Riemann Zeta Function of 4/Proof 2

Theorem

The Riemann zeta function of $4$ is given by:

$\ds \map \zeta 4$

$=$

$\ds \dfrac 1 {1^4} + \dfrac 1 {2^4} + \dfrac 1 {3^4} + \dfrac 1 {4^4} + \cdots$

$\ds $

$=$

$\ds \dfrac {\pi^4} {90}$

$\ds $

$\approx$

$\ds 1 \cdotp 08232 \, 3 \ldots$

By Fourier Series of x squared, for $x \in \closedint {-\pi} \pi$:

$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$

Hence:

$\ds \frac 1 \pi \int_{-\pi}^\pi x^4 \rd x$

$=$

$\ds \frac 1 2 \paren {\frac {2 \pi^2} 3}^2 + \sum_{n \mathop = 1}^\infty \paren {\frac {4 \paren {-1}^n} {n^2} }^2$

$\ds \leadsto \ \ $

$\ds \frac 2 \pi \int_0^\pi x^4 \rd x$

$=$

$\ds \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}$

$\ds \leadsto \ \ $

$\ds \frac {2 \pi^4} 5$

$=$

$\ds \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}$

$\ds \leadsto \ \ $

$\ds \frac {8 \pi^4} {45}$

$=$

$\ds \sum_{n \mathop = 1}^\infty \frac {16} {n^4}$

$\ds \leadsto \ \ $

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^4}$

$=$

$\ds \frac {\pi^4} {90}$

$\blacksquare$