# Riemann Hypothesis/Introduction to the Zeta function

Definition 1
${\displaystyle \zeta (s):=\sum _{n=1}^{\infty }n^{-s}\,\forall s\in \mathbb {C} :\Re (s)>1}$
Theorem 1
${\displaystyle \zeta (s)=\prod _{p|{\text{prime}}}{\frac {1}{1-p^{-s}}}}$

Where ${\displaystyle \prod _{p|{\text{prime}}}}$ denotes a product over the primes.

Proof

Note that,

${\displaystyle {\frac {1}{2^{s}}}\zeta (s)={\frac {1}{2^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{6^{s}}}\ldots }$

And hence,

${\displaystyle \left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}\ldots }$

One will notice that every other term has been removed. It then follows that,

${\displaystyle {\frac {1}{3^{s}}}\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)={\frac {1}{3^{s}}}+{\frac {1}{9^{s}}}+{\frac {1}{15^{s}}}\ldots }$

Subtracting,

${\displaystyle \left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+{\frac {1}{11^{s}}}\ldots }$

One may notice that this process sieves the RHS, meaning that,

${\displaystyle \left(\prod _{p|{\text{prime}}}1-p^{-s}\right)\zeta (s)=1}$

It therefore follows that,

${\displaystyle \zeta (s)=\prod _{p|{\text{prime}}}{\frac {1}{1-p^{-s}}}}$ ${\displaystyle \blacksquare }$
Theorem 2
${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {t^{s-1}}{e^{t}-1}}\mathrm {d} t}$

Multiplying the integrand through ${\displaystyle {\frac {e^{-t}}{e^{-t}}}}$,

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {e^{-t}t^{s-1}}{1-e^{-t}}}\mathrm {d} t}$

Writing ${\displaystyle (1-e^{-t})^{-1}}$ as a power series,

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }e^{-t}t^{s-1}\sum _{n=0}^{\infty }e^{-nt}\mathrm {d} t}$
${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }\sum _{n=1}^{\infty }e^{-nt}t^{s-1}\mathrm {d} t}$

Using the substitution ${\displaystyle \xi =nt\implies {\frac {1}{n}}\mathrm {d} \xi =\mathrm {d} t}$

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }\sum _{n=1}^{\infty }e^{-\xi }\left({\frac {\xi }{n}}\right)^{s-1}{\frac {1}{n}}\mathrm {d} \xi }$
${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}e^{-\xi }\xi ^{s-1}\mathrm {d} \xi }$

Using properties of infinite series',

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}\int _{0}^{\infty }e^{-\xi }\xi ^{s-1}\mathrm {d} \xi }$

By the definition of the ${\displaystyle \Gamma }$ function,

${\displaystyle \implies \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}$

Which is true by definition. ${\displaystyle \blacksquare }$