Zeta summation: Basel problem

For $s= 2$ , the zeta series is:

$\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}= 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots$

We would like to evaluate this series using the representation presented the previous post in we obtain

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

For $s = 2$ ,

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

since $\Gamma(1) = 1$ . We need to evaluate

$\displaystyle \xi(2) = \int_0^\infty \frac{\vartheta(u) - 1}{2} \, du.$

We split the integral as follows:

$\displaystyle \int_0^1 \frac{\vartheta(u) - 1}{2} \, du + \int_1^\infty \frac{\vartheta(u) - 1}{2} \, du$

For the second part, we make the change of variable $u = 1/v$ , so $du = -dv/v^2$ . Using the functional equation of the theta function presented in a previous post.

$\displaystyle \vartheta(u) = u^{-1/2} \vartheta(1/u), \qquad u>0.$

This can be rewritten as

$\displaystyle \vartheta(1/v) = \sqrt{v} \, \vartheta(v)$

one can show (after standard calculations) that the two parts combine to yield a convergent integral. The final evaluation, which relies on the Poisson summation formula applied to the Gaussian, gives

$\displaystyle \xi(2) = \frac{\pi}{6}$

We obtain

$\displaystyle \zeta(2) = \pi \cdot \xi(2) = \pi \cdot \frac{\pi}{6} = \frac{\pi^2}{6}$

Finally

$\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots = \frac{\pi^2}{6}$

The Basel problem was sloved by Euler in 1735, who showed that the sum equals $\displaystyle \frac{\pi^2}{6}$ .

Zeta summation: Basel problem

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