TITLE: Why is the Gamma function shifted from the factorial by 1?
QUESTION [148 upvotes]: I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty  t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty  t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the multiplication formula is simpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$
instead of
$$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$
the infinite product definitions reduce from
$$\begin{align}
\Gamma(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} 
= \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\
\end{align}$$
to
$$\begin{align}
\Pi(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)} 
= \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Pi(z) &= e^{-\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\
\end{align}$$
 and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$
I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

REPLY [5 votes]: I'm going to elaborate on Pietro Majer's answer a bit.
Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$
\begin{align}
\Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt]
\Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m),
\end{align}
where $n,m$ are positive real numbers. Then
$$
\Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m).
$$
A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.
Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by
\begin{align}
\Pr(T_n \in A) & = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} \\[10pt]
& = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u.
\end{align}