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


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

Without loss of generality, the Riemann Double Sum of this multi-variable function can be set such that $0<xy<1$.

By Geometric Series:


 * $\displaystyle\int_0^1\int_0^1\frac{1}{1-xy}\mathrm d A=\int_0^1\int_0^1\sum_{n\mathop=0}^\infty(xy)^n\mathrm d A$

By Tonelli's Theorem and Linear Combination of Integrals:


 * $\displaystyle\int_0^1\int_0^1\frac{1}{1-xy}\mathrm d A=\sum_{n\mathop=0}^\infty\left(\int_0^1 x^n\mathrm d x\int_0^1 y^n\mathrm d y\right)$

Calculating both integrals via Fundamental Theorem of Calculus:


 * $\displaystyle\int_0^1\int_0^1\frac{1}{1-xy}\mathrm d A=\sum_{n\mathop=0}^\infty\left(\frac{1}{n+1}\frac{1}{n+1}\right)$

Correcting the sum shows the relationship between the first expression and the value to calculate.


 * $\displaystyle\zeta\left({2}\right)=\sum_{n\mathop=1}^\infty\frac{1}{n^2}=\int_0^1\int_0^1\frac{1}{1-xy}\mathrm d A$

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

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

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


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

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

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

and letting


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

allows us to make a Substitution into each integral, giving us


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

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


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

Let


 * $\displaystyle\theta=\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)$

Then


 * $\displaystyle\tan^2\theta=\frac{1-u}{1+u}$

and from Sum of Squares of Sine and Cosine/Corollary 1, we have that


 * $\displaystyle\sec^2\theta=\frac{2}{1+u}$

We solve for $u$, and by the Double Angle Formulas/Cosine/Corollary 1 we have that


 * $\displaystyle\cos2\theta=u$

Taking arccosines at both sides, and using Sine of Complement equals Cosine we finally have


 * $\theta=\displaystyle\frac{1}{2}\arccos u=\frac{\pi}{4}-\frac{\arcsin u}{2}$

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


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

Just substituting


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

into the arcsines, and splitting the second integral allows us to see that


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