Approximaths

Zeta summation V

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In a previous post we have computed the meromorphic continuation into the entire complex plane of the Zeta function:

\displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}

This analytic continuation allows us to sum the Zeta function in the whole complex plane. For special (‘known’) values of the Xi function we can calculate the Zeta function. For example for

s = 2 we have \xi(2) = \frac{\pi}{6} and \Gamma(2/2) = \Gamma(1) = 1 and therefore:

\displaystyle \zeta(2) = \frac{\pi^2}{6}
\displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}

For s = 4 we have \xi(4) = \frac{\pi^2}{90} and \Gamma(4/2) = \Gamma(2) = 1:

\displaystyle \zeta(4) = \frac{\pi^4}{90}
\displaystyle \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + ... = \frac{\pi^4}{90}

For s = 6 we have \xi(6) = \frac{2\pi^3}{945} and \Gamma(6/2) = \Gamma(3) = 2:

\displaystyle \zeta(6) = \frac{\pi^6}{945}
\displaystyle \frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ... = \frac{\pi^6}{945}

So, Zeta of all the positive even integers has the form

\displaystyle\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!}

Where B_{2n} are Bernoulli numbers.

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