Basel Problem/Proof 8

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Theorem

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.


Proof

By Fourier Series of Identity Function over $-\pi$ to $\pi$, for $x \in \openint {-\pi} \pi$:

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


Hence:

\(\displaystyle \frac 1 \pi \int_{-\pi}^\pi x^2 \rd x\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {\frac{2 \paren {-1}^{n + 1} } n}^2\) Parseval's Theorem
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 2 \pi \int_0^\pi x^2 \rd x\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\) Definite Integral of Even Function
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {2 \pi^2} 3\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$