Basel Problem/Proof 9

Proof
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:


 * $\map f x = \begin {cases}

\paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:


 * $\map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$

Setting $x = 0$: