Basel Problem/Proof 8

Theorem

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

where $\zeta$ denotes the Riemann zeta function.

Proof

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

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

Hence:

\(\ds \frac 1 \pi \int_{-\pi}^\pi x^2 \rd x\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \paren {\frac{2 \paren {-1}^{n + 1} } n}^2\)

Parseval's Theorem

\(\ds \leadsto \ \ \)

\(\ds \frac 2 \pi \int_0^\pi x^2 \rd x\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\)

Definite Integral of Even Function

\(\ds \leadsto \ \ \)

\(\ds \frac {2 \pi^2} 3\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)

\(=\)

\(\ds \frac {\pi^2} 6\)

$\blacksquare$