Math 411: Honours Complex Variables - University of Alberta

12.2. THE GAMMA FUNCTION 81
12.2 The Gamma Function
For Re(z) > 0, define
Γ+(z) :=
� ∞
0 +
e −t t z−1 dt,
where the integration is performed along the positive real axis. Then Γ+ is holomorphic
in the right half plane {z ∈ C : Re(z) > 0}. A single integration by parts yields
the following recurrence relation
� ∞
Γ+(z +1) = e (12.2)
−t t z dt = −e −t t z
�∞
�
� ∞
�
� +z e −t t z−1 dt
0 +
= zΓ+(z).
Since Γ+(1) = � ∞
0 e−t dt = 1 = 0!, we see that Γ+(n+1) = n! for n ∈ N0. Continuing
in this manner we find that Γ+(z+n) = (z+n−1)...(z+1)zΓ+(z). On rearranging
this formula,
Γ+(z) =
0
0 +
Γ+(z +n)
z(z +1)...(z +n−1) ,
it is possible to analytically continue the function to the left-half plane:
⎧
⎨Γ+(z)
Re(z) > 0,
Γ(z) := Γ+(z +n)
⎩
−n < Re(z) ≤ −n+1,z �= −n+1,n = 1,2,3,...
z(z +1)...(z +n−1)
The resulting function Γ(z) is holomorphic in the complex plane except at z =
0,−1,−2,..., where it has simple poles. The graph of Γ(x) for x ∈ R is shown
in Figure 12.1 and an interactive three-dimensional plot of the surface Γ(z) for z ∈ C
is shown in Figure 12.2.
We proceed to derive a few useful relationships involving the Γ function.
• For α ∈ (0,1) we have
Γ(α) =
which leads to
�
Γ(α)Γ(1−α) = 2
� ∞
0 +
e −t t α−1 � ∞
dt = 2
0 +
= 4
� ∞
0 +
� ∞
0 +
� ∞
0 +
e −(x2 +y2 )
� π/2 � ∞
e −y2
y 2α−1 dy (letting t = y2 ),
e −y2
y 2α−1 ��
dy 2
= 4 tan
0
2α−1 θ
� π/2
0
= 2 tan 2α−1 θdθ.
0
�
y
x
� ∞
0 +
�2α−1 e −x2
x 1−2α �
dx
dxdy
e −r2
rdrdθ