Basel Problem/Proof 1

Theorem

$\displaystyle \zeta \left({2}\right) = \sum_{n = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.

Proof

From Double Integral of Power of Product:

$\displaystyle \frac 1 {n^2} = \int_0^1 \int_0^1 (xy)^{n-1} \ dx \ dy$

Hence:

$\displaystyle \sum_{n = 1}^\infty {\frac 1 {n^2}} = \sum_{n = 1}^\infty \int_0^1 \int_0^1 (xy)^{n-1} \ dx \ dy$

By the Monotone Convergence Theorem we have:

$\displaystyle \sum_{n = 1}^\infty \frac 1 {n^2} = \int_0^1 \int_0^1 \left({\sum_{n = 1}^\infty \left({x y}\right)^{n-1} }\right) dx \ dy = \int_0^1 \int_0^1 \frac {dx \ dy} {1-xy}$

Let $\displaystyle (u,v) = \left({\frac{x+y} 2, \frac{y-x} 2}\right)$ so that $(x,y) = (u-v, \ u+v)$.

Then:

$\displaystyle \zeta \left({2}\right) = 2 \iint_S \frac{du \ dv} {1 - u^2 + v^2}$

where $S$ is the square defined by the coordinates $\displaystyle (0,0), \ \left({\frac 1 2, -\frac 1 2}\right), \ (1,0), \ \left({\frac 1 2, \frac 1 2}\right)$.

Exploiting the symmetry of the square we have:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \int_0^u \frac {dv \ du} {1-u^2 + v^2} + \int_{\frac 1 2}^1 \int_0^{1-u} \frac {dv \ du} {1 - u^2 + v^2}}\right)$

After some straightforward calculations we get:

$\displaystyle \zeta \left({2}\right) = 4 \left( \int_{0}^{\frac{1}{2}}{\frac{\arcsin u}{\sqrt{1-u^{2}}}\,du} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt{1-u^2}} \left( \frac{\pi } 4-\frac{\arcsin u} 2 \right)\,du} \right) = 4 \left( \frac{\pi^2}{72}+\frac{\pi^2} {36} \right) = \frac {\pi^2} 6$

$\blacksquare$