Riemann Zeta Function at Even Integers/Proof 2

Proof
Let $k \in \N$.

Let $\map S x$ be equal to $x^{2 k}$ on $\closedint {-\pi} \pi$ and be periodic with period $2 \pi$.

Let $\ds \map I {2 m, n} = \int_0^\pi x^{2 m} \cos n x \rd x$.

Let $\map A {2 m, n} = \dfrac {\pi^{2 m - 1} \paren {-1}^n 2 m} {n^2}$.

Let $\map B {2 m, n} = -\dfrac {2 m \paren {2 m - 1} } {n^2}$.

By Fourier Series for Even Function over Symmetric Range:

We have:

Thus:

From the above:
 * $\map \zeta 2 = \dfrac {\pi^2} 6$

which serves as our base case.

Assume the induction hypothesis that, for $1 \le m \le k - 1$:
 * $\map \zeta {2 m} = \paren {-1}^{m + 1} \dfrac {B_{2 m} 2^{2 m - 1} \pi^{2 m} } {\paren {2 m}!}$

Then:

which completes the induction step.

Thus by Proof by Mathematical Induction, for all $n \ge 1$:
 * $\map \zeta {2 n} = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}$