2. Gamma function :
Gamma function is define by the improper interval
𝑒−𝑥
∞
0
𝑥 𝑛−1 𝑑𝑥 , 𝑛 > 0
And it is denoted by Γ n.
Alternate form of gamma function ,
Γ n = 𝑒−𝑥2∞
0
𝑥2𝑛−1
𝑑𝑥
3. Proof by definition
Γ n = 𝑒−𝑥∞
0
. 𝑥 𝑛−1
𝑑𝑥
Let x=𝑡2
and 𝑑𝑥 = 2tdt
Γ n= 𝑒−𝑡2∞
0
. 𝑡2𝑛−2
𝑑𝑡
=2 𝑒−𝑡2∞
0
. 𝑡2𝑡−1
𝑑𝑡
Now changing the variable t to x,
Γ n=2 𝑒−𝑥2∞
0
𝑥2𝑛−1
𝑑𝑥
5. Proof of the property :
1] Γ n+1 = 𝑒−𝑥∞
0
. 𝑥 𝑛
𝑑𝑥,
Inter changing by parts ,
Γ n+1 = | 𝑒−𝑥
. 𝑥 𝑛
|0
∞
- 𝑒−𝑥∞
0
. 𝑥 𝑛−1
𝑑𝑥
= 𝑛 𝑒−𝑥∞
0
. 𝑥 𝑛−1
𝑑𝑥
=n Γ n
Hence Γn+1 = n Γ n proved.
This is known as recurrence reduction formula for gamma function.
6. NOTE :
Γ n+1 = n! if n is a positive integer
Γn+1 = n Γn if n is a real number .
Γn = Γ
𝑛+1
𝑛
if n is negative fraction.
Γn Γ1-n =
Π
sin 𝑛 Π
7. (2) Γ
1
2
= Π
PROOF :
By alternate form of gamma function ,
Γ
1
2
= 2 𝑒−𝑥2∞
0
. 𝑥2(
1
2
)−1
𝑑𝑥
= 𝑒−𝑥2
. 𝑑𝑥
∞
0
Γ
1
2
Γ
1
2
= 𝑒−𝑥2
𝑑𝑥
∞
0
. 𝑒−𝑦2
. 𝑑𝑦
∞
0
changing to polar coordinates x= 𝑟 cos 𝜃 ,
𝑦 = 𝑟 sin 𝜃
∴ 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑎𝑑𝜃
Limits of x x=0 to x → ∞
Limits of y y=0 to y → ∞
8. This shows that the region of integration is the first quadrant.
Draw the elementary radius vector in the region which starts from
the pole extend up to ∞.
Limits of r r=0 to r → ∞
Limits of 𝜃 𝜃 =0 to 𝜃 →
𝜋
2
.
Γ
1
2
Γ
1
2
= 4 𝑒−𝑟2∞
0
𝜋
2
0
. 𝑟𝑑𝑟𝑑𝜃
= 4 𝑑𝜃 (
−1
2
) 𝑒−𝑟2∞
0
𝜋
2
0
. (−2𝑟) 𝑑𝑟
=
4
−2
|𝜃|0
𝜋
2
|𝑒−𝑥2
|0
∞
[∵ 𝑒 𝑓 𝑟 𝑓′ 𝑟 𝜃𝑟 = 𝑒 𝑓 𝑟 ]
= -2.
𝜋
2
(0 − 1)
=𝜋
Γ
1
2
= 𝜋