# Stirling's formula and a Gamma function relation

I am trying to understand a paper by by A. Booker on poles of Artin $$L$$-functions where in one of the lemmas he uses the following identity, derived using Stirling's formula: $$\frac{\Gamma(s/2)^2}{2^{-s}\Gamma(s-1/2)} = \sqrt{8\pi}\left( 1+\frac{c_1}{s}+\ldots \frac{c_n}{s^n}+O\left(\frac{1}{s^{n+1}}\right)\right),\qquad \Re s\geq 1$$ (for certain numbers $$c_k$$).

I don't quite understand how they got this, can anyone help?

• Write this as $\exp(2\log \Gamma(s/2) + s \log 2 - \log \Gamma(s - 1/2))$ and use Stirling's formula in the form $\log \Gamma(s) = (s - 1/2) \log s - s + (1/2) \log 2\pi + \cdots$, where there are lower order terms involving powers of $1/s$. Then use the fact that $\exp(x) = 1 + x + x^2/2! + \cdots$. Feb 23, 2022 at 0:34

Denote your fraction by $$R(s)$$. By the duplication formula, $$\Gamma\! \left( {s - \frac{1}{2}} \right) = \Gamma \!\left( {2\left( {\frac{s}{2} - \frac{1}{4}} \right)} \right) = \frac{{2^{s - 3/2} }}{{\sqrt \pi }}\Gamma\! \left( {\frac{s}{2} - \frac{1}{4}} \right)\Gamma \!\left( {\frac{s}{2} + \frac{1}{4}} \right).$$ Thus, $$R(s) = \sqrt {8\pi } \frac{{\Gamma ^2 \!\left( {\frac{s}{2}} \right)}}{{\Gamma\! \left( {\frac{s}{2} - \frac{1}{4}} \right)\Gamma\! \left( {\frac{s}{2} + \frac{1}{4}} \right)}}.$$ Now for any fixed complex $$h$$, we have $$\log \Gamma (z + h) \sim \left( {z + h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) + \sum\limits_{n = 1}^\infty {\frac{(-1)^{n+1}{B_{n + 1} (h)}}{{n(n + 1)z^n }}}$$ as $$z\to \infty$$ in $$|\arg z|\le \pi-\delta<\pi$$. Here $$B_n(h)$$ denotes the Bernoulli polynomials. Thus, \begin{align*} \log \frac{{R(s)}}{{\sqrt {8\pi } }} & \sim \sum\limits_{n = 1}^\infty {\frac{{(-1)^{n + 1}2^n (2B_{n + 1} (0) - B_{n + 1} (-\frac{1}{4}) - B_{n + 1} ( \frac{1}{4}))}}{{n(n + 1)s^n }}} \\ & = - \frac{1}{{8s}} - \frac{1}{{8s^2 }} - \frac{{17}}{{192s^3 }} - \ldots \end{align*} as $$s\to \infty$$ in $$|\arg s|\le \pi-\delta<\pi$$. Finally, by taking the exponential of each side and expanding the right-hand side in negative powers of $$s$$, we deduce $$R(s) \sim \sqrt {8\pi } \left( {1 - \frac{1}{{8s}} - \frac{{15}}{{128s^2 }} - \frac{{75}}{{1024s^3 }} - \ldots } \right)$$ as $$s\to \infty$$ in $$|\arg s|\le \pi-\delta<\pi$$.