Proof of Γ(1/2)
The gamma function is defined as
∞
Z
xα−1 e−x dx.
Γ(α) =
0
Making the substitution x = u2 gives the equivalent expression
Z ∞
2
Γ(α) = 2
u2α−1 e−u du
0
A special value of the gamma function can be derived when 2α − 1 = 0 (α = 12 ). When α = 12 , Γ 12
simplifies as
Z ∞
2
1
Γ
e−u du
=2
2
0
1
To derive the value for Γ 2 , the following steps are used. First, the value of Γ 21 is squared. Second,
the squared value is rewritten as a double integral. Third, the double integral is evaluated by
transforming to polar coordinates.
Fourth, the Γ 12 is explicitly solved for.
First, square the value for Γ 12 and rewrite as a double integral. Hence,
2 1
1
1
Γ
= Γ
Γ
2
2
2
∞
Z
= 2
0
∞
Z
∞
Z
2
e−u du 2
2
e−v dv
0
∞
Z
e−(u
=4
0
2
+v 2 )
dvdu
(1)
0
The region R which defines the first quadrant, is the region of integration for the integral in (1). The
bivariate transformation u = r cos θ, v = r sin θ will transform the integral problem from cartesian
coordinates to polar coordinates, (r, θ). These new variables will range from 0 6 r 6 ∞ and 0 6 θ 6 π2
for the first quadrant. The Jacobian of the transformation is
∂u ∂u ∂r ∂θ cos θ −r sin θ
= r cos2 θ + r sin2 θ = r
|J| = ∂v ∂v = sin θ r cos θ ∂r
∂θ
Hence, (1) can be written as
2
Z ∞Z ∞
Z π2 Z ∞
2
2
2
1
Γ
e−r rdrdθ
=4
e−(u +v ) dvdu = 4
2
0
0
0
0
"Z
=4
#"Z
π
2
dθ
0
#
∞
e
0
−r
2
rdr
u = −r 2
du = −2rdr
h π i 1 Z −∞
u
=4
−
e du
2
2 0
= −π [0 − 1]
=π
2
Finally, since e−u > 0 for all u > 0, then Γ 12 > 0. Hence,
2
√
1
1
Γ
= π =⇒ Γ
= π
2
2