A Note on Explicit Evaluations of Extended Euler Sums

We study the extended Euler sums and the alternating extended Euler sums and establish their explicit expressions in terms of Riemann zeta functions and Hurwitz zeta functions. Comparing with the existing results, ours are simpler and thus yield significantly better accuracy when Matlab is used for numerical calculation.


Introduction
We study the extended Euler sums and the alternating extended Euler sums (1.2) where [x] is the largest integer ≤ x.For k = 1, there have been many results [1][2][3] about the evaluations of these sums.For k > 1, p = 1, Mark W. Coffey [4] has given a detailed study about these sums.For k > 1, Wang and Eie [5] have developed a systematic way for establishing explicit evaluations of these sums.Our goal is to use a different way to establish expressions of these sums and improve the results by Wang and Eie.In section 3, we compare our results with Wang and Eie's and show the improvement when Matlab is used for their numerical calculations.
Theorem A(Theorem 1 [5]) For an odd weight w = p + q, q, p > 1, Theorem B(Theorem 5 [5]) For positive integers m, n, nonnegative integer k, and w = 2m + 2n + 1, (1.11) In (1.8)-(1.11),C n (θ) and S n (θ) are defined by which are the cosine and the sine parts of the periodic zeta function defined by (1.14) The above results are clearly very beautiful, but the calculation of C n (θ) and S n (θ) is quite difficult.In particular, the involvement of trigonometric functions makes it very hard to ensure the accuracy of numerical calculation.On the other hand, Theorem B does not include E and the case when k is even.To improve the results, we prove the following theorems.
Theorem 1 For an odd weight w = p + q with p, q > 1, and k ≥ 1, we have (1.16) Theorem 2 For an odd weight w = p + q with p, q > 1, and k ≥ 1, we have where η(0) = 1 2 and η(s, a) is defined by the formula Theorem 3 For an odd weight w = p + q with p, q > 1, and k ≥ 1, we have Theorem 4 For an odd weight w = p + q with p, q > 1, and k ≥ 1, we have (1.20) Theorem 5 For the integer k ≥ 1 and m ≥ 1, we have (1.24) Remark 6 (1.9) requires k to be odd.Theorems 2-4 have released this requirement.Theorem B is about . (1.25) Remark 7 Comparing Theorems A-C and Theorems 1-5, we may obtain a group of identities some of which may be new.For example, Theorems A and 1 give the following identity.

Proof of Theorems
We use the classical residue method used by [1,2].The following lemma plays an important role in the proof as it does in the calculation of all other Euler sums.
Lemma 8 (Cauchy, Lindelöf [2]) Let ξ(s) be a kernel function and r(s) be a rational function which is O(s −2 ) at infinity.Then where S is the set of poles of r(s), O is the set of the poles of r(s) that are not poles of ξ(s) and Res(h(s))| s=λ denotes the residue of h(s) at λ.
The key of using the residue method is the wise selection of kernel functions.Here, we make an essential use of kernels , cot πs, and csc πs. (2. 2) The poles of ψ ± (k, s) are at the integers n and fractions n + j k with j = 1, 2, • • • , k − 1 and n = 0, 1, 2, • • • .The power series expansions of ψ ± (k, s) at these poles are, respectively, where 3), we have The kernels π cot πs and π csc πs have the following expansions, respectively, Consequently, the residues of these functions are to base function r(s) = s −q respectively, we can obtain theorems 1-4 by using Lemma 1 and results (2.8) and (2.9).Similarly, applying kernel functions ξ(s) = π cot πsψ ± (k, s), π csc πs ψ ± (k, s) to the base function r(s) = s −q respectively, one may obtain theorem 5.

Simplification of Theorems A-C and Numerical Comparisons
As mentioned in section 1, the involvement of trigonometric functions in Theorems A-C causes larger error in numerical calculation.To compare our results with Theorems A-C, we first simplify Theorems A-C and try to eliminate the trigonometric functions.We will use the following standard notations for ς ± (q, x) and η ± (q,x).
It is well-known that the Hurwitz zeta function and the trigonometric functions are related by the identities: where j k is the Stirling subset number, and sin(2πp/q) cos(2y) − cos(2πp/q) .By using Fourier expansion, we know We can rewrite C n (θ) as Similarly we may write S n (θ) and Using the properties of the Hurwitz zeta function and (3.7) Substituting (3.2) and (3.7) into (1.9),(1.11) and (1.14), we get  p,q by using Matlab.

.i
Let y j,k = cot jπ k and z j,k = csc jπ k .By (1.7), we haveJ = 0 for i > l , a (l) 0 = l! for l ≥ 0, a (l) i = (l − 2i + 2)a (l−1) i−1 + (l − 2i)a (l−1) i for 0 < i ≤ l,and b(l) 0 = l! for l ≥ 0, b (l) i = 0 for l ≤ i, and b (l) i = (l−2i+1)b (l−1) i−1 +(l−2i)b (l−1) i for 0 < i < l.The following table shows the numerical results and the errors for the calculation of E (k) .10) Now, we compare the efficiency of the expressions given by Theorems A-C, Theorems 1-5 and those obtained in this section.The following table shows the numerical results and the errors for the calculation of E It is clear that the errors generated by (1.21) and (3.10) are significantly smaller than those generated by(1.14)Theorems1-4 need more precise expressions for J