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Again, order of growth problems.

Show that the function $(s-1)\zeta(s)$ is an entire function of growth order $1$; or equivalently, $$|(s-1)\zeta(s)| \leq A_{\epsilon} \; \exp \left(a_{\epsilon}|s|^{1+\epsilon} \right).$$

Of course, $\zeta (\cdot)$ denotes the Riemann zeta function.

Many thanks in advance.

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It's not quite true that those statements are equivalent. The first implies the second (assuming an implicit universal quantifier for $\epsilon\gt0$, existential quantifiers for $a_\epsilon$ and $A_\epsilon$, and a universal quantifier for $s$), but it further implies that there are no such $a_\epsilon$ and $A_\epsilon$ for $\epsilon\lt0$. –  joriki Dec 11 '11 at 9:56

3 Answers 3

up vote 2 down vote accepted
+50

Another approach is through the Laurent series of the Riemann zeta function at $s=1$,

$$\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n\;,$$

where the $\gamma_n$ are the Stieltjes constants. Multiplying by $s-1$ yields

$$(s-1)\zeta(s)=1+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^{n+1}\;.$$

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

$$f(z)=\sum_{n=0}^\infty a_nz^n$$

in terms of the coefficients:

$$\mu=\limsup_{n\to\infty}\frac{n\log n}{\log (1/|a_n|)}\;.$$

To evaluate the limit superior, we can use bounds found by Matsuoka: For all $n\ge10$,

$$|\gamma_n|\le\frac{\exp(n\log\log n)}{10000}\;,$$

and for infinitely many n

$$|\gamma_n|\gt\exp(n\log\log n-n\epsilon)\;.$$

By substituting Stirling's approximation for the factorial,

$$\log n!\sim n\log n -n\;,$$

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

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This is an application of Phragmen-Lindelof, Laplace-Stirling asymptotics of Gamma (or less), and the functional equation. The completed zeta function $\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$ has the functional equation $\xi(1-s)=\xi(s)$, and simple poles at $s=1,0$. Thus, $s(s-1)\xi(s)$ is entire.

From the resulting $\zeta(1-s)=\pi^{s/2}\Gamma(s/2)\zeta(s)/\pi^{-(1-s)/2}\Gamma((1-s)/2))$, multiplying through by $s(s-1)$ to eliminate the poles, and from asymptotics of $\Gamma$, we find that $\zeta(s)$ is of vertical growth of the order ${1\over 2}+|\Re(s)|$ to the left of $0$, with as explicit (crude) constants as one wants. Applying Phragmen-Lindelof gives bounded order of growth in the critical strip. (The details of this latter concern Lindelof and RH, but are irrelevant for the present coarser statement.)

This sort of consideration (at very modest detail) is sufficient to invoke Hadamard-product business.

The Hadamard-product stuff is treated in detail in Ahlfors' "Complex Analysis", as are asymptotics for $\Gamma$. The functional equation of $\zeta$ is treated many places. The argument I sketched is also given many places, such as Lang's "Algebraic Number Theory". It should be viewed as "standard".

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I think this is done in any serious treatment of the zeta function, but is a bit long to write out here. Try Davenport's Multiplicative Number Theory.

EDIT: or try these notes by my colleague, William Chen. He defines $$\xi(s)=(1/2)s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$$ Theorem 6C gives a product formula for $\xi$. Section 6.3 introduces functions of order 1. Theorem 6K relates functions of order 1 to products over zeros. All that goes into Section 6.4, where Chen proves $$\xi(s)=O_{\alpha}\left(e^{|s|^{\alpha}}\right)$$

MORE EDIT: Perhaps the discussion in Bateman and Diamond, Analytic Number Theory, would help. Theorem 8.1 gives the "asymmetric functional equation," $$\zeta(s)=2^s\pi^{s-1}\Gamma(1-s)\zeta(1-s)$$ Then Section 8.2 gives various estimates for $\zeta$. With $s=\sigma+it$, they note $$\zeta(s)=1+O(2^{-\sigma})$$ as $\sigma\to\infty$. By Theorem 8.1 and Stirling they get $$\zeta(\sigma+it)=O(|t|^{(1/2)-\sigma})$$ for $-a\le\sigma\le-b\lt0$, $|t|\ge1$. Then they get (Lemma 8.4) $$\zeta(s)=O(\log t)$$ for $\sigma\ge1$, $t\ge2$, and $$\zeta(s)=O_{\delta}(t^{1-\delta})$$ for $0\lt\delta\lt1$, $\sigma\ge\delta$, $t\ge2$.

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Davenport is kind of vague. I guess the inequality is to be deduced from the functional equation $\zeta(s)=\pi^{s-1/2} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \zeta(1-s)$, but I don't see it yet... –  Anna Dec 11 '11 at 20:12
    
Ok, I understand the proof that $\xi$ has order at most $1$. But how does that imply that $(s-1)\zeta$ has order $1$ (given the function eqn)? –  Anna Dec 11 '11 at 23:44
    
I suppose you have to know something about the Gamma function. Perhaps Stirling's formula is enough. –  Gerry Myerson Dec 12 '11 at 0:38
    
@Anna Where did you find the proof that $\xi$ has order at most 1? Can you comment the book/weblink you used? I've been looking around for that proof, but I have not been able to find it. Also, it can be shown that the other functions in $\zeta$'s functional equation have order of growth 1, hence (s-1)$\zeta(s)$ has order of growth 1. –  nick Jul 9 at 7:15

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