Why is the Gamma function shifted from the factorial by 1?

Question

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 $\Pi(z)$?

Answer 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."

Answer 2

I would argue against the OP's opinion. The definition $\Gamma(z)$ becomes very natural if you write it as $\Gamma(z) = \int_0^\infty t^{z} e^{-t} d^\times t$, where $d^\times t=dt/t$ is the Haar measure on the multiplicative group of positive numbers. Moreover, $t^z$ is a character of this group, hence the definition is an instance of the Fourier transform on locally compact abelian groups, in this case called the Mellin transform. In fact, this is why this version works well for the Riemann zeta function and in fact for any automorphic $L$-function: $\pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s)$ is invariant under $s\to 1-s$. Of course, one might say that $\zeta(s)$ is not normalized in the right way, but in terms of the Dirichlet coefficients of $\zeta(s)$, or more generally in terms of the Langlands parameters of an automorphic $L$-function, the current normalization is the right one (cf. Ramanujan conjecture)!

EDIT: I just realized this is an elaboration of a comment Emerton made earlier.

Answer 3

It was so that Legendre could do with the gamma function what the Catholic church did 170 years later: He put a simple pole at the origin.

Answer 4

They both are are equally "good". Unlike conventional calculus in discrete calculus there are two equally valid differentiation operators with little reason to prefere one over the other - forward difference $\Delta f(x)$ and backward difference $\nabla f(x)$. In discrete multiplicative calculus there are also two similar operators - discrete multiplicative forward difference $\frac{f(x+1)}{f(x)}=\exp(\Delta \ln f(x))$ and discrete multiplicative backward difference $\frac{f(x)}{f(x-1)}=\exp(\nabla \ln f(x))$. They both have their respective inverse operators - forward discrete multiplicative integral and backward discrete multiplicative integral. So the $\Gamma(x)$ is the forward discrete multiplicative integral of $f(x)=x$ and $\Gamma(x+1)=x!$ is the backward discrete multiplicative integral of the same function.

In the scientific applications there is a preference to using forward difference rather than backward difference I think because if is possible to find forward differences of arbitrary order of a function defined on only positive integers. Among other considerations, it allows to represent in the form of Newton series a function which is defined only on natural numbers (Newton series with backward difference would require the function to be defined on negative integers).

Answer 5

I would also go for $\Pi(t)$ or $t!$, but a possible reason to prefer the shifted version, $\Gamma(t)$, is the following. The gamma densities $\gamma_t$, $t\in \mathbb{R}$ defined as

$$\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)},$$

are a convolution semigroup, so that $\Gamma(t)$ appears naturally as the normalization factor of $\gamma_t$. (And, of course, the semigroup relation

$$\gamma_t*\gamma_s=\gamma_{t+s}$$

would be destroyed shifting from $t-1$ to $t$ in the definition of $\gamma_t$)

Also note that the expression of the Beta function

$$B(t,s):=\int_0^1 x^{t-1}(1-x)^{s-1} \, dx$$

in terms of the $\Gamma$ function, if shifted, would also loose the useful form

$$B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)}.$$

(incidentally note that this relation follows plainly from the semigroup property since as a general fact, the integral of a convolution of two functions is the product of their integrals).

Answer 6

The formula for the volume of the unit $n$-ball is nicer in terms of $\Pi$, instead of $\Gamma$, since it is $\frac{\pi^{\frac{n}{2}}}{\Pi(\frac{n}{2})}$, as opposed to $\frac{\pi^{\frac{n}{2}}}{\Gamma(1 + \frac{n}{2})}$.

Answer 7

One advantage to the conventional definition of the Beta function $B(s,t)$ is that a random variable whose probability distribution is the Beta distribution with probability distribution proportional to $x^{s-1}(1-x)^{t-1}\ dx$ for $0\le x\le 1$ has expected value $s/(s+t)$.

Answer 8

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}

Answer 9

This drives me crazy, too. However, Andrews, Askey and Roy give their reasons for preferring the Legendre definition in their book on special functions. Click on the link to page 6, http://books.google.com/books?id=kGshpCa3eYwC&lpg=PP1&dq=andrews%20askey%20roy&pg=PA6#v=onepage&q=legendre&f=false

I'm not sure exactly what they are referring to in their section 1.10, but this may have something to do with it. Click on the link to page 39, http://books.google.com/books?id=kGshpCa3eYwC&lpg=PP1&dq=andrews%20askey%20roy&pg=PA39#v=onepage&q=jacobi&f=false

Answer 10

If you look at the $L$ function of a modular form, it is $\Gamma$ that comes naturally for the functional equation. It is also $\Gamma$ that tells you what the order of the zeros of the $L$ function are (and these orders have a meaning).

Answer 11

Whatever the real reason might have been (or if indeed it was even a matter of conscious choice to begin with) I'm afraid we may never truly know. However, I do find it a fortunate coincidence, inasmuch as it enables many mathematical results to be expressed in terms of the $\Gamma$ and $\zeta$ or $\eta$ functions of the same argument. Here are but a few beautiful examples:

$$\int_0^\infty\frac{x^z}{e^x-1}dx=\Gamma(z+1)~\zeta(z+1),\qquad\qquad\qquad\qquad\Re(z)>1.$$

$$\int_0^\infty\frac{x^z}{e^x+1}dx=\Gamma(z+1)~\eta(z+1),\qquad\qquad\qquad\qquad\Re(z)>1.$$

$$\oint_\gamma\frac{(-x)^z}{e^x-1}dx=2i~\Gamma(z+1)~\zeta(z+1)\sin(z\pi),\qquad\qquad\qquad z\in\mathbb C,$$

$\qquad\qquad\qquad\qquad\qquad\qquad\quad$ where $\gamma=R~e^{i\alpha}$, with $\alpha\in(0,2\pi)$ and $R\to\infty$.

$$\int_0^\infty\ln\Big(1-e^{-x^{\Large a}}\Big)dx=-\Gamma\bigg(\frac1a+1\bigg)~\zeta\bigg(\frac1a+1\bigg),\qquad\qquad a>0.$$

$$\int_0^\infty\ln\Big(1+e^{-x^{\Large a}}\Big)dx=\quad\Gamma\bigg(\frac1a+1\bigg)~\eta\bigg(\frac1a+1\bigg),\qquad\qquad a>0.$$

$$\ln\Gamma(x)=\zeta'(0,x)-\zeta'(0),\qquad\qquad\qquad\qquad\qquad\qquad x\in\mathbb R^\star\setminus\mathbb Z^-.$$