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.
Why is the Gamma function off by 1 from the factorial?
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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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(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
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...