gamma function

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Introduction

The gamma function can be thought of asthe natural way to generalize the concept of the factorialto non-integer arguments.

Leonhard Euler (http://planetmath.org/EulerLeonhard)came up with a formula for such a generalization in 1729.At around the same time,James Stirling independently arrived at a different formula,but was unable to show that it always converged.In 1900, Charles Hermite showed that the formula given by Stirling does work,and that it defines the same function as .

Definitions

$$\mathrm{\Gamma }(z)=\underset{n\to \mathrm{\infty }}{lim}\frac{n^{z}n!}{{\prod }_{k=0}^n(z+k)}.$$

However, it is now more commonly defined by

$$\mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{z-1}𝑑t$$

for $z\in ℂ$ with $\mathrm{Re}(z)>0$,and by analytic continuation for the rest of the complex plane,except for the non-positive integers (where it has simple poles).

Another equivalent definition is

$$\mathrm{\Gamma }(z)=\frac{e^{-\gamma z}}{z}\prod _{n=1}^{\mathrm{\infty }}{\left(1+\frac{z}{n}\right)}^{-1}{e}^{z/n},$$

where $\gamma$ is Euler’s constant.

Functional equations

The gamma function satisfies the functional equation

$$\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$$

except when $z$ is a non-positive integer.As $\mathrm{\Gamma }(1)=1$, it follows by induction that

$$\mathrm{\Gamma }(n)=(n-1)!$$

for positive integer values of $n$.

Another functional equation satisfied by the gamma function is

$$\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)=\frac{\pi }{\sin \pi z}$$

for non-integer values of $z$.

Approximate values

The gamma function for real $z$ looks like this:

(generated by GNU Octave and gnuplot)

It can be shown that $\mathrm{\Gamma }(1/2)=\sqrt{\pi }$.Approximate values of $\mathrm{\Gamma }(x)$ for some other $x\in (0,1)$ are:

If the value of $\mathrm{\Gamma }(x)$ is known for some $x\in (0,1)$,then one may calculate the value of $\mathrm{\Gamma }(n+x)$ for any integer $n$by making use of the formula $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$.We have

$\mathrm{\Gamma }(n+x)$$=$$(n+x-1)\mathrm{\Gamma }(n+x-1)$

$=$$(n+x-1)(n+x-2)\mathrm{\Gamma }(n+x-2)$

$\mathrm{⋮}$

$=$$(n+x-1)(n+x-2)\mathrm{\cdots }(x)\mathrm{\Gamma }(x)$

which is easy to calculate if we know $\mathrm{\Gamma }(x)$.