Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


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?

share|cite|improve this question
The accepted answer on that MO question doesn't really reflect community consensus; I am much more willing to support Emerton's comment. – Qiaochu Yuan Jan 20 '11 at 22:12
@Nate Thanks for the link. I checked Wikipedia and the archives on this site, but didn't look at MathOverflow. – Mark Eichenlaub Jan 20 '11 at 22:17
This thread is ancient. But I've added a new answer. See below. – Michael Hardy Sep 10 '12 at 18:51
up vote 13 down vote accepted

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

share|cite|improve this answer
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. – t.b. Jan 21 '11 at 1:14

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.

share|cite|improve this answer

$$ \Gamma(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x}\,dx. $$ 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.

share|cite|improve this answer
(I know this is old) Couldn't you just write the probability function as (using the "pi" function) this? $$\frac{x^{\alpha}e^{-x}}{\Pi(\alpha)}$$ – NotNotLogical May 17 '14 at 23:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.