Evaluation of Inﬁnite Series by Integrals

: We examine a large class of inﬁnite triple series and establish a general summation formula. This is done by expressing the triple series in terms of deﬁnite integrals involving arctangent function that are evaluated in turn in closed forms. Numerous explicit formulae are tabulated for the triple series whose values result in elegant expressions as π , ln2 and the Catalan constant G .

Their tensor product with zeta function and extensions lead to intensive investigations on Euler-Zagier sums and multiple zeta values (see for example [11][12][13]). These works suggest the authors to examine naturally, for λ ∈ Z and µ, ν ∈ N 0 , the following triple series The aim of this article is to evaluate explicitly this series. Firstly, it is trivial to see that there holds the following symmetry: Then for m ∈ N 0 and n ∈ Z, recall the binomial identity −λ n = m ∑ k=0 (−1) m−k m k k − λ n + m , which can be explained by the finite differences (i.e., the mth difference of a polynomial of degree m + n). When m ≤ min{µ, ν}, the series S λ (µ, ν) can be manipulated as follows: where the lower limit m of the sum with respect to n is replaced by 0 because Hence, we get the following recurrence relation.
Particularly for m = 1, the following simplified recurrence holds: By writing the two inner sums as definite integrals where exchanging orders of summation and integration is justified by Lebesgue's dominated convergence theorem ( [14], §11.32), we can express S λ (µ, ν) as Evaluating the inner sum by the binomial theorem leads us to the following double integral representation.
Based on Lemmas 1 and 2, we deduce the following preliminary facts.
• λ, µ, ν with λ = 0 In this case, the corresponding double integral becomes a product of two single integrals Both integrals are computable since their integrands are simple rational functions. • λ, µ, ν with λ < 0 According to the binomial theorem, by expanding (1 + x 2 y 2 ) −λ in Lemma 2, we can express S λ (µ, ν) in terms of S 0 (µ , ν ). • λ, µ, ν with λ > 0 In view of Lemma 1, we can write S λ (µ, ν) in terms of S λ (µ , 0) or S λ (0, ν ). Taking account of symmetry, we only need to evaluate S λ (µ, 0) for λ ∈ Z and µ ∈ N 0 . By making use of the algebraic relation we can express Therefore, the evaluation of series S λ (µ, ν) is simplified to analyzing series S λ (0, 0) and T λ (µ). They will be treated separately in the next two sections. Finally, the paper will end in Section 4, where a conclusive theorem will be presented together with several tabulated sample formulae.
Throughout the paper, we shall utilize the following notations. For an indeterminate x and n ∈ N 0 , the rising and falling factorials are defined by (x) 0 = x 0 = 1 and for n ∈ N.

Evaluation of S λ (0, 0)
In view of the integral expression in Lemma 2, consider the difference By applying the equation and then the symmetry, we can manipulate the double integral By employing the integration by parts we get the following recurrence relation Iterating this relation λ − 1 times (under the condition λ ≥ 1), we find that By substitution, we can proceed further where the first one is done as follows: Then we can express the difference Denote the last sum by ∇, we can manipulate it as follows: Therefore, we have established the following recurrence relation.

Evaluation of T λ (µ)
Now we are going to evaluate Recalling the partial fraction decomposition This reduces T λ (µ) to a single integral where the integrand is given by Observing that we find the recurrence relation where the non-homogeneous term ∆(λ, µ) reads explicitly as The integral ∆(λ, µ) will be evaluated in Lemma 6. According to the recurrence (11), we infer that as long as T λ (0) are known, we can deduce all T λ (µ) for µ > 0. Analogously, we claim that as long as T 1 (µ) are known, we can deduce all T λ (µ) for λ ≤ 0.

∆(λ, µ)
In order to evaluate ∆(λ, µ) explicitly, we have to do that for the above two integrals, that are treated in the next two separate lemmas.
Firstly, it is not hard to evaluate the arctan-integral below.

Iterating times this equation gives that
By assigning = m and = n − 1 for n > m and n ≤ m, respectively, we get from the above equation the following recurrent formulae.
In the above lemma, the two integrals on the right are evaluated explicitly below: Finally, we find the following expression for the non-homogeneous term.
For λ = 1, it is not difficult to check that This is justified by combining two integrals Interestingly, there is a counterpart integral evaluation In general for λ > 1, integrating with respect to x gives Recalling Lemma 3, we can compute the rightmost integral This leads us to the general formula When λ ≤ 0, it is almost routine to proceed with Summing up, we have proved the following general result.
According to Proposition 2, it is possible to compute T λ (µ) when λ ∈ Z and µ ∈ N 0 are specified by small values. For −3 ≤ λ ≤ 3 and 0 ≤ µ ≤ 3, the corresponding values of T λ (µ) are recorded in the following
In view of the algebraic relation we can proceed further with By substitution, we arrive at the following conclusive theorem.