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arXiv:1202.3933v1 [math.HO] 16 Feb 2012 h i fti oei oso htteei neda aywyt dedu to way f least easy at an interesting, indeed be is to there appears that that show fact a to (2), is from note directly this of aim The ocnetr bu osbeatmt yReant n e ways new find to by attempts ζ possible about conjecture to otedsoeyo h ucinleuto ftezt funct zeta the of equation the of discovery the to ausof values n hn uliguo hsrpeetto,h defines he representation, this upon building then, and ieetpof o hsseilcs aebe olce yR Cha R. by collected been have mentio [1]. case also We note special unpublished ones. this previous for f of [2] proofs to references different refer useful we bibliography, for the and overload proof to not order In known. here] that elhv enti rbe fderiving of problem this been have well ainof uation where motnei h relation the is importance (2 n nhsbo nteReanszt ucin[,p 2,Po.Ewrso Edwards Prof. 12], p. [3, zeta Riemann’s the on book his In eakbewl-nw eut rtotie yElri 75 sthe is 1735, in Euler by obtained first result, well-known remarkable A nhsfmu ae 4,Reanitoue h nerlrepresent the introduces Riemann [4], paper famous his In Let ,snems ftecretykonmtosbcm oua much popular became methods known currently the of most since ), Teei oes a oddc hsfmu oml fEuler’s of formula famous this deduce to way easy no is “There rmReansitga oml for formula integral Riemann’s from B ζ i ( s s ζ o ol imn evaluate Riemann would How sthe is eteReanzt ucindfie,for defined, function zeta Riemann the be ) ( yuigtecnoritga equation integral contour the using by s o vnitgrvle of values integer even for ) i t enul ubr pca aeo atclrhistorical particular of case special A number. Bernoulli -th ζ sin(2 2 ( s = ) ζ Γ( πs (2 1 ζ s n )Γ( 2 = (2) ) = ) Z s ζ ac Dalai Marco 0 ) ( ∞ ζ (2 s ( := ) s e π π x = ) x ) 2 s 2 − − 1 / n k X e.()here] (4) [eq. ,frwihtdysvrlpof are proofs several today which for 6, s 1 2(2 ∞ ( =1 1 i htis that , − Z dx, 1) ζ ∞ n k 1 ( ∞ )! s s n . euto 3 here] (3) [equation ) +1 ( − e B x x 2 − ℜ ) n s ( − , 1 s nwwihldRiemann led which anew 1 ) ℜ dx. > ( s ζ ζ 1 ion ) ( , > (2 s o l complex all for ) .I sntunfair not is It ”. ,b h sum the by 1, o historical a rom n htfourteen that n rarcn new recent a or )? mni an in pman n tmay it and oevaluate to ation e.(4) [eq. bserves e(4) ce later. eval- (2) (3) (4) (1) point of view, but doesn’t seem to have been previously observed to the best of the author’s knowledge. Note that, even if this proof relies on in the complex domain, no use is made of the residue theorem and only Cauchy’s theorem is used. Thus, even in the special case s = 2, this proof is different from all those presented in [1]. First note that, by a change of variable x =2τ in the integral appearing on the right hand side of (2), and then using the identity 2 1 1 = − , (5) e2τ − 1 eτ − 1 eτ +1 we obtain, after simple manipulations,

∞ s−1 τ − 1−s τ dτ = (1 2 )Γ(s)ζ(s). (6) Z0 e +1 Now, for complex z = x + iy and integer s, let f(z) = zs−1/(ez − 1) and, given R> 0, let ΓR be the rectangular contour of vertices 0, R, R + iπ and iπ in the complex plane. By Cauchy’s theorem, we have that

f(z)dz =0. (7) ZΓR It is easy to see that the integral over the right vertical side of the contour tends to zero when R → ∞. So, when R → ∞, equation (7) gives

∞ s−1 ∞ s−1 π s−1 x − (x + iπ) (iy) x dx x+iπ dx = i iy dy. (8) Z0 e − 1 Z0 e − 1 Z0 e − 1 For the particular case s = 2, using (2) and (6), this equation reads ∞ 1 1 π y i π y sin y ζ(2) + ζ(2) + iπ x dx = dy + dy. 2 Z0 e +1 Z0 2 2 Z0 1 − cos y Taking the real part of both sides one finds the desired result ζ(2) = π2/6. Similarly, in the case of even integer s = 2n, using (2) and (6), after some manipulations the real part of (8) gives

n−1 2n − π Γ(2n)ζ(2n)+ α(n, k)ζ(2n − 2k) = (−1)n 1 (9) 4n Xk=0 where − 2n − 1 α(n, k)= 1 − 21 2n+2k (−π2)k Γ(2n − 2k). (10)  2k   This allows us to obtain the values ζ(2n) recursively, the solution being then expressible in closed form with the use of Bernoulli numbers. It is reasonable to speculate that Riemann could have noticed this procedure while working out the analytic extension of ζ(s) by means of (3) using (2) as

2 a starting point. It is worth pointing out that, as a side result, taking the imaginary part of (8) for even s or the real part for odd s, one obtains integral and representations for the values ζ(2k+1). This also suggests an intuitive interpretation of the different nature of ζ(2n) and ζ(2n+1), being the quantities associated respectively with the imaginary and the real parts of the function f(z)= z/(ez − 1) on the imaginary axis.

References

[1] R. Chapman, Evaluating ζ(2) (1999), available at http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

[2] E. De Amo, M. D´ıaz Carrillo and J. Fern´andez-S´anchez, Another proof of Euler’s formula for ζ(2k), Proc. Amer. Math. Soc., 139 (2011) 1441-1444. [3] H. M. Edwards,Riemann’s Zeta Function. Academic Press, New York, 1974. [4] G. F. B. Riemann, Uber¨ die anzahl der primzahlen unter einer gegebenen gr¨osse, Monatsber. Berlin. Akad., (1859) 671-680.

Department of Information Engineering, University of Brescia, Italy [email protected]

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