Analytic Continuation of Riemann Zeta Function

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Theorem

The Riemann zeta function is meromorphic on $\C$.


Proof

The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\set {s: \map \Re s > 0}$ is given by Integral Representation of Riemann Zeta Function in terms of Fractional Part:

$(1): \quad \ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$

where $\fractpart x$ is the fractional part of $x$.

Let $\map \Re s \le 0$.

Then the value of $\map \zeta s$ can be computed from the relation:

$\map \Gamma {\dfrac s 2} \pi^{-s/2} \map \zeta s = \map \Gamma {\dfrac {s - 1} 2} \pi^{\frac {1 - s} 2} \map \zeta {1 - s}$

That is:

$\map \xi s = \map \xi {1 - s}$

where $\xi$ is the completed zeta function.


First we show that $(1)$ is analytic for $\map \Re s > 0$.

For $n \ge 1$, let:

\(\ds a_n\) \(=\) \(\ds s \int_n^{n + 1} \fractpart x x^{-s - 1}\rd x\)
\(\ds \) \(\ll\) \(\ds s \int_n^{n + 1} x^{-s - 1} \rd x\)
\(\ds \) \(=\) \(\ds \bigintlimits {-x^{-s} } n {n + 1}\)
\(\ds \) \(=\) \(\ds \frac 1 {n^s} - \frac 1 {\paren {n + 1}^s}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1}^s - n^s} {n^s \paren {n + 1}^s}\)

Here $\ll$ is the order notation.

By the Mean Value Theorem, for some $n \le \theta \le n + 1$:

$\paren {n + 1}^s - n^s = s \theta^{s - 1} \le s \paren {n + 1}^{s-1}$

Thus if $s = \sigma + i t$:

$\size {a_n} \le \size {\dfrac s {n^{s + 1} } } = \dfrac {\sigma^2 + t^2} {n^{\sigma + 1} }$

Since:

$\ds \map \zeta s = \frac s {s - 1} - \sum_{n \mathop \ge 1} a_n$

it follows that this representation converges absolutely uniformly on $\map \Re s > 0$.

Thus by Uniform Limit of Analytic Functions is Analytic $\map \zeta s$ is analytic for $\map \Re s > 0$ and $s \ne 1$.


For all $s$ with $\map \Re s < \dfrac 1 2$, $\map \zeta s$ is simply the reflection of $\zeta$ in the upper half-plane.

Therefore, $\zeta$ is also analytic for all $s$ with $\map \Re s < 0$.

$\blacksquare$