# The $\Gamma$ function

## Idea and Definition

Leonhard Euler solved the problem of finding a function of a continuous variable $x$ which for integer values of $x=n$ agrees with the factoriel function $n↦n!$. In fact, gamma function is a shift by one of the solution of this problem.

For a complex variable $x\ne -1,-2,\dots$, we define $\Gamma \left(x\right)$ by the formula

$\Gamma \left(x\right)=\underset{k\to \infty }{\mathrm{lim}}\frac{k!{k}^{x-1}}{\left(x{\right)}_{k}}$\Gamma(x) = \lim_{k\to \infty} \frac{k! k^{x-1}}{(x)_k}

where $\left(x{\right)}_{0}=1$ and for positive integer $k=1,2,\dots$,

$\left(x{\right)}_{k}=x\left(x+1\right)\left(x+2\right)\cdots \left(x+k-1\right)$(x)_k = x (x+1) (x+2)\cdots (x+ k-1)

is the shifted factorial. It easily follows that $\Gamma \left(n+1\right)=n!\mathrm{for}\mathrm{natural}\mathrm{numbers}$n = 0, 1, 2, \ldots\$.

## Properties

As a function of a complex variable, the Gamma function $\Gamma \left(x\right)$ is a meromorphic function with simple poles at $x=0,-1,-2,\dots$.

Extending the recursive definition of the ordinary factorial function, the Gamma function satisfies the following translation formula:

$\Gamma \left(x+1\right)=x\Gamma \left(x\right)$\Gamma(x+1) = x\Gamma(x)

away from $x=0,-1,-2,\dots$.

It also satisfies a reflection formula, due to Euler:

$\Gamma \left(x\right)\Gamma \left(1-x\right)=\frac{\pi }{\mathrm{sin}\left(\pi x\right)}.$\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}.

Quite remarkably, the Gamma function (this time as a function of a real variable) is uniquely characterized in the following theorem:

###### Theorem (Bohr-Mollerup)

The restriction of the Gamma function to the interval $\left(0,\infty \right)$ is the unique function $f$ such that $f\left(x+1\right)=xf\left(x\right)$, $f\left(1\right)=1$, and $\mathrm{log}f$ is convex.

A number of other representations of the Gamma function are known and frequently utilized, e.g.,

• Product representation:

$\frac{1}{\Gamma \left(x\right)}=x{e}^{\gamma x}\prod _{n=1}^{\infty }\left(1+\frac{x}{n}\right){e}^{-x/n}$\frac1{\Gamma(x)} = x e^{\gamma x} \prod_{n=1}^{\infty} (1 + \frac{x}{n})e^{-x/n}

where $\gamma$ is Euler's constant?.

• Integral representation:

$\Gamma \left(x\right)={\int }_{0}^{\infty }{t}^{x}{e}^{-t}\frac{dt}{t}.$\Gamma(x) = \int_{0}^{\infty} t^x e^{-t} \frac{d t}{t}.

## References

• George Andrews, Richard Askey, Ranjan Roy, Special Functions. Encyclopedia of Mathematics and Its Applications 71, Cambridge University Press, 1999.

Revised on October 23, 2011 01:06:59 by Toby Bartels (67.141.101.96)