Talk:Gamma function

Proof of Weierstrass definition

In the proof of the Weierstrass definition, ${\textstyle \Gamma (s)=\lim _{n\to \infty }{\frac {n!}{s\left(s+1\right)...\left(s+n\right)}}n^{s}}$ is stated without proof. While it is obvious (to me) that this is true when s is a positive integer, I see no way to prove this for non-integers without invoking one of the previous definitions of the gamma function. In particular, I think that this equality needs to be assumed as part of the Weierstrass definition. If I am right, I think we need to say this explicitly. —Quantling (talk | contribs) 14:26, 23 May 2023 (UTC)

I made some edits to address this and some other issues. Please take a look. —Quantling (talk | contribs) 15:56, 23 May 2023 (UTC)

The integral definition gives the Weierstrass product. Then the Weierstrass product gives

$\Gamma (z)=\lim _{n\to \infty }{\frac {n!}{z\left(z+1\right)\cdots \left(z+n\right)}}n^{z}.$ A1E6 (talk) 17:42, 23 May 2023 (UTC)

If you would replace the current proof of the Weierstrass definition with one that is explicitly based upon the integral, that'd be great! (As it is, we are instead assuming about the asymptotic behavior of Γ(z) as the real part of z goes to infinity.) We could then have the Euler's infinite product definition be based upon, and subsequent to, the Weierstrass definition. (Unless, the proof that starts with the integral goes via the infinite product, in which case, maybe we'd keep the current order.) Thanks —Quantling (talk | contribs) 18:23, 23 May 2023 (UTC)

Page formatting in Mobile Site

In the initial description at the start of the page the short description is in the middle of the formula of the gamma function in the mobile version of the site. This does not seem to be the case in the desktop site. Unanimous350 (talk) 18:45, 27 May 2023 (UTC)

Proof of equivalence of the three definitions

Is there value in using this instead or additionally?:

$${\begin{aligned}&\int _{0}^{\infty }t^{z-1}e^{-t}dt\\&=\lim _{n\rightarrow \infty }\int _{0}^{n+1}t^{z-1}\left(1-{\frac {t}{n+1}}\right)^{n}dt\\&=\lim _{n\rightarrow \infty }{\frac {1}{z(z+1)\ldots (z+n-1)}}{\frac {n!}{(n+1)^{n}}}\int _{0}^{n+1}t^{z+n-1}\left(1-{\frac {t}{n+1}}\right)^{0}dt\\&=\lim _{n\rightarrow \infty }{\frac {n!}{z(z+1)\ldots (z+n-1)}}{\frac {1}{(n+1)^{n}}}{\frac {(n+1)^{z+n}}{z+n}}\\&=\lim _{n\rightarrow \infty }{\frac {n!}{z(z+1)\ldots (z+n)}}(n+1)^{z}\\&={\frac {1}{z}}\prod _{n=1}^{\infty }{\frac {(1+1/n)^{z}}{1+z/n}}\end{aligned}}$$

where the second equality is integration by parts n times. —Quantling (talk | contribs) 22:42, 27 May 2023 (UTC)

This proves it only for $\Re (z)>0$. And as noted by Plusjeremy, that alone is insufficient for proving the reflection formula. Can you show that the infinite product is analytic for $z\in \mathbb {C} \setminus \mathbb {Z} _{0}^{-}$? (We're assuming the integral definition together with $\Gamma (z+1)=z\Gamma (z)$). A1E6 (talk) 09:18, 28 May 2023 (UTC)

I suggest adding it. Hawkeye7 (discuss) 00:02, 28 May 2023 (UTC)

are all these definitions shown?

are all these defenitions of $\Gamma (x)$ shown?

$${\begin{aligned}\Gamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}dt\\\Gamma (x)=\int _{0}^{1}(-\ln(t))^{x-1}dt\\\Gamma (x)=\lim _{n\to \infty }n^{x-1}\prod \limits _{k=1}^{n}{\frac {1}{k+x-1}}\end{aligned}}$$

Erikgobrrrr (talk) 21:50, 1 October 2023 (UTC)

Euler's definition as an infinite product

The reference for Euler's infinite product says that the limit goes to n!, not 1. 104.187.53.82 (talk) 16:21, 6 November 2023 (UTC)

I believe that you are talking about two different expressions. Both of these are true:

$$\lim _{n\to \infty }{\frac {n!\,\left(n+1\right)^{z}}{(n+z)!}}=1$$

$$\lim _{n\to \infty }{\frac {1}{z}}\prod _{k=1}^{n}\left[{\frac {1}{1+{\frac {z}{k}}}}\left(1+{\frac {1}{k}}\right)^{z}\right]=\Gamma (z)$$

—Quantling (talk | contribs) 16:41, 6 November 2023 (UTC)