View source for Riemann Hypothesis/Introduction to the Zeta function

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;Definition 1

:<math>\zeta(s) := \sum_{n=1}^\infty n^{-s} \, \forall s\in\mathbb{C}:\Re(s) > 1</math>
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;Theorem 1

:<math>\zeta(s) = \prod_{p|\text{prime}} \frac{1}{1-p^{-s}}</math>

Where <math>\prod_{p|\text{prime}}</math> denotes a product over the primes. 

;Proof

Note that, 

:<math>\frac{1}{2^s}\zeta(s) = \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} \ldots</math>

And hence, 

:<math>\left(1-\frac{1}{2^s}\right)\zeta(s) = 1 + \frac{1}{3^s} + \frac{1}{5^s} \ldots</math>

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

:<math>\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</math>

Subtracting, 

:<math>\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</math>

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

:<math>\left(\prod_{p|\text{prime}}1-p^{-s}\right)\zeta(s) = 1</math>

It therefore follows that, 

:<math>\zeta(s) = \prod_{p|\text{prime}} \frac{1}{1-p^{-s}}</math> <math>\blacksquare</math> 
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;Theorem 2
:<math>\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^{t}-1} \mathrm{d}t</math>

Multiplying the integrand through <math>\frac{e^{-t}}{e^{-t}}</math>, 

:<math>\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{e^{-t}t^{s-1}}{1-e^{-t}}\mathrm{d}t</math>

Writing <math>(1-e^{-t})^{-1}</math> as a power series, 

:<math>\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty e^{-t}t^{s-1} \sum_{n=0}^\infty e^{-nt} \mathrm{d}t</math>
:<math>\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=1}^\infty e^{-nt}t^{s-1}\mathrm{d}t</math>

Using the substitution <math>\xi = nt \implies \frac{1}{n}\mathrm{d}\xi = \mathrm{d}t</math>

:<math>\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</math>
:<math>\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</math>

Using properties of infinite series', 

:<math>\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</math>

By the definition of the <math>\Gamma</math> function, 

:<math>\implies \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}</math>

Which is true by definition. <math>\blacksquare</math>
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