|The Gamma function is defined as, Γ(x)||=|
|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.
We shall integrate Γ(x + 1) by parts (integrating et and differentiating tx):
|Γ(x + 1)||=|
|tx et dt||=||[tx et]|
|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)|
|tx1 et dt||Γ(1)||=|
|= (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.