## Riemann zeta function

### Function

 definition $\zeta: \mathbb{C}\setminus\{1\} \to \mathbb C$ definition $\zeta(s) := \begin{cases} \sum_{n=1}^\infty\, n^{-s} &\hspace{.5cm} \mathrm{if}\hspace{.5cm} \mathfrak{R}(s)>1 \\\\ \text{analytic continuation}\hspace{.5cm} &\hspace{.5cm} \mathrm{else} \end{cases}$

“$\text{analytic continuation}$”

#### Theorems

Euler product:

$\zeta(s)=\prod_{p\in\text{primes}}\frac{1}{1-p^{-s}}$

##### Representations
$\zeta(s) = \dfrac{\int_0^\infty\frac{x^{s}}{e^x-1}\frac{{\mathrm d}x}{x}}{\int_0^\infty \frac{x^{s}}{e^x-0}\frac{{\mathrm d}x}{x}}=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s}}{e^x-1}\frac{{\mathrm d}x}{x}$
$\log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x^s-1}\frac{{\mathrm d}x}{x}$

where $\pi$ is the prime-counting function.

##### Functional equation

Tells you most values:

$\zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s)$
##### Specific values
$\zeta(-2m)=0$
$\zeta(2m)=(-1)^{m+1}\frac{(2\pi)^{2m}}{2(2m)!} B_{2m}$

so that

$\zeta(1-2m) = \dfrac{2(2m-1)!}{(4\pi^2)^m}\cos(m\pi)\zeta(2m)$

$\zeta(1-2m)=(-1)^{m+1}\frac{1}{2m}B_{2m}$

E.g.

$\zeta(2)=\pi^2/6$
$\zeta(4)=\pi^4/90$

Also

$\zeta(-1)=-1/12$
##### Expansions

Fix an $N$ and split the infinite sum into a finite part ($\sum_{n=1}^N\, n^{-s}$ is also the subject of Faulhaber's formula) and $\sum_{n=N}^\infty\, n^{-s}$. We can learn about its divergence via the following approximation

$\sum_{n=N}^\infty\, n^{-s}\approx \lim_{\varepsilon\to 0}\int_N^{N/\varepsilon}n^{-s}$

$=\dfrac{N^{-s+1}}{-s+1}\left(\lim_{\varepsilon\to 0}\dfrac{1}{\varepsilon^{-s+1}}-1\right)$

$=\dfrac{1}{s-1}\dfrac{N}{N^s}\left(1-\lim_{\varepsilon\to 0}\varepsilon^{{\mathrm{Re}}(s)-1}{\mathrm e}^{i\,{\mathrm{Im}}(s)\log(\varepsilon)}\right)$

Due to the term $\frac{1}{s-1}$, the value $s=1$ is a proper singularity and this is also true for the analytical continuation of the sum, i.e. the zeta function. On the other hand, the term $\lim_{\varepsilon\to 0}\varepsilon^{{\mathrm{Re}}(s)-1}$ is responsible for the sum being undefined for $\mathrm{Re}(s)\le 1$.

##### 1+2+3+...

(maybe delete this)

Sum[k, {k, 0, n}]
Binomial[n + 1, 2]
Plot[Binomial[n, 2], {n, -1, 4}, ImageSize -> 200]

Integrate[Binomial[n, 2], {n, s, s + d}]/d // Simplify
Solve[% == -1/12, d]
Table[{s, %}, {s, 0, 1, 1/2}] // TableForm
##### On Riemanns paper

In Riemanns original paper, Riemann observed that a integration variable substitution x→n·x in the definition of the Gamma function

$\Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x$

let's you write

${\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x$

and thus

$\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \sum_{n=1}^\infty \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x$

Now consider the geometric series

$\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$

(check signs)

The above sum over the integral is convergent for $s>1$, while the expression

$\frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x$

(check signs)

works for all complex $s\ne 1$. We thus found the analytical continuation.

The integrand $\frac{1} { {\mathrm e}^x-1}$ diverges periodically in steps of $2\pi\,i$. He discovers that the function obeys a reflection formula