Γ=∫0∞tx−1e−tdt
For 0<x<1, it is an improper integral because
tx−1 becomes infinite at the lower limit.
But, it is convergent for x>0. For
x≤0 it is not defined because the integral diverges.
For
x≤0,
Γ(x) has not so far been defined. BUT, let's define it by the recursion relation
Γ(x+1)=xΓx. Which is what you're onto.
Γ(x)=x1Γ(x+1) defines
Γ(x) for x<0. .........
[1]
Here is an example:
Γ(−0.5)=−0.51Γ(0.5)≈−3.545....
Γ(−1.5)=−1.51⋅Γ(0.5)1≈2.363.... and so on.
Since
Γ(1)=1, we see that
Γ(x)=xΓ(x+1)→∞ as x→0
From this and successive uses of [1], we see that
Γ(x) becomes infinite not only at 0 but also at all negative
integers.
For positive x,
Γ(x) is a continuous function passing through the points
x=n, Γ(x)=(n−1)!.
For negative x,
Γ(x) is discontinuous at the negative integers. In the intervals between the integers it alternates between positive and negative: negative from 0 to -1, positive from -1 to -2, and so on.
We can also use the identity, which I will not bother deriving,
Γ(−x)=x⋅Γ(x)sin(πx)−π
As can be seen, if
x∈Z, then the denominator is 0 because sin is 0 for integer multiples of Pi.
Was that a nice tutorial?. I have always been interested in the gamma and beta functions. They are cool and can be used to evaluate integrals which are otherwise difficult using elementary methods.