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This question already has an answer here:

Why didn't they define it as $$ \tilde \Gamma(x) = \int_0^\infty t^x e^{-t} \, dt ?$$ Then the definition would have two less characters than the standard definition of $\Gamma(x)$, and we would have $\tilde \Gamma(n) = n!$ for $n$ a non-negative integer. And this would save a lot of confusion.

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marked as duplicate by J. M. is a poor mathematician, Jonas Meyer, Dietrich Burde, Mark Viola, user147263 Jul 15 '15 at 22:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This was asked in Mathoverflow some time ago, and got nice answers. You should look there. $\endgroup$ – Mariano Suárez-Álvarez Jul 15 '15 at 20:07
  • $\begingroup$ I think this is a notational "mistake" that Laplace (?) made that we are just stuck with. There have been attempts to fix it but the sheer volume of math that would need fixing as a result is quite daunting. David's answer is the best, by far. It's the way I think of it. $\endgroup$ – Cameron Williams Jul 15 '15 at 20:07
  • $\begingroup$ Related: math.stackexchange.com/questions/18356/… $\endgroup$ – HDE 226868 Jul 15 '15 at 20:11
  • $\begingroup$ Thanks everyone. I did look around to see if it was a duplicate, but I used the incorrect search criteria! The answers are very helpful. Seeing as most people like David Ullrich's answer, I'll accept that one. (It is also the same as Emerton's answer on the MO site.) Honestly, I don't think it is a very good answer. But its the best a posteriori explanation for what I think is otherwise a mistake of history. $\endgroup$ – Stephen Montgomery-Smith Jul 15 '15 at 21:15
  • $\begingroup$ Here is the link for the MO question. What do you think of Detlev Gronau's idea, because of the nicer relation to the beta function ? $\endgroup$ – Dietrich Burde Jul 16 '15 at 14:29
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My opinion is that it's because the "right" way to write the standard definition is $\Gamma(x)=\int_0^\infty t^xe^{-t}\frac{dt}{t}$. Putting in that multiplicative Haar measure makes a lot of other things easier to get straight. Same for a lot of integrals with $t$ to some exponent with a curious $-1$ attached...

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  • $\begingroup$ In other words, $\Gamma(x) = \int_{-\infty}^\infty e^{xs} e^{-e^s} \, ds $ after the substitution $s = \log t$. $\endgroup$ – Stephen Montgomery-Smith Jul 15 '15 at 20:54
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    $\begingroup$ I guess. I didn't think that was my point but you may well be right. My point was maybe just that when we have an integral $\int_0^\infty$ the most natural way to write it is in terms of $\frac{dt}t$ instead of $dt$... $\endgroup$ – David C. Ullrich Jul 15 '15 at 21:04
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    $\begingroup$ But $\Gamma(x+1) = \int_0^1 (-\ln t)^x \, dt$ looks much nicer to me. $\endgroup$ – Stephen Montgomery-Smith Jul 15 '15 at 21:08
  • $\begingroup$ But also, I do see your point. $\endgroup$ – Stephen Montgomery-Smith Jul 15 '15 at 21:09
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(Copied from https://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1)

From Riemann's Zeta Function, by H. M. Edwards, available as a Dover paperback, footnote on page 8: "Unfortunately, Legendre subsequently introduced the notation $\Gamma(s)$ for $\Pi(s-1).$Legendre's reasons for considering $(n-1)!$ instead of $n!$ are obscure (perhaps he felt it was more natural to have the first pole at $s=0$ rather than at $s = -1$) but, whatever the reason, this notation prevailed in France and, by the end of the nineteenth century, in the rest of the world as well. Gauss's original notation appears to me to be much more natural and Riemann's use of it gives me a welcome opportunity to reintroduce it."

The book in question: https://books.google.com/books?id=5uLAoued_dIC&pg=PA7&source=gbs_toc_r&cad=4#v=onepage&q&f=false

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Actually, this question has been considered by many authors. The article Why is the gamma function so as it is? gives some explanations.

Edit: From a theoretical to a practical answer: It seems to be due to Euler, who influenced Legendre involving the nice relation with the beta function $$ B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}. $$ see identity $(14)$. The answer of Pietro Majer at the MO question is in this spirit.

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  • $\begingroup$ I have edited. Could you remove the downvote then ? $\endgroup$ – Dietrich Burde Jul 16 '15 at 14:26

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