35
$\begingroup$

As a physics student, I have occasionally run across the gamma function

$$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$

when we want to generalize the concept of a factorial. Why not define the gamma function so that

$$\Gamma(n) = n!$$

instead?

I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option. Are there some areas of mathematics where the accepted definition looks more natural? Are there some formulas that work out more cleanly with the accepted definition?

$\endgroup$
5
  • 9
    $\begingroup$ mathoverflow.net/questions/20960/… $\endgroup$
    – Nate Eldredge
    Jan 20, 2011 at 21:46
  • 1
    $\begingroup$ The accepted answer on that MO question doesn't really reflect community consensus; I am much more willing to support Emerton's comment. $\endgroup$
    – Qiaochu Yuan
    Jan 20, 2011 at 22:12
  • $\begingroup$ @Nate Thanks for the link. I checked Wikipedia and the archives on this site, but didn't look at MathOverflow. $\endgroup$
    – Mark Eichenlaub
    Jan 20, 2011 at 22:17
  • $\begingroup$ This thread is ancient. But I've added a new answer. See below. $\endgroup$
    – Michael Hardy
    Sep 10, 2012 at 18:51
  • 1
    $\begingroup$ With the $\Pi$ function, there is something gratifying about writing the volume of an $n$ ellipsoid with radii $r_1, \ldots, r_n$ as $$\frac{\pi^{n/2}}{\Pi(n/2)}\prod_{k=1}^n r_k$$ Look at all that pi! $\endgroup$
    – user123641
    Mar 2, 2018 at 19:51

3 Answers 3

Reset to default
20
$\begingroup$

I find it more illuminating to see what that extra $ t^{-1} $ does to the integral. As a generalization of the factorial, $ \Gamma $ is inherently multiplicative*. On the other hand, integration, which is essentially a sum, is inherently additive. Thinking about it this way it seems a bit odd that an integral could give an appropriate generalization. However, there is a simple function that takes the (positive) multiplicative reals to the reals under addition: the logarithm. Thinking of gamma as $$ \Gamma(s) = \int_0^{\infty} t^s e^{-t} \frac{dt}{t} $$ we see that the natural log arises naturally (what's the first thing to come to mind when you see $ \frac{dt}{t} \: $?) in this context (and there are no pesky $s-1$'s left).

I haven't done much with integration theory (so I'm not sure my terminology is correct), but I believe this intuitive argument may be made rigorous by considering the integral formula for $ \Gamma(s) \: $ as an integral over the positive reals under multiplication (i.e. the interval $ (0,+\infty) $ ) with respect to the multiplicative Haar measure.

*note that by "multiplicative" I just mean that it's realized as a product, not that it's a multiplicative arithmetic function or anything like that

$\endgroup$
1
  • 11
    $\begingroup$ Concerning the last paragraph. Yes, that's precisely right. The measure $\frac{dt}{t}$ is the Haar measure on the multiplicative reals: by the substitution rule we have $\frac{d(at)}{at} = \frac{dt}{t}$ so that the measure is invariant under multiplication. Note that this is precisely Emerton's point in the MO-link in one of the comments above. $\endgroup$
    – t.b.
    Jan 21, 2011 at 1:14
7
$\begingroup$

$$ \Gamma(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x}\,dx. $$ Why $\alpha-1$ instead of $\alpha$? Here's one answer; there are probably others. Consider the probability density function $$ f_\alpha(x)=\begin{cases} \dfrac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)} & \text{for }x>0 \\[12pt] 0 & \text{for }x<0 \end{cases} $$ The use of $\alpha-1$ instead of $\alpha$ makes the family $\{f_\alpha : \alpha > 0\}$ a "convolution semigroup": $$ f_\alpha * f_\beta = f_{\alpha+\beta} $$ where the asterisk represents convolution.

$\endgroup$
4
  • $\begingroup$ (I know this is old) Couldn't you just write the probability function as (using the "pi" function) this? $$\frac{x^{\alpha}e^{-x}}{\Pi(\alpha)}$$ $\endgroup$
    – user142299
    May 17, 2014 at 23:12
  • $\begingroup$ @NotNotLogical : If you write it that way, then you don't have a convolution semigroup. $\endgroup$
    – Michael Hardy
    Apr 21, 2017 at 3:36
  • $\begingroup$ Your PDF is undefined at $0$. I think you wanted one of the cases to be $\ge$ or $\le$ so that $f_\alpha(0)=0$? $\endgroup$
    – Simply Beautiful Art
    Aug 20, 2017 at 22:49
  • $\begingroup$ @SimplyBeautifulArt : PDFs are not defined pointwise. What is of interest about them is what you get when you integrated them over a set. So defining it at an isolated point just isn't a thing. $\endgroup$
    – Michael Hardy
    Mar 4, 2021 at 0:36
6
$\begingroup$

A completely simple explanation is provided in the first section here.

EDIT: Consider a general gamma density function. That is, a function $f(x;c,\lambda)$, $x>0$, of the form $f(x;c,\lambda) = \lambda^c x^{c-1}e^{-\lambda x} / \Gamma(c)$, where $c$ and $\lambda$ are arbitrary positive constants. The counterpart with respect to the alternative definition, $\tilde \Gamma (p) := \int_0^\infty {t^p e^{ - t} \,{\rm d}t}$, is a function $f(x;c,\lambda)$, $x>0$, of the form $f(x;c,\lambda) = \lambda^{c+1} x^{c}e^{-\lambda x} / \tilde \Gamma(c)$, where $c > -1$ and $\lambda > 0$. Obviously, the former form is preferable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .