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?


4 Answers 4


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'.

  • $\begingroup$ @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...). $\endgroup$ Mar 11, 2012 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, 2012 at 21:47
  • $\begingroup$ @FUZxxl: no problem and fine reading! $\endgroup$ Mar 11, 2012 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, $\endgroup$ Jun 23, 2016 at 22:31

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

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.

  • $\begingroup$ I personally enjoy this approach very much :-) $\endgroup$ Aug 20, 2017 at 22:43
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    $\begingroup$ 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$$\int_0^\infty x^ne^{-x}~\mathrm dx=n!$$ $\endgroup$ Aug 20, 2017 at 22:45

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

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


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

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.

  • 2
    $\begingroup$ 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, 2012 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, 2012 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, 2012 at 21:57
  • 1
    $\begingroup$ @FUZxxl: I have added a paragraph. See the edit. $\endgroup$
    – Aryabhata
    Mar 11, 2012 at 21:58

In Leonhard Euler's Integral: A Historical Profile of the Gamma Function: In Memoriam: Milton Abramowitz by Philip J. Davis in The American Mathematical Monthly , Dec., 1959: Apparently, Euler, experimenting with infinite products of numbers, chanced to notice that if n is a positive integer,

$$\small n!=\Bigg [ \left(\frac 2 1 \right)^n \frac1{n+1}\Bigg ]\,\Bigg [ \left(\frac 3 2 \right)^n \frac2{n+2}\Bigg ]\,\Bigg [ \left(\frac 4 3 \right)^n \frac3{n+3}\Bigg ]\cdots$$

and succeeding in transforming this infinity product into an integral, extending the factorial beyond integers, upon noticing that for certain values the infinite product yielded $\pi, $ suggesting areas of a circle.

But there is a really neat intuition already expressed in one of the answers, and beautifully presented by Robert Andrew Martin here, simply expanding the integral part in the integration by parts of a polynomial modulated by an exponential. In essence, the counterpart of the factorials in Taylor series.

For instance for $x^4$ (leaving constant of integration out):

$$\begin{align}\small \int x^4\; e^{-x} \; dx &\small= -x^4\;e^{-x} +\int 4\,x^3\;e^{-x}\;dx\\ &\small = -x^4 \; e^{-x}-4\,x^3\;e^{-x} +\int_0^\infty 4\cdot 3\,x^2\;e^{-x}\;dx\\ &\small = -x^4 \; e^{-x} -4\,x^3\;e^{-x} -4\cdot 3\,x^2\;e^{-x} +\int 4\cdot 3\cdot 2\,x\;e^{-x}\;dx\\ &\small =-x^4 \; e^{-x} -4\,x^3\;e^{-x} -4\cdot 3\,x^2\;e^{-x} -4\cdot 3\cdot2\,x\;e^{-x} - \underbrace{4\cdot 3\cdot 2\cdot 1}_{4!}\;e^{-x} \end{align}$$

Generalizing and integrating from $0$ to $\infty:$

$$\small\int_0^\infty x^n\; e^{-x} \; dx=-x^n \; e^{-x} -n\,x^{n-1}\;e^{-x} \cdots - \underbrace{n\cdot (n-1)\cdots 3\cdot 2\cdot 1}_{n!}\;e^{-x}\;\;\Bigr|_{x=0}^\infty=n!$$

which can immediately be extended beyond integers as $\displaystyle \small x! = \int_0^\infty t^x\; e^{-t} \; dt$

essentially the gamma function, except for the accepted slightly different definition: $$\Gamma(x)=\int_0^\infty t^{x-1}\; e^{-t} \; dt$$

that makes $\small (x-1)!=\Gamma(x).$


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