Talk:Gamma function

Proof of equivalence of the three definitions
Is there value in using this instead or additionally?: $$\begin{align} & \int_0^\infty t^{z-1} e^{-t} dt \\ & = \lim_{n\rightarrow\infty} \int_0^{n+1} t^{z-1} \left(1-\frac{t}{n+1}\right)^n dt \\ & = \lim_{n\rightarrow\infty} \frac{1}{z(z+1)\ldots(z+n-1)}\frac{n!}{(n+1)^n} \int_0^{n+1} t^{z+n-1} \left(1-\frac{t}{n+1}\right)^0 dt \\ & = \lim_{n\rightarrow\infty} \frac{n!}{z(z+1)\ldots(z+n-1)}\frac{1}{(n+1)^n} \frac{(n+1)^{z+n}}{z+n} \\ & = \lim_{n\rightarrow\infty} \frac{n!}{z(z+1)\ldots(z+n)} (n+1)^{z} \\ & = \frac{1}{z} \prod_{n=1}^{\infty} \frac{(1+1/n)^z}{1 + z/n} \end{align}$$ where the second equality is integration by parts $n$ times. — Q uantling (talk &#124; 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? $$\begin{align} \Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt\\ \Gamma(x)=\int_{0}^{1}(-\ln(t))^{x-1}dt\\ \Gamma(x)=\lim_{n\to\infty}n^{x-1}\prod\limits_{k=1}^{n}\frac{1}{k+x-1} \end{align}$$ 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:
 * $$\lim_{n \to \infty} \frac{n! \, \left(n+1\right)^z}{(n+z)!} = 1$$
 * $$\lim_{n \to \infty} \frac{1}{z} \prod_{k=1}^n \left[ \frac{1}{1+\frac{z}{k}} \left(1 + \frac{1}{k}\right)^z \right] = \Gamma(z)$$ — Q uantling (talk &#124; contribs) 16:41, 6 November 2023 (UTC)

log vs ln
I don't care much about which, but the article should be consistent. Do we want to use "$log$" or "$ln$" to indicate a natural logarithm? We could use one or the other throughout. Or, if we mostly go with "$ln$", we could nonetheless use "$log$" in those fewer cases where the expression works regardless of the base of the logarithm. What do you prefer? — Q uantling (talk &#124; contribs) 17:06, 10 June 2024 (UTC)


 * I've made it consistent. Since I had to choose one, I chose log, with a notation on first use that it is the natural logarithm. I think this is the more common style for mathematical analysis and analytic number theory, the main subfields of mathematics for this topic. —David Eppstein (talk) 18:18, 10 June 2024 (UTC)

Equation appears wrong
The equation immediately following the words "Laplace transform" appears wrong.

The number log(π) has a positive sign in the previous equation, on the right of the equals sign.

So when it is brought to the left of the equals sign as in the equation following "Laplace transform", it should have a negative sign.

But it does not.

If this observation is correct, I hope someone familiar with this subject can fix this.