Riemann Zeta Function at Even Integers

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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:

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

where $\zeta$ denotes the Riemann zeta function.


We also have:

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


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

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

$\blacksquare$


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

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:

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


We have:

\(\ds \map I {0, n}\) \(=\) \(\ds \int_0^\pi \cos n x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {\sin n x} n} 0 \pi\)
\(\ds \) \(=\) \(\ds \frac {\sin n \pi} n - \frac {\sin 0} n\)
\(\ds \) \(=\) \(\ds 0 - 0\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \map I {2 k, n}\) \(=\) \(\ds \int_0^\pi x^{2 k} \cos n x \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {\frac {x^{2 k} \sin n x} n} 0 \pi - \frac {2 k} n \int_0^\pi x^{2 k - 1} \cos n x \rd x\) 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 presented 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 $r = 2 n$, $n \ge 1$.


It can also be expressed using the archaic form of the Bernoulli numbers as:

\(\ds \map \zeta {2 n}\) \(=\) \(\ds \dfrac {2^{2 n - 1} \pi^{2 n} {B_n}^*} {\paren {2 n}!}\)


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:

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


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

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