Riemann Zeta Function at Even Integers/Proof 2

Theorem

The Riemann $\zeta$ function can be calculated for even integers as follows:

 $\displaystyle \map \zeta {2 n}$ $=$ $\displaystyle \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1^{2 n} } + \frac 1 {2^{2 n} } + \frac 1 {3^{2 n} } + \frac 1 {4^{2 n} } + \cdots$

where:

$B_n$ are the Bernoulli numbers
$n$ is a positive integer.

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

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

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

 $\displaystyle \map S x$ $=$ $\displaystyle \frac {\frac 2 \pi \int_0^\pi \map S x \rd x} 2 + \sum_{n \mathop = 1}^\infty \cos n x \frac 2 \pi \int_0^\pi \map S x \cos n x \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {\int_0^\pi x^{2 k} \rd x} \pi + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n x \int_0^\pi x^{2 k} \cos n x \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n x \map I {2 k, n}$ $\displaystyle \leadsto \ \$ $\displaystyle \map S \pi$ $=$ $\displaystyle \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n \pi \map I {2 k, n}$ $\displaystyle \leadsto \ \$ $\displaystyle \pi^{2 k}$ $=$ $\displaystyle \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {2 k \pi^{2 k} } {2 k + 1}$ $=$ $\displaystyle \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}$

We have:

 $\displaystyle \map I {0, n}$ $=$ $\displaystyle \int_0^\pi \cos n x \rd x$ $\displaystyle$ $=$ $\displaystyle \bigintlimits {\frac {\sin n x} n} 0 \pi$ $\displaystyle$ $=$ $\displaystyle \frac {\sin n \pi} n - \frac {\sin 0} n$ $\displaystyle$ $=$ $\displaystyle 0 - 0$ $\displaystyle$ $=$ $\displaystyle 0$ $\displaystyle \map I {2 k, n}$ $=$ $\displaystyle \int_0^\pi x^{2 k} \cos n x \rd x$ $\displaystyle$ $=$ $\displaystyle \intlimits {\frac {x^{2 k} \sin n x} n} 0 \pi - \frac {2 k} n \int_0^\pi x^{2 k - 1} \cos n x \rd x$ Integration by Parts $\displaystyle$ $=$ $\displaystyle 0 - 0 - \frac {2 k} n \int_0^\pi x^{2 k - 1} \sin n x \rd x$ $\displaystyle$ $=$ $\displaystyle -\frac {2 k} n \int_0^\pi x^{2 k - 1} \sin n x \rd x$ $\displaystyle$ $=$ $\displaystyle -\frac {2 k} n \paren {\intlimits {\frac {-x^{2 k - 1} \cos n x} n} 0 \pi + \frac {2 k - 1} n \int_0^\pi x^{2 k - 2} \cos n x \rd x}$ Integration by Parts $\displaystyle$ $=$ $\displaystyle -\frac {2 k} n \paren {\frac {\pi^{2 k - 1} \paren {-1}^{n + 1} } n + \frac{2 k - 1} n \map I {2 k - 2, n} }$ $\displaystyle$ $=$ $\displaystyle \frac {\pi^{2 k - 1} \paren {-1}^n 2 k} {n^2} - \frac {2 k \paren {2 k - 1} } {n^2} \map I {2 k - 2, n}$ $\displaystyle$ $=$ $\displaystyle \map A {2 k, n} + \map B {2 k, n} \map I {2 k - 2, n}$ $\displaystyle$ $=$ $\displaystyle \map A {2 k, n} + \map B {2 k, n} \map A {2 k - 2, n} + \map B {2 k, n} \map B {2 k - 2, n} \map I {2 k - 4, n}$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop = 0}^{k - 1} \map A {2 k - 2 m, n} \prod_{j \mathop = 0}^{m - 1} \map B {2 k - 2 j, n}$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^n \paren {2 k - 2 m} } {n^2} \prod_{j \mathop = 0}^{m - 1} -\frac {\paren {2 k - 2 j} \paren {2 k - 2 j - 1} } {n^2}$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^{n + m} \paren {2 k - 2 m} } {n^{2 \paren {m + 1} } } \prod_{j \mathop = 0}^{m - 1} \paren {2 k - 2 j} \paren {2 k - 2 j - 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^{n + m} \paren {2 k}!} {n^{2 \paren {m + 1} } \paren {2 k - 2 m - 1}!}$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2m + 1} \paren {-1}^{n + m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!}$

Thus:

 $\displaystyle \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}$ $=$ $\displaystyle \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2 m + 1} \paren {-1}^{n+ m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!}$ $\displaystyle$ $=$ $\displaystyle 2 \sum_{n \mathop = 1}^\infty \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!}$ $\displaystyle$ $=$ $\displaystyle 2 \sum_{m \mathop = 1}^k \sum_{n \mathop = 1}^\infty \frac {\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!}$ $\displaystyle$ $=$ $\displaystyle 2 \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \sum_{n \mathop = 1}^\infty \frac 1 {n^{2 \paren m} }$ $\displaystyle$ $=$ $\displaystyle 2 \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}$ Definition of Riemann Zeta Function $\displaystyle \leadsto \ \$ $\displaystyle \frac{ 2 k \pi^{2 k} } {2 k + 1}$ $=$ $\displaystyle 2 \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}$ $\displaystyle \leadsto \ \$ $\displaystyle 2 \paren {2 k}! \paren {-1}^{k - 1} \map \zeta {2 k}$ $=$ $\displaystyle \frac{ 2 k \pi^{2 k} } {2 k + 1} - 2 \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}$ $\displaystyle \leadsto \ \$ $\displaystyle \map \zeta {2 k}$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} - \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}$

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:

 $\displaystyle \map \zeta {2 k}$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} - \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} - \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \paren {-1}^{m + 1} \dfrac {B_{2 m} 2^{2 m - 1} \pi^{2 m} } {\paren {2 m}!}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} + \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k} \paren {-1}^k B_{2 m} 2^{2 m - 1} } {\paren {2 k - 2 m + 1}! \paren {2 m}!}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} + \frac{\pi^{2 k} \paren {-1}^k} {2 \paren {2 k + 1}!} \sum_{m \mathop = 1}^{k - 1} \binom {2 k + 1}{2m} B_{2 m} 2^{2 m}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} + \frac{\pi^{2 k} \paren {-1}^k} {2 \paren {2 k + 1}!} \paren {2 k - 2^{2 k} B_{2 k} \paren {2 k + 1} }$ Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} + \frac{k \pi^{2 k} \paren {-1}^k} {\paren {2 k + 1}!} + \frac{B_{2 k} 2^{2 k - 1}\pi^{2 k} \paren {-1}^{k + 1} } {\paren {2 k}!}$ $\displaystyle$ $=$ $\displaystyle \frac {B_{2 k} 2^{2 k - 1}\pi^{2 k} \paren {-1}^{k + 1} } {\paren {2 k}!}$

which completes the induction step.

Thus by Proof by Mathematical Induction, for all $n \ge 1$:

$\map \zeta {2 n} = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}$

$\blacksquare$