25
$\begingroup$

It always puzzles me, how the Gamma function's inventor came up with its definition $$\Gamma(x+1)=\int_0^1(-\ln t)^x\;\mathrm dt=\int_0^\infty t^xe^{-t}\;\mathrm dt$$ Is there a nice derivation of this generalization of the factorial?

$\endgroup$
18
$\begingroup$

Here is a nice paper of Detlef Gronau Why is the gamma function so as it is?.
Concerning alternative possible definitions see Is the Gamma function mis-defined? providing another resume of the story Interpolating the natural factorial n! .

Concerning Euler's work Ed Sandifer's articles 'How Euler did it' are of value too, in this case 'Gamma the function'.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @FUZxxl: Well it provides the letters from Euler where this generalization appears (Euler usually didn't give names to his functions, Gamma and +1 came later...). $\endgroup$ – Raymond Manzoni Mar 11 '12 at 21:38
  • $\begingroup$ Sorry. I found out that the article actually contains an answer to my question, thus I removed my post. $\endgroup$ – FUZxxl Mar 11 '12 at 21:47
  • $\begingroup$ @FUZxxl: no problem and fine reading! $\endgroup$ – Raymond Manzoni Mar 11 '12 at 21:52
  • $\begingroup$ @DanielG: Thanks for the update! Concerning the last paragraph we may use 'How Euler did it' and 'Gamma the function' from the irreplaceable "The Euler Archive". Cheers, $\endgroup$ – Raymond Manzoni Jun 23 '16 at 22:31
8
$\begingroup$

I guess you can say this is yet another application of the power of integration by parts (and I am guessing that is how the integral formula "was come up with" initially).

If you are trying to find the antiderivative of $P(t) e^t$, where $P(t)$ is a polynomial, integration by parts arises naturally and I would say it(integral of $P(t) e^t$) is quite natural to encounter during ones study of mathematics. And if you actually work it out, you notice the factorial like recursion. We can rid of the "non-integral" parts of the integration by parts formula by using the limits $0$ and $\infty$.

If $I_n = \int_{0}^{\infty} t^n e^{-t} \text{dt}$ then integration by parts gives us

$$I_n = -e^{-t}t^n|_0^{\infty} + n\int_{0}^{\infty} t^{n-1} e^{-t} = nI_{n-1}$$

so if

$f(x) = \int_{0}^{\infty} t^x e^{-t} \text{dt}, \quad x \ge 0$

then

$f(x) = x f(x-1), \quad x \ge 1$.

Also, we have that $f(0) = 1$, thus the integral definition agrees with the factorial function at the non-negative integers and can serve as a real extension for factorial.

Using Analytic continuation its domain can be extended further.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you for that answer although you don't point out how one get's to $\int_0^\infty t^ne^{-1}\;\mathrm dt$. $\endgroup$ – FUZxxl Mar 11 '12 at 21:48
  • $\begingroup$ @FUZxxl: I don't understand. I interpreted your question as "how did one come up with the integral". Or as you asking for why $\Gamma(x+1) = x!$ and not $\Gamma(x) = x!$? $\endgroup$ – Aryabhata Mar 11 '12 at 21:50
  • 1
    $\begingroup$ YOur first interpretion is right. But your answer shows only that the integral satisfies the recurrence relation $\Gamma(x+1)=x\Gamma(x)$ and does not show how one can derive that integral. $\endgroup$ – FUZxxl Mar 11 '12 at 21:57
  • 1
    $\begingroup$ @FUZxxl: I have added a paragraph. See the edit. $\endgroup$ – Aryabhata Mar 11 '12 at 21:58
6
$\begingroup$

$$ \int e^{ax} dx = \frac{1}{a} e^{ax} + c $$

Take $\left .\frac{d}{da}\right |_{a=1}$ on both sides $n$ times, and algebra to get rid of $(-1)^n$, you'll have an integral equal to $n!$.

This is an intuitive way to get the Gamma function. You've shown that for integers it holds from this simple derivation.

Mathematicians then went through a great deal of work to show that it holds true for allot more than just the integer case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I personally enjoy this approach very much :-) $\endgroup$ – Simply Beautiful Art Aug 20 '17 at 22:43
  • 5
    $\begingroup$ To be more precise, we have$$\int_0^\infty e^{ax}~\mathrm dx=-\frac1a\quad \forall a<0$$Differentiate both sides $n$ times to get$$\int_0^\infty x^ne^{ax}~\mathrm dx=\frac{(-1)^{n+1}n!}{a^{n+1}}$$Now sub in $a=-1$ to get$$\int_0^\infty x^ne^{-x}~\mathrm dx=n!$$ $\endgroup$ – Simply Beautiful Art Aug 20 '17 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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