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

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 $\Pi(z)$?

• overlapping edits, it seems. Does anyone object if I put the formulas in equation mode rather than inline? Some of us are making do with small screens and failing eyesight – Yemon Choi Apr 10 '10 at 22:33
• Note: the integral for the $\Gamma$ function is not so unreasonable, if you think of it as an integral of $t^z e^{-t}\frac{dt}{t}$, i.e. an integral transform over $\mathbb R^{\times}_{>0}$ with respect to multiplicative Haar measure. – Emerton Apr 10 '10 at 23:15
• Emerton, that's exactly the explanation my number theory professor just gave me yesterday. Still, I don't think it's very satisfying given everything else. – Kevin Casto Apr 10 '10 at 23:23
• I would argue that both functional equations you listed are really arguments in favor of $\Gamma$ over $\Pi$, especially since the symmetric form of the functional equation for $\zeta$ (using the $\xi$ function) is even more compact. Emerton's Haar measure argument is quite strong when viewed in the broader context of Tate's thesis, where you need to consider homogeneous measures at each place. Other points in favor of $\Gamma$ include the beta function definition, and the Mellin transform formula for $\zeta$ in terms of $\theta$. – S. Carnahan Apr 11 '10 at 3:25
• For what it's worth, I always felt that the answer to this question would be embedded somehow in Tate's thesis and the theory of motives and general gamma factors at infinity, but having tried to get on top of this stuff the only thing I convinced myself of was that there is no "one correct gamma function"---one uses $\Gamma(s/2)$ and $\Gamma((s+1)/2)$ and $\Gamma(s)$ and it's not clear to me why any of these are more fundamental than any other. – Kevin Buzzard Apr 11 '10 at 8:25

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."

• Yeah, this is basically what I assumed. It is nice to have someone directly address the issue instead of sweeping it under the rug. And for those interested, the book is on Google Books - books.google.com/books?id=5uLAoued_dIC - and pages 8-10 have most of the identities listed above. – Kevin Casto Apr 10 '10 at 23:27
• "Legendre's reasons...are obscure". The main issue in my opinion is slightly different. Let's recall that 200 years ago (but even more recently) standardizing the notation was much more a difficult problem, and less urgent at the same time -no internet, no common scientific language, no or very few international meetings. In such a situation it was quite natural and harmless even introducing a notation ad hoc in any new memoir. Note that Weierstrass' one was the "factorial function", Fc(z), the reciprocal of the Gamma function. – Pietro Majer May 28 '10 at 17:03

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.

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.

• Ba-dom-pom-pssht. – Ketil Tveiten Oct 18 '10 at 13:10
• Sorry, I couldn't resist. :-) – Greg Kuperberg Oct 18 '10 at 13:38
• the Church put a pole at the origin thousand years earlier – Bob Apr 23 '11 at 9:41

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).

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).

• You would define $\tilde B(t,s) = \int_0^1 x^t (1-x)^s \, dx$. Then $\tilde B(m,n) = 1\big/ \binom{m+n}m$. – Stephen Montgomery-Smith Jul 15 '15 at 21:00
• $\int_0^z \frac{x^{t-1}}{(t-1)!} \frac{(z-x)^{s-1}}{(s-1)!} \, dx = D_{z}^{-s} \frac{z^{t-1}}{(t-1)!} = \frac{z^{s+t-1}}{(s+t-1)!}$ is easy to remember if one can remember to associate $s=1$ with integration and $s=0$ with the identity or Dirac delta function. – Tom Copeland Jul 16 '15 at 1:53
• @Stephen Montgomery-Smith. In fact the formula you wrote is wrong, corroborating my claim that $\Gamma(t)\Gamma(s)/\Gamma(t+s)$ is easier to deal with ;) – Pietro Majer Jan 30 '16 at 18:55

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 $\frac{\pi^{\frac{n}{2}}}{\Gamma(1 + \frac{n}{2})}$.

• On the other hand, the surface area of the unit hypersphere in $\mathbb{R}^n$ can be written as $\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}$ instead of $\frac{2\pi^{n/2}}{\Pi(\frac{n}{2}-1)}$. – S. Carnahan Sep 16 '14 at 10:38

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 $s/(s+t)$.

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

• The thing is, the Beta function is itself (to me) incorrectly defined; it should be the integral of t^x*(1-t)^y. Likewise, there doesn't seem to be any reason why, in the generalization to characters, we can't have J be the analog of B' (where B'(x,y) = B(x+1,y+1)) and g the analog of Pi. – Kevin Casto Apr 11 '10 at 3:10
• But not only two of them. There are much more factorial generalizations which interpolate integer factorials. End even have not poles! For example the Hadamard generalization of factorials, look on Wolfram. – Sergei Apr 14 '14 at 15:58

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).

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 $R\to\infty$.

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

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

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

• You can replace some of the $z+1$ terms with $z$ by replacing $dx$ with the natural scale-invariant measure $d^\times \! x = \frac{dx}{x}$. – S. Carnahan Sep 16 '14 at 10:26

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 $$\Pr(T_n \in A) = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u.$$