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There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^t dt$ is "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students.

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Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there. –  Noldorin Aug 4 '10 at 15:05
    
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this. –  Larry Wang Aug 4 '10 at 16:28
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It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues. –  pbrooks Aug 4 '10 at 17:06
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9 Answers 9

up vote 72 down vote accepted

The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties

  • $f(1)=1$;
  • $f(x+1)=xf(x)$ for every $x\geq 0$;
  • $\log f$ is a convex function.

The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.


By the way, the gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

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Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to). –  Larry Wang Aug 4 '10 at 16:59
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The logarithm $\log \Gamma(c)$ can be thought of as a "continuous sum" $\log \Gamma(x) = \sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex? –  Greg Graviton Aug 4 '10 at 17:31
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Is there a function that satisfies the first 2 properties but is not convex? –  Dan Apr 18 '12 at 1:26
    
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1. –  Charles Sep 25 '12 at 20:05
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Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.

Euler's Gamma Function
Euler's Gamma Function

Hadamard's Gamma function
Hadamard's Gamma function

Luschny's factorial function
Luschny's factorial function


See here for more information.

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For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that

$\displaystyle \int_{0}^1 t^a (1 - t)^b \, dt = \frac{a! b!}{(a+b+1)!}.$

For (non-negative?) real values of $a$ and $b$ the correct generalization is

$\displaystyle \int_0^1 t^a (1 - t)^b \, dt = \frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}.$

And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.

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Why is that the "correct generalization"? –  Neil G Aug 28 '10 at 9:30
    
Also, please see my first question -- I guess you know the answer! –  Neil G Aug 28 '10 at 9:32
    
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one. –  Qiaochu Yuan Aug 28 '10 at 12:22
    
I think your formula should be $\displaystyle \int_0^1 t^a (1 - t)^b \, dt = \frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}$. –  John Bentin Mar 23 '12 at 12:37
    
@John: whoops! Corrected. –  Qiaochu Yuan Mar 23 '12 at 16:26
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This is a comment posted as an answer for lack of reputation.

Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)

This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.

EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.

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Answers are generally accepted for comment this long anyway –  Casebash Aug 5 '10 at 11:41
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Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:

  • $f(1)=1$
  • $f(z+1)=zf(z)$ for all $z>0$
  • $f(z)$ is analytic for $\operatorname{Re}z>0$
  • $f(z)$ is bounded for $1\leq \operatorname{Re}z\leq 2$

(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)

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Looking for a difference that makes a difference.

Flipping the gamma function and looking at Newton interpolation provides another angle on the question:

Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $\bigtriangledown^{s-1}_{n}c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s-1}{n}c_n$, Newton interpolation gives $$\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!}=\frac{x^{s-1}}{(s-1)!}$$

for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.

Then with the sequence $a_j=1=\int_0^\infty e^{-x} \frac{x^{j}}{j!}dx$,

$\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j}a_j=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j}1=1=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j}\int_0^\infty e^{-x} \frac{x^{j}}{j!}dx=\int_0^\infty e^{-x} \frac{x^{s-1}}{(s-1)!}dx$.

This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.

This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)

From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.

To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:

Consider the fractional integro-derivative

$\displaystyle\frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=FP\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz=FP\displaystyle\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz$

$=\displaystyle\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}$

where FP denotes a Hadamard-type finite part, $x>0$, and $\alpha$ and $\beta$ are real.

For $\alpha>0$ and $\beta<0$, the finite part is not required for the beta integral, and it can be written as

$\displaystyle\int_{0}^{1}\frac{(1-z)^{\alpha}}{\alpha!}\frac{z^{-\beta-1}}{(-\beta-1)!} dz=\sum_{n=0}^{\infty } (-1)^n \binom{\alpha }{n}\frac{1}{n-\beta}\frac{1}{\alpha!}\frac{1}{(-\beta-1)!}$

$=\displaystyle\sum_{n=0}^{\infty }\frac{\beta!}{(\alpha-n)! n!}\frac{\sin (\pi (\beta -n))}{\pi (\beta -n)}=\frac{1}{(\alpha-\beta)!}, $ or

$$\displaystyle\sum_{n=0}^{\infty }\frac{1}{(\alpha-n)! n!}\frac{\sin (\pi (\beta -n))}{\pi (\beta -n)}=\frac{1}{(\alpha-\beta)!\beta!},$$

where use has been made of $\frac{\sin (\pi \beta)}{\pi \beta}=\frac{1}{\beta!(-\beta)!}$, and $\alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(\alpha)>-1$ and all complex $\beta$.

Euler's motivation (update July 2014):

R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially

$\displaystyle\frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}$.

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The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts. –  Tom Copeland May 2 '12 at 21:21
    
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus. –  Tom Copeland May 10 '12 at 1:27
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Comment on Jens: You can take any function on $[0,1)$ and extend to $\mathbb{R}$ using this functional equation. You have three examples above. These can obviously be made continuous. More surprising is that there are several ways of getting a convex function. However if you insist on log-convex there is a unique function.

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Comments should not be made until you have gained enough rep to make a comment rather than an answer (Its only 50 rep, so it shouldn't take very long to obtain). –  Casebash Aug 4 '10 at 21:12
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True, and I wasn't expecting to be upvoted. I don't think that is a reason to downvote. –  BWW Aug 4 '10 at 21:19
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For me another argument is the convincing one.

Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.

Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.

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Doesn't one choose the Gamma Function because it satisfies the functional equation $\Gamma(x) = (x-1)\Gamma(x-1)$ for all positive $x$ and therefore looks a lot like the factorial?

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The other Gamma functions mentioned here also satisfy that functional equation. –  Michael Lugo Aug 4 '10 at 17:53
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Just looking at their graphs, it is apparent that they don't, since it would entail going to infinity as $x$ goes to zero. –  Robin Chapman Aug 5 '10 at 6:25
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