4)− ζ (3) ln 2+ ζ (2) ln2 2− ln4 2−2Li . n3 4 4 2 12 4 2 n≥1 By the integral deﬁnition of harmonic numbers 1 (3) Z 2 n X Hn 1 ln x X (1 − x ) = dx n (2n − 3) 2 (1 − x) n (2n − 3) n≥1 0 n≥1 1 Z 2 1 ln x 3 −1 √ = 2x + 2 ln 2 − 2 − ln (1 − x) − 2x 2 tanh x dx 2 (1 − x) 0 1 Z 2 2 1 2 1 ln x 3 −1 √ (18) = ζ (3) ln 2 + ζ (4) − − x 2 tanh x dx. 3 12 3 3 (1 − x) 0 Now we apply a Taylor series expansion to the integrand so that, 1 1 Z 3 −1 √ Z x 2 tanh ( x) 2 X 2n−1 + 3 2 X 2hn ln x dx = hn x 2 2 ln x dx = , (1 − x) (n + 2)3 0 0 n≥1 n≥1 22 Anthony Sofo

1 where hn = H2n − 2 Hn. By a change of index and using (4), (7) and (8) we have 1 3 −1 √ Z x 2 tanh ( x) 35 8 49 448 38 ln2 x dx = ζ (4) + ζ (3) + ζ (2) − ln 2 − − 8L (3) (1 − x) 4 3 9 27 27 0 and substituting into (18) we obtain (17). Finally 1 Z 2 2 Q (2, 1, 2, 1, 2) = x χ1 (x) Li2 (−x) ln (x) dx 0 307 23 40 7 89 1 = − ζ (2) − ln 2 − ζ (2) ln 2 + ζ (4) + ln2 2 324 108 81 27 144 27 5 1 7 5 1 − ζ (3) + ζ (2) ζ (3) − ζ (3) ln 2 − ζ (5) − L (3) . 72 8 18 24 3 6. Let a = 2, b = 1, m = q = 3, p = 0, then 1 Z 2 3 Q (2, 1, 3, 0, 3) = x χ0 (x) Li3 (−x) ln (x) dx 0  n+1 (4)  3 X (−1) H n = ζ (4) η (3) − 2 8  n3  n≥1 135 483 15 = ζ (4) ζ (3) − ζ (7) − ζ (5) ζ (2) 32 128 16 21 9 15 + ζ (4) + ζ (3) + ζ (2) − 3. 32 16 16 7. Let a = 0, b = 4, m = q = 1, p = 0, then 1 Z 4 Q (0, 4, 1, 0, 1) = χ0 (x) Li1 −x ln (x) dx 0   n+1 (2) 1 X (−1) H = ζ (2) η (1) − 2n 4  n2  n≥1 1 1 9 = πG − ζ (3) + ζ (2) ln 2. 8 2 32 8. Let a = −1, b = 1, q = 0, then for any even weight (m + p), we can explicitly evaluate 1 Z −1 m Q (−1, 1, m, p, 0) = x χp (x) Li0 (−x) ln (x) dx 0  (m+1) (m+1)  m 2m+1H − 2H (−1) m! X 2n−1 n− 1 = 2 − 2mη (m + 1) λ (p) , 2m+1  (2n − 1)p  n≥1 Familes of Legendre Chi Function Integrals 23 when m = 2, p = 6 we have

1 Z −1 2 x χ6 (x) Li0 (−x) ln (x) dx 0 28105 3207 675 189 = ζ (9) − ζ (7) ζ (2) − ζ (5) ζ (4) − ζ (6) ζ (3) . 512 128 64 128

9. Let a = 0, b = 2, m = 1, p = 1, q = 0 then

1 Z 2 Q (0, 2, 1, 1, 0) = χ1 (x) Li0 −x ln (x) dx 0  6ζ(2) + 16 ln 2 + 8Hn  (2n+1)2 (2n+1)3 (2n+1)3 1 X n+1 = (−1)   4   n≥1 − 2 ζ (2) − H(2) (2n+1)2 n 3π3 1 π = − G ln 2 − ζ (2) + ln2 2. + ln 2 − 2W (3) . 64 4 16

1 10. Consider the degenerate case a = 2, b = 2, δ = 2 , p = q = 0, then

1 1 Z 1 Z x5 x2 χ (x) Li − x2 lnm (x) dx = lnm (x) dx 0 0 2 x4 − x2 + 2 0 0 1 Z X Jn X = − x2n+3 lnm (x) dx 2n n≥1 0 n≥1 where

(n−1) n [ 2 ] [ 2 ] X n − 1 − r X 2rn n − r J = 2r = n r n − r r r=1 r=1

1 n n are the Jacobsthal numbers, or in Binet form, Jn = 3 (2 − (−1) ) . Therefore

1 Z 2 1 2 m m+1 X Jn x χ0 (x) Li0 − x ln (x) dx = (−1) m! 2 2n+m+1 (n + 2)m+1 0 n≥1 m+1 n (−1) m! X (2n − (−1) ) = 3 · 2m+1 n m+1 n≥1 2 (n + 2) 24 Anthony Sofo

  m+1 n+1 (−1) m! X 1 X (−1) =  +  3 · 2m+1 m+1 n m+1 n≥1 (n + 2) n≥1 2 (n + 2) (−1)m+1 m! 1 1 1 = ζ (m + 1) − 1 − + Φ − , m + 1, 3 3 · 2m+1 2m+1 2 2 (−1)m+1 m! 1 = ζ (m + 1) − 3 − 4Li − . 3 · 2m+1 m+1 2

3 11. Finally consider the instructive degenerate case a = − 2 , b = 2, m ∈ N, δ = 1 t for 1 ≤ t ≤ −1, p = q = 0, then by expansion

1 1 Z Z 3 − 3 1 2 m x 2 m J (m, t) = x 2 χ (x) Li − x ln (x) dx = ln (x) dx 0 0 t (x2 − 1) (x2 + t) 0 0

1 n j Z n m+j m+1 X X (−1) 2n− 1 m X X (−1) m!2 = x 2 ln (x) dx = tj tj (4n + 1)m+1 n≥1 j=1 0 n≥1 j=1 m n (−1) m! X 1 (−1) = − 1 (1 + t) 1 m+1 tn n≥1 2n + 2 (−1)m m! 1 1 1 5 = 22m+2 − ζ m + 1, − Φ − , m + 1, . 2m+1 (1 + t) 4 t t 4

m+1 1 (−1) (m) 1 The Hurwitz zeta function ζ m + 1, 4 = m! ψ 4 , so that in terms of Euler and Bernoulli numbers, see [17], we have. For m = 2u, u ∈ N 1 1 1 22u−1 ζ 2u + 1, = − ψ(2u) = (2u)!22u+1λ (2u + 1) + π2u+1 |E | 4 (2u)! 4 (2u)! 2u where λ (·) is the dirichlet lamba function defined by (3) and E2u are the Euler numbers defined by X |E2u| π sec (z) = z2n; |z| < . (2n)! 2 n≥0 For m = 2u − 1, u ∈ N 1 1 1 24u−2 ζ 2u + 1, = ψ(2u−1) = π2u 22u − 1 |B | + 2 (2u)!β (2u) , 4 (2u − 1)! 4 (2u)! 2u where β (2u) is the Dirichlet beta function, defined by (2) and B2u are the Bernoulli numbers defined by

2n 2n X 2 2 − 1 |B2u| π tan (z) = z2n−1; |z| < . (2n)! 2 n≥0 Familes of Legendre Chi Function Integrals 25

Finally

1 Z − 3 1 2 2u J (2u, t) = x 2 χ (x) Li − x ln (x) dx 0 0 t 0   (2u)!22u+1λ (2u + 1)   4u+2 22u−1  2 − (2u)!    (2u)!  2u+1  =  +π |E2u|  , 22u+1 (1 + t)     1 1 5 − t Φ − t , 2u + 1, 4 and 1 Z − 3 1 2 2u−1 J (2u − 1, t) = x 2 χ (x) Li − x ln (x) dx 0 0 t 0 (2u − 1)! 24u−2 1 1 5 = − 24u − π2u 22u − 1 |B | + 2 (2u)!β (2u) − Φ − , 2u, . 22u (1 + t) (2u)! 2u t t 4

1 Z − 3 1 2 5 J (5, 2) = x 2 χ (x) Li − x ln (x) dx 0 0 2 0 5 1 5 = Φ − , 6, − 2560 + 1280β (6) + 1260ζ (6) , 16 2 4 and

1 Z − 3 1 2 4 J (4, 5) = x 2 χ (x) Li − x ln (x) dx 0 0 5 0 5 1 1 5 = 128 − π5 − Φ − , 5, − 62ζ (7) . 25 40 3 4

Concluding Remarks: We have carried out a systematic study of the product of the Legendre Chi function, polylogarithm function and log function in terms of Euler sums. We believe most of our results are new in the literature and given many examples which most are not amenable to a mathematical computer package.

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Anthony Sofo (Received 06. 05. 2020.) Victoria University, (Revised 12. 05. 2021.) Melbourne, Australia Email:[email protected]