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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?

mathseeker
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    At first glance, this just seems like $\Gamma(x+a)=\Gamma(x)\cdot(x+1)\cdot(x+2)\cdots(x+a)$, and since $x\gg a$, we would have $x+a\approx x$, which would lead to the approximation you posted. – Raad Shaikh Mar 26 '22 at 15:35
  • It says there in the article that it's valid for $\alpha\in\mathbb{C}$. – G Tony Jacobs Mar 26 '22 at 15:35
  • @RaadShaikh Thank you. That makes sense. – mathseeker Mar 26 '22 at 15:45
  • @GTonyJacobs I am interested to know if it works when both $x$ and $\alpha$ are real – mathseeker Mar 26 '22 at 15:47
  • You can use the Stirling's approximation: $\Gamma(x+1)=\sqrt{2\pi x}\big(\frac{x}{e}\big)^x\Big(1+O\big(\frac{1}{x}\big)\Big)$. $$\lim_{x\to\infty}\frac{\Gamma(x+a)}{\Gamma(x),x^a}=\lim_{x\to\infty}\sqrt\frac{2\pi(x+a-1)}{2\pi(x-1)}\frac{\big(\frac{x+a-1}{e}\big)^{x+a-1}}{\big(\frac{x-1}{e}\big)^{x-1}x^a}$$ $$=\lim_{x\to\infty}\big(\frac{x+a-1}{e}\big)^{a}\big(1+\frac{a}{x-1}\big)^{x-1}\frac{1}{x^a}=\lim_{x\to\infty}\frac{(x+a-1)^a}{x^a,e^{a}}e^{(x-1)\ln(1+\frac{a}{x-1})}=1$$ – Svyatoslav Mar 26 '22 at 16:12
  • @mathseeker Real numbers are also part of $\mathbb{C}$. If it works for complex $\alpha$, then it works for real $\alpha$, because $\mathbb{R}\subset\mathbb{C}$. Additionally, $x$ is real in the formula, as stated in the article where the formula is presented. – G Tony Jacobs Mar 27 '22 at 13:03

2 Answers2

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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 $$ \Gamma(x+\alpha) \underset{x\rightarrow +\infty}{\sim} \sqrt{2\pi x}\left(\frac{x}{e}\right)^x x^{\alpha}\underset{x\rightarrow +\infty}{\sim}\Gamma(x)x^{\alpha} $$

Tuvasbien
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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.

Claude Leibovici
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