Asymptotic for the gamma function on vertical lines

401 Views Asked by At

On page 135 of Joerg Bruedern's "Einfuehrung in die analytische Zahlentheorie" he claims that Stirling's formula implies for fixed $\sigma <0$, any $t\geq 1$, and some constant $C$ (I assume depending on $\sigma $ but not on $t$)

$$\Delta (\sigma +it)=Ce^{-it\log t+it(1+\log 2\pi )}t^{1/2-\sigma }+\mathcal O(t^{-1/2-\sigma }),$$

where $\Delta (s)$ is "the factor in the functional equation for the Riemann Zeta Function", which I assume means

$$\Delta (s)=\frac {\pi ^{s-1/2}\Gamma ((1-s)/2)}{\Gamma (s/2)}.$$

Stirling's formula says that for fixed $\delta >0$ and $\text {arg}(z)\leq \pi -\delta $

$$\log \Gamma (z)=(z-1/2)\log z-z+\frac {\log 2\pi }{2}+\mathcal O(1/|z|).$$

How does the claim follow from this? I've tried to expand $z^{z-1/2}$ by writing it in terms of the real and imaginary parts of $z$ and using the definition of complex powers, but my answer always seems far from a simple formula like above. In particular I always have terms of the form

$$e^{i\text { arg}(z)}$$

which I don't know what I should do with.