Relation between $\Gamma(1/2-x)$ and $\Gamma(-x)$ for small $x$

Question

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.

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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 $\Gamma(x+1)=x\Gamma(x)$.

Answer 1

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...

Answer 2

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 $(2)$.