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I want to find the relation between $\Gamma(\frac{1}{2}-x)$ and $\Gamma(-x)$ for small $x$.

For large $x$, we can use asymptotic expansion, i.e., $\Gamma(x+a) \sim \Gamma(x) x^{a}$, here I am considering small $x$ limit.

For integer difference i.e., $\Gamma(x+n)$ and $\Gamma(x)$, there is a nice relation

\begin{align} \frac{\Gamma\left(x+n\right)}{\Gamma\left(x\right)}&=\prod_{k=0}^{n-1}{\left(x+k\right)} \end{align} which comes from the properties of $\Gamma(x+1)=x\Gamma(x)$.

Gary
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2 Answers2

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For small values of $x$ $$\Gamma \left(\frac{1}{2}-x\right)=\sqrt{\pi }+\sqrt{\pi } (\gamma +2\log (2))x+O\left(x^2\right)\tag 1$$ $$\Gamma \left(-x\right)=-\frac{1}{x}-\gamma -\frac{6 \gamma ^2+\pi ^2}{12} x+O\left(x^2\right)\tag2 $$

Solve $(1)$ for $x$ and plug in $(2)$.

Claude Leibovici
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    You are fast Claude (I was editing the expansion of $;\displaystyle f(x):=\frac {\Gamma(1-x);\Gamma\left(\frac 12\right)}{\Gamma\left(\frac 12-x\right)}$ starting with $1-\log(4)x+\cdots$ :-) Cheers, – Raymond Manzoni May 25 '23 at 09:05
  • @RaymondManzoni. Hi Raymond ! How are you ? – Claude Leibovici May 25 '23 at 09:08
  • In vacation and fine thank you! And you? – Raymond Manzoni May 25 '23 at 09:09
  • @RaymondManzoni. Good to know ! I shall be on the billard on coming tuesday; the question is : where next ? (I enjoy dark humor but who knows ?). Cheers :-) – Claude Leibovici May 25 '23 at 09:18
  • Sorry, wishing you all the best for your operation. And I know you'll use mathematics to be where you wish... Cheers, – Raymond Manzoni May 25 '23 at 09:22
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    @RaymondManzoni. Once upon a time, I was in my room at hosptial and the surgeon came in telling : let(s go, it is time ! I replied : five more minuts, I have to look at a result on my laptop. This is not a joke and everybody waited for me. – Claude Leibovici May 25 '23 at 09:25
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    I'll trust you on this and else! There is a well known story about Hardy who had to take a boat to Denmark in stormy weather. He sent a postcard claiming that he had proved the Riemann hypothesis reasoning that God would not allow him to die which such a fame! :-) – Raymond Manzoni May 25 '23 at 09:34
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We may start with $\;\displaystyle f(x):=-x\frac {\Gamma\left(\frac 12\right)\,\Gamma(-x)}{\Gamma\left(\frac 12-x\right)}\;$ (with $\Gamma\left(\frac 12\right)=\sqrt{\pi}\;$) and obtain the expansion : $$f(x)=1-2\log(2)\,x-\left(\zeta(2)-2\log(2)^2\right)x^2+\left(\pi^2\log(2)-4\log(2)^3-6\zeta(3)\right)\frac {x^3}3+O\left(x^4\right)$$

To get more terms you may use $\;\displaystyle f(x)=-x\,\operatorname{B}\left(-x,\frac 12\right)\;$ with $\operatorname{B}$ the beta function (see for example this Alpha expansion). There is also a formula by Wolfram functions $\left(a=-x,b=\frac 12\right)$

A simpler relation linking directly $\Gamma\left(\frac 12-x\right)$ and $\Gamma\left(-x\right)\,$ appears unlikely from the known relations in A&S.
The duplication formula $6.1.18$ allows us to write their product as $\;\displaystyle\sqrt{\pi}\;2^{2x+1}\;\Gamma(-2x)\;$ and you may obtain simple expansions by considering their fraction with one of them squared and so on...

Raymond Manzoni
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