Inverse of the Gamma Function

Question

I've come across an equation that needed use of the gamma function's functional inverse, and I know that you can't represent the functional inverse of the gamma function by any combination of elementary functions. I thought that there might be an approximation out there that I could use. However, I couldn't find any useful approximations of the inverse. Is there any chance someone might have one under their belt?

Answer 1

I tried approaching it different ways, and one of them did, in fact, work out. For the first method, I tried inverting Stirling's approximation, but that didn't work. For the second, I tried rearranging terms in Stirling approximation and then inverting it, but that also turned up nothing. For method #3, I used a representation of $Γ(x+\frac12)$ and substituted all the gamma functions for Stirling's approximation and then tried to invert it, which worked.

METHOD #3: $Γ(x+\frac12)=\frac{Γ(2x+1)\sqrt{\pi}}{4^xΓ(x+1)}$

$Γ(x+1)\approx\sqrt{2\pi x}(\frac{x}{e})^x$

$Γ(x+\frac12)\approx\frac{\sqrt{4\pi x}{(\frac{2x}{e})}^{2x}\sqrt{\pi}}{4^x\sqrt{2\pi x}(\frac{x}{e})^x}=\frac{{2}^{2x}\sqrt{2\pi}{(\frac{x}{e})}^{2x}}{4^x(\frac{x}{e})^x}=\sqrt{2\pi}(\frac{x}{e})^x$

So $Γ(x+\frac12)\approx\sqrt{2\pi}(\frac{x}{e})^x$, which means that $Γ(x)\approx\sqrt{2\pi}{(\frac{x-\frac12}{e})}^{x-\frac12}$

Inverting...

$x=\sqrt{2\pi}{(\frac{y-\frac12}{e})}^{y-\frac12}$

$\frac{\ln(\frac{x}{\sqrt{2\pi}})}{e}=\frac{y-\frac12}{e}\ln(\frac{y-\frac12}{e})$

$W(\frac{\ln(\frac{x}{\sqrt{2\pi}})}{e})=\ln(\frac{y-\frac12}{e})$

${e}^{W(\frac{\ln(\frac{x}{\sqrt{2\pi}})}{e})+1}=y-\frac12$

${e}^{W(\frac{\ln(\frac{x}{\sqrt{2\pi}})}{e})+1}+\frac12=y$

So, an approximation for the functional inverse of the gamma function is ${e}^{W(\frac{\ln(\frac{x}{\sqrt{2\pi}})}{e})+1}+\frac12$, where $W(x)$ is the Lambert W Function.