Repeated Integration and Explicit Formula for the N-Th Integral of Xm(Lnx)

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In particular, the simplification formulas for definite recurrent integrals will involve partitions. Hence, exploiting this connection between recurrent integrals and partitions, we derive some partition identities. ′ A particular repeated integral in which we are interested is the n-th integral of xm(ln x)m . ′ We are interested in the repeated integral of xm(ln x)m as developing a formula for such an integral will require the use of all three main theorems (the variation formula, inversion formula, and reduction formula) of [1]. Hence, we present simplification formulae for this integral as an application to that cited article. This integral is surprisingly linked to the harmonic sum as well as to a more general form called the multiple harmonic star sum (which is a particular case ′ of recurrent sums [1]). In fact, the reduction formula for the repeated integral of xm(ln x)m is expressed in terms of a recurrent sum (in particular, a shifted multiple harmonic star sum). The connection between repeated integrals and recurrent sum structures has been previously explored. In 2018, Ce Xu [10] illustrated the connection between similar integrals involving logarithms and a particular recurrent sum (the multiple zeta star function or multiple harmonic star sum). In this article, we further contribute to establishing this connection between logarithmic integrals and multiple harmonic star sums. The multiple harmonic star sum (MHSS) [11, 12, 13] is defined as follows: 1 ζ⋆(s ,s ,...,s )= . n 1 2 k N s1 N s2 ··· N sk 1≤N1≤N2≤···≤N ≤n 1 2 k X k For the purpose of this article, let us also define the following more general notation: 1 ζ⋆ (s ,s ,...,s )= . q,n 1 2 k N s1 N s2 ··· N sk q≤N ≤N ≤···≤N ≤n 1 2 k 1 X2 k Also, let (s1,...,sk)=({p}k) represent (s1,...,sk)=(p,...,p). The connection goes both ways: In the same way multiple harmonic star sums help in simplifying such integrals, these integrals can be used to simplify certain harmonic sums. Study [3], presents several formulae for simplifying harmonic sums, repeated harmonic sums as well as a variant involving binomial coefficients (that is referred to as the binomial-harmonic sum). Exploiting the connection established between logarithmic integrals and harmonic sums, these relations will be utilized to derive formulae for the repeated harmonic sum and the repeated binomial-harmonic sum.

In Section 2, we derive reduction formulas for indefinite repeated integrals as well as indefinite recurrent integrals which allow us to express such integrals in terms of simple (non- repetitive) integrals. Then, we present a generalization of the fundamental theorem of calculus for repeated integrals and recurrent integrals. These formulae will then be used to express the definite repeated and recurrent integrals in terms of definite single integrals. Partition identities will also be derived from the recurrent integral formulae. In Section 3, we propose explicit ′ formulas for the n-th order repeated integral of xm(ln x)m . These formulas will include n, m, and m′ as parameters in the expression. They will also include certain number theoretical concepts such as MHSS and partitions. In Section 4, the relation between this integral and the harmonic sum will be exploited to produce results related to the harmonic sum such as an alternating sum representation of the harmonic sum. We also utilize the relations derived in [3] to derive a similar alternating sum representation for the repeated harmonic sum as well as for the modified form of the repeated harmonic sum (the repeated binomial-harmonic sum).

2. Repeated integration In this section, we consider two techniques of repeated integration: repeated integrals and recurrent integrals. We begin by deﬁning these types of repetitive integrals and presenting some

2 notation.

Definition. We define the indefinite repeated integral of order n of the function f(x), denoted In[f(x)](x) or simply In, by the following structure:

n In = ··· f(x) dx ··· dx = ··· f(x) dx . (1) Z Z Z Z It can also be deﬁned by the following recurrent relation:

In+1 = In dx, (I0 = Rf(x). Definition. We define the n-th order definite repeated integral of the function f(x), denoted In,a,b[f(x)] or simply In,a,b, by the following structure:

b x3 x2 In,a,b = ··· f(x1) dx1dx2 ··· dxn. (2) Za Za Za Definition. We define the indefinite recurrent integral of order n of the function f(x), denoted Jn[f(x)](x) or simply Jn, by the following recurrent relation:

Jn+1 = f(x)Jn dx, (J0 =1R. Explicitly, it can also be expressed as follows:

n Jn = f(x) f(x) ··· f(x) dx . (3) Z Z Z Definition. We define the n-th order definite recurrent integral of the function f(x), denoted Jn,a,b[f(x)] or simply Jn,a,b, by the following structure:

b xn x2 Jn,a,b = f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn. (4) Za Za Za Remark. As we can see, repeated integrals are analogous to repeated sums [3] while recurrent integrals are analogous to recurrent sums [1].

Likewise, before we proceed, we need to generalize the concept of constant of integration: Let us deﬁne Cn(x) as − n 1 xi C (x)= c n i i! i=0 X where the ci’s are constants of integration.

Remark. If we are using Cn(x) as a function of x, for simplicity, we will denote it as Cn.

2.1. Repeated integrals In this section, we begin by providing a way to compute indefinite repeated integrals in terms of simple indefinite integrals. Then we generalize the concept of primitive and present a generalization of the fundamental theorem of calculus. Finally, we present a formula for computing definite repeated integrals in terms of simple definite integrals.

3 2.1.1. Indefinite repeated integrals We begin by proving a formula for expressing indefinite repeated integrals in terms of single integrals. We refer to this theorem as the reduction formula for indefinite repeated integrals. Theorem 2.1. Let f(x) be a function of x, for any n ∈ N∗, we have that

n xk−1 xn−k ··· f(x)dxn = (−1)n−k f(x)dx + C . (k − 1)! (n − k)! n k Z Z X=1 Z Remark. By substituting k by n − k and performing a few manipulations, this theorem can be rewritten as follows:

n− xn−1 1 n − 1 (−1)k ··· f(x)dxn = xkf(x)dx + C . (n − 1)! k xk n k Z Z X=0 Z Proof. 1. Base case: verify true for n = 1.

1 xk−1 x1−k (−1)1−k f(x)dx + C = f(x) dx + C . (k − 1)! (1 − k)! 1 1 k X=1 Z Z 2. Induction hypothesis: assume the statement is true until n.

n xk−1 xn−k ··· f(x)dxn = (−1)n−k f(x)dx + C . (k − 1)! (n − k)! n k Z Z X=1 Z 3. Induction step: we will show that this statement is true for (n + 1). We have to show the following statement to be true:

n+1 xk−1 xn−k+1 ··· f(x)dxn+1 = (−1)n−k+1 f(x)dx + C . (k − 1)! (n − k + 1)! n+1 k Z Z X=1 Z

To be concise, we denote the left hand side term by I. By applying the induction hypothesis,

I = ··· f(x) dxn dx Z Z Z n xk−1 xn−k = (−1)n−k f(x)dx + C dx (k − 1)! (n − k)! n Z k=1 Z ! X − n xk−1 xn−k n 1 xi = (−1)n−k f(x)dx dx + c dx + C (k − 1)! (n − k)! i i! Z k=1 Z ! Z i=0 ! X − X n xk−1 xn−k n 1 xi+1 = (−1)n−k f(x)dx dx + c + C. (k − 1)! (n − k)! i (i + 1)! k i=0 X=1 Z Z X xk−1 xn−k Setting du = (k−1)! dx and v = (n−k)! f(x) dx, then applying integration by parts, we have

xk−1 xn−k R xk xn−k xk xn−k f(x)dx dx = f(x) dx − f(x) dx (k − 1)! (n − k)! k! (n − k)! k! (n − k)! Z Z Z Z xk xn−k n xn = f(x) dx − f(x) dx. k! (n − k)! k n! Z Z 4 Substituting back, we get

− n xk xn−k n n xn n 1 xi+1 I = (−1)n−k f(x) dx − (−1)n−k f(x) dx + c + C k! (n − k)! k n! i (i + 1)! k=1 Z k=1 Z i=0 Xn k n−k X n n X n i n−k x x n x k n x = (−1) f(x) dx − (−1) f(x) dx (−1) + c − + C. k! (n − k)! n! k i 1 i! k k i=1 X=1 Z Z X=1 X n k n Knowing that k=0 (−1) k = 0, P n n n n (−1)k = (−1)k − 1= −1. k k k k X=1 X=0 Also, let us consider the set of constants bi such that b0 = C and, for i ≥ 1, bi = ci−1. Hence, n xk xn−k xn n xi I = (−1)n−k f(x) dx +(−1)n f(x) dx + b k! (n − k)! n! i i! k i=0 X=1 Z Z X n+1 xk−1 xn−k+1 xn n xi = (−1)n−k+1 f(x) dx +(−1)n f(x) dx + b (k − 1)! (n − k + 1)! n! i i! k i=0 X=2 Z Z X n+1 xk−1 xn−k+1 n xi = (−1)n−k+1 f(x) dx + b . (k − 1)! (n − k + 1)! i i! k i=0 X=1 Z X n xi Noting that i=0 bi i! represents Cn+1, the theorem is proven by induction. Hence, toP find the n-th integral of f(x) according to Theorem 2.1, it is enough to find an expression for the integral of xkf(x) in terms of k. Thus, the problem is reduced from computing an n-th integral to computing a simple integral. Using this theorem, we find two propositions that will be useful later on. Proposition 2.1. For f(x) = 1, using Theorem 2.1, we get that xn ··· 1 dxn = + C . n! n Z Z Proof. Applying Theorem 2.1 for f(x)= 1, we get

n− xn−1 1 n − 1 (−1)k ··· 1 dxn = xkdx + C (n − 1)! k xk n Z Z k=0 Z Xn− xn 1 n − 1 (−1)k = + C (n − 1)! k k +1 n k=0 − X xn n 1 n = (−1)k + C n! k +1 n k=0 xn X = + C . n! n

Proposition 2.2. For f(x) = ln x, using Theorem 2.1, we get that

n 1 xn n (−1)k ··· ln x dxn − C = ··· dxn+1 − C = ln x + . n x n+1 n! k k " k # Z Z Z Z X=1 5 Remark. Theorem 2.1 can also be generalized to non-integer orders to obtain:

∞ xα−1 α − 1 (−1)k I = xkf(x) dx α Γ(α) k xk k X=0 Z where α − 1 (α − 1) ··· (α − k) = . k k! 2.1.2. Relation between definite and indefinite repeated integrals The fundamental theorem of calculus relates the definite integral of a function to its primitive. In this section, we develop a generalization of the fundamental theorem of calculus that relates the definite repeated integral of a function to its repeated primitives. We begin by defining and generalizing the concept of primitive.

Deﬁnition. We denote by F (x) the primitive of f(x). We use primitive to mean the expression of the indeﬁnite integral without the constant of integration (In other words, the case where the constant of integration is zero).

Definition. We define the primitive of order n of the function f(x), denoted Fn[f(x)](x) or Fn(x), as the expression of the n-th indefinite integral of f(x) excluding the constants of integration (In other words, the case where the constants of integration are zero).

The following theorem illustrates how to compute the n-th primitive of a function f(x) in terms of a simple primitive, speciﬁcally, the primitive of xkf(x).

k Theorem 2.2. Let Φk(x) be the primitive of x f(x). The n-th primitive of f(x) can be expressed as follows:

n k−1 n−1 n−1 x Φ − (x) x n − 1 Φ (x) F (x)= (−1)n−k n k = k . n (k − 1)! (n − k)! (n − 1)! k (−x)k k k X=1 X=0 Proof. Using the deﬁnition of the n-th primitive presented as well as the notation presented, we can see, from Theorem 2.1, that the n-th primitive of f(x) can be expressed as stated by this theorem.

Example 2.1. The n-th primitive of f(x)=1 is xn F (x)= . (5) n n! Example 2.2. The n-th primitive of f(x) = ln x is

n 1 xn n (−1)k F [ln x](x)= F (x)= ln x + . (6) n n+1 x n! k k " k # X=1 In the following lemma, we present a modiﬁed version of the fundamental theorem of calculus which is needed to derive a generalization of the fundamental theorem of calculus for repeated integrals.

Lemma 2.1. We have that

b b Fn(x)dx = Fn+1(x)|a= Fn+1(b) − Fn+1(a). Za 6 Proof. Fn+1(x) is the primitive of Fn(x). Hence, from the fundamental theorem of calculus, we have the lemma. In what follows, we prove a generalization of the fundamental theorem of calculus. Theorem 2.3. Let f(x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have n b x3 x2 (b − a)n−k ··· f(x )dx dx ··· dx = F (b) − F (a). 1 1 2 n n (n − k)! k a a a k Z Z Z X=1 Remark. By applying the binomial theorem, this theorem can be expressed as follows:

n n−k b x3 x2 bn−k n − k a ℓ ··· f(x )dx dx ··· dx = F (b) − F (a) (−1)ℓ . 1 1 2 n n k (n − k)! ℓ b a a a k ℓ Z Z Z X=1 X=0 Proof. 1. Base case: verify true for n = 1.

1 (b − a)1−k b F (b) − F (a)= F (b) − F (a)= f(x )dx . 1 (1 − k)! k 1 1 1 1 k a X=1 Z 2. Induction hypothesis: assume the statement is true until n. n b x3 x2 (b − a)n−k ··· f(x )dx dx ··· dx = F (b) − F (a). 1 1 2 n n (n − k)! k a a a k Z Z Z X=1 3. Induction step: we will show that this statement is true for (n + 1). We have to show the following statement to be true:

n b x3 x2 +1 (b − a)n−k+1 ··· f(x )dx dx ··· dx = F (b) − F (a). 1 1 2 n+1 n+1 (n − k + 1)! k a a a k Z Z Z X=1

b x3 x2 b xn+1 x3 x2 ··· f(x1)dx1dx2 ··· dxn+1 = ··· f(x1)dx1dx2 ··· dxn dxn+1. Za Za Za Za Za Za Za Letting t = xn+1 and applying the induction hypothesis, n b x3 x2 b (x − a)n−k ··· f(x )dx dx ··· dx = F (x ) − n+1 F (a) dx 1 1 2 n+1 n n+1 (n − k)! k n+1 Za Za Za Za " k=1 # n X b b (t − a)n−k = F (t)dt − F (a) dt. n k (n − k)! a k a Z X=1 Z Let us note that, letting u =(t − a) and, hence, du = dt, we have

b b (t − a)n−k b un−k un−k+1 (b − a)n−k+1 dt = du = = . (n − k)! (n − k)! (n − k + 1)! (n − k + 1)! Za Za a Using Lemma 2.1, we get n b x3 x2 (b − a)n−k+1 ··· f(x )dx dx ··· dx = F (b) − F (a) − F (a) 1 1 2 n+1 n+1 n+1 (n − k + 1)! k Za Za Za k=1 n X +1 (b − a)n−k+1 = F (b) − F (a). n+1 (n − k + 1)! k k X=1 The case for (n + 1) is proven. Hence, the theorem is proven by induction.

7 Proposition 2.3. For f(x) = 1, from Theorem 2.3, we have that b x3 x2 (b − a)n ··· 1 dx dx ··· dx = . 1 2 n n! Za Za Za Proof. By applying Theorem 2.3 for f(x) = 1, we have that n b x3 x2 bn (b − a)n−k ak ··· 1 dx dx ··· dx = − 1 2 n n! (n − k)! k! Za Za Za k=1 X n bn 1 n = − (b − a)n−kak n! n! k k X=1 bn (b − a + a)n (b − a)n = − − n! n! n! (b − a)n = . n!

2.1.3. Definite repeated integrals In this section, we develop a reduction formula for definite repeated integrals which expresses such integrals in terms of definite single integrals. According to Theorem 2.4, it is enough to determine an expression for the definite integral of xkf(x) in function of k to compute the n-th definite integral of f(x). Hence, the problem is reduced from computing an n-th integral to computing a single integral. Theorem 2.4. Let f(x) be a function defined in the interval [a, b], for any n ∈ N∗, we have n− b x3 x2 bn−1 1 n − 1 (−1)k b ··· f(x )dx dx ··· dx = xkf(x) dx. 1 1 2 n (n − 1)! k bk a a a k a Z Z Z X=0 Z Proof. 1. Base case: verify true for n = 1. − b1−1 1 1 1 − 1 (−1)k b b xkf(x) dx = f(x) dx. (1 − 1)! k bk k a a X=0 Z Z 2. Induction hypothesis: assume the statement is true until n. n− b x3 x2 bn−1 1 n − 1 (−1)k b ··· f(x )dx dx ··· dx = xkf(x) dx. 1 1 2 n (n − 1)! k bk a a a k a Z Z Z X=0 Z 3. Induction step: we will show that this statement is true for (n + 1). We have to show the following statement to be true: n b x3 x2 bn n (−1)k b ··· f(x )dx dx ··· dx = xkf(x) dx. 1 1 2 n+1 n! k bk a a a k a Z Z Z X=0 Z To be concise, we denote the left hand side term by I. By applying the induction hypothesis,

b xn+1 x3 x2 I = ··· f(x1)dx1dx2 ··· dxn dxn+1 Za Za Za Za b n−1 n−1 k xn+1 (xn+1) n − 1 (−1) k = k x f(x) dx dxn+1 (n − 1)! k (xn+1) Za " k=0 Za # n−1 X 1 n − 1 b xn+1 = (−1)k (x )n−k−1 xkf(x) dxdx . (n − 1)! k n+1 n+1 k a a X=0 Z Z 8 n−k−1 t k Let t = xn+1. Setting du = t dt and v = a x f(x) dx, then applying integration by parts, we have R b b tn−k t t=b b tn−k vdu = [uv]b − udv = xkf(x) dx − [tkf(t)] dt a n − k n − k Za Za Za t=a Za bn−k b b tn = xkf(x) dx − f(t) dt. n − k n − k Za Za Hence, substituting back, we get

n− n− 1 1 n − 1 bn−k b 1 1 n − 1 b tn I= (−1)k xkf(x) dx − (−1)k f(t) dt (n − 1)! k n − k (n − 1)! k n − k k=0 Za k=0 Za n− X n−X bn 1 n (−1)k b b tn 1 n = xkf(x) dx − f(t) dt (−1)k. n! k bk n! k k a a k X=0 Z Z X=0 Noticing that − n 1 n n n (−1)k = (−1)k − (−1)n = −(−1)n k k k k X=0 X=0 and that n bn n (−1)k b b tn xkf(x) dx =(−1)n f(t) dt , n! k bk n! k n a a X= Z Z then substituting back, we obtain the theorem.

Remark. Using Theorem 2.4, we can provide an additional proof for Proposition 2.3:

n− b x3 x2 bn−1 1 n (−1)k ··· 1 dx dx ··· dx = [bk+1 − ak+1] 1 2 n n! k +1 bk Za Za Za k=0 Xn n bn n n a k = (−1)k+1 − (−1)k+1 n! k k b "k=1 k=1 # nX X 1 n = (−1)kbn−kak n! k k X=0 (b − a)n = . n! Remark. Theorem 2.4 can also be generalized to non-integer orders to obtain:

∞ bα−1 α − 1 (−1)k b I = xkf(x) dx α,a,b Γ(α) k bk k a X=0 Z where α − 1 (α − 1) ··· (α − k) = . k k! 2.2. Recurrent integrals In this section, we present formulas for computing indefinite and definite recurrent integrals of a function in terms of the primitive of this same function. We also introduce the concept of recurrent primitive and present a generalization of the fundamental theorem of calculus for recurrent integrals. We finish by deriving a few partition identities from this type of integrals.

9 2.2.1. Indefinite recurrent integrals Now we prove a formula for expressing indefinite recurrent integrals in terms of single integrals. We can call this the reduction formula for indefinite recurrent integrals.

Theorem 2.5. Let f(x) be a function of x, for any n ∈ N∗, we have that

n− 1 n 1 1 i f(x) ··· f(x) f(x) dxn = f(x)dx + c f(x)dx . n! i i! i=0 Z Z Z Z X Z Remark. Using the notation introduced, this theorem can be rewritten as 1 f(x) ··· f(x) f(x) dxn = (F (x))n + C (F (x)). n! n Z Z Z Proof. Let X = f(x) dx and dX = f(x) dx, we can rewrite the recurrent integral as follows: R f(x) ··· f(x) f(x) dxn = ··· 1 dXn. Z Z Z Z Z Z Using Proposition 2.1, we get

n− Xn 1 n 1 1 i f(x) ··· f(x) f(x) dxn = + C (X)= f(x)dx + c f(x)dx . n! n n! i i! i=0 Z Z Z Z X Z

Example 2.3. For the recurrent integral of order 3 of sin x, we have

sin x sin x sin x dxdxdx = sin x sin x (− cos x + c2) dxdx Z Z Z Z Z cos2 x = sin x − c cos x + c dx 2 2 1 Z cos3 x cos2 x = − + c − c cos x + c 6 2 2 1 0 1 3 2 1 i = sin x dx + c sin x dx . 3! i i! i=0 Z X Z Example 2.4. For the recurrent integral of order 3 of ex, we have

x x x x x x e e e dxdxdx = e e (e + c2) dxdx Z Z Z Z Z e2x = ex + c ex + c dx 2 2 1 Z e3x e2x = + c + c ex + c 6 2 2 1 0 1 3 2 1 i = ex dx + c ex dx . 3! i i! i=0 Z X Z

10 2.2.2. Relation between deﬁnite and indeﬁnite recurrent integrals We begin by introducing the concept of recurrent primitive.

Definition. We define the recurrent primitive of order n of the function f(x), denoted ρn(x), as the resulting expression of the n-th indefinite recurrent integral of f(x) excluding the constants of integration (In other words, the case where the constants of integration are zero).

Theorem 2.6. The n-th recurrent primitive of f(x) can be expressed as follows: 1 1 ρ (x)= (ρ (x))n = (F (x))n . n n! 1 n! Proof. Using the deﬁnition of the n-th recurrent primitive presented as well as the notation presented, we can see, from Theorem 2.5, that the n-th recurrent primitive of f(x) can be expressed as follows: 1 ρ (x)= (F (x))n . n n!

Noticing that ρ1(x) is simply F (x), the proof is complete.

This definition of ρn(x) will be used in the next section to derive a reduction formula for the definite recurrent integral of a given function f(x). Now we prove a variant of the fundamental theorem of calculus for recurrent integration. This is needed to prove an expression for definite recurrent integrals in terms of indefinite recurrent integrals.

Lemma 2.2. We have that

b f(x)ρn(x) dx = ρn+1(b) − ρn+1(a). Za

Proof. By definition of a recurrent integral, ρn+1(x) = f(x)ρn(x) dx. Hence, ρn+1(x) is the primitive of f(x)ρ (x). Therefore, by the fundamental theorem of calculus, we obtain the n R lemma. Before we can proceed, we need to present the concept of partition of an integer as partitions are involved in the generalized fundamental theorem of calculus for recurrent integrals. As defined by the author in [1, 2], a partition is defined as follows:

Deﬁnition. A partition of a non-negative integer m is a set of positive integers whose sum equals m. We can represent a partition of m as an ordered set (yk,1,...,yk,m) that veriﬁes

yk,1 +2yk,2 + ··· + myk,m = iyk,i = m. (7) i=1 X The coeﬃcient yk,i is the multiplicity of the integer i in the k-th partition of m. Note that 0 ≤ yk,i ≤ m while 1 ≤ i ≤ m. Also note that the number of partitions of an integer m is given by the partition function denoted p(n). Another way of representing a partition of an integer is with a set (λ1,...,λℓ) that satisﬁes:

λ + ··· + λ = ℓ λ = m, 1 ℓ i=1 i (8) λ ≤ λ . ( i i+1 P Note that the second condition is necessary to ensure a unique representation for each partition of the integer. The λi’s are called the parts of this partition and ℓ is called the length of this

11 given partition. Note that 1 ≤ λi ≤ m and that the λi’s are not necessarily all distinct. Also note that the partitions of an integer do not all have the same length. In terms of the ﬁrst m deﬁnition, ℓ is the sum of the multiplicities: ℓ = i=1 yi = y1 + ··· + ym. As partitions are not a fundamental part of this article, we will not go into more details. However, for readers interested in a detailed explanation of partitions, seeP [14]. Now that the concept of partition of an integer has been presented, we introduce the generalized fundamental theorem of calculus for recurrent integrals.

Theorem 2.7. Let f(x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have

b xn x2 f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn Za Za Za n n−k yi yi = (ρk(b) − ρk(a)) (−1) (ρi(a)) . k − i=1 X=1 P iyXi=n k Y Remark. This theorem can also be rewritten using the second partition notation:

b xn x2 n ℓ ℓ f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn = (ρk(b) − ρk(a))(−1) ρλi (a). a a a Z Z Z k=1 P λi=n−k i=1 X λiX≤λi+1 Y

This sum is over all partitions (λ1,...,λℓ) of all integers n − k with 1 ≤ k ≤ n. It is important to note that it is not equivalent to a sum over all partitions (k,λ1,...,λℓ) of the integer n as k does not necessarily satisfy k ≤ λ1. Proof. 1. Base case: verify true for n =1

1 1−k yi yi (ρk(b) − ρk(a)) (−1) (ρi(a)) = ρ1(b) − ρ1(a). k − i=1 X=1 P iyXi=1 k Y Likewise, from Lemma 2.2,

b b f(x1) dx1 = f(x1)ρ0(x1) dx1 = ρ1(b) − ρ1(a). Za Za 2. Induction hypothesis: assume the statement is true until n.

n n−k yi yi Jn,a,b = (ρk(b) − ρk(a)) (−1) (ρi(a)) . k − i=1 X=1 P iyXi=n k Y 3. Induction step: we will show that this statement is true for (n + 1). We have to show the following statement to be true:

n+1 n−k+1 yi yi Jn+1,a,b = (ρk(b) − ρk(a)) (−1) (ρi(a)) . k − i=1 X=1 P iyiX=n k+1 Y

12 Using the induction hypothesis,

b xn+1 x2 Jn+1,a,b = f(xn+1) f(xn) ··· f(x1) dx1 ··· dxn dxn+1 Za Za Za b n n−k = f(x ) (ρ (x ) − ρ (a)) (−1)yi (ρ (a))yi dx n+1  k n+1 k i  n+1 a k − i=1 Z X=1 P iyXi=n k Y n b  n−k  yi yi = f(xn+1)ρk(xn+1)dxn+1 (−1) (ρi(a)) k a − i=1 X=1 Z P iyXi=n k Y n b n−k yi yi − ρk(a) f(xn+1)dxn+1 (−1) (ρi(a)) . k a − i=1 X=1 Z P iyXi=n k Y From Lemma 2.2, b f(xn+1)ρk(xn+1) dxn+1 = ρk+1(b) − ρk+1(a), Za b b f(xn+1) dxn+1 = f(xn+1)ρ0(xn+1) dxn+1 = ρ1(b) − ρ1(a). Za Za Hence,

n n−k yi yi Jn+1,a,b = (ρk+1(b) − ρk+1(a)) (−1) (ρi(a)) k − i=1 X=1 P iyXi=n k Y n n−k yi yi − ρk(a)(ρ1(b) − ρ1(a)) (−1) (ρi(a)) k − i=1 X=1 P iyXi=n k Y n+1 n−k+1 yi yi = (ρk(b) − ρk(a)) (−1) (ρi(a)) k − i=1 X=2 P iyiX=n k+1 Y n n−k yi yi +(ρ1(b) − ρ1(a)) (−1)ρk(a) (−1) (ρi(a)) . k − i=1 X=1 P iyXi=n k Y

Let us deﬁne a partition (Y1,...,Yn) such that Yk = yk +1 and, for i =6 k, if 1 ≤ i ≤ n − k, Yi = yi, otherwise, Yi = 0. Noticing that iYi = iyi + k = n, hence, (Y1,...,Yn) is a partition of n for all k. Thus, we have P P n n−k yi yi (ρ1(b) − ρ1(a)) (−1)ρk(a) (−1) (ρi(a))

k=1 P iyi=n−k i=1 X X n Y Yi Yi =(ρ1(b) − ρ1(a)) (−1) (ρi(a)) i=1 PXiYi=n Y 1 n−k+1 Yi Yi = (ρk(b) − ρk(a)) (−1) (ρi(a)) . k − i=1 X=1 P iYiX=n k+1 Y Hence, substituting back, we get the (n + 1) case and the theorem is proven.

13 2.2.3. Definite recurrent integrals In this section, we develop two reduction formulae for definite recurrent integrals: the first involves a sum over partitions while the second does not. Then, in the later section, by equating the expressions from these two formulae, we derive partition identities.

Theorem 2.8. Let f(x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have

b xn x2 f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn a a a Z n Z Z n−k (F (b))k − (F (a))k (−1)yi = (F (a))n−k . k! i!yi k − i=1 X=1 P iyXi=n k Y Remark. This theorem can also be expressed using the second partition notation:

b xn x2 f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn Za Za Za n ℓ (F (b))k − (F (a))k 1 = (F (a))n−k(−1)ℓ . k! λi! k=1 P λi=n−k i=1 X λiX≤λi+1 Y

Proof. To be concise, we will represent the recurrent integral by Jn,a,b. From Theorem 2.7,

n n−k yi yi Jn,a,b = (ρk(b) − ρk(a)) (−1) (ρi(a)) . k − i=1 X=1 P iyXi=n k Y Applying Theorem 2.6, we get

n n−k k k iyi (ρ1(b)) − (ρ1(a)) yi (ρ1(a)) Jn,a,b = (−1) k! i!yi k − i=1 X=1 P iyXi=n k Y n n−k k k yi (ρ1(b)) − (ρ1(a)) n−k (−1) = (ρ1(a)) . k! i!yi k − i=1 X=1 P iyXi=n k Y

Noticing that ρ1(x) is simply the primitive F (x) of f(x), we get the theorem. Theorem 2.9. Let f(x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have

n b xn x2 1 b (F (b) − F (a))n f(x ) f(x − ) ··· f(x ) dx ··· dx = f(x) dx = . n n 1 1 1 n n! n! Za Za Za Za

Proof. Let Xi = f(xi) dxi = F (xi) and dXi = f(xi) dxi, we can rewrite the deﬁnite recurrent integral as follows (Notice that the limits of integration should also be modiﬁed accordingly): R

b xn x2 F (b) Xn X2 f(xn) f(xn−1) ··· f(x1) dx1 ··· dxn = ··· 1 dX1 ··· dXn. Za Za Za ZF (a) ZF (a) ZF (a) Using Proposition 2.3, we get the theorem.

14 2.2.4. Partition identities derived from recurrent integrals By equation the expressions of the deﬁnite recurrent integral from Theorem 2.8 and Theorem 2.9, we get the following theorem.

Theorem 2.10. Let F (x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have

n n−k n (F (b))k − (F (a))k (−1)yi (F (b) − F (a)) (F (a))n−k = . k! i!yi n! k − i=1 X=1 P iyXi=n k Y Equivalently, we have

n ℓ (F (b))k − (F (a))k 1 (F (b) − F (a))n (F (a))n−k(−1)ℓ = . k! λi! n! k=1 P λi=n−k i=1 X λiX≤λi+1 Y

From this general partition identity, we derive some special partition identities.

Corollary 2.1. For any n ∈ N∗, we have that

n n−k n 1 (−1)yi (−1) +1 = . k! i!yi n! k − i=1 X=1 P iyXi=n k Y Equivalently, we have (−1)ℓ+1 (−1)n = . k! λ1! ··· λℓ! n! k+λ1···+λℓ=n λiX≤λi+1 Proof. Letting F (x) be such that F (a) = 1 and F (b) = 0, Theorem 2.10 reduces to this corollary. The second equality is obtained by using the second partition notation and combining the two sums into one.

Corollary 2.2. For any n ∈ N∗ and any a, b ∈ C, we have that

n n−k bkan−k (−1)yi (b − a)n − (−a)n = . k! i!yi n! k − i=1 X=1 P iyXi=n k Y Equivalently, we have

bkan−k (−1)ℓ bk (−1)ℓaλ1+···+λℓ (b − a)n − (−a)n = = . k! λ1! ··· λℓ! k! λ1! ··· λℓ! n! k+λ1···+λℓ=n k+λ1···+λℓ=n λiX≤λi+1 λiX≤λi+1

Proof. Letting F (x) be such that F (a)= a and F (b)= b, from Theorem 2.10, we have

n n−k n n−k n bkan−k (−1)yi 1 (−1)yi (b − a) − an = . k! i!yi k! i!yi n! k − i=1 k − i=1 X=1 P iyXi=n k Y X=1 P iyXi=n k Y Applying Corollary 2.1, we obtain this corollary. The second equality is obtained by using the second partition notation and combining the two sums into one.

15 Corollary 2.3. For any n ∈ N∗ and any b ∈ C, we have that

n n−k bk (−1)yi (b − 1)n − (−1)n = . k! i!yi n! k − i=1 X=1 P iyXi=n k Y Equivalently, we have

bk (−1)ℓ (b − 1)n − (−1)n = . k! λ1! ··· λℓ! n! k+λ1···+λℓ=n λiX≤λi+1

Proof. By setting a = 1 in Corollary 2.2, we obtain this corollary.

2.2.5. Generalized recurrent integrals The recurrent integral structure is one where all functions are the same. It is analogous to the simple recurrent sum (presented in [1]) where all sequences are the same. However, that last article also presents a more general sum structure where all sequences are distinct. Similarly, we can introduce a generalized recurrent integral structure where all functions are distinct. Definition. We define the general indefinite recurrent integral of order n and denote it as:

n Jn[f1,...,fn](x)= fn(x) fn−1(x) ··· f1(x) dx . (9) Z Z Z Definition. We define the n-th order general definite recurrent integral and denote it as:

b xn x2 Jn,a,b[f1,...,fn]= fn(xn) fn−1(xn−1) ··· f1(x1) dx1 ··· dxn. (10) Za Za Za In what follows, we conjecture formulas for simplifying each of the indefinite and definite general recurrent integrals. But first, we define the follow notation: The permutation group Sn is the set of all permutations of the set {1,...,n}. Let σ ∈ Sn be a permutation of the set {1,...,n} and let σ(i) represent the i-th element of this given permutation. Let Cn,k be the set of all combinations (without repetition) of k elements from the set {1,...,n}. Let ε ∈ Cn,k be a combination of k elements from the set {1,...,n} and let ε(i) represent the i-th element of this given combination. Conjecture 2.1. Let f(x) be a function of x, for any n ∈ N∗, we have that

n fσ(n)(x) fσ(n−1)(x) ··· fσ(1)(x) dx σ∈S Xn Z Z Z n n−1 k

= fi(x) dx + ck,ε(1),...,ε(k) fε(i)(x)dx i=1 k ε∈C i=1 Y Z X=0 Xn,k Y Z where the ck,ε(1),...,ε(k) are constants of integration.

x Example 2.5. For f1(x)= e and f2(x) = sin x, we have

sin x ex dx2 + ex sin x dx2 Z Z Z Z x x = e dx sin x dx + c1,1 sin x dx + c1,2 e dx + c0

Zx Z x Z Z = −e cos x − c1,1 cos x + c1,2e + c0.

16 Conjecture 2.2. Let f(x) be a function deﬁned in the interval [a, b], for any n ∈ N∗, we have

b xn x2 n b fσ(n)(xn) fσ(n−1)(xn−1) ··· fσ(1)(x1) dx1 ··· dxn = fi(x) dx . σ∈S a a a i=1 a Xn Z Z Z Y Z x Example 2.6. For f1(x)= e and f2(x) = sin x, we have

b x2 b x2 x1 x sin x2 e dx1dx2 + e2 sin x1 dx1dx2 a a a a Z b Z b Z Z = ex dx sin x dx = −eb cos b − ea cos a + eb cos a + ea cos b. Za Za

′ 3. Explicit formula for the n-th integral of xm(ln x)m An integral table can be found at the back of any calculus textbook. Such integral tables contain general solutions to some general integrals. However, one integral that cannot be found ′ is the n-th order integral of xm(ln x)m ,

′ xm(ln x)m dxn. Z This integral is of great importance due to its connection with multiple harmonic star sums (MHSSs). In this section, we develop an explicit formula for this general integral. Through the formulas we derive in this section, we further contribute to strengthening the connection between logarithmic integrals and multiple harmonic star sums. We derive two expressions for this integral: the first in terms of shifted MHSSs and the second in terms of a sum over partitions of shifted harmonic sums. Hence, further highlighting the number theoretical nature of this integral. But, before we begin, for the sake of conciseness, let us define the following notation: f(x) dxn = f(x) dnx = ··· f(x) dx ··· dx . Z Z Z Z n n | {z } 3.1. Lemmas | {z } Before we proceed to prove the needed lemmas, we need to present the gamma function. The gamma function [15, 16], denoted Γ(z), is an extension of the factorial to complex numbers excluding the non-positive integers as it satisfies the following property: Γ(z +1) = zΓ(z). It was developed by Euler and is defined by the following expression:

∞ Γ(z)= xz−1e−x dx. Z0 If z ∈ N, the gamma function reduces back to the following factorial: Γ(z)=(z − 1)! . We will denote by Z− the set of negative integers. Hence, Γ(z + 1) is deﬁned for z ∈ C \ Z−.

Now that we have introduced the gamma function, we proceed in proving the lemmas. The ﬁrst lemma consists of proving an expression for a simpler version of the integral.

Lemma 3.1. For any m ∈ C \ {−1} and for any m′ ∈ N, we have that

′ m+1 m k m m′ ′ x m′−k (ln x) 1 x (ln x) dx = m ! (−1) ′ + C . m +1 k! (m + 1)m −k 1 k Z X=0 17 Proof. 1. Base case: verify true for m′ = 0.

xm+1 0 (ln x)k 1 xm+1 0! (−1)−k = = xm dx. m +1 k! (m + 1)−k m +1 k X=0 Z 2. Induction hypothesis: assume the statement is true until m′.

′ m+1 m k m m′ ′ x m′−k (ln x) 1 x (ln x) dx = m ! (−1) ′ . m +1 k! (m + 1)m −k k Z X=0 3. Induction step: we will show that this statement is true for (m′ + 1). We have to show the following statement to be true:

′ m+1 m +1 k m m′+1 ′ x m′−k+1 (ln x) 1 x (ln x) dx =(m + 1)! (−1) ′ . m +1 k! (m + 1)m −k+1 k Z X=0

m′+1 ′ 1 m′ u = (ln x) , du =(m + 1) x (ln x) dx We set m+1 v = x , dv = xm dx. ( m+1 Applying integration by parts, we get

m+1 ′ ′ x ′ m +1 ′ xm(ln x)m +1 dx = (ln x)m +1 − xm(ln x)m dx. m +1 m +1 Z Z Applying the induction hypothesis,

′ xm(ln x)m +1 dx

Z ′ m+1 ′ m+1 m k x m′+1 m +1 ′ x m′−k (ln x) 1 = (ln x) − m ! (−1) ′ m +1 m +1 m +1 k! (m + 1)m −k k=0 X′ m+1 m+1 m k x m′+1 ′ x m′−k+1 (ln x) 1 = (ln x) +(m + 1)! (−1) ′ m +1 m +1 k! (m + 1)m −k+1 k=0 ′ X m+1 m +1 k ′ x m′−k+1 (ln x) 1 =(m + 1)! (−1) ′ . m +1 k! (m + 1)m −k+1 k X=0 Hence, the lemma is proven by induction. In [1], the author proved the following theorem (called the variation formula for recurrent sums),

n+1 N3 N2 m n N3 N2 m−k ··· aNm ··· aN2 aN1 = (an+1) ··· aNk ··· aN2 aN1 . (11) N =q N2=q N1=q k N =q N2=q N1=q ! Xm X X X=0 Xk X X This relation needs to be adapted to a certain form in order for it to be used in the proof of Theorem 3.1. This needed adaptation is illustrated by the following lemma. This lemma consists of a multiple harmonic star sum identity expressing a MHSS in terms of lower order MHSSs.

18 Lemma 3.2. We have that

n m N N + +1 3 2 1 1 1 ··· ··· Nm′−i N2 N1 N ′ m N m N m m −Xi=1+ 2X=1+ 1X=1+ m′ n m N N 1 + 3 2 1 1 1 = k−i ··· ··· . (m + n + 1) Nm′−k N2 N1 k i N ′ =1+m N2=1+m N1=1+m X= m −Xk X X Remark. Using the notation, we can write the lemma as follows:

m′ ⋆ ′ ⋆ ζ1+m,n+m({1}m −k) ζ ({1} ′− )= . 1+m,n+m+1 m i (n + m + 1)k−i k i X= ′ 1 Proof. From equation (11) with m substituted by m − i, with aN = , with q =1+ m, and k Nk with n substituted by n + m, we have that

n m N N + +1 3 2 1 1 1 ··· ··· Nm′−i N2 N1 N ′ =1+m N2=1+m N1=1+m m −Xi X X m′−i n m N N 1 + 3 2 1 1 1 = m′−k−i ··· ··· . Nk N N k (m + n + 1) N m N m N m 2 1 ! X=0 kX=1+ 2X=1+ 1X=1+ For the sake of conciseness, let us denote the ﬁrst term as S. Substituting k by m′ − k, we get

n m N N 0 1 + 3 2 1 1 1 S = ··· ··· . k−i ′ ′ ′  Nm −k N2 N1  m −k m −i (m + n + 1) N ′ m N m N m X= m −Xk=1+ 2X=1+ 1X=1+   Noticing that the interval m′ − i ≤ m′ − k ≤ 0 is equivalent to the interval i ≤ k ≤ m′,

m′ n m N N 1 + 3 2 1 1 1 S = k−i ··· ··· . m n  Nm′−k N2 N1  k i ( + + 1) N ′ =1+m N2=1+m N1=1+m X= m −Xk X X   The relation is proven.

3.2. Theorems Now that the needed lemmas have been proven, we can develop the following explicit ex- ′ pression for the n-th integral of xm(ln x)m in terms of a shifted multiple harmonic star sum. Theorem 3.1. For any m ∈ C \ Z−, any m′ ∈ N, and any n ∈ N∗, we have that

′ n+m m k m m′ n ′ x m′−k (ln x) ⋆ x (ln x) d x =(m )!Γ(m +1) (−1) ζ ({1} ′− ) + C Γ(n + m + 1) k! 1+m,n+m m k n "k # Z X=0 where n m N N + 3 2 1 1 1 ⋆ ′ ζ1+m,n+m({1}m −k)= ··· ··· . Nm′−k N2 N1 N ′ m N m N m m −Xk=1+ 2X=1+ 1X=1+ Remark. If m ∈ N, we replace the gamma function by the corresponding factorial.

′ n+m m k m m′ n ′ x m′−k (ln x) ⋆ x (ln x) d x =(m)!(m )! (−1) ζ ({1} ′− ) + C . (n + m)! k! 1+m,n+m m k n "k # Z X=0 19 Remark. If m ∈ Z− but (n + m) < 0, by taking the limit of the gamma functions ratio,

′ n+m m k m m′ n ′ x m′−k (ln x) ⋆ x (ln x) d x =(m )! (−1) ζ ({1} ′− ) + C . (m + 1) ··· (n + m) k! 1+m,n+m m k n "k # Z X=0 Remark. As the reason for the constants of integration represented by Cn is trivial, we will neglect this term in the proof of the theorem for simplicity. Proof. 1. Base case: verify true for n = 1.

′ 1+m m k ′ x m′−k (ln x) ⋆ Γ(m + 1)(m )! (−1) ζ ({1} ′− ) Γ(2 + m) k! 1+m,1+m m k "k=0 # ′ X m+1 m k ′ x m′−k (ln x) 1 = m ! (−1) ′ . m +1 k! (m + 1)m −k k X=0 Likewise, from Lemma 3.1,

′ m+1 m k m m′ ′ x m′−k (ln x) 1 x (ln x) dx = m ! (−1) ′ . m +1 k! (m + 1)m −k k Z X=0 2. Induction hypothesis: assume the statement is true until n.

′ n+m m k m m′ n ′ x m′−k (ln x) ⋆ x (ln x) d x = Γ(m + 1)(m )! (−1) ζ ({1} ′− ) . Γ(n + m + 1) k! 1+m,n+m m k "k # Z X=0 3. Induction step: we will show that this statement is true for (n + 1). We have to show the following statement to be true

′ n+m+1 m k m m′ n+1 ′ x m′−k (ln x) ⋆ x (ln x) d x = Γ(m +1)(m )! (−1) ζ ({1} ′− ) . Γ(n + m + 2) k! 1+m,n+m+1 m k "k # Z X=0

Using the induction hypothesis,

′ ′ xm(ln x)m dn+1x = xm(ln x)m dnx dx

Z Z Z ′ ′ n+m m k (m )! x m′−k (ln x) ⋆ = Γ(m + 1) (−1) ζ ({1} ′− ) dx Γ(n + m + 1) k! 1+m,n+m m k Z "k=0 # ′ X ′ m m′−k Γ(m + 1)(m )! (−1) ⋆ n+m k = ζ ({1} ′− ) x (ln x) dx . Γ(n + m + 1) k! 1+m,n+m m k k X=0 Z Using Lemma 3.1, we have

k xm+n+1 (ln x)i 1 xn+m(ln x)k dx = k! (−1)k−i . m + n +1 i! (m + n + 1)k−i i=0 Z X Substituting back and simplifying, we get

′ ′ m+n+1 m k i ⋆ ′ m !Γ(m + 1)x ′ (ln x) ζ ({1}m′−k) xm(ln x)m dx = (−1)m −i 1+m,n+m . Γ(m + n + 2) i! (m + n + 1)k−i k i=0 Z X=0 X 20 Now we need to invert the order of summation. From Theorem 3.1 of [1], we have

m′ k m′ m′

bk ai = ai bk. k i=0 i=0 k i X=0 X X X= Let i m′−i (ln x) 1 1 ⋆ a =(−1) , b = ζ ({1} ′− ). i i! (m + n + 1)−i k (m + n + 1)k 1+m,n+m m k Hence, applying this formula to interchange the order of summation, we get

′ m k i ⋆ ′ (ln x) ζ ({1}m′−k) (−1)m −i 1+m,n+m i! (m + n + 1)k−i k=0 i=0 X X′ ′ m i m ⋆ ′ (ln x) 1 ζ ({1}m′−k) = (−1)m −i 1+m,n+m i! (m + n + 1)−i (m + n + 1)k i=0 k=i X′ ′ X m i m ⋆ ′ (ln x) ζ ({1}m′−k) = (−1)m −i 1+m,n+m . i! (m + n + 1)k−i i=0 k i X X= Applying Lemma 3.2, we get

m′ k m′ i ⋆ ′ i m′−i (ln x) ζ1+m,n+m({1}m −k) m′−i (ln x) ⋆ (−1) = (−1) ζ ({1} ′− ) . i! (m + n + 1)k−i i! 1+m,n+m+1 m i k=0 i=0 i=0 X X X Substituting back into the expression of the integral, we get

′ n+m+1 m i m m′ n+1 ′ x m′−i (ln x) ⋆ x (ln x) d x = Γ(m + 1)(m )! (−1) ζ ({1} ′− ) . Γ(n + m + 2) i! 1+m,n+m+1 m i " i=0 # Z X Hence, the case for (n + 1) is proven. Thus, the relation is proven by induction. Before we proceed to prove a second expression for this logarithmic integral, we present a few corollaries that will be useful later. Corollary 3.1. For any m ∈ C \ {−1}, we have that xm+1 1 xm ln x dx = ln x − + C . m +1 m +1 1 Z Corollary 3.2. For any m ∈ C \ Z− and for any n ∈ N∗, we have that

n m xn+m + 1 xm ln xdnx = Γ(m + 1) ln x − + C . Γ(n + m + 1) N n " N=1+m # Z X Remark. If m ∈ N, we replace the gamma function by the corresponding factorial.

n m xn+m + 1 xm ln xdnx = m! ln x − + C . (n + m)! N n " N=1+m # Z X Remark. If m ∈ Z− but (n + m) < 0, by taking the limit of the gamma functions ratio,

n m xn+m + 1 xm ln xdnx = ln x − + C . (n + m) ··· (m + 1) N n " N=1+m # Z X 21 Using the reduction theorem for recurrent sums presented in [1], we can develop the following ′ explicit expression for the n-th integral of xm(ln x)m in terms of a sum over partitions. Theorem 3.2. For any m ∈ C \ Z−, any m′ ∈ N, and any n ∈ N∗, we have that

′ xm(ln x)m dnx Z ′ ′ y ′ n+m m k m −k n+m k,i (m )!Γ(m + 1)x m′−k (ln x) 1 1 =  (−1)  +Cn. Γ(n + m + 1) k! y ! iyk,i N i i k,i ! k=0 k ′ =1 N=1+m X P iyk,iX=m −k Y X      Remark. If m ∈ N, we replace the gamma function by the corresponding factorial.

′ xm(ln x)m dnx Z ′ ′ y n+m m k m −k n+m k,i ′ x m′−k (ln x) 1 1 =(m)!(m )!  (−1) y i  + Cn. (n + m)! k! yk,i! i k,i N ! k=0 k ′ i=1 N=1+m X P iyk,iX=m −k Y X      Remark. If m ∈ Z− but (n + m) < 0, by taking the limit of the gamma functions ratio,

′ xm(ln x)m dnx Z ′ ′ y ′ n+m m k m −k n+m k,i (m )! x m′−k (ln x) 1 1 =  (−1)  + Cn. (n + m) ··· (m + 1) k! y ! iyk,i N i i k,i ! k=0 k ′ =1 N=1+m X P iyk,iX=m −k Y X      Proof. In [1], the author proved the following reduction formula,

y n N2 m n k,i 1 1 i ··· aNm ··· aN1 = (aN ) . (yk,i)! i Nm=q N1=q k i=1 N=q ! X X P i.yXk,i=m Y X

Applying this formula (with m substituted by m′, n substituted by n + m, q = 1+ m, and 1 aN = ) to Theorem 3.1, we obtain this theorem. k Nk Logarithms and logarithmic integrals have always held a special place in number theory. The greatest symbol of this connection is the prime number theorem. This connection is further highlighted here as a logarithmic integral was shown to be related to two major number theoretical concepts: MHSSs in Theorem 3.1 and partitions in Theorem 3.2.

Although it is not of major importance, we can use the theorems previously developed for ′ deﬁnite repeated integrals to develop some formulae for the n-th deﬁnite integral of xm(ln x)m .

m m′ Deﬁnition. We deﬁne Ln,m,m′ (x) as the n-th primitive of x (ln x) whose expression can be seen from Theorem 3.1 or Theorem 3.2. Theorem 3.3. For any m ∈ C \ Z−, any m′ ∈ N, and any n ∈ N∗, we have that

b x3 x2 n ′ − m m ′ n k ′ ··· x1 (ln(x1)) dx1dx2 ··· dxn = Ln,m,m (b) − (b − a) Lk,m,m (a). a a a k Z Z Z X=1 22 ′ Proof. By applying Theorem 2.1 for f(x)= xm(ln x)m , we get this theorem. Theorem 3.4. For any m ∈ C \ Z−, any m′ ∈ N, and any n ∈ N∗, we have that

b x3 x2 n−1 n−1 k b ′ b n − 1 (−1) ′ ··· xm(ln(x ))m dx dx ··· dx = xm+k(ln x)m dx. 1 1 1 2 n (n − 1)! k bk a a a k a Z Z Z X=0 Z ′ Proof. By applying Theorem 2.4 for f(x)= xm(ln x)m , we get this theorem. Theorem 3.5. For any m ∈ C \ Z−, any m′ ∈ N, and any n ∈ N∗, we have that

b x3 x2 m m′ ··· x1 (ln(x1)) dx1dx2 ··· dxn a a a Z Z Z ′ n−1 n−1 k m m′−j ′ b n − 1 (−1) (−1) m ! m+k+1 j m+k+1 j = ′ b (ln b) − a (ln a) . (n − 1)! k bk (m + k + 1)m −j+1 j! k=0 j=0 X X Proof. By applying Lemma 3.1 to Theorem 3.4, we get this theorem.

3.3. Particular cases In what follows, we illustrate some special particular cases of Theorem 3.1. • For n = 1:

′ m+1 m k m m′ ′ x m′−k (ln x) 1 x (ln x) dx = m ! (−1) ′ + C . (12) m +1 k! (m + 1)m −k 1 k Z X=0 • For n = 1 and m = 0:

′ m k ′ ′ (ln x) (ln x)m dx = m′! x (−1)m −k + C . (13) k! 1 k Z X=0 • For m = 0:

′ n m k m′ n ′ x m′−k (ln x) ⋆ (ln x) d x = m ! (−1) ζ ({1} ′− )+ C , (14) n! k! n m k n k Z X=0 where n N N 3 2 1 1 1 ⋆ ′ ζn({1}m −k)= ··· ··· . Nm′−k N2 N1 N ′ =1 N2=1 N1=1 mX−k X X • For m′ = 1: n m xn+m + 1 xm(ln x) dnx = Γ(m + 1) ln x − + C . (15) Γ(n + m + 1) N n " N m # Z X=1+ • For n = 1 and m′ = 1: xm+1 1 xm(ln x) dx = ln x − + C . (16) m +1 m +1 1 Z • For m′ = 0: xn+m xm dnx = + C . (17) Γ(n + m + 1) n Z 23 4. Application to the expression of harmonic sums In this section, we use the connection between logarithmic integrals and harmonic sums established in the previous section to prove some results related to the harmonic sum, repeated harmonic sum, and binomial-harmonic sum.

4.1. An alternating expression for the harmonic sum In this section, we derive a new expression for the harmonic sum as an alternating sum. We present a particularly original proof for this expression using the logarithmic integral from Corollary 3.2. We begin by proving the following lemma. Lemma 4.1. For any n ∈ N∗ and any m ∈ N, we have that

n dn (n + m)! (−1)k+1 m + n (xn+m ln x)= xm ln x + n! xm . dxn m! k m + k k X=1 Proof. Let u = xn+m, v = ln x. From Leibniz’s theorem, we get

dn n n n n xn+m ln x =(uv)(n) = u(n−k)v(k) = [xn+m](n−k)[ln x](k). dxn k k k=0 k=0 X X n n+m (n−k) (k) (n+m)! m For k = 0, k [x ] [ln x] = m! x ln x. Hence,

n dn (n + m)! n xn+m ln x = xm ln x + [xn+m](n−k)[ln x](k). dxn m! k k=1 X We also know the following: (m + n)! (m + n)! ∀z ∈ N, [xn+m](z) = xm+n−z, hence, [xn+m](n−k) = xm+k. (m + n − z)! (m + k)!

(−1)k+1(k − 1)! ∀k ∈ N∗, v(k) = . xk Hence, by substituting back, we get

n dn (n + m)! n (m + n)! (−1)k+1(k − 1)! xn+m ln x = xm ln x + xm+k dxn m! k (m + k)! xk k=1 X n (n + m)! (−1)k+1 (m + n)! n! = xm ln x + xm m! k (m + k)!(n − k)! k=1 X n (n + m)! (−1)k+1 m + n = xm ln x + n! xm . m! k m + k k X=1 The formula is proven. Using the lemma, we develop an alternating sum expression for the harmonic sum. Theorem 4.1. For any n ∈ N∗ and any m ∈ N, we have that

n m n m + n + 1 (−1)k+1 m + n = . m k k m + k k m k =1+X X=1 24 Proof. From Corollary 3.2, we have that n m xn+m + 1 xm ln x dxn = m! ln x − + C . (n + m)! N n " N m # Z X=1+ We rearrange the terms as follows, n m + 1 (n + m)! xn+m = xn+m ln x − xm ln x dxn + C . N m! n N=1+m X Z By diﬀerentiating n times, we get n m + 1 dn dn (n + m)! dn dn (xn+m)= (xn+m ln x) − xm ln x dxn + (C ). N dxn dxn m! dxn dxn n N m ! =1+X Z We know the following: dn dn (n + m)! dn (C )=0, (xn+m)= xm, xm ln x dxn = xm ln x, dxn n dxn m! dxn Z and, from Lemma 4.1, n dn (n + m)! (−1)k+1 m + n (xn+m ln x)= xm ln x + n! xm . dxn m! k m + k k X=1 Hence, substituting back, we get n m n + 1 (n + m)! (n + m)! (−1)k+1 m + n (n + m)! xm = xm ln x + n! xm − xm ln x N m! m! k m + k m! N=1+m ! k=1 X n X (−1)k+1 m + n = n! xm . k m + k k X=1 Dividing both sides by n! xm, we get the theorem. Corollary 4.1. For m =0, Theorem 4.1 becomes n n 1 (−1)k+1 n = . k k k k k X=1 X=1 4.2. Applications of the alternating expression 4.2.1. Expression of the repeated harmonic sum in terms of an alternating sum In [3], the author showed how the repeated harmonic sum can be expressed in terms of a simple shifted harmonic sum. In this section, we express the repeated harmonic sum as an alternating sum. Theorem 4.2. For any n ∈ N∗ and any m ∈ N, we have that

n k k n 3 2 1 (−1)k+1 m + n ··· = . k k m + k k k k 1 k mX+1=1 X2=1 X1=1 X=1 Proof. In [3], the following formula was proven,

n k n m 2 1 n + m + 1 ··· = . k m i k k 1 i=1+m mX+1=1 X1=1 X By applying Theorem 4.1, we obtain the theorem.

25 4.2.2. Expression of the repeated binomial-harmonic sum in terms of an alternating sum Similarly, in [3], the author also showed that the binomial-harmonic sum can be expressed as a shifted harmonic sum. In this section, we develop a new expression in terms of an alternating sum. Theorem 4.3. For any n, k ∈ N∗ and any m ∈ N, we have that

n N N m n 2 N + m 1+ 1 (−1)i+1 m + k + n ··· 1 = . m i i m + k + i N =1 N1=1 " i=1+m # i=1 Xk X X X Proof. In [3], the following formula was proven,

n N N m n m k 2 N + m 1+ 1 n + m + k + + 1 ··· 1 = . m i m + k i N =1 N1=1 " i=1+m # i m k Xk X X =1+X+ Applying Theorem 4.1, we obtain the theorem.

4.2.3. Application to the expression of the n-th integral of xm ln x In this section, we present an alternative expression for the n-th integral of xm ln x. Theorem 4.4. For any n ∈ N∗ and any m ∈ N, we have that

n+m n+m n k+1 m n x x (−1) m + n x ln x dx = ln x − + Cn. n n+m n+m 2 k m + k ! m n! k Z m X=1 Proof. From Corollary 3.2, we have

n+m n+m m n x 1 x ln x dx = ln x − + Cn. n n+m k ! m " k m # Z =1+X Hence, applying Theorem 4.1, we obtain the theorem. Corollary 4.2. For m =0, Theorem 4.4 becomes n xn xn (−1)k+1 n ln x dxn = ln x − + C . n! n! k k n k Z X=1 4.2.4. Divergence of the alternating series The harmonic series was proven to diverge by several mathematicians including Nicole Oresme [17], Pietro Mengoli [18], Johann Bernoulli [19], and Jacob Bernoulli [20, 21]. Twenty additional proofs can also be found in [22]. In this section, we use this fact to proof the divergence of the alternating series presented in this section. Theorem 4.5. For any m ∈ N, we have that

n (−1)k+1 m + n lim = ∞. n→∞ k m + k "k # X=1 Proof. n + m (n + m)! 1 lim = lim = lim (n + m) ··· (n +1)= ∞. n→∞ m n→∞ m! n! m! n→∞ n m n m m ∞ m + 1 + 1 1 1 1 lim = lim − = − = ∞. n→∞ k n→∞ k k k k k m " k k # k k =1+X X=1 X=1 X=1 X=1 Hence, by using Theorem 4.1, we prove that this alternating series diverges.

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