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:


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

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:

By Unsymmetric Functional Equation for Riemann Zeta Function:


 * $\map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\dfrac 1 2 s \pi} \map \Gamma s \map \zeta s$

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

By Complex Exponential Function is Entire, the factor $\ds 2^{1 - t} \pi^{-t}$ 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 $\zeta$.