The Gamma function, $Î“$, is a curious mathematical creature that pops up in many, seemingly unrelated places. Yet it defies naive intuition;

- Over the positive integers, it simplifies to the factorial function, shifted by $1$.
- Over the negative integers, it diverges.
- At $0$, it diverges.
- It has even more interesting behavior over non-integer fractions and complex numbers.

What makes $Î“$ so interesting?

The Gamma function is defined as follows:

$Î“:CÎ“(s)â€‹â†’C=0âˆ«âˆžâ€‹t_{sâˆ’1}e_{âˆ’t}dtâ€‹$It converges for all complex numbers $s$ with positive real part (i.e. $Re(s)>0$). It also converges for complex numbers with negative real part (i.e. $Re(s)<0$) except where $Re(s)$ is a negative integer or $0$.

When $Re(s)$ is zero or a negative integer and the imaginary part of $s$ is $0$, then $Î“(s)$ diverges and is undefined.

Note that the integers are embedded in the complex numbers as the subset obtained by setting the imaginary part to $0$ and restricting the real part accordingly.

Over the positive integers, $Î“$ has two general properties, outlined below.

### Special Case

Consider $Î“(n)$ for $n=1$. The integral has a surprisingly simple simplification:

$Î“(1)â€‹=0âˆ«âˆžâ€‹t_{1âˆ’1}e_{âˆ’t}dt=0âˆ«âˆžâ€‹e_{âˆ’t}dt(sinceÂt_{0}=1)=[âˆ’e_{âˆ’t}]_{0}=0âˆ’(âˆ’1)=1â€‹$### General Case: Positive Integers

Take $n$ to be an arbitrary positive integer, i.e. $nâˆˆN$ and $n>1$. To prove that $Î“$ is indeed equivalent to the factorial function (albeit shifted by $1$), we need to show that $Î“(n+1)=nÎ“(n)$.

Revisit the definition of $Î“$, applied to $n+1$:

$Î“(n+1)â€‹=0âˆ«âˆžâ€‹t_{n}e_{âˆ’t}dtâ€‹$This integral can be simplified using the rule for integration by parts:

$âˆ«udv=uvâˆ’âˆ«vduâ€‹$Take $u=t_{n}$ and $dv=e_{âˆ’t}dt$; then:

$duvThus,Î“(n+1)â€‹=dtdâ€‹t_{n}=nt_{nâˆ’1}=âˆ«e_{âˆ’t}dt=âˆ’e_{âˆ’t}=0âˆ«âˆžâ€‹t_{n}e_{âˆ’t}dt=0âˆ«âˆžâ€‹ut_{n}â€‹â€‹â‹…dve_{âˆ’t}â€‹â€‹dt=uv[âˆ’t_{n}e_{âˆ’t}]_{0}â€‹â€‹âˆ’0âˆ«âˆžâ€‹vnt_{nâˆ’1}â€‹â€‹â‹…du(âˆ’e_{âˆ’t})â€‹â€‹dt=(0âˆ’0)âˆ’0âˆ«âˆžâ€‹âˆ’nt_{nâˆ’1}e_{âˆ’t}dt=n0âˆ«âˆžâ€‹t_{nâˆ’1}e_{âˆ’t}dt=nâ‹…Î“(n)â€‹$This is an important identity, usually expressed in any of the following two forms:

$Î“(n+1)Î“(n)â€‹=nÎ“(n)=n1â€‹Î“(n+1)â€‹$### Proof by Induction

So far, we have seen the following results:

$Î“(1)Î“(n+1)â€‹=1â‰¡0!=nÎ“(n)forÂnâˆˆN_{>1}â€‹$By induction on $n$, suppose we know that $Î“(n)=(nâˆ’1)!$ (with the base-case being $Î“(1)=0!$). We can compute $Î“(n+1)$ as follows:

$Î“(n+1)â€‹=nâ‹…Î“(n)=nâ‹…(nâˆ’1)!=n!â€‹$$Î“$ was not intended to be a generalization of the factorial function. The result was a somewhat accidental one, albeit useful.

By restricting $Î“$ to the positive integers, we discovered that:

$Î“(n)â€‹=(nâˆ’1)!forÂnâˆˆNâ€‹$But what happens when we plug in a fraction, say $23â€‹$?

$Î“(23â€‹)â€‹=Î“(21â€‹+1)=21â€‹â‹…Î“(21â€‹)=21â€‹0âˆ«âˆžâ€‹t_{âˆ’21â€‹}e_{âˆ’t}dtâ€‹$To simplify the integral, let's try to get rid of the $t_{âˆ’21â€‹}$ term by substituting $t=x_{2}$:

$Î“(23â€‹)â€‹=21â€‹0âˆ«âˆžâ€‹t_{âˆ’21â€‹}e_{âˆ’x_{2}}â‹…2t_{21â€‹}dx=0âˆ«âˆžâ€‹e_{âˆ’x_{2}}dxâ€‹$This is a familiar integral, known as the Gaussian integral:

$I(a)I(âˆž)I_{2}(âˆž)â€‹=aâˆ«aâ€‹e_{âˆ’x_{2}}dx=âˆžâˆ«âˆžâ€‹e_{âˆ’x_{2}}dx=â€‹âˆ’âˆžâˆ«âˆžâ€‹e_{âˆ’x_{2}}dxâ€‹â‹…â€‹âˆ’âˆžâˆ«âˆžâ€‹e_{âˆ’y_{2}}dyâ€‹=âˆ’âˆžâˆ«âˆžâ€‹âˆ’âˆžâˆ«âˆžâ€‹e_{âˆ’(x_{2}+y_{2})}dxdy=âˆ’âˆžâˆ«âˆžâ€‹âˆ’âˆžâˆ«âˆžâ€‹e_{âˆ’r_{2}}rdrdÎ¸=0âˆ«2Ï€â€‹0âˆ«âˆžâ€‹e_{âˆ’r_{2}}rdrdÎ¸=2Ï€0âˆ«âˆžâ€‹e_{âˆ’r_{2}}rdr=2Ï€âˆ’âˆžâˆ«0â€‹21â€‹e_{u}du(substitutingÂu=âˆ’r_{2})=Ï€[e_{u}]_{âˆ’âˆž}=Ï€(e_{0}âˆ’e_{âˆ’âˆž})=Ï€(1âˆ’0)=Ï€â€‹$Since $Î“(23â€‹)$ restricts the integration to $[0,âˆž)$ and the function is even:

$Î“(23â€‹)â€‹=21â€‹âˆ’âˆžâˆ«âˆžâ€‹e_{âˆ’x_{2}}dx=21â€‹â‹…I_{2}(âˆž)â€‹=2Ï€â€‹â€‹â€‹$Other fractions of the form $(n+21â€‹),nâˆˆN$ have similar results simplifications:

$Î“(21â€‹)Î“(23â€‹)Î“(25â€‹)Î“(27â€‹)Î“(29â€‹)â‹®â€‹=Ï€â€‹=21â€‹Ï€â€‹=43â€‹Ï€â€‹=815â€‹Ï€â€‹=16105â€‹Ï€â€‹â€‹$In general,

$Î“(n1â€‹)âˆ¼nâˆ’Î³â€‹$where $Î³$ is the Euler-Mascheroni constant and $âˆ¼$ denotes asymptotic equivalence.

but unfortunately the integral has to be computed or simplified directly for each case.

$Î“$ has a meromorphic extension to the complex numbers, with simple poles at the non-positive integers and $0$. It is defined with the same rules and the relation

$Î“(s+1)=sÎ“(s)$holds for all complex numbers $s$. For example;

$Î“(i)â€‹=(âˆ’1+i)!â‰ˆâˆ’0.1549âˆ’0.4980iâ€‹$Various values of $Î“$ are tabulated here.

For some useful applications of $Î“$ in the Riemann zeta function, see this post.