$\frac{\Gamma(\frac{1}{2}s)}{\Gamma(\frac{1-s}{2})} = O(|t|^{\sigma - \frac{1}{2}})$

Question

I want to prove this formula using Stirling's approximation or otherwise:$\frac{\Gamma(\frac{1}{2}s)}{\Gamma(\frac{1-s}{2})} = O(|t|^{\sigma - \frac{1}{2}})$ wher $s=\sigma+it$.

Can someone please answer here or give me a reference for the above formula so that I can study its proof?

edit In this answer (Stirling formula on $\frac{\gamma(\frac{1}{2}s)}{\gamma(\frac{1-s}{2})} = O(|t|^{\sigma - \frac{1}{2}})$) we assume that $\sigma$ is bounded. What if $\sigma$ is only bounded from above, so what if $\sigma<0$?

Answer 1

Ok, you need the following points:

1. $\Gamma(\frac{1}{2}+z)\Gamma(\frac{1}{2}-z)=\frac{\pi}{\cos(\pi z)}$;
2. $\cos(iz)= \cosh(z)$;
3. $\Gamma(2z)=\frac{2^{(2z-1)}}{\sqrt(\pi)} \Gamma(z) \Gamma(z+\frac{1}{2})$ - Legendre's duplication formula;
4. Stirling's formula $\Gamma(z)\sim\sqrt{\frac{2\pi}{z}}e^{z(\ln(z)-1)}$, $|\arg z| < \pi$;
5. $\ln(\sigma + it) = \ln\sqrt{\sigma^2 + t^2} + i\,{\arg}(\sigma + it) \sim \ln|t| + \frac{\pi }{2} i\,{\rm sign}(t)$.

So, $$\frac{\Gamma(s/2)}{\Gamma(1/2-s/2)} = \frac{2^{1-s}}{\sqrt{\pi}}\cos(\pi s/2) \Gamma(s) = O\left[\cosh\left(\frac{\pi}{2}(t-i\sigma)\right) \Gamma(\sigma + it)\right] = O \left( e^{\pi |t|/2}\Gamma(\sigma + it)\right) \\ = O\left[ \exp\left( \pi |t|/2 + \left(\ln|t|+\frac{\pi i}{2}{\rm sign}(t)-1\right)(\sigma+it) - |t|/2\right) \right] = O(e^{(\ln|t|-1)\sigma-|t|/2}) = O(|t|^{\sigma-1/2}).$$, where $t\to\pm\infty$, $\sigma$ fixed.

It looks it holds for any $\sigma$. It follows from the fact that $\,{\arg}(\sigma + it) \sim \frac{\pi }{2} i\,{\rm sign}(t)$ which does not depend on $\sigma$.

In the common case when $\sigma \to \pm\infty$, $t\to \pm\infty$ we have again:

$$\cos(\pi/2(\sigma+it)) = \cos(\pi\sigma/2)\cosh(\pi t/2) - i \sin(\pi\sigma/2) \sinh(\pi t/2) = O(e^{\pi|t|/2}),$$ where $t\to \pm\infty$ and any $\sigma$.

But now $$\Gamma(s)=\Gamma(\sigma+it)= O\left[\exp(s(\ln s-1)-\frac{1}{2}\ln s)\right] = O\left[\exp\left( (\sigma+ it)(\ln \sqrt{\sigma^2+t^2}+i\arg(\sigma+it)-1)-\frac{1}{2}\ln\sqrt{\sigma^2+t^2} \right)\right] = O\left[\exp\left((\sigma - \frac{1}{2})\ln\sqrt{\sigma^2+t^2}-\sigma-t\arg(\sigma+it)\right)\right].$$

Multiplying by $2^{1-s}=O(2^{-\sigma})=O(e^{-\sigma\ln2})$ we have

$$ \frac{\Gamma(s/2)}{\Gamma(1/2-s/2)} = O\left((\sigma^2 +t^2)^{\frac{\sigma-1/2}{2}} e^{\pi|t|/2-t\arg(\sigma+it)-\sigma(1+\ln2)}\right).$$

From here we obtain

1. $\sigma$ is fixed, $t\to \pm \infty$ (result already obtained above): $$O(|t|^{\sigma-1/2});$$

2. $t\ne0$ is fixed, $\sigma\to \pm \infty$: $$O(|\sigma|^{\sigma-1/2}e^{-\sigma(1+\ln2)}).$$