Riemann Zeta Function at Even Integers

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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.


Corollary

\(\displaystyle B_{2 n}\) \(=\) \(\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \sum_{k \mathop = 1}^\infty \frac 1 {k^{2 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{n + 1} \dfrac {\left({2 n}\right)!} {2^{2 n - 1} \pi^{2 n} } \left({1 + \frac 1 {2^{2 n} } + \frac 1 {3^{2 n} } + \cdots}\right)\)


Proof 1

Lemma

Let $x \in \R$ be such that $\left\lvert{x}\right\rvert < 1$.

Then:

$\displaystyle \pi x \cot {\pi x} = 1 - 2 \sum_{n \mathop = 1}^\infty \zeta \left({2 n}\right) x^{2 n}$

where $\zeta$ denotes the Riemann zeta function.


We also have:

\(\displaystyle \pi x \cot {\pi x}\) \(=\) \(\displaystyle i \pi x \frac {e^{i \pi x} + e^{- i \pi x} } {e^{i \pi x} - e^{- i \pi x} }\) Cotangent Exponential Formulation
\(\displaystyle \) \(=\) \(\displaystyle i \pi x \frac {e^{2 i \pi x} + 1 } {e^{2 i \pi x} - 1 }\)
\(\displaystyle \) \(=\) \(\displaystyle i \pi x \left({1 + \frac 2 {e^{2 i \pi x} - 1} } \right)\)
\(\displaystyle \) \(=\) \(\displaystyle i \pi x + \frac {2 i \pi x} {e^{2 i \pi x} - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle i \pi x + \sum_{n \mathop = 0}^\infty \frac {B_n \left({2 i \pi x}\right)^n} {n!}\) Definition of Bernoulli Numbers
\(\displaystyle \) \(=\) \(\displaystyle 1 + \sum_{n \mathop = 2}^\infty \frac {B_n \left({2 i \pi x}\right)^n} {n!}\) as $B_0 = 1$ and $B_1 = - \dfrac 1 2$
\(\displaystyle \) \(=\) \(\displaystyle 1 - 2 \sum_{n \mathop = 2}^\infty \left({-\frac 1 2}\right) \frac {B_n \left({2 i \pi x}\right)^n} {n!}\)
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle 1 - 2 \sum_{n \mathop = 1}^\infty \left({-\frac 1 2}\right) \frac {B_{2n} \left({2 i \pi x}\right)^{2n} } {\left({2n}\right)!}\) 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 $S \left({x}\right)$ be equal to $x^{2k}$ on $\left[{-\pi \,.\,.\, \pi}\right]$ and be periodic with period $2 \pi$.

Let $\displaystyle I \left({2 m, n}\right) = \int_0^\pi x^{ 2m} \cos \left({n x}\right) \rd x$.

Let $A \left({2 m, n}\right) = \dfrac {\pi^{2 m - 1} \left({-1}\right)^n 2 m} {n^2}$.

Let $B \left({2 m, n}\right) = -\dfrac {2 m \left({2 m - 1}\right)} {n^2}$.


By Fourier Series for Even Function over Symmetric Range:

\(\displaystyle S \left({x}\right)\) \(=\) \(\displaystyle \frac {\frac 2 \pi \int_0^\pi S\left(x\right) \rd x} 2 + \sum_{n \mathop = 1}^\infty \cos\left( n x\right) \frac 2 \pi \int_0^\pi S\left(x\right) \cos\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac { \int_0^\pi x^{2k} \rd x} \pi + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos\left( n x\right) \int_0^\pi x^{2k} \cos\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac { \pi^{2k} } {2k+1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos\left( n x\right) I\left(2k, n \right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle S\left(\pi\right)\) \(=\) \(\displaystyle \frac { \pi^{2k} } {2k+1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos\left( n \pi\right) I\left(2k, n \right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \pi^{2k}\) \(=\) \(\displaystyle \frac { \pi^{2k} } {2k+1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \left( -1\right)^n I\left(2k, n \right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac{ 2k \pi^{2k} } {2k+1}\) \(=\) \(\displaystyle \frac 2 \pi \sum_{n \mathop = 1}^\infty \left( -1\right)^n I\left(2k, n \right)\)


We have:

\(\displaystyle I\left(0, n \right)\) \(=\) \(\displaystyle \int_0^\pi \cos\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \left[ \frac {\sin\left( n x\right)} n \right]_0^\pi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sin\left( n \pi \right)} n - \frac {\sin\left(0 \right)} n\)
\(\displaystyle \) \(=\) \(\displaystyle 0 - 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)
\(\displaystyle I \left(2k, n \right)\) \(=\) \(\displaystyle \int_0^\pi x^{2k} \cos\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \left[ \frac {x^{2k} \sin\left( n x\right)} {n} \right]_0^\pi - \frac{2k}{n} \int_0^\pi x^{2k-1} \cos\left( n x\right) \rd x\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle 0 - 0 - \frac{2k}{n} \int_0^\pi x^{2k-1} \sin\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac{2k}{n} \int_0^\pi x^{2k-1} \sin\left( n x\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac{2k}{n} \left[ \left[ \frac {-x^{2k-1} \cos\left( n x\right)} n \right]_0^\pi + \frac{2k-1}{n} \int_0^\pi x^{2k-2} \cos\left( n x\right) \rd x \right]\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle -\frac{2k}{n} \left[ \frac {\pi^{2k-1} \left(-1\right)^{n+1} } n + \frac{2k-1}{n} I\left(2k-2, n \right) \right]\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^{2k-1} \left(-1\right)^{n} 2k} {n^2} - \frac{2k\left(2k-1\right)}{n^2} I\left(2k-2, n \right)\)
\(\displaystyle \) \(=\) \(\displaystyle A \left(2k, n\right) + B\left(2k, n\right) I\left(2k-2, n \right)\)
\(\displaystyle \) \(=\) \(\displaystyle A \left(2k, n\right) + B\left(2k, n\right) A\left(2k-2, n\right) + B\left(2k, n\right) B\left(2k-2, n\right) I\left(2k-4, n \right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^{k-1} A\left(2k-2m, n\right) \prod_{j\mathop=0}^{m-1} B\left(2k-2j, n\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^{k-1} \frac {\pi^{2k-2m-1} \left(-1\right)^{n} \left(2k-2m\right)} {n^2} \prod_{j\mathop=0}^{m-1} -\frac{\left(2k-2j\right)\left(2k-2j-1\right)}{n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^{k-1} \frac {\pi^{2k-2m-1} \left(-1\right)^{n+m} \left(2k-2m\right)} {n^{2\left(m+1\right)} } \prod_{j\mathop=0}^{m-1} \left(2k-2j\right)\left(2k-2j-1\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^{k-1} \frac {\pi^{2k-2m-1} \left(-1\right)^{n+m} \left(2k\right)!} {n^{2\left(m+1\right)} \left(2k-2m-1\right)!}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 1}^{k} \frac {\pi^{2k-2m+1} \left(-1\right)^{n+m-1} \left(2k\right)!} {n^{2\left(m\right)} \left(2k-2m+1\right)!}\)


Thus:

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


From the above:

$\zeta \left({2}\right) = \dfrac {\pi^2} 6$

which serves as our base case.


Assume, for induction, that for $1 \le m \le k-1$

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


Then:

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

which completes the Induction Hypothesis.


Thus, for all $n \ge 1$:

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

$\blacksquare$


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:

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


Riemann Zeta Function of $4$

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

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


Riemann Zeta Function of $6$

$\displaystyle \map \zeta 6 = \sum_{n \mathop = 1}^{\infty} \frac 1 {n^6} = \frac {\pi^6} {945}$

where $\zeta$ denotes the Riemann zeta function.


Riemann Zeta Function of $8$

$\displaystyle \map \zeta 8 = \sum_{n \mathop = 1}^{\infty} \frac 1 {n^8} = \frac {\pi^8} {9450}$

where $\zeta$ denotes the Riemann zeta function.


Riemann Zeta Function of $26$

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

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


Also see


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.


Sources