# Riemann Zeta Function at Even Integers

## Theorem

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

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

where:

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

### Corollary

 $\ds B_{2 n}$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }$ $\ds$ $=$ $\ds \paren {-1}^{n + 1} \dfrac {\paren {2 n}!} {2^{2 n - 1} \pi^{2 n} } \paren {1 + \frac 1 {2^{2 n} } + \frac 1 {3^{2 n} } + \cdots}$

## Proof 1

### Lemma

Let $x \in \R$ be such that $\size x < 1$.

Then:

$\displaystyle \pi x \cot {\pi x} = 1 - 2 \sum_{n \mathop = 1}^\infty \map \zeta {2 n} x^{2 n}$

where $\zeta$ denotes the Riemann zeta function.

We also have:

 $\ds \pi x \cot {\pi x}$ $=$ $\ds i \pi x \frac {e^{i \pi x} + e^{- i \pi x} } {e^{i \pi x} - e^{- i \pi x} }$ Cotangent Exponential Formulation $\ds$ $=$ $\ds i \pi x \frac {e^{2 i \pi x} + 1 } {e^{2 i \pi x} - 1 }$ $\ds$ $=$ $\ds i \pi x \left({1 + \frac 2 {e^{2 i \pi x} - 1} } \right)$ $\ds$ $=$ $\ds i \pi x + \frac {2 i \pi x} {e^{2 i \pi x} - 1}$ $\ds$ $=$ $\ds i \pi x + \sum_{n \mathop = 0}^\infty \frac {B_n \left({2 i \pi x}\right)^n} {n!}$ Definition of Bernoulli Numbers $\ds$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n \left({2 i \pi x}\right)^n} {n!}$ as $B_0 = 1$ and $B_1 = - \dfrac 1 2$ $\ds$ $=$ $\ds 1 - 2 \sum_{n \mathop = 2}^\infty \left({-\frac 1 2}\right) \frac {B_n \left({2 i \pi x}\right)^n} {n!}$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds 1 - 2 \sum_{n \mathop = 1}^\infty \left({-\frac 1 2}\right) \frac {B_{2n} \left({2 i \pi x}\right)^{2n} } {\left({2n}\right)!}$ Odd Bernoulli Numbers Vanish

Equating the coefficients of $(1)$ with the expression given in the lemma:

$\zeta \left({2 n}\right) = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n}} {\left({2 n}\right)!}$

$\blacksquare$

## Proof 2

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

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

We have:

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

Thus:

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

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:

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

which completes the induction step.

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

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

$\blacksquare$

## Also rendered as

This can also be seen rendered in the elegant form:

$\map \zeta r = \dfrac 1 2 \size {B_r} \dfrac {\paren {2 \pi}^r} {r!}$

for $r = 2 n$, $n \ge 1$.

## Examples

### Riemann Zeta Function of $2$

The Riemann zeta function of $2$ is given by:

 $\ds \map \zeta 2$ $=$ $\ds \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^2} 6$ $\ds$ $\approx$ $\ds 1 \cdotp 64493 \, 4066 \ldots$

### Riemann Zeta Function of $4$

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

 $\ds \map \zeta 4$ $=$ $\ds \dfrac 1 {1^4} + \dfrac 1 {2^4} + \dfrac 1 {3^4} + \dfrac 1 {4^4} + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^4} {90}$ $\ds$ $\approx$ $\ds 1 \cdotp 08232 \, 3 \ldots$

### Riemann Zeta Function of $6$

The Riemann zeta function of $6$ is given by:

 $\ds \map \zeta 6$ $=$ $\ds \dfrac 1 {1^6} + \dfrac 1 {2^6} + \dfrac 1 {3^6} + \dfrac 1 {4^6} + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^6} {945}$ $\ds$ $\approx$ $\ds 1 \cdotp 01734 \, 3 \ldots$

### Riemann Zeta Function of $8$

The Riemann zeta function of $8$ is given by:

 $\ds \map \zeta 8$ $=$ $\ds \dfrac 1 {1^8} + \dfrac 1 {2^8} + \dfrac 1 {3^8} + \dfrac 1 {4^8} + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^8} {9450}$ $\ds$ $\approx$ $\ds 1 \cdotp 00408 \, 3 \ldots$

### Riemann Zeta Function of $26$

The Riemann zeta function of $26$ is given by:

 $\ds \map \zeta {26}$ $=$ $\ds \dfrac 1 {1^{26} } + \dfrac 1 {2^{26} } + \dfrac 1 {3^{26} } + \dfrac 1 {4^{26} } + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^{26} \times 2^{24} \times 76 \, 977 \, 927} {27!}$

## Historical Note

Leonhard Paul Euler calculated the sums of all the even powers of the reciprocals of the integers, all the way up to the $26$th power.