# How to come up with the gamma function?

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?

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'.

• @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...). – Raymond Manzoni Mar 11 '12 at 21:38
• Sorry. I found out that the article actually contains an answer to my question, thus I removed my post. – FUZxxl Mar 11 '12 at 21:47
• @FUZxxl: no problem and fine reading! – Raymond Manzoni Mar 11 '12 at 21:52
• @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, – Raymond Manzoni Jun 23 '16 at 22:31

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.

• Thank you for that answer although you don't point out how one get's to $\int_0^\infty t^ne^{-1}\;\mathrm dt$. – FUZxxl Mar 11 '12 at 21:48
• @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 $\Gamma(x) = x!$? – Aryabhata Mar 11 '12 at 21:50
• 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. – FUZxxl Mar 11 '12 at 21:57
• @FUZxxl: I have added a paragraph. See the edit. – Aryabhata Mar 11 '12 at 21:58

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

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.

• I personally enjoy this approach very much :-) – Simply Beautiful Art Aug 20 '17 at 22:43
• 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$$\int_0^\infty x^ne^{-x}~\mathrm dx=n!$$ – Simply Beautiful Art Aug 20 '17 at 22:45