Basel Problem/Proof 1

Proof
By Riemann Zeta Function as a Multiple Integral,


 * $\displaystyle \map \zeta 2 = \int_0^1 \int_0^1 \frac 1 {1 - x y} \rd A$

Let $\tuple {u, v} = \tuple {\dfrac {x + y} 2, \dfrac{y - x} 2}$ so that:
 * $\tuple {x, y} = \tuple {u - v, u + v}$

Let:
 * $\size J = \size {\dfrac {\partial \tuple {x, y} } {\partial \tuple {u, v} } } = 2$

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


 * $\map \zeta 2 = \displaystyle 2 \iint \limits_S \frac {\d u \rd v} {1 - u^2 + v^2}$

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

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

Factoring $1 - u^2$ gives us:
 * $\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} }$

and letting:


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

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


 * $\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac u {\sqrt {1 - u^2} } } \rd u + \int_{\frac 1 2}^1 \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac {1 - u} {\sqrt {1 - u^2} } } \rd u}$

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

Applying Pythagoras's Theorem:


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

Let:

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


 * $\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt {1 - u^2} } \rd u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt {1 - u^2} } \paren {\frac \pi 4 - \frac {\arcsin u} 2} \rd u} }$

Substituting:


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

into the arcsines, and splitting the second integral:


 * $\map \zeta 2 = 4 \paren {\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} } = \dfrac {\pi^2} 6$