Why isn't the gamma function defined so that $\Gamma(n) = n! $?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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