## General Factorial

#### Problem

The Gamma function is defined as, Γ(`x`) | = | | | `t`^{x1} `e`^{t} `dt`. |

Given that Γ(`x`) is continuous for `x` 0 prove that Γ(`x` + 1) = `x`! for all positive integer values, and consequently is a candidate for extending the factorial function to non-integer positive values.

#### Solution

We shall integrate Γ(`x` + 1) by parts (integrating `e`^{t} and differentiating `t`^{x}):

Γ(`x` + 1) | = | | | `t`^{x} `e`^{t} `dt` | = | [`t`^{x} `e`^{t}] | | + `x` | | | `t`^{x1} `e`^{t} `dt` |

For `x` 0, we can see that `t`^{x} = 0 when `t` = 0 and `e`^{t} = 0 when `t` = , hence we get Γ(`x` + 1) = `x`Γ(`x`).

Therefore for integers, `n` 1, we get:

Γ(`n` + 1) | = | `n`Γ(`n`) |

| = | `n`(`n`1)Γ(`n`1) |

| = | `n`(`n`1)(`n`2)Γ(`n`2) |

| = | ... |

| = | `n`(`n`1)(`n`2)(`n`3)...321Γ(1) |

Γ(`x`) | = | | | `t`^{x1} `e`^{t} `dt` | | Γ(1) | = | | | `e`^{t} `dt` | = | [`e`^{t}] | | = (01) = 1 |

Therefore Γ(`n` + 1) = `n`! for `n` 1.

Although we are not permitted to substitute `n` = 0 into the equation, we are pleased to see that Γ(1) = 0! = 1, as expected.

As Γ(`x` + 1) maps to all the expected values of `x`! for integer values and is continuous for `x` 0, it is a suitable candidate for non-integer values.

It is interesting to note that for `x` 0 the function can also be shown to be holomorphic (everywhere differentiable) and logarithmic convex. In fact, it turns out to be the ONLY function which produces a smooth curve through all the points of the graph `y` = `x`! for `x` 0.

Problem ID: 251 (02 Dec 2005) Difficulty: 4 Star

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