For \(s \in \mathbb{C}\) with \(\textrm{Re } s > 1\), the zeta function is defined as

\[ \begin{aligned} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \end{aligned} \]

We can readily see \(\zeta(s)\) converges absolutely in \(\{\textrm{Re } s > 1\}\). What's interesting is the convergence is uniform in \(\{\textrm{Re } s > r\}\) for any \(r > 1\), which can be used to show that \(\zeta(s)\) is analytic in \(\{\textrm{Re } s > 1\}\).

Riemann was able to show that \(\zeta(s)\) has an analytic continuation into \(\mathbb{C} \setminus \{1\}\), and this continuation satisfies that \(\zeta(s) \rightarrow \infty\) as \(s \rightarrow 1\).

Some interesting facts:

• The only zeros of the zeta function (ie when \(\zeta(s) = 0\)) outside the strip \(\{0 \le \textrm{Re } s \le 1\}\) are at the negative even integers, \(-2, -4, -6, \cdots\), which are called the "trivial zeros".
• Zeta has no zeros on the line \(\{ \textrm{Re } s = 1\}\), nor the line \(\{ \textrm{Re } s = 0\}\).

Riemann Hypothesis

    In the strip \(\{0 \le \textrm{Re } s \le 1\}\), all zeros of \(\zeta\) are on the line \(\displaystyle \{ \textrm{Re } s = \frac{1}{2} \}\)

which has significant implication on the distribution of prime numbers and the growth of many important arithmetic functions, but remains unproved.

Amazingly,

\[ \begin{aligned} \zeta(s) = \prod_p \frac{1}{1 - p^{-s}} \end{aligned} \]

where the product is over all primes ! Why ? Observe

\[ \begin{aligned} \zeta(s) &= \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \\ &= \big(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{9^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{5^s} + \frac{1}{25^s} + \cdots\big) \cdots \\ \end{aligned} \]

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 10:23 AM