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If you ask someone to evaluate $\left(-\frac{1}{2}\right)!$ they would probably provide you with $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$. Similarly, $\left(-\frac{3}{4}\right)!$ would get called $\Gamma\left(\frac{1}{4}\right)=\sqrt[4]{\pi}\sqrt{\sqrt{2}-1}\frac{\zeta\left(\frac{3}{4}\right)}{\zeta\left(\frac{1}{4}\right)}$. Each instance uses the principal root.

It's generally accepted that the gamma function is "the" continuation of the factorial. But why? Analytic continuation doesn't really work uniquely for sets with disjoint elements. If $f:\mathbb{Z}\to\mathbb{R}$ is given by $f(x)=0$, then two different analytic functions that agree with $f$ on $\mathbb{Z}$ are $g:\mathbb{R}\to\mathbb{R},\;g(x)=0$ and $h:\mathbb{R}\to\mathbb{R},\;h(x)=\sin(\pi x)$

So if we took $\tilde{\Gamma}(x)=\Gamma(x)\operatorname{cis}(2\pi x)$, where $\operatorname{cis}(t)=\cos(t)+i\sin(t)$, then this would agree (in an offset way) with the factorial at all the non-negative integers, and is analytic, but wouldn't be considered as a substitute for the gamma function. People wouldn't think $\left(-\frac{3}{4}\right)!=\tilde{\Gamma}\left(\frac{1}{4}\right)$ ever.

Is it because the gamma function is the only analytic function which agrees with the factorial and is also real along the real line (save the select poles)? $\tilde{\Gamma}$ isn't real along all of $\mathbb{R}$, but that's just one example. Is there a way to prove this, or disprove it? Or is it just that the integral definition got popularised, and it's just a byproduct of tradition?

Darmani V
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  • Tradition and popularity, where by popularity I mean the function appeared in many other places as well. – SK19 Oct 28 '24 at 18:12
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    As wikipedia notes, $e^{13.3\sin(7\pi x)}\Gamma(x)$ would be meromorphic, real on $\Bbb R\setminus{\text{poles}}$ and still satisfy $g(x+1)=xg(x)$ and $g(1)=1$. – Sassatelli Giulio Oct 28 '24 at 18:21
  • @BrianTung that comment is answer-worthy imo :) – SK19 Oct 28 '24 at 18:34
  • @SK19: OK, I've converted my comments into an answer, thanks for the suggestion. – Brian Tung Oct 28 '24 at 18:42
  • Just a remark to support SK19 answer. Tradition and popularity could feel like superfluous properties, but the fact is that it is popular because its useful. – jjagmath Oct 28 '24 at 18:56
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    @MartinR: Yeah, I just saw that dupe. I'm afraid I'm going to vote to close as duplicate, although my answer below has some extra information I think is somewhat interesting (the link to the article, notably). – Brian Tung Oct 28 '24 at 18:57

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One factor is the Bohr-Mollerup theorem, which states that this $\Gamma(\cdot)$ (due to Daniel Bernoulli) is the only function that is positive, equals $1$ at $x = 1$, satisfies $x\Gamma(x) = \Gamma(x+1)$, and notably, is logarithmically convex (that is, its logarithm is also convex).

Otherwise, as SK19 wrote in the comments, it is partly tradition and popularity. There is also Hadamard's gamma function, which has some appealing properties—it's entire, for example—but is not convex.

Philip Davis's article from 1959, "Leonhard Euler's Integral: A Historical Profile of the Gamma Function", provides some interesting history.

On a side note: Harald Bohr (of the Bohr-Mollerup theorem, above) was the brother of physicist Niels Bohr, and also an Olympic silver medal-winning footballer.

Brian Tung
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    I've voted to close this question as a duplicate of this one, although I'll leave this answer here for the link to the Davis article. – Brian Tung Oct 28 '24 at 19:01
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    Alternatively you could add that information as an answer to the other Q&A, so that all relevant information are at one place. – Martin R Oct 28 '24 at 19:03