In a recent paper [
Citation: Junjie Quan. Explicit formulas of alternating multiple zeta star values ζ⋆(ˉ1,{1}m−1,ˉ1) and ζ⋆(2,{1}m−1,ˉ1)[J]. AIMS Mathematics, 2022, 7(1): 288-293. doi: 10.3934/math.2022019
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In a recent paper [
We begin with some basic notation. Let s:=(s1,…,sk)∈Nk, the (alternating) multiple harmonic (star) sums are defined by
ζn(s)≡ζn(s1,…,sk):=∑n≥n1>⋯>nk>0k∏j=1n−|sj|jsgn(sj)nj, | (1.1) |
ζ⋆n(s)≡ζ⋆n(s1,…,sk):=∑n≥n1≥⋯≥nk≥1k∏j=1n−|sj|jsgn(sj)nj, | (1.2) |
where sj stands for non-zero integer, and
sgn(sj):={1,sj>0,−1,sj<0. |
We may compactly indicate the presence of an alternating sign. When sgn(sj)=−1, by placing a bar over the corresponding integer exponent sj. For example,
ζn(ˉ2,3,ˉ1,4)=ζn(−2,3,−1,4)=∑n≥n1>n2>n3>n4≥1(−1)n1+n3n21n32n3n44. |
Moreover, for convenience, by {s1,…,sj}d we denote the sequence of depth dj with d repetitions of {s1,…,sj}. For example,
ζn(2,3,{1}2)=ζn(2,3,1,1),ζ⋆n(5,2,{1}3)=ζ⋆n(5,2,1,1,1). |
The sums of types (1.1) and (1.2) (one of more the sj barred) are called the alternating multiple harmonic sums and alternating multiple harmonic star sums, respectively. Conventionally, we call dep(s)=k the depth and w≡wt(s):=s1+⋯+sk the weight.
Obviously, the limit cases of alternating multiple harmonic (star) sums give rise to alternating multiple zeta (star) values, for example,
ζ(ˉ2,3,ˉ1,4)=limn→∞ζn(ˉ2,3,ˉ1,4). |
The systematic study of multiple zeta values began in the early 1990s with the works of Hoffman [2] and Zagier [6]. After that it has been attracted a lot of research on them in the last three decades (see, for example, the book of Zhao [7]).
The main purpose of the present paper is to obtain the explicit formulas of alternating multiple zeta star values ζ⋆(ˉ1,{1}m−1,ˉ1) and ζ⋆(2,{1}m−1,ˉ1).
In this section we first give a lemma, which will be useful in the development of the main results of this paper.
Lemma 2.1. (see [3,4]) For n,m∈N and x∈[−1,1), the following relation holds:
x∫0tn−1lnm(1−t)dt=1nlnm(1−x)(xn−1)+m!(−1)mnζ⋆n({1}m;x)−1nm−1∑i=1(−1)i−1i!(mi)lnm−i(1−x){ζ⋆n({1}i;x)−ζ⋆n({1}i)}, | (2.1) |
where the parametric multiple harmonic star sum ζ⋆n(s1,⋯,sk−1,sk;x) is defined by
ζ⋆n(s1,⋯,sk−1,sk;x):=∑n≥n1≥⋯≥nk≥1xnkns11⋯nsk−1k−1nskk, |
and ζ⋆n(∅;x):=xn.
We note the fact that ∫x0=∫10+∫x1 and ∫10tn−1lnm(1−t)dt=m!(−1)mnζ⋆n({1}m), then (2.1) can be rewritten as
x∫1tn−1lnm(1−t)dt=1nm∑i=0(−1)ii!(mi)lnm−i(1−x){ζ⋆n({1}i;x)−ζ⋆n({1}i)}. | (2.2) |
From (2.2), by a direct calculation, we can find that
ζ⋆n({1}m;x)=nm!m∑i=0(−1)i(mi)lnm−i(1−x)x∫1tn−1lni(1−t)dt+ζ⋆n({1}m)=n(−1)m−1m!1∫xtn−1lnm(1−t1−x)dt+ζ⋆n({1}m). | (2.3) |
Theorem 2.2. For positive integers k,m and real x∈[−1,1),
∞∑n=1ζn−1({1}k−1)ζ⋆n({1}m;x)n2=m∑j−0(k+jj)lnm−j(1−x)(m−j)!Li2,{1}k+j−1(x)−m−1∑j=0(k+jj)lnm−j(1−x)(m−j)!ζ(k+j+1), | (2.4) |
where Lis1,s2,⋯,sk(x) is multiple polylogarithm function defined by
Lis1,s2,⋯,sk(x):=∑n1>n2>⋯>nk≥1xn1ns11ns22⋯nskk,x∈[−1,1). | (2.5) |
Proof. Multiplying (2.3) by ζn−1({1}k−1)n2, then summing both sides of it over n yields
∞∑n=1ζn−1({1}k−1)ζ⋆n({1}m;x)n2−∞∑n=1ζn−1({1}k−1)ζ⋆n({1}m)n2=(−1)k+m−1k!m!1∫xlnm(1−t1−x)lnk(1−t)tdt=(−1)k−1k!m!m∑j=0(mj)(−1)jlnm−j(1−x)1∫xlnj+k(1−x)xdx. | (2.6) |
From [3,Eq (2.24)], an elementary calculation shows that
x∫0lnk(1−t)tdt=(−1)kk!Li2,{1}k−1(x). |
From [1,3], we have the duality relation ζ(m+1,{1}k−1)=ζ(k+1,{1}m−1). Hence, letting m=1 yields ζ(2,{1}k−1)=ζ(k+1). Thus, the formula (2.4) holds.
Similarly, we can also give the following results of alternating zeta type values:
Theorem 2.3. For positive integer m and x∈[−1,1],
∞∑n=1ζ⋆n({1}m;x)n(−1)n=Lim+1(1−x2)−Lim+1(12). | (2.7) |
Proof. We note that the formula (2.3) can be rewritten as
ζ⋆n({1}m;x)=nm!m∑i=0(−1)i(mi)lnm−i(1−x)x∫0tn−1lni(1−t)dt−m−1∑i=0ζ⋆n({1}i)(m−i)!lnm−i(1−x). | (2.8) |
Multiplying (2.8) by (−1)nn and summing with respect to n, we obtain
∞∑n=1ζ⋆n({1}m;x)n(−1)n=−1m!m∑i=0(−1)i(mi)lnm−i(1−x)x∫0lni(1−t)1+tdt−m−1∑i=0ζ⋆(ˉ1,{1}i)(m−i)!lnm−i(1−x). | (2.9) |
In [5], Xu proved the result
x∫0lnm(1−t)1+tdt=(−1)mm!Lim+1(12)+lnm(1−x)ln(1+x2)+m∑l=1(−1)l+1lnm−l(1−x)l!(ml)Lil+1(1−x2). | (2.10) |
Combining (2.9) with (2.10), by an elementary calculation, we have the result (2.7).
Corollary 2.4. For a positive integer m,
ζ⋆(ˉ1,{1}m−1,ˉ1)=ζ(m+1)−Lim+1(12), | (2.11) |
ζ⋆(2,{1}m−1,ˉ1)=m∑i=0i+1(m−i)!lnm−i(2)ζ(ˉ2,{1}i)−m−1∑i=0i+1(m−i)!ζ(i+2)lnm−i(2), | (2.12) |
where ([1])
ζ(ˉ2,{1}m−1)=(−1)m(m+1)!lnm+1(2)+(−1)m(ζ(m+1)−Lim+1(12))−(−1)mm∑j=1lnm+1−j(2)(m+1−j)!Lij(12). |
Proof. Letting k=1,x=−1 in (2.4) and x=−1 in (2.7) and noting the facts that (according to the definitions of alternating multiple zeta star values)
ζ⋆(ˉ1,{1}m−1,ˉ1)=∞∑n=1ζ⋆n({1}m−1,ˉ1)n(−1)nandζ⋆(2,{1}m−1,ˉ1)=∞∑n=1ζ⋆n({1}m−1,ˉ1)n2 |
yield the desired results.
Hence, from Corollary 2.4, we obtain the explicit evaluations of alternating multiple zeta values ζ⋆(ˉ1,{1}m−1,ˉ1) and ζ⋆(2,{1}m−1,ˉ1).
In this paper, we use the integrals of logarithmic function to establish explicit formulas of the two (alternating) multiple zeta type values involving multiple harmonic star sum and parametric multiple harmonic star sum
∞∑n=1ζn−1({1}k−1)ζ⋆n({1}m;x)n2and∞∑n=1ζ⋆n({1}m;x)n(−1)n |
in terms of zeta values and multiple polylogarithms. Further, by applying these formulas obtained, we obtain the explicit evaluations of alternating multiple zeta values ζ⋆(ˉ1,{1}m−1,ˉ1) and ζ⋆(2,{1}m−1,ˉ1).
It is possible that some similar sums involving (parametric) multiple harmonic star sum can be computed by the methods and techniques given in this paper. For example, we can get the following theorem:
Theorem 3.1. For positive integers m,k and x∈[−1,1),
∞∑n=1(ζ⋆n({1}m;x)−ζ⋆n({1}m))ζn(ˉ1,{1}k−1)n(−1)n=(−1)k−1m!k!m∑j=0(mj)(−1)jlnm−j(1−x)1∫xlnj+k(1−t)tdt−(−1)k−1m!k!m∑j=0(mj)(−1)jlnm−j(1−x)1∫xlnj+k(1−t)1+tdt. | (3.1) |
Proof. The proof of Theorem 3.1 is similar as the proof of Theorem 2.3. Multiplying (2.3) by ζn(ˉ1,{1}k−1)n(−1)n and summing with respect to n, and applying the identity
lnk(1−t)1+t=(−1)kk!∞∑n=1ζn(ˉ1,{1}k−1)(−t)n(k∈N, t∈(−1,1)), |
we have
∞∑n=1(ζ⋆n({1}m;x)−ζ⋆n({1}m))ζn(ˉ1,{1}k−1)n(−1)n=(−1)m+k−1m!k!∫1xlnm(1−t1−x)lnk(1−t)t(1+t)dt. |
Hence, we may easily deduce the desired result by a direct calculation.
In particular, setting x=0 in (3.1) yields
∞∑n=1ζ⋆n({1}m)ζn(ˉ1,{1}k−1)n(−1)n=(m+kk)(ζ(m+k+1)+ζ(ˉ1,ˉ1,{1}m+k−1)). | (3.2) |
We thank the anonymous referee for suggestions which led to improvements in the exposition. The author is supported by the National Natural Science Foundation of China (Grant No. 12101008), the Natural Science Foundation of Anhui Province (Grant No. 2108085QA01) and the University Natural Science Research Project of Anhui Province (Grant No. KJ2020A0057).
The author declares no conflict of interest in this paper.
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