Poles of Riemann Zeta Function

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Theorem

Let $\zeta$ be the Riemann zeta function.

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


Proof

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

$\displaystyle \zeta \left({s}\right) = \frac s {s-1} - s \int_1^\infty \left\{ {x}\right\} x^{-s-1} \, \mathrm d x$

is meromorphic for $\Re \left({s}\right) > 0$, 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:

\(\displaystyle \operatorname{Res} \left({\zeta, 1}\right)\) \(=\) \(\displaystyle \lim_{s \to 1} \left({s - 1}\right) \left({\frac s {s-1} - s \int_1^\infty \left\{ {x}\right\} x^{-s-1} \, \mathrm d x}\right)\) Residue at Simple Pole
\(\displaystyle \) \(=\) \(\displaystyle 1\)

By Unsymmetric Functional Equation for Riemann Zeta Function:

$\displaystyle \zeta \left({1 - s}\right) = 2^{1 - s} \pi^{-s} \cos \left({\frac 1 2 s \pi}\right) \Gamma \left({s}\right) \zeta \left({s}\right)$

Therefore, for $\Re \left({s}\right) \le 0$:

\(\displaystyle \zeta \left({s}\right)\) \(=\) \(\displaystyle \zeta \left({1 - t}\right)\) for some $t$ with $\Re \left({t}\right) \ge 1$
\(\displaystyle \) \(=\) \(\displaystyle 2^{1-t}\pi^{-t} \cos \left({\frac 1 2 t \pi}\right)\Gamma \left({t}\right) \zeta \left({t}\right)\)

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

By Poles of Gamma Function, $\Gamma \left({t}\right)$ has no poles when $\Re \left({t}\right) \ge 1$.

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

Therefore, the only possible pole is a simple pole at $t = 1$ from the factor $\zeta \left({t}\right)$.

But at this point:

$\cos \left({\dfrac 1 2 t \pi}\right) = \cos \left({\dfrac \pi 2}\right) = 0$

which cancels the pole of $\zeta$.

$\blacksquare$