Poles of Riemann Zeta Function

From ProofWiki
Jump to navigation Jump to search


Let $\zeta$ be the Riemann zeta function.

Then $\zeta$ has a simple pole at $s = 1$ with residue $1$, and no other poles.


By Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part:

$\displaystyle \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$

is meromorphic for $\map \Re s > 0$, and the integral converges to a finite value for fixed $s$ in this region.

Therefore in this region the only pole of $\zeta$ is at $s = 1$, with residue:

\(\ds \Res \zeta 1\) \(=\) \(\ds \lim_{s \mathop \to 1} \paren {s - 1} \paren {\frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x}\) Residue at Simple Pole
\(\ds \) \(=\) \(\ds 1\)

By Unsymmetric Functional Equation for Riemann Zeta Function:

$\displaystyle \map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\frac 1 2 s \pi} \, \map \Gamma s \, \map \zeta s$

Therefore, for $\map \Re s \le 0$:

\(\ds \map \zeta s\) \(=\) \(\ds \map \zeta {1 - t}\) for some $t$ with $\map \Re t \ge 1$
\(\ds \) \(=\) \(\ds 2^{1 - t} \pi^{-t} \map \cos {\frac 1 2 t \pi} ,\ \map \Gamma t \,\map \zeta t\)

By Complex Exponential Function is Entire, the factor $\displaystyle 2^{1 - t} \pi^{-t}$ has no poles when $\map \Re t \ge 1$.

By Poles of Gamma Function, $\map \Gamma t$ has no poles when $\map \Re t \ge 1$.

By Complex Cosine Function is Entire, $\map \cos {\dfrac 1 2 t \pi}$ also has no poles in this region.

Therefore, the only possible pole is a simple pole at $t = 1$ from the factor $\map \zeta t$.

But at this point:

$\map \cos {\dfrac 1 2 t \pi} = \map \cos {\dfrac \pi 2} = 0$

which cancels the pole of $\zeta$.