Completed Riemann Zeta Function has Order One

Theorem
The completed Riemann zeta function $\xi$ has order at most $1$.

Proof
We are required to prove that:


 * $\xi \left({s}\right) = \dfrac 1 2 s \left({s - 1}\right) \pi^{-s/2} \Gamma \left({\dfrac s 2}\right) \zeta \left({s}\right) \ll \exp \left({|s|^\beta}\right)$

for all $\beta > 1$, where $\ll$ is the order notation.

Note that by the Functional Equation for Riemann Zeta Function, it is sufficient to check this for $\Re \left({s}\right) \ge \dfrac 1 2$.

We simply check this fact for each factor.

Evidently:

for some $c_1 > 0$.

For the gamma factor, we have Stirling's Formula for Gamma Function:


 * $\displaystyle \log \Gamma \left({s}\right) = \left({s - \frac 1 2}\right)\log s - s + \frac {\log 2 \pi} 2 + \sum_{n \mathop = 1}^{d-1} \frac{B_{2 n}}{ 2 n \left({2 n - 1}\right)s^{2 n-1}} + \mathcal O \left({s^{1 - 2 d} }\right)$

This is valid only away from the poles of $\Gamma$ at $s = 0, -1, -2, \ldots$

However, it is assumed that $\Re \left({s}\right) \ge \dfrac 1 2$, so this is not a problem.

The error term $\mathcal O \left({s^{1 - 2d} }\right)$ is small for large $s$.

More generally, the largest contribution is the term $\left({s - \dfrac 1 2}\right)\log s$, so we have:


 * $\log \Gamma\left({\dfrac s 2}\right) \ll \left\vert{s}\right\vert \log \left\vert{s}\right\vert$

That is:


 * $\Gamma \left({\dfrac s 2}\right) \ll \exp \left({\left\vert{s}\right\vert \log \left\vert{s}\right\vert}\right)$

Finally, from Integral Representation of Riemann Zeta Function in terms of Fractional Part, for $\Re \left({s}\right) > \dfrac 1 2$:


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

It is seen that for $\Re \left({s}\right) > \dfrac 1 2$, the integral is bounded, and therefore:


 * $\left({1 - s}\right) \zeta \left({s}\right) \ll \mathcal O (\left\vert{s}\right\vert^2) \ll \exp \left({\left\vert{s}\right\vert}\right)$

Combining these facts, and using that $\log s \ll s^\epsilon$ for all $\epsilon > 0$ (shown by Upper Bound of Natural Logarithm), we have:


 * $\left|{\xi \left({s}\right)}\right| \ll \exp\left({\left\vert{s}\right\vert^{1 + \epsilon} }\right)$

for all $\epsilon > 0$, and the proof is complete.