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 the Riemann Zeta Function,


 * $\displaystyle \zeta(s) = \frac s {s-1} - s \int_1^\infty \{ x\} x^{-s-1}\ dx$

is meromorphic for $\Re(s) > 0$, and 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:


 * $\displaystyle \zeta(1-s) = 2^{1-s}\pi^{-s} \cos\left(\frac12 s\pi \right)\Gamma(s)\zeta(s)$

Therefore, for $\Re(s) \le 0$,

By Exponential is Entire, the factor $\displaystyle2^{1-t}\pi^{-t}$ has no poles when $\Re(t) \ge 1$.

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

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 $\zeta(t)$.

But at this point:
 * $\cos \left({\dfrac 1 2 t \pi}\right) = \cos \left({\dfrac \pi 2}\right) = 0$

which cancels the pole of $\zeta$.