# Basel Problem

## Theorem

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

where $\zeta$ denotes the Riemann zeta function.

## Proof 1

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

Hence:

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

By the Monotone Convergence Theorem (Real Analysis) we have:

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

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:

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

$\blacksquare$

## Proof 2

Let

$\displaystyle f(x) = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots$.

We note that

$\displaystyle f(0)=1-\frac{0}{3!}+\frac{0}{5!}-\frac{0}{7!}+\frac{0}{9!}-\cdots = 1$.

Now for all $x\neq 0$ $\frac{x}{x}=1$. Therefore,

$\displaystyle f(x) = \frac{x}{x}f(x)=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots}{x}$.

From the analytic definition of sine:

$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.

Therefore, for all $x\neq 0$

$\displaystyle f(x)=\frac{\sin x}{x}$.

From Euler's formula for the sine function we know that

$\displaystyle f(x) = \left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right) \left( 1 - \frac{x}{2\pi} \right) \left( 1 + \frac{x}{2\pi} \right) \left( 1 - \frac{x}{3\pi} \right) \left( 1 + \frac{x}{3\pi} \right) \cdots = \left[ 1 - \frac{x^2}{\pi^2} \right] \left[ 1 - \frac{x^2}{4\pi^2} \right] \left[ 1 - \frac{x^2}{9\pi^2} \right] \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots$.

Expanding this infinite product we get

$\displaystyle \left[ 1 - \frac{x^2}{\pi^2} \right] \left[ 1 - \frac{x^2}{4\pi^2} \right] \left[ 1 - \frac{x^2}{9\pi^2} \right] \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots$.

From this

$\displaystyle 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots$.

Now the first element in each infinite sum is equal to the first element in the second and so on. From this we know that

$\displaystyle \frac{x^2}{3!} = \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2$.

Dividing each side of this equation by $x^2$ and factoring out $\frac{1}{\pi^2}$ on the right side of the equation we obtain

$\displaystyle \frac{1}{6} = \frac{1}{\pi^2}\left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Multiplying each side of this equation by $\pi^2$ we obtain

$\displaystyle \frac{\pi^2}{6} = \left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Therefore,

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

$\blacksquare$

## Historical Note

Although often credited to Euler, it has been suggested that it was in fact solved by Nicolaus I Bernoulli.