The Gamma function, $Γ$, 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 $Γ$ so interesting?

The Gamma function is defined as follows:

$Γ:CΓ(s) →C=0∫∞ t_{s−1}e_{−t}dt $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 $Γ(s)$ 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, $Γ$ has two general properties, outlined below.

### Special Case

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

$Γ(1) =0∫∞ t_{1−1}e_{−t}dt=0∫∞ e_{−t}dt(sincet_{0}=1)=[−e_{−t}]_{0}=0−(−1)=1 $### General Case: Positive Integers

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

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

$Γ(n+1) =0∫∞ t_{n}e_{−t}dt $This integral can be simplified using the rule for integration by parts:

$∫udv=uv−∫vdu $Take $u=t_{n}$ and $dv=e_{−t}dt$; then:

$duvThus,Γ(n+1) =dtd t_{n}=nt_{n−1}=∫e_{−t}dt=−e_{−t}=0∫∞ t_{n}e_{−t}dt=0∫∞ ut_{n} ⋅dve_{−t} dt=uv[−t_{n}e_{−t}]_{0} −0∫∞ vnt_{n−1} ⋅du(−e_{−t}) dt=(0−0)−0∫∞ −nt_{n−1}e_{−t}dt=n0∫∞ t_{n−1}e_{−t}dt=n⋅Γ(n) $This is an important identity, usually expressed in any of the following two forms:

$Γ(n+1)Γ(n) =nΓ(n)=n1 Γ(n+1) $### Proof by Induction

So far, we have seen the following results:

$Γ(1)Γ(n+1) =1≡0!=nΓ(n)forn∈N_{>1} $By induction on $n$, suppose we know that $Γ(n)=(n−1)!$ (with the base-case being $Γ(1)=0!$). We can compute $Γ(n+1)$ as follows:

$Γ(n+1) =n⋅Γ(n)=n⋅(n−1)!=n! $$Γ$ was not intended to be a generalization of the factorial function. The result was a somewhat accidental one, albeit useful.

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

$Γ(n) =(n−1)!forn∈N $But what happens when we plug in a fraction, say $23 $?

$Γ(23 ) =Γ(21 +1)=21 ⋅Γ(21 )=21 0∫∞ t_{−21}e_{−t}dt $To simplify the integral, let's try to get rid of the $t_{−21}$ term by substituting $t=x_{2}$:

$Γ(23 ) =21 0∫∞ t_{−21}e_{−x_{2}}⋅2t_{21}dx=0∫∞ e_{−x_{2}}dx $This is a familiar integral, known as the Gaussian integral:

$I(a)I(∞)I_{2}(∞) =a∫a e_{−x_{2}}dx=∞∫∞ e_{−x_{2}}dx= −∞∫∞ e_{−x_{2}}dx ⋅ −∞∫∞ e_{−y_{2}}dy =−∞∫∞ −∞∫∞ e_{−(x_{2}+y_{2})}dxdy=−∞∫∞ −∞∫∞ e_{−r_{2}}rdrdθ=0∫2π 0∫∞ e_{−r_{2}}rdrdθ=2π0∫∞ e_{−r_{2}}rdr=2π−∞∫0 21 e_{u}du(substitutingu=−r_{2})=π[e_{u}]_{−∞}=π(e_{0}−e_{−∞})=π(1−0)=π $Since $Γ(23 )$ restricts the integration to $[0,∞)$ and the function is even:

$Γ(23 ) =21 −∞∫∞ e_{−x_{2}}dx=21 ⋅I_{2}(∞) =2π $Other fractions of the form $(n+21 ),n∈N$ have similar results simplifications:

$Γ(21 )Γ(23 )Γ(25 )Γ(27 )Γ(29 )⋮ =π =21 π =43 π =815 π =16105 π $In general,

$Γ(n1 )∼n−γ $where $γ$ is the Euler-Mascheroni constant and $∼$ denotes asymptotic equivalence.

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

$Γ$ 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

$Γ(s+1)=sΓ(s)$holds for all complex numbers $s$. For example;

$Γ(i) =(−1+i)!≈−0.1549−0.4980i $Various values of $Γ$ are tabulated here.

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