Basel Problem/Proof 1

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
Consider the expression:


 * $\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x y} \ \mathrm d A$

WLOG, the Riemann Double Sum of this multi-variable function can be set such that $0 < x y < 1$.

Then:

This shows the relationship between the first expression and the value to calculate.

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

Let:
 * $\left|{J}\right| = \left\vert{\dfrac{\partial \left({x, y}\right)} {\partial \left({u, v}\right)} }\right\vert = 2$

Then, by Change of Variables Theorem (Multivariable Calculus):


 * $\displaystyle \zeta \left({2}\right) = 2 \iint \limits_S \frac{\mathrm d u \ \mathrm d v} {1 - u^2 + v^2}$

where $S$ is the square defined by the coordinates:
 * $\left({0, 0}\right), \ \left({\dfrac 1 2, -\dfrac 1 2}\right), \ \left({1, 0}\right), \ \left({\dfrac 1 2, \dfrac 1 2}\right)$

Exploiting the symmetry of the square and the function over the $u$-axis, we have:
 * $\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2}}\right)$

Factoring $1-u^2$ gives us:
 * $\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \ \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1}}\right)$

and letting:


 * $s = \dfrac v {\sqrt{1 - u^2} }, \mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

allows us to make a substitution into each integral, giving:


 * $\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac u {\sqrt{1 - u^2} } }\right) \mathrm d u + \int_{\frac 1 2}^1 \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac{1 - u} {\sqrt{1 - u^2} } }\right) \mathrm d u}\right)$

Consider the right triangle with sides $1$, $x$ and $\sqrt{1 - x^2}$.

From it arises the identity:


 * $\arcsin x = \arctan \dfrac x {\sqrt{1 - x^2} }$

Let:

This allows us to convert the arctans from the integrals into arcsines:


 * $\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt{1 - u^2} }\, \mathrm d u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt{1 - u^2} } \left({\frac \pi 4 - \frac {\arcsin u} 2}\right)\,\mathrm d u} }\right)$

Substituting:


 * $s = \arcsin u$, $\mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

into the arcsines, and splitting the second integral:


 * $\zeta \left({2}\right) = 4 \left({\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} }\right) = \dfrac {\pi^2} 6$