Some asymptotics for zeta function. 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.
 A: 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 $\sigma < 0$.
