Smk =(-1)k+1 k = four ,andSmk =2k = -1 .(61)For the following instance
Smk =(-1)k+1 k = four ,andSmk =2k = -1 .(61)For the following instance, it’s easy to try to remember the definition with the Riemann zeta function (s) and some formulae for its analytic continuation. The Riemann zeta function can be defined by Formula [89] (s) = 1 , ns n =(62)Mathematics 2021, 9,14 ofwhere n goes by means of all integers (this formula is as a consequence of Euler and Riemann, who’ve regarded genuine and complicated values for the variable s, respectively [47]), or by the expression (s) =p1-1 ps-,(63)where p goes via all prime Methyl jasmonate manufacturer number [89]. When s C, i.e., s = Re(s) + i Im(s), the Dirichlet series [60,96] is convergent for the half-plane Re(s) 1, and at any finite area in which Re(s) 1 + , 0, it can be uniformly convergent. Thus, it’s attainable to define (s) as an analytic function, frequent for Re(s) 1. The infinite item present inside the second definition can also be totally convergent for the half-plane Re(s) 1. These two forms from the Riemann zeta function may be seen as analytic equivalents from the basic theorem of arithmetic, which uniquely expresses an integer as a solution of primes, and is revealing of your significance with the Riemann zeta function (s) inside the theory of prime numbers [89]. More about the Riemann zeta function (including historical aspects) is reported in [89,979]. The Riemann zeta function (s) admits analytic continuation and is standard for all s except for any uncomplicated pole at s = 1, with GNF6702 Description residue 1. Strategies to obtain such analytic continuation can be noticed in [22,47,89]. Titchmarsh [89] discussed the analytic continuation for the Riemann zeta function using the following functional equation: (s) = 2s s-1 sin 1 s (1 – s) (1 – s) , 2 (64)getting an approximation close to s = 1 which can be obtained by (s) = 1 + + O(|s – 1|) . s-1 (65)A technique due to Riemann [89] makes use of the fundamental formula (s) = 1 (s)x s -1 dx ex -( Re(s) 1)(66)z s -1 dz, exactly where C is definitely an adequate line contour that excludes the z C e -1 poles, to show that for Re(s) 1, it holds that plus the integral I (s) = (s) = e-1s (1 – s) 2i z s -1 dx . ez – 1 (67)CThis expression defines an analytic continuation of (s) over the entire s-plane. The very simple pole in s = 1 may be the unique feasible singularity because the integral I (s) (convergent for any infinite region) vanishes within the singularities of (1 – s) [89]. An analytic expansion for the Riemann zeta function, obtained throughout the EMSF, is often identified in [22,47]. One more analytic continuation for the Riemann zeta function is provided by Tao [9], in the 1 1 context of smoothed sums. For 1 n ks , the asymptotic expansion yields k=k =k 1 = (s) + C,-s n1-s + O 1/n , n ks(68)for complex quantity (s). It truly is significant to remember that such number (s) doesn’t rely from the selected cutoff . It follows that (s) = lim 1 k – C,-s n1-s s k n k =n(69)Mathematics 2021, 9,15 ofcan be interpreted as a brand new definition for (s) inside the half-plane R(s) 1 [9]. Observing n that C,-s n1-s = 1 x -s ( x/n) dx – 1/(s – 1) holds for n substantial enough, Tao proposed (s) = 1 + lim s – 1 n k 1 – ks n k =nx -sx dx n(70)as a single version of analytical continuation for the Riemann zeta function (s), valid for Re(s) 1 and for Re(s) 1 [9]. Hence, Tao employed the idea of smoothed sums to present a new definition of (s) that holds on the complex plane and that recovers the asymptotic expansion presented in (65) close to s = 1. Considering the analytic continuations for Riemann zeta function (s) and evaluating them in the context in the smooth.
