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General Factorial

Problem

The Gamma function is defined as, Γ(x) = 
infinity


0
txminus1 eminust dt.

Given that Γ(x) is continuous for x greater than 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 eminust and differentiating tx):

Γ(x + 1) = 
infinity


0
tx eminust dt =  [minustx eminust]
infinity

0
+ x
infinity


0
txminus1 eminust dt

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

Therefore for integers, n greater than or equal 1, we get:

Γ(n + 1) = nΓ(n)
  = n(nminus1)Γ(nminus1)
  = n(nminus1)(nminus2)Γ(nminus2)
  = ...
  = n(nminus1)(nminus2)(nminus3)times...times3times2times1timesΓ(1)

Γ(x) = 
infinity


0
txminus1 eminust dt  implies  Γ(1) = 
infinity


0
eminust dt = minus[eminust]
infinity

0
 =  minus(0minus1) = 1

Therefore Γ(n + 1) = n! for n greater than or equal 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 greater than 0, it is a suitable candidate for non-integer values.

It is interesting to note that for x greater than 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 greater than 0.

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

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