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General Factorial
Problem
| The Gamma function is defined as, Γ(x) | = |  |
| tx 1 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) | = | |
| tx et dt | = | [tx et] |
| + x | |
| 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) | = | |
| tx1 et dt |  | Γ(1) | = | |
| et dt | = | [et] |
| = (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.
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