Inspired by a problem of my book wich is :
Let $0<x$ then we have :
$$f(x)=\left(x!\right)\left(\frac{1}{x}\right)!\geq 1$$
I have tried several function (log,LambertW,$x^a/x+1$,...) but now I found it
I have tried to find a Stirling Formula for $f(x)$ and accordingly to Wolfram alpha we have :
$$\lim_{x\to\infty}\left(\left(x!\right)\left(\frac{1}{x}\right)!\right)^{\frac{1}{x}}-\left(e^{-1}x+\sinh^{-1}\left(x\right)e^{-\left(1+\ln\left(2\right)\right)}+\frac{\ln\left(\pi\right)}{2e}\right)=0$$
Where $\sinh^{-1}$ is the inverse function of the more common $\sinh(x)$ e.g $$\sinh^{-1}(\sinh(x))=x$$
I have some questions :
1.How to check the limit ?
2.Is it new or just trivial ?
3.Have you any reference ?
5.Can we add some other term like in a power series ?
6.Is it better than the Stirling formula ?
Thanks !
Ps:I don't want to show the first inequality I just speak about a possible new Stirling formula for this product .
For large values of $x$ $$F=\frac 1 x \log\Bigg[\left(x!\right)\left(\frac{1}{x}\right)!\Bigg]=\log(x)-1+\frac {\log(2 x)+\log(\pi)}{2x}+O\left(\frac{1}{x^2}\right)$$ is a very good approximation even for small values of $x$.
On the other side,
$$G=\log\left(e^{-1}x+\sinh^{-1}\left(x\right)e^{-\left(1+\log \left(2\right)\right)}+\frac{\ln\left(\pi\right)}{2e}\right)=\log(x)-1+\frac{\log(2x)+2e \log \left(\frac{\pi }{2 e}\right)}{2x}+O\left(\frac{1}{x^2}\right)$$
$$F-G=\frac{\log(\pi)-2e \log \left(\frac{\pi }{2 e}\right) }{2x}+O\left(\frac{1}{x^2}\right)$$
Computing exactly for $x=123$, $F=3.839184921$, $G=3.822391606$ then $F-G=0.016793$ while the above gives $F-G=0.016773$
Edit
A more detailed expansion of $F$ is $$F=\log \left(\frac{x}{e}\right)+\frac {\log(2 \pi x)}{2x}-\frac{12\gamma-1}{12x^2}+\frac {\pi^2}{12x^3}-\frac{120\zeta(3)+1}{360x^4}+\frac {\pi^4}{360x^5}+O\left(\frac{1}{x^6}\right)$$