Poles of Riemann Zeta Function

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:

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

By Exponential 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 Cosine 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 $\zeta$.