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I need to use the functional equation for $\zeta(s)$ and Stirling's formula, to show that for $s=\sigma +it$ , with $\sigma <0$:

$$ |\zeta(s)| << \left(\frac{t}{2\pi}\right)^{1/2-\sigma}$$

as $t\rightarrow \infty$ (where $\sigma$ is fixed), i.e, $\,\displaystyle{\frac{|\zeta(s)|}{\left(\frac{t}{2\pi}\right)^{1/2-\sigma}}}$ is bounded by some constant that depends on $\sigma$ as $t\rightarrow \infty$.

Any reference or the solution itself?

I tried use the fact that

$$\frac{\zeta(s)}{\zeta(1-s)} = \frac{\pi^{s-1/2}\, \Gamma\left(\frac{1-s}{2}\right)}{\Gamma(s/2)}$$, where $\zeta(1-s)$ is well defined, doesn't vanish and bounded I guess this is the constant that depends on $\sigma$, my problem is how to evaluate the fraction with the $\Gamma$'s.

Thanks in advance.

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You may want to use the function equation $$\zeta(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)$$ this comes from the fact that $\Gamma(1-s) \Gamma(s) = \dfrac{\pi}{\sin(\pi s)}$. But I tried in vain using this. – user17762 Jun 2 '12 at 0:56
Can you tell the source of the problem? and why you are interested in the asymptotic of $\zeta$ outside the critical strip? – user17762 Jun 2 '12 at 17:38
The question is an assignment I got, here: question 3, I haven't seen this assertion in textbook, do you happen to know how to solve this or even just a good reference to read? :-) – MathematicalPhysicist Jun 3 '12 at 4:35
I think I have now answered your question. A minor correction in the link you sent. In the Stirling's formula, the right hand side must read $\exp \left( (s-1/2) \log(s) -s + \frac12 \log(2 \pi) + \mathcal{O}(1/|s|)\right)$ i.e. the right hand side must be exponentiated. – user17762 Jun 3 '12 at 6:35
up vote 5 down vote accepted

From the function equational for $\zeta(s)$, we have that $$\zeta(s) = \pi^{(s-1/2)}\zeta(1-s) \dfrac{\Gamma \left( \dfrac{1-s}{2} \right)}{\Gamma \left( \dfrac{s}{2} \right)}$$ $s = \sigma+it$, where $\sigma < 0$. Note that here we fix $\sigma < 0$ and let $t \rightarrow \infty$.

First note that, $\zeta(1-s)$ is trivially bounded by $\zeta(1-\sigma)$. (Remember $\sigma < 0$)

Also, $\pi^{(s-1/2)}$ is bounded by $\pi^{-(-\sigma+1/2)} = \dfrac1{\pi^{-\sigma+1/2}}$. From Stirling's formula, we have that $$\Gamma(s+1) \sim \sqrt{2 \pi s} \left(\dfrac{s}{e} \right)^s$$for a fixed $\sigma < 0$ and large $|t|$. Note that Stirling's formula is valid when the imaginary part of $s$ is large, fixing the real part i.e. when $t$ is sufficiently bounded away from $0$ for a fixed $\sigma < 0$.

Now lets turn our attention to the ratio of $\Gamma$ functions. We, from now on, assume that we fix $\sigma < 0$ and let $t \rightarrow \infty$. This is what we mean by $\lvert s \rvert \rightarrow \infty$. $$\Gamma \left( \dfrac{1-s}{2}\right) = \Gamma \left( - \left(\dfrac{1+s}{2} \right) + 1\right) \sim \sqrt{-2 \pi \left( \dfrac{1+s}{2}\right)} \times \left( - \left(\dfrac{1+s}{2e} \right) \right)^{- \left(\dfrac{1+s}{2} \right)}$$

$$\Gamma \left( \dfrac{s}{2}\right) = \Gamma \left( \left(\dfrac{s}{2} -1\right) + 1\right) \sim \sqrt{2 \pi \left( \dfrac{s}{2} - 1\right)} \times \left( \dfrac{s/2-1}{e}\right)^{s/2-1}$$

$$\dfrac{\Gamma \left( \dfrac{1-s}{2}\right)}{\Gamma \left( \dfrac{s}{2}\right)} \sim \sqrt{\dfrac{1+s}{2-s}} \dfrac{\left( - \left(\dfrac{1+s}{2e} \right) \right)^{- \left(\dfrac{1+s}{2} \right)}}{\left( \dfrac{s-2}{2e}\right)^{s/2-1}}$$

Hence, $$\left \lvert \dfrac{\Gamma \left( \dfrac{1-s}{2}\right)}{\Gamma \left( \dfrac{s}{2}\right)} \right \rvert \sim \left \lvert \sqrt{\dfrac{1+s}{2-s}} \dfrac{\left( - \left(\dfrac{1+s}{2e} \right) \right)^{- \left(\dfrac{1+s}{2} \right)}}{\left( \dfrac{s-2}{2e}\right)^{s/2-1}} \right \rvert$$

For a fixed $\sigma < 0$, where $s = \sigma + it$, as $t \rightarrow \infty$, we have that $$\left \lvert \sqrt{\dfrac{1+s}{2-s}} \right \rvert \rightarrow 1$$ $$\left \lvert -\left(\dfrac{1+s}{2e} \right)\right \rvert \sim \left \lvert \left(\dfrac{s}{2e} \right)\right \rvert$$ $$\left \lvert \left(\dfrac{s-2}{2e} \right)\right \rvert \sim \left \lvert \left(\dfrac{s}{2e} \right)\right \rvert$$ Hence, we get that $$\left \lvert \dfrac{\Gamma \left( \dfrac{1-s}{2}\right)}{\Gamma \left( \dfrac{s}{2}\right)} \right \rvert \sim \left \lvert \dfrac{\left \lvert\dfrac{s}{2e} \right \rvert^{-(1+s)/2}}{\left \lvert\dfrac{s}{2e} \right \rvert^{s/2-1}} \right \rvert \sim \left \lvert \left \lvert \dfrac{s}{2e}\right \rvert ^{\left(1/2 - s \right)} \right \rvert \sim \left \lvert \dfrac{s}{2e}\right \rvert ^{\left(1/2 - \sigma \right)} \sim \left \lvert \dfrac{t}{2e}\right \rvert ^{\left(1/2 - \sigma \right)}$$

Hence, putting all this together, we get that $$\left \lvert \zeta(s) \right \rvert = \left \lvert \pi^{(s-1/2)}\zeta(1-s) \dfrac{\Gamma \left( \dfrac{1-s}{2} \right)}{\Gamma \left( \dfrac{s}{2} \right)} \right \rvert = \left \lvert \pi^{(s-1/2)} \right \rvert \left \lvert \zeta(1-s) \right \rvert \left \lvert \dfrac{\Gamma \left( \dfrac{1-s}{2} \right)}{\Gamma \left( \dfrac{s}{2} \right)} \right \rvert$$ Assembling all together, we get that $$\left \lvert \zeta(s) \right \rvert \ll \pi^{\sigma - 1/2} \zeta(1 - \sigma) \left \lvert \dfrac{t}{2e}\right \rvert ^{\left(1/2 - \sigma \right)} = \zeta(1 - \sigma) \left \lvert \dfrac{t}{2 \pi e}\right \rvert ^{\left(1/2 - \sigma \right)}$$

This is what you wanted to prove. You also get some idea of how the constants grow with $\sigma$, where $\sigma < 0$.

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You mean $|s| \rightarrow \infty$, right? cause when $t\rightarrow \infty$ and $\sigma$ is finite, $|s|\rightarrow \infty$. – MathematicalPhysicist Jun 3 '12 at 10:05
@MathematicalPhysicist Yes :). It was a typo. I have changed it now. – user17762 Jun 3 '12 at 14:09

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