Replacing xby −x, the equation (12.82) is sometimes represented
Γ(1 −x) = −xΓ(−x) (12 .83)
Using the recurrence relation (12.81) one can show that
Γ(n) =(n−1)Γ(n−1)
Γ(n−1) =(n−2)Γ(n−2)
Γ(n−2) =(n−3)Γ(n−3)
..
. =
..
.
Γ(3) =2 Γ(2)
Γ(2) =1 Γ(1) = 1
(12 .84)
The equations (12.81) and (12.84) demonstrate that
Γ(n+ 1) = n(n−1)(n−2)··· 3 · 2 ·1 = n! (12 .85)
Observe that when n= 0 the equation (12.85) becomes Γ(1) = 0!, but we know
Γ(1) = 1, hence this is one of the reasons for the convention of defining 0! as 1.
Figure 12-6.
The Gamma function Γ(x)and 1 /Γ(x).
Write equation (12.81) in the form Γ(n) =
Γ(n+ 1)
n to show that for n= 0,−^1 ,−^2 ,...
the function Γ(n)becomes infinite. The function Γ(x)and 1 /Γ(x)are illustrated in
the figure 12-6.
