# Deriving a particular form of Stirling's Approximation of the Gamma function?

I am working from some notes concerning bounds on Dedekind zeta functions and am trying to derive a supposed version of Stirling's approximation contained therein:

Let $$\sigma$$ be fixed and $$|t|\rightarrow \infty$$, then a version of Stirling's approximation formula yields $$|\Gamma(\sigma+it)| \sim |t|^{\sigma-\tfrac{1}{2}}e^{-\tfrac{\pi}{2}|t|}\ll |t|^{\sigma-\tfrac{1}{2}}e^{-|t|}$$ where we define $$f(z)\ll g(z) \text{ iff } \exists c\in \mathbb{R}: \exists z_0\in \mathbb{C}: \forall |z|>|z_0|: |f(z)|

I started working from what appears to be the standard rendering of Stirling: $$\Gamma(s)=\frac{\sqrt{2\pi}s^{s-\frac{1}{2}}}{e^s}\big(1+O(\tfrac{1}{s})\big) \qquad \underset{\text{Since }O\big(\tfrac{1}{s}\big)\ll \tfrac{1}{s} \text{ by definition}}{\Rightarrow} \qquad \Gamma(s)\ll \Big|\frac{\sqrt{2\pi}s^{s-\frac{1}{2}}}{e^s}\Big|\cdot (1+|\tfrac{1}{s}|)$$ But making the relevant substitutions, shuffling things around, and ignoring the negligible contribution of $$\sigma$$ to the modulus when applicable, I have only been able to find $$|\Gamma(\sigma+it)|\ll |t|^{\sigma-\tfrac{1}{2}}+|t|^{\sigma-\tfrac{3}{2}}$$

• There should indeed be a decay behavior for large $t$, for example $|\Gamma(10+100i)|$ is less than $10^{-48}$. Intuitively this is caused by cancellations in the integral defining $\Gamma(\sigma+it)$ when $|t| \gg \sigma$. – Ian Jan 4 '20 at 0:06
• @Ian I see what you mean and have edited the question. I will update if I find the answer you hint at. – Gary D Jan 4 '20 at 0:29
• It comes from $z^z=\exp(z \log(z))$. When $|t/\sigma|$ is large, $\log(z)$ has imaginary part close to $\pm \pi/2$ (with the same sign as $t$ has), so $z^z$ has a modulus not much bigger than $e^{-\pi/2 |t|}$. – Ian Jan 4 '20 at 0:51
• @Ian Ah, thank you! If you care to write your comment in answer form I will mark it as accepted and upvote. – Gary D Jan 4 '20 at 0:55

The decay comes from $$z^z=\exp(z \log(z))$$. When $$|t/\sigma|$$ is large, $$\log(z)$$ has imaginary part close to $$\operatorname{sign}(t) \pi/2$$, so that the real part of the exponent behaves like $$\sigma \log(\sigma) - \pi/2 |t|$$ which is dominated by the second term, giving the decay.