Proove that
gamma (1/2) = √π
Answers
The gamma function is defined as
Γ(α) = Z ∞
0
x
α−1
e
−x
dx.
Making the substitution x = u
2 gives the equivalent expression
Γ(α) = 2 Z ∞
0
u
2α−1
e
−u
2
du
A special value of the gamma function can be derived when 2α − 1 = 0 (α =
1
2
). When α =
1
2
, Γ
1
2
simplifies as
Γ
1
2
= 2 Z ∞
0
e
−u
2
du
To derive the value for Γ
1
2
, the following steps are used. First, the value of Γ
1
2
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
1
2
is explicitly solved for.
First, square the value for Γ
1
2
and rewrite as a double integral. Hence,
Γ
1
2
2
=
Γ
1
2
Γ
1
2
=
2
Z ∞
0
e
−u
2
du 2
Z ∞
0
e
−v
2
dv
= 4 Z ∞
0
Z ∞
0
e
−(u
2+v
2
)
dvdu (1)
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
π
2
for the first quadrant. The Jacobian of the transformation is
|J| =
∂u
∂r
∂u
∂θ
∂v
∂r
∂v
∂θ
=
cos θ −r sin θ
sin θ r cos θ
= r cos2
θ + r sin2
θ = r
Hence, (1) can be written as
Γ
1
2
2
= 4 Z ∞
0
Z ∞
0
e
−(u
2+v
2
)
dvdu = 4 Z π
2
0
Z ∞
0
e
−r
2
rdrdθ
= 4 "Z π
2
0
dθ# " Z ∞
0
e
−r
2
rdr
u = −r
2
du = −2rdr
#
= 4 h
π
2
i
−
1
2
Z −∞
0
e
u
du
= −π [0 − 1]
= π
Finally, since e
−u
2
> 0 for all u > 0, then Γ
1
2
> 0. Hence,
Γ
1
2
2
= π =⇒ Γ
1
2
=