On A Central Binomial Series Related to ζ(4).

Transcription

1 On A Central Binomial Series Related to ζ. Vivek Kaushik November 8 arxiv:8.67v [math.ca] 5 Nov 8 Abstract n n n 7π 3 by We prove a classical binomial coefficient series identity n evaluating a double integral equal to this sum in two ways. The latter way will lead us to evaluating a sum of polylogarithmic integrals, whose values are linear combinations of ζ n n π 6 and ζ n n π 9. Introduction We denote the central binomial series C n n. n n We show that C 7π /3 using standard techniques from real-analysis. We evaluate the double integral I log +y y dy d in two ways. First, we will change variables u + v/,y u v/ and integrate with respect to v. Upon converting the resulting integrand into a binomial series, echanging integral and series, and performing term-by-term integration with respect to u, we will obtain I C. On the other hand, reconsidering the original definition of I and integrating with respect to y, we will obtain a sum of 3 different polylogarithmic integrals. These integrals have values that are linear combinations of ζ n n π 6, ζ n π 9. n We assume the knowledge of the values of these two series throughout the paper. Thus, our goal is to prove C 7 ζ 36 We first recall the definition of the polylogarithm function and then evaluate a certain series that prominently appears in the computation of one of the three aforementioned polylogarithmic integrals. Then we prove and eplain why is important in generalizing Apostol s classical ζ proof in []. In the literature, C is a classical result from the theories of both central binomial sums and of log-sine integrals. See [] for a combinatorial treatment and evaluation of C. On the other hand, see all of[,3,5,6] for the evaluation involving Clausen functions and other sophisticated tools from Fourier Analysis and Comple Variables.

2 Some Preliminaries We recall the polylogarithm of order k N, It is easy to see Li k z n z n nk, z C. Li ζ, Li ζ. We state without proof the crucial differentiation and integration identities for the polylogarithm. See [7] for a detailed treatment of polylogarithms. and z d dz Li kz Li k z z Li k t t dt Li k+ z. The number Li / will prominently appear in our later calculations. We state its value and proof. Proof. Consider the double integral Li / ζ log. +z z +yz Integrating with respect to y, we get is equal to +z z +yz dy dz log z zz + dy dz. dz log z log z dz + dz z z + logt ζ+ dt 3 t ζ+ logt t/ dt ζ+ logtt/ n dt ζ+ n n logtt/ n dt ζ Li /, in which the second term of 3 follows from substituting z t to the second term.

3 Ontheotherhand, wereversetheorderofintegrationinandintegratewithrespect to z to see y dz dy +z z +yz y y z + dz dy y z + log y y dy log t dt 5 +t log t +t + log +t in which 5 follows from substituting y t. Hence, dt 6 Li /+log, 7 ζ Li / Li /+log, and the desired result follows from rearranging terms. Evaluating C Recall the series in question the double integral from the introduction: I I C. C n n, 8 n n log +y y Proof. We change variables u+v u v,y to see that I u n n n log u dv du u u +v 8tan u u log u du u 8sin u log u u n dy d. 9 du u n log u n n du n u n log u n n du n n 3 n n n n n C, n 3

4 where in, we used the identity n n n n n sin, <, 3 and in, we used integration by parts. Finally, follows from the simplification of the summand recalling the definition of the binomial coefficient n n! k n k!k! Now we seek the actual value of I, which is the purpose of this paper. I log+ log Li + + log Li Li 3 + Proof. This follows from integrating the inner integral in 9 with respect to y. With Mathematica, we obtained the antiderivative log +y y dy log y + log +y Li +y + log+y + d. Li +y which may be analytically confirmed by differentiating the right hand side with respect to y. Now, let ψy be the right hand side of 5. Taking ψ ψ gives the desired integrand. Hence, the desired result follows from writing I ψ ψ d., We evaluate I by splitting it into the three integrals: log+ log I d, 6 Li log Li + I 3 + d, 7 Li 3 + I 3 d. 8 We first evaluate I, which is the easiest of the three. I 7 6 ζ.

5 Proof. From Mathematica, we obtain the antiderivative identity log+ log d Li +Li log Li 3 log. 9 Evaluating the right hand side at the end points and and taking the difference, we see I Li n n n n + n n n + n n n + n n ζ 7 6 ζ, n where follows from observing the odd terms in the Li sum are negative while the even terms are positive. Then follows from splitting the series representation of ζ into sums of the even and odd terms: ζ n n + n, n n n Remark. An alternative proof of the previous theorem is to recall log+ n,. n n By putting the series on the right hand side of in place of the log+ epression in the integrand of 6, echanging sum and integral, and finally integrating term by term using integration by parts, we may obtain Li /. The result follows from repeating the final steps of the previous proof. The evaluations of I and I 3 are much harder. The reason is that the antiderivatives of both integrals, upon evaluating at the endpoints and, possess a large number of constants that are not rational multiples of π. Such constants do turn out cancelling out with one another in the final result. Hence, to simplify the process, we only search for the following terms: Li ζ, Li / ζ log, Li ζ, Li 7 8 ζ. 3 All the terms in 3 possess rational multiples of π. 5

6 I 7 8 ζ. Proof. We make the substitution u to 7 to see + I Li ulog u u Li3 u u u Li ulog u u u u du log ulog u u u log 3 u u du du Li 3 u u u du logulog u u Li 3 u uu du + log ulog u u du Li 3 u uu du where follows from integrating by parts on the first term with the differentiating function being Li u, and 5 follows from epanding the integrand in the first term. Using Mathematica, we obtain 5 log 3 u du 3Li u+k u 6 u logulog u u du Li u+li +K u 7 u u log ulog u du K 3 u 8 u Li 3u uu du Li u +K u, 9 where K u,...,k u are linear combinations of miscellaneous functions. It turns out that π log + log + K i / K i. Note the first two terms are part of Li /. Upon summing the antiderivatives, evaluating at the endpoints u / and u and taking the difference, we see i I 3Li +Li +Li + Li ζ ζ in which 3 follows from the identity 7 ζ, 3 8 ζ 5 ζ. 6

7 Now we evaluate I 3. I 3 C + 35 ζ. Proof. Changing variables u on 8, we get + Li u 3 I 3 u u du nu n n 3 u du n n n nu n n 3 u du n β n 3 n, 3 where βu,v Now, we recall the well known identity: where and the facts u v d. βu,v ΓuΓv Γu+v, Γu u e d, Γ/ π, Γn n! for all n N. Then epanding 3 into even and odd terms, we see I 3 n n n β 3, + n n n 3β, n n n n + 7π 6 n n C C + 7π ζ where the first term in 3 follows from simplifying the summand with the recalled facts and the second term is a summation identity n n n 3β n, 3 7π 6, 33 7

8 which we obtained using Mathematica. At this moment, we do not know of a proof for 33, but we suspect the sum on the left hand side is equal to the integral πlog π 6 πlog sinθ dθ. Combining our obtained values for I,I,I 3, we see which implies upon rearranging terms, C I +I +I ζ 7 C ζ ζ 7 C ζ+ 8 C 3 Hence, we have proved the desired result. 7 8 ζ 7 7π ζ Relation to Apostol s Method We conclude by mentioning an interesting connection to Apostol s method to prove ζ π /6 in []. Apostol evaluates the double integral Putting y n y dy d. 3 y n, y <, in place of the integrand in 3, echanging sum and integral, and integrating term by term with respect to y and then gives ζ. On the other hand, 3 is equal to π /6, upon making the change of variables u+v u v,y, which transforms 3 into the sum of arctangent integrals u u dv du+ u +v u In particular, the first integral in 35 is equal to u u dv du u +v 8 +u dv du. 35 u +v sin u u du n n 36 n n π 8,

9 where 36 follows from replacing the integrand with the series in 3, echanging the sum and integral, and integrating term by term. Hence, we see the first integral in 35 is equal to a central binomial series, namely 36, which fortunately we could easily evaluate with elementary calculus. With a similar geometric series argument, we can show that the integral yzw dw dz dy d 37 is equal to ζ. But if we change variables u+v u v,y on 37, split it into a sum of two integrals over two different regions as in 35, we will get one of the integrals to be u u u wz +v wz wz sin dv du dw dz dw dz wz n n 38 n n C wz where 38 can beobtained by replacing sin epression with theantiderivative of the series from 3 evaluated at wz, and then performing the usual steps of echanging sum and integral and integrating term by term twice with respect to w and z. Thus, to generalize Apostol s method to find ζ, we would have to evaluate C, but this is unfortunately difficult to do if we don t assume the value of ζ in the first place. References [] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Epansions, Springer Netherlands,. [] Junesang Choi and Young Joon Cho and H. M. Srivastava, Log-Sine integrals involving series associated with the zeta function and polylogarithms, Mathematica Scandinavica 5 9, no., [3] Jonathan M. and Straub Borwein Armin, Special Values of Generalized Log-sine Integrals, ISSAC, ACM, New York, NY, USA,. [] Tom M. Apostol, A Proof that Euler Missed: Evaluating ζ the Easy Way, Springer New York, New York, NY,. [5] Alfred J. van der Poorten, Some wonderful formulas...an introduction to polylogarithms, Queen s Papers in Pure and Appl. Math., vol. 5, Queen s Univ., Kingston, Ont., 98. [6] David Borwein and Jonathan M. Borwein, On an Intriguing Integral and Some Series Related to ζ, Proceedings of the American Mathematical Society 3 995, no., [7] L. Lewin, Polylogarithms and associated functions, North Holland, 98. 9