By drawing a graph of the geometric derivative of $x!$, $e^{\left(\frac{\text{d}ln(x!)}{\text{d}x}\right)}$, i guessed that $e^{\left(\frac{\text{d}ln(x!)}{\text{d}x}\right)}\sim_{+\infty}(x+1/2)$.
Then I applied the reverse operation on the equivalent ($x+\frac{1}2$) found above:
geometric integral of $(x+1/2)$ : $e^{\int ln(x+1/2) \text{d}x}=\frac{(x+1/2)^{\left(x+1/2\right)}}{e^{\left(x+1/2\right)}}$
We can then prove using Stirling's approximation $x!\sim_{+\infty}\left(\frac{x}{e}\right)^x\sqrt{2\pi x}$ that $\frac{(x+1/2)^{\left(x+1/2\right)}}{e^{\left(x+1/2\right)}}\sqrt{2\pi}\sim_{+\infty}x!$.
However, i would like to show that $e^{\left(\frac{\text{d}ln(x!)}{\text{d}x}\right)}\sim_{+\infty}(x+1/2)$ with a mathematical proof and not just graphical observations.
How could I prove it ?
Knowing the asymptotic properties of the function digamma, the proof is rather easy :
Ref.: Eq.(16) in http://mathworld.wolfram.com/DigammaFunction.html