I am looking for an "intuition pump" (Daniel Dennett's phrase) for the gamma function, showing why $\Gamma(n + 1)$ is the same as $\int_0^\infty x^n e^{-x} \, dx$.

For instance, is the fact that $e^{x}$ can be expressed as ${1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\cdots}$ be used to provide a non-rigorous proof ?

This means that $n!$ can also be expressed as $\int_0^\infty \frac {x^n} {1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\cdots} \,dx$

Given that exponents, factorials and the natural logarithm are such simple building blocks in maths, you'd think there would be an intuitive proof, or even a 'proof without words', but I haven't found anything on the internet.


closed as unclear what you're asking by Calvin Khor, gammatester, Xander Henderson, Jyrki Lahtonen, user99914 Sep 10 '18 at 5:10

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  • $\begingroup$ $\Gamma (s)$ is a continuous version of $n!$ [the factorial of natural numbers]. $\endgroup$ – xbh Sep 5 '18 at 13:41
  • $\begingroup$ what is $x$ in $x^3/(x^3/3!)$ and where does this expression come from? $\endgroup$ – Calvin Khor Sep 5 '18 at 13:42
  • $\begingroup$ @CalvinKhor -- Look at the second line. $\endgroup$ – mr_e_man Sep 5 '18 at 13:50
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    $\begingroup$ @mr_e_man No, the OP claims that the area under the graph is $x^3/(x^3/3!)$ when $n=3$, not $\int_a^b x^3/(x^3/3!) dx$. His expression is correct for any $x\neq 0$ but i could say its also equal to $$x^{100} sin(x)/(3!x^{100} sin(x))$$ yet I don't expect this to mean anything $\endgroup$ – Calvin Khor Sep 5 '18 at 13:56
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    $\begingroup$ Re your edit, the value of the maximiser of the integrand can sometimes be used to approximate the integral, though I don't believe its via the above mentioned expressions. Have a look at Laplace's method en.wikipedia.org/wiki/… and also the method of steepest descent $\endgroup$ – Calvin Khor Sep 5 '18 at 14:22

This is my attempt at non-rigorous intuitive approach to the factorial properties of the Gamma Function, limited to building on the simple recurrence relation property of the Gamma Function; $\Gamma(n+1)=n\,\Gamma(n)$.

Lets start by integrating a simple function $\frac{1}{\sqrt{x}}$ or $x^{-1/2}$ say.

$$I_1=\int x^{-1/2} \, dx=\frac{x^{1/2}}{\frac{1}{2}}$$ Integrating twice in succession we have $$I_2=\int\int x^{-1/2} \, dx\, dx=\frac{x^{3/2}}{ \frac{1}{2}\frac{3}{2} }$$ and integrating $n$ times successively we have $$I_{n}=\int_n...\int x^{-1/2} \, dx^n=\frac{x^{n-1+\frac{1}{2}}}{ \frac{1}{2}\frac{3}{2}...(n-1+\frac{1}{2}) }$$

To generalise slightly let $q$ be any rational number between $0$ and $1$ and we have $$I_n=\int_n...\int x^{-q} \, dx^n=\frac{x^{n-1+q}}{ q (1+q)...(n-1+q) }$$

where $n$ is any positive non-zero integer.

Then we could naively define the continuous version of the factorial as

$$[n-1+q]!=q \,(1+q)...(n-1+q)=\prod_{k=0}^{n-1}(k+q)$$

This version interpolates values between (n-1)! and (n)! using a simple straight line approximation. This is what I am calling the naïve version of the "continuous factorial".

The naïve version is normally by convention called the rising factorial and can be written in terms of the Gamma Function by building on the simple recurrence relation given earlier i.e.


or slightly rearranged we have

$$\Gamma(s)=\Gamma(n+q)=\Gamma(q)\prod_{k=0}^{n-1}(k+q)$$ where $s$ is any positive rational number $s=n+q$

So rather than a series of ugly discontinuous straight line interpolations between integer factorial values we have now defined what turns out to be a smooth function between positive integer factorial values ($1!$, $2!$ etc) which we can mathematically connect to our earlier naïve approach. The Gamma Function can be further generalised for use with negative numbers and complex numbers as well.

  • $\begingroup$ The OP was asking for an explanation of $$\int\frac{x^n}{\sum_k x^k/k!}dx = \frac{x^n}{x^n/n!}$$ $\endgroup$ – mr_e_man Sep 5 '18 at 16:12
  • $\begingroup$ @mr_e_man: I was attempting to answer the question in the middle of the post "Even so, is there some way of using this to create an intuitive explanation of the gamma function for someone with only basic calculus (me) ?" since I could not answer the other question. $\endgroup$ – James Arathoon Sep 5 '18 at 16:32
  • $\begingroup$ But what is that "this"? I think it's the equation above. $\endgroup$ – mr_e_man Sep 5 '18 at 16:33
  • $\begingroup$ @mr_e_man: Ok my mistake. $\endgroup$ – James Arathoon Sep 5 '18 at 16:39

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