## General Factorial

#### Problem

The Gamma function is defined as, Γ(x) =
 0
tx1 et 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 et and differentiating tx):

Γ(x + 1) =
 0
tx et dt =  [tx et]
 0
+ x
 0
tx1 et dt

For x 0, we can see that tx = 0 when t = 0 and et = 0 when t = , hence we get Γ(x + 1) = xΓ(x).

Therefore for integers, n 1, we get:

 Γ(n + 1) = nΓ(n) = n(n1)Γ(n1) = n(n1)(n2)Γ(n2) = ... = n(n1)(n2)(n3)...321Γ(1)

Γ(x) =
 0
tx1 et dt    Γ(1) =
 0
et dt = [et]
 0
=  (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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