P1 1 π2 Another Proof of n=1 n2 = 6 Sourangshu Ghosh∗1 1Department of Civil Engineering, Indian Institute of Technology, Kharagpur 2020 Mathematics Subject Classification: 11M06. Keywords: Zeta Function, Pi, Basel Problem. Abstract In this paper we present a new solution to the Basel problem using a complex line log(1+z) integral of z . 1 Introduction The Basel problem is a famous problem in number theory, first posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1735. The Problem remained open for 90 years, π2 until Euler found the exact sum to be 6 , in 1734. He would eventually propose three P1 1 separate solutions to the problem during his lifetime for n=1 n2 . The Basel problem asks for the sum of the series: 1 X 1 1 1 1 = + + + :: n2 12 22 32 n=1 The first proof which was given by Euler used the Weierstrass factorization theorem to rep- resent the sine function as an infinite product of linear factors given by its roots [13]. This same approach can be used to find out values of zeta function at even values using formulas obtained from elementary symmetric polynomials [22]. A solution in similar lines was given by Kortram [19]. Many solutions of the Basel problem use double integral representations of ζ(2) [8][24][25]. Proofs using this method are presented in [3][4][1][12][14]. Choe [9] used the power series expansion of the inverse sine function to find a solution to the problem. Many solutions to the problem use extensive trigonometric inequalities and identities, as presented in [6][2][20][15]. Many other textbook methods using Fourier analysis, complex analysis, and multivariable calculus for this problem are listed in [7][10][11]. A very strange solution was given by Pace [21]. They used probabilistic methods using random variables that are independent and have Cauchy Distribution to give a solution to the Basel Problem. This inspired the study of relationship between independent Cauchy ∗[email protected] 1 variables and ζ(2k), which is still an active area of research. This method was later ex- tended by [18] to other harmonic sums. Another solution to the Basel Problem was given by [23] which utilizes well-known identities of the Fejer kernel. [5] gave another solution to P1 (−1)n π the Basel Problem by squaring the well known Gregory formula n=0 2n+1 = 4 Another solution coming from an unexpected direction was stated by [16]. He used the Sum of Squares Function formula to find the number of representations of a positive integer as a sum of four squares, all of which are themselves integer. This Sum of Squares Function formula for sum of eight, sixteen, and higher no of squares can be similarly be used to find values of ζ(2k).A good survey of such solution to the Basel Problem as mentioned above that mathematicians have discovered can be found in [7][10][11]. In this paper we present a simple solution to the Basel problem using the Cauchy integral theorem applied to the log(1+z) complex line integral of z . 2 Proof We first take a note of the Taylor series expansion of log(1 + x) which is as follows: x2 x3 x4 log(1 + x) = x − + − + ::: 2 3 4 The series converges for jxj < 1. Now if we divide the expression by x we get log(1 + x) x x2 x3 = 1 − + − + ::: (2.1) x 2 3 4 The domain of convergence is 0 < jxj < 1.Integrating both sides of (1) from −1 to 1, we get the following equality: 1 Z 1 log(1 + x) Z 1 x x2 x3 X 1 dx = (1 − + − + :::) = 2 ∗ (2.2) x 2 3 4 (2n + 1)2 −1 −1 n=0 The integrand above has a removable discontinuity at 0 and the radius of convergence of the series in (2.1) is 1. Therefore it converges uniformly on compact subintervals of [−1; 1] and hence can be integrated term-wise, as in (2.2). Further note that 1 1 1 1 1 X 1 X 1 X 1 1 X 1 3 X 1 = − = (1 − ) = (2.3) (2n + 1)2 n2 (2n)2 4 n2 4 n2 n=0 n=1 n=1 n=1 n=1 P1 1 R 1 log(1+x) Therefore we can get the sum of n=1 n2 by knowing the definite integral −1 x . Let us solve this integral by using complex variables. For our convenience we change the variable of integration as z. Therefore our only task remaining is to compute the integral Z 1 log(1 + x) dx (2.4) −1 x Notice that the integrand in (2.4) has a removable singularity at 0 as we have discussed earlier. 2 Figure 1: Contour Integral Path Consider the contour C1 + C2 + C3 , as in the above figure.. Here C2 is a straight line from -1 + to 1, C1 and C3 are portions of circles of radii 1 and , centered at 0 and -1, respectively. Z log(1 + z) Z log(1 + z) Z log(1 + z) Z log(1 + z) dz = dz + dz + dz (2.5) C z C1 z C2 z C3 z Here the direction of the integration in the arc is anticlockwise. Now we shall use the Cauchy integral theorem. Before using that let us first state the theorem. Theorem 1. Let U ⊆ C be an open set, and let f : U ! C be a holomorphic function. Let γ :[a; b] ! U be a smooth closed curve. If γ is homotopic to a constant curve, Then we have Z f(z)dz = 0 γ Let us put f(z) = log(1 + z)=z . Using the Cauchy integral theorem we hence write Z log(1 + z) dz (2.6) C z We can then write using equation (2.5) and (2.6) that Z 1 log(1 + z) Z log(1 + z) Z log(1 + z) Z log(1 + z) = lim dz = − lim dz − lim dz !0 !0 !0 −1 z C2 z C1 z C3 z (2.7) it Parameterizing C3 as −1 + e : 0 ≥ t ≥ t0, for some t0, we obtain Z Z t0 it Z t0 log(1 + z) log(e ) it log() + it it dz = i it e dt = i it e dt C3 z 0 −1 + e 0 −1 + e Z t0 eit Z t0 teit = i log() it dt − it dt 0 −1 + e 0 −1 + e Both the integrals are bounded on upper because we know by ML-inequality Z t0 it Z t0 Z t0 it Z t0 2 e 1 t0 te t0 t0 j it dtj ≤ dt = ; it dt ≤ dt = 0 −1 + e 0 1 − 1 − 0 −1 + e 0 1 − 1 − The first integral is multiplied with log() and the second one is multiplied with . Both of them have tends to zero as tends to zero. Hence R log(1+z) dz also tends to zeros as tends C3 z to zero. 3 Also note that C1 tends to the complete outer circle, as ! 0. Therefore Z log(1 + z) Z π log(1 + eiθ) Z π lim dz = ieiθdθ = i log(1 + eiθ)dθ !0 iθ C1 z 0 e 0 Z π Z π θ i θ θ θ = i log(2 cos e 2 )dθ = i (i + log(2 cos ))dθ 0 2 0 2 2 −π2 Z π θ −π2 = + i log(2 cos )dθ = + iI (say) (2.8) 4 0 2 4 Note that π π π Z θ Z 2 Z 2 I = log(2 cos )dθ = 2 log(2 cos u)du = 2 log(2 sin u)du 0 2 0 0 Further we obtain π π Z 2 Z 2 2I = 2 [log(2 sin v) + log(2 cos v)]dv = 2 log(2 sin 2u)du 0 0 π Z π Z 2 = log(2 sin v)dv = 2 log(2 sin v)dv = I 0 0 This concludes that I = 0. Hence from (2.7) and (2.8) we obtain Z 1 log(1 + x) −π2 π2 dx = −( + iI) = −1 x 4 4 Applying (2.3) and (2.2), we conclude that 1 1 X 1 4 X 1 2 Z 1 log(1 + x) π2 = = dx = n2 3 (2n + 1)2 3 x 6 n=1 n=1 −1 References [1] T. Apostol , A proof that Euler missed: Evaluating ζ(2) the easy way, Math. Intelli- gencer, 5 (1983) 59–60. [2] T. Apostol, Mathematical Analysis , Addison-Wesley (1974). [3] F. Beukers; E. Calabi; J.A.C. Kolk, Sums of generalized harmonic series and volumes, Nieuw Arch. Wisk., 11 (1993) 217-224. [4] F. Beukers, A note on the irrationality of ζ(2) and ζ(3), Bull. London Math. Soc., 1 (1979) 268–272. [5] J. M. Borwein and P. B. Borwein, Pi and the AGM , Canadian Mathematical Society Series of Monographs and Advanced Texts, No. 4 , John Wiley Sons, New York (1998). [6] A.L. Cauchy , Cours d’Analyse, Note VIII (1821) [7] R. Chapman, Evaluating ζ(2), Preprint, http://www.maths.ex.ac.uk/rjc/etc/zeta2.pdf, (2003). 4 [8] D. F. Connon, Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers (Volume I), arXiv:0710.4028 [math.HO], (2008). P1 1 π2 [9] B.R. Choe, An Elementary Proof of n=1 n2 = 6 , Amer. Math. Monthly, 94 No. 1, (1987) 662-663. [10] S. Ghosh, The Basel Problem, arXiv:2010.03953 [math.GM] (2020).