Gamma, Beta, and Stirling
1
Gamma function
The gamma function can be defined as an improper definite integral:
Z ∞
Γ(α) =
tα−1 e−t dt
0
To get its most notable property, let’s do integration by parts on Γ(α + 1) (it’s
a little easier to add 1) using u = tα , dv = e−t , so that du = αtα−1 , v = −e−t
Z ∞
tα e−t dt
(1)
Γ(α + 1) =
0
Z ∞
= tα (−e−t )|∞
−
(−e−t )(αtα−1 dt)
(2)
0
0
Z ∞
= (0 − 0) + α
tα−1 e−t dt
(3)
0
= αΓ(α)
(4)
R ∞ −t
Since we can do Γ(1) = 0 e dt = 1 easily, then we get Γ(2) = 1×Γ(1) = 1,
and then Γ(3) = 2Γ(2) = 2, and then the more general Γ(k + 1) = k! for any
integer k > 0. Handling zero and negative values, where Γ(x) goes to ∞, requires
more tools.
√
One case that requires a different approach is Γ(1/2) = π.
The behavior for large values is described by Stirling’s formula:
√
Γ(x + 1)
= 2π
x+1/2
−x
x→∞ x
e
lim
2
Rescale Gamma Integral
If we rescale the variable in the integral we can change the variable of integration
to simplify. Here change with s = at, so ds = adt, leading to the following:
Z ∞
Z ∞
Z ∞
tα−1 e−at dt =
(s/a)α−1 e−s (ds/a) = a−α
sα−1 e−s ds = Γ(α)/aα
0
3
0
0
Beta integral
The Beta integral begins with two gamma integrals in variables s and t, and
then changing variables: s + t = u on (0, ∞] and s/(s + t) = v on (0, 1). The
inverse function has then s = uv and t = u(1 − v), and the Jacobian of the
transformation has determinant −u.
ds/du ds/dv
v
u
J=
=
dt/du dt/dv
1 − v −u
1
The sequence of integrals then becomes:
Z
Z ∞
sα−1 e−s ds ×
Γ(α)Γ(β) =
tβ−1 e−t dt
(5)
(uv)α−1 e−uv (u(1 − v))β−1 e−u(1−v) | − u|dudv
Z 1 Z ∞
uα+β−1 e−u du v α−1 (1 − v)β−1 dv
=
(6)
0
0
Z
1
Z
=
0
0
∞
∞
0
(7)
0
Z
=
1
v α−1 (1 − v)β−1 dv × Γ(α + β)
0
From this last result, we can view the following as the beta integral:
Z 1
Γ(α)Γ(β)
v α−1 (1 − v)α−1 dv =
Γ(α + β)
0
jfm, 02 July 2015
2
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