Establishing order of the $\xi(s)$, the Riemann xi function

Question

Clearly,

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

Now from various literature, I understand that the order of the $\xi(s)=1$. The problem remains how to show it. The polynomials have the order of $0$, so I believe it is enough to show that the order of $\Gamma\left(\frac{s}{2}\right)$ and $\zeta(s)$ is $1$ for each of the two? I am really struggling with this and I need to establish this for my dissertation.

I know there exists a formula:

An entire function $f$ is said to be of finite order if there exist numbers $a$,$r>0$ such that $$|f(z)|\leq exp(|z|^a)$$ for all $|z|>r$. The infimum of all numbers $a$ for which this inequality holds is called the function order of $f$, denoted $\lambda=\lambda(f)$.

But I have no idea how to use this definition for the gamma or the zeta functions. Thank you for help

Answer 1

$\xi(s)=\xi(1-s)$ and $\xi$ is an entire function. By the integral representation for the $\zeta$ function over the half-plane $\text{Re}(s)>0$ it is clear that $\zeta(s)\ll e^{|s|}$. $s(s-1)$ has order $0$ and $\pi^{s/2}$ has order $1$. By the Weierstrass (or Euler's) product for the $\Gamma$ function, or just by Stirling's inequality, $\Gamma(s)\ll e^{|s|^{1+\varepsilon}}$ over $\text{Re}(s)>0$. It follows that the order of $\xi$ is $\leq 1$. By studying $\xi(s)$ over $\mathbb{R}^+$, it is straightforward to deduce that the order of $\xi$ is exactly $1$, since $\zeta(s)\to 1$ as $s\to +\infty$.

Answer 2

Another approach to Jack's idea above is through the Laurent series of the Riemann zeta function evaluated at $s=1$,

\begin{equation} \zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n \end{equation} where each of the $\gamma_n$ are Stieltjes constants. Multiplying this by $s-1$ gives

\begin{equation} (s-1)\zeta(s)=1+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^{n+1} \end{equation}

Now, examining the Theorem 2.2.2 of Entire Functions by Ralph Philip Boas expresses the order $\mu$ of an entire function given by a power series

\begin{equation} f(z)=\sum_{n=0}^\infty a_nz^n \end{equation}

in terms of the coefficients:

\begin{equation} \mu=\limsup_{n\to\infty}\frac{n\log n}{\log (1/ \lvert a_n \rvert)} \end{equation} The limit superior can be evaluated by bounds found by Matsuoka: that is, $\forall\, n\ge 10$,

\begin{equation} \lvert\gamma_n \rvert \le\frac{\exp(n\log\log n)}{10000} \end{equation} and for infinitely many n

\begin{equation} \lvert\gamma_n \rvert \gt\exp(n\log\log n-n\epsilon) \end{equation}

By substituting Stirling's approximation for the factorial,

\begin{equation} \log n!\sim n\log n -n\ \end{equation}

we can see that the limit superior is $1$: Then the upper bound on $\gamma_n$ dictates that this is eventually below $1+\delta$, and the lower bound dictates that it is also infinitely often above $1$.