Stirling's Approximation of Gamma Function

Question

In the wikipedia page of Gamma Function here, it is stated that, when $x \to \infty $,

$$\Gamma(x+\alpha) = \Gamma(x)x^\alpha$$

1. Is it valid for both real $x$ and real $\alpha$ ?
2. Could you please provide the proof of this approximation?
3. Is there any book/ paper from here I can cite this particular result?

Thank you very much

Update: Thank you for your answers. Is it also possible to get the proof for lower and upper bound for this approximation?

Answer 1

This is a consequence of Stirling approximation : $$ \begin{aligned} \Gamma(x+\alpha) &\underset{x\rightarrow +\infty}{\sim}\sqrt{2\pi(x+\alpha)}\left(\frac{x+\alpha}{e}\right)^{x+\alpha} \\ &\underset{x\rightarrow +\infty}{\sim}\sqrt{2\pi x}\left(\frac{x+\alpha}{e}\right)^x x^{\alpha}e^{-\alpha} \\ \end{aligned} $$ Moreover, $$ (x+\alpha)^x e^{-\alpha}=x^x\exp\left(x\log\left(1+\frac{\alpha}{x}\right)-\alpha\right)=x^x e^{o(1)}\underset{x\rightarrow +\infty}{\sim}x^x $$ Thus $$ \Gamma(x+\alpha) \underset{x\rightarrow +\infty}{\sim} \sqrt{2\pi x}\left(\frac{x}{e}\right)^x x^{\alpha}\underset{x\rightarrow +\infty}{\sim}\Gamma(x)x^{\alpha} $$

Answer 2

Using Stirling approximation $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \left(\log \left(\frac{1}{p}\right)+\log (2 \pi )\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ make $p=(x+\alpha)$ and continue with Taylor series for large values of $x$. This would give $$\log (\Gamma (x+\alpha))-\log (\Gamma (x))=\alpha \log (x)+\frac{(\alpha -1) \alpha }{2 x}-\frac{(\alpha -1) \alpha (2 \alpha -1)}{12 x^2}+O\left(\frac{1}{x^3}\right)$$ Now, using $$\frac{ \Gamma (x+\alpha)} {\Gamma (x) }=e^{\log (\Gamma (x+\alpha))-\log (\Gamma (x))}$$ $$\frac{ \Gamma (x+\alpha)} {\Gamma (x) }=x^{\alpha } \left(1+\frac{(\alpha -1) \alpha }{2 x}+\frac{(\alpha -2) (\alpha -1) \alpha (3 \alpha -1)}{24 x^2}+O\left(\frac{1}{x^3}\right)\right)$$ and then the very first approximation that you wrote.