Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Pedagogically speaking, I see two problems with defining $\Gamma(z)$ (at least for real $z$) by the limit $$\Gamma(z)=\lim_{m\to\infty}\frac{m! m^z}{\prod_{i=0}^m (z+i)}$$ as compared with the formula $$\Gamma(z)=\lim_{m\to\infty}\frac{\lfloor m+z\rfloor! (z+m+1)^{\{z\}}}{\prod_{i=0}^m (z+i)}$$ (which one easily sees gives equivalent values in the limit). (The exponent $\{z\}$ means the fractional part of $z$.) Of the two, only the second formula yields $\Gamma(z+1)=z!$ for non-negative integers $z$ via trivial limits of constant sequences. And so far as I can see only the second formula arises immediately from a simple one-sentence story that a student might use to discover the formula as an exercise: march $\Gamma(z)$ to $\Gamma(z+m)$ using $\Gamma(z+1)=z\Gamma(z)$, then estimate $\Gamma(z+m)$ as a weighted geometric mean of the nearest factorials.

Questions: does the traditional limit support an equally compelling narrative? Ought one prefer it on other grounds? Neither limit serves well for numerical computation, but curiously, for a given $m$, they give errors of approximately equal magnitude but opposite sign. Does this have a conceptual explanation?

share|improve this question
Do you want to avoid the standard approach, using integration by parts and analytic continuation? –  Frank Thorne Dec 26 '10 at 22:54
@Frank Thorne I see the same problem - that the teacher writes down a formula, in your case $\int_0^\infty x^{n-1} e^{-x} dx$, does a calculation to see that the formula gives factorials at integer arguments and on the basis of that declares the proper interpolation found. Perhaps the teacher later mumbles "Bohr–Mollerup theorem" but figures that the proof would be a distraction in an introductory course. My formula -- and sentence -- explains how to find the interpolation and why it's the right one, and anticipates the proof of Bohr-Mollerup right in the definition. –  David Feldman Dec 26 '10 at 23:32
As far as the motivation for the integral formula, see math.stackexchange.com/questions/3444/… . –  Qiaochu Yuan Dec 27 '10 at 4:53

3 Answers 3

I guess it depends on whether the objective is (a) to introduce and motivate the Gamma function, and only the Gamma function, in as efficient a manner as possible, or (b) to present some useful mathematics which includes (but is not restricted to) a motivation for the Gamma function.

The Euler limit formulation can be motivated by starting with the asymptotic

$\binom{n+m}{m} \approx \frac{m^n}{n!}$ (1)

when $n$ is fixed and $m \to \infty$. Ostensibly, this fact is not directly related to the Gamma function, but is a useful asymptotic to know nonetheless (e.g. in relating differential calculus to difference calculus). But once one observes that the binomial $\binom{n+m}{m} = \frac{(n+1) \ldots (n+m)}{m!}$ can be defined even for non-integer $n$ (keeping $m$ integer, of course), this asymptotic can then be used to define $n!$ for non-integer n, which soon leads to Euler's limit definition for the Gamma function.

Now, one could certainly make a shorter route to the Gamma function by not presenting the asymptotic (1); this would be better at achieving the "local" objective (a), but perhaps not the "global" objective (b).

EDIT: In my experience, the Gamma function tends to enter into mathematics not so much as "an extension of the factorial function to the non-integer case", but rather as "normalisation constants obtained whenever integrating exponentials (additive characters) against monomials (multiplicative characters)"; the fact that the latter often happens to be expressible in terms of factorials when certain exponents are integer seems more of a secondary feature than a fundamental raison d'etre for Gamma. It's similar to how the Riemann zeta function is interesting in its own right, and not primarily as an extension of the Bernoulli numbers to the case of non-integer subscripts.

share|improve this answer
Maybe I'm missing something, but since essentially your asymptotic will immediately produce Euler's definition from my second formula, I don't see anything here sacrificed on the alter of efficiency. That said, I find your EDIT very compelling, namely the idea that pedagogically, solving the interpolation problem is actually a kind of distraction. So may I ask, please, can one axiomate the formal properties of Gamma that makes it work as "normalization constants..." –  David Feldman Dec 27 '10 at 6:00
Your analogy to the Riemann zeta function doesn't help me much, because I don't understand what nature was thinking when she burdened the zeta function with both Bernoulli number extension and prime distribution asymptotics. Years ago I once asked Stephen Lichtenbaum, following a talk he gave, a question along that line. He told me that Howard Garland had an idea for a K-theory indexed by arbitrary complex numbers, with the hope that it would supply the missing link for both the theory of special values of L-functions and the Hilbert-Polya attack on RH. I don't know what became of the idea. –  David Feldman Dec 27 '10 at 6:08
My point regarding the asymptotic (1) is that it can be used as an alternate way to define Gamma, namely as the unique extension of the factorial function (shifted by one, of course) that makes (1) valid even when n is not an integer. The advantage of doing things this way is that (1) is independently interesting even if one does not directly care about Gamma. As regarding the axiomatisation of Gamma as a normalisation constant, only one axiom is required: $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\ dt$, interpreting the integral using distributions or analytic continuation as necessary. –  Terry Tao Dec 27 '10 at 7:25

I see many reasons that support introducing Euler's limit as a first definition for the Gamma function. I agree with you, that if all you want is a function interpolating the factorial, you might as well use the second limit which makes the jump from discrete to continuous a little more clear. However, because the Gamma function pops up in so many different contexts, one wants to understand more of its properties, and we see that the second limit is not quite natural because floor functions and fractional parts are not "calculus friendly".

A first example is that Euler's limit is one step away from the Weierstrass product form $$\Gamma(z)=\lim_{m\to \infty}\frac{m!m^z}{z(z+1)\cdots(z+m)}=\frac{1}{z}\lim_{m\to \infty}\prod_{k=1}^m \frac{(1+\frac{1}{k})^z}{(1+\frac{z}{k})}=\frac{e^{-\gamma z}}{z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{-1}e^{z/k}$$

Second, it makes it easier to introduce the digamma function (logarithmic derivative of $\Gamma$) $$\psi=\frac{d}{dz}\ln \Gamma(z)=\frac{d}{dz}\lim_{m\to \infty}\bigg(\ln m!+z\ln m-\ln z-\ln(z+1)-\cdots \ln(z+m)\bigg)$$ $$=\lim_{m\to \infty} \left(\ln m-\frac{1}{z}-\frac{1}{z+1}\cdots-\frac{1}{z+m}\right)=-\gamma+\sum_{k=1}^{\infty}\frac{z-1}{k(k+z-1)}$$ It also seems like deriving the integral forms of $\Gamma$ from the second limit would be harder.

share|improve this answer

For many years I used to finish the course of analysis for the first year's undergraduate students in Physics giving an elementary yet rigorous introduction of the (real) Gamma function. These students use it in other courses, including laboratory, and on the other hand this last short chapter gave me the opportunity of a review of most of the material of the year (real and complex numbers, differential calculus, Riemann integral, power series, uniform convergence, scalar ODE).

I understand your perplexity in choosing the best definition. The Eulerian integral, or the Weierstrass infinite product, or the limit? Although each of these has its point of strength, I prefer not to emphasize on a particular representation. Rather, after giving some motivations from analysis and probability, I introduced the Gamma function via the elegant Artin-Bohr-Mollerup characterization: the unique (up to a multiplicative constant) log-convex solution of the functional equation. By means of a logarithmic derivative (convex functions are differentiable up to at most countably many points) the problem is reduced to the existence and uniqueness of the solution to the linear functional equation for the digamma function, which is a very nice elementary exercise on series.

The point in my opinion is not, choosing the more general representation, or the most practical to compute, or the easiest to derive. It is, rather finding the most central, the one richer of consequences, and able to establish connections more naturally.

Artin's definition has a central position in the theory: with the uniqueness result in hand, all representations and all identities and asymptotic formulas are easily deduced by checking the hypotheses for the characterizations (one example for all: the multiplicative formula $\Gamma(x)=\dots$ defines a functions on the RHS, which is clearly log-convex and satisfies the functional equation; the multiplicative constant is then found comparing the asymptotic expansions of both sides).

Physics students are of course practical people, but this doesn't mean they can't appreciate abstraction, when it is useful!

share|improve this answer

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.