Gamma Function

1. Gamma Function
Dr. Anjali Devi JS
Guest Faculty
School of Chemical Sciences
M G University

2. Factorial – A business of positive integers only?
n! =n (n-1) (n-2) (n-3)……3.2.1
1! =1
=2
2!
=6
3!
1 2 3
1
2
6
●
●
●
We want a function
(1) f(1) =1
=n (n-1)!
(2) f(n) =n (f(n-1))
Solution: Gamma Function
“Generalised Factorial” is
Gamma function.

3. Gamma Function
Γ 𝑛 = 0
∞
𝑒−𝑥𝑥(𝑛−1)𝑑𝑥 n>0
Put n=1
Γ 1 = 0
∞
𝑒−𝑥𝑥(1−1)𝑑𝑥 = 0
∞
𝑒−𝑥𝑑𝑥 =
𝑒−𝑥
−1 0
∞
= 1
The gamma function , (Γ(n)) is defined to be an extension of the
factorial to complex and real number arguments.
Properties of Gamma function
Γ 1 = 1
(1)
Put n=0
(2)
∞
𝛤 0 = Γ 0 = ∞

4. (3) Reduction formula for Γ(n)
Γ 𝑛 + 1 = 0
∞
𝑒−𝑥𝑥(𝑛+1−1)𝑑𝑥 = 0
∞
𝑒−𝑥 𝑥(𝑛)𝑑𝑥
1st fn.
2nd fn.
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 by parts
𝑢𝑣𝑑𝑥 = 𝑢 𝑣 𝑑𝑥 − 𝑢′
( 𝑣 𝑑𝑥)𝑑𝑥
0
∞
𝑥(𝑛)𝑒−𝑥𝑑𝑥 = 𝑥(𝑛)
0
∞
𝑒−𝑥𝑑𝑥 − 0
∞
𝑛𝑥𝑛−1
0
∞
𝑒−𝑥𝑑𝑥 𝑑𝑥
Put n=n+1
=n 0
∞
𝑒−𝑥𝑥𝑛−1𝑑𝑥 = 𝑛Γ 𝑛
Γ 𝑛 + 1 = n Γ 𝑛 for n>0
Properties

5. (4) Value of Γ(n) in terms of factorial
Γ 𝑛 + 1 = n Γ 𝑛
Γ(2) = Γ(1 + 1) = 1. Γ(1)= 1 = 1!
Γ 3 = Γ 2 + 1 =
Γ 4 = Γ 3 + 1 = 3Γ 3 =
2Γ 2 = 2 = 2!
6= 3!
Γ 𝑛 = 𝑛 − 1 !
Remember: Here n should be a positive integer.
Properties

6. 𝜞
𝟏
𝟐
= 𝝅
Γ
1
2
=
0
∞
𝑒−𝑥
𝑥(𝑛−1)
𝑑𝑥
Put x=u2,
At x=0, u=0
x= ∞, u= ∞
dx=2udu
Γ
1
2
=
0
∞
𝑒−𝑢2
𝑢(−1)(2𝑢)𝑑𝑢
= 2
0
∞
𝑒−𝑢2
𝑑𝑢
Square both sides
Γ
1
2
2
=
4
0
∞
𝑒−𝑢2
𝑑𝑢
0
∞
𝑒−𝑣2
𝑑𝑣
= 4
0
∞
0
∞
𝑒−(𝑢2+𝑣2)𝑑𝑢𝑑𝑣
Proof
Properties (5)

7. Convert to polar coordinates using u=r Cos𝜃 𝑣 = 𝑟𝑆𝑖𝑛𝜃
𝜞
𝟏
𝟐
= 𝝅
Proof (Contd..)
Γ
1
2
2
=4 0
𝜋
2
0
∞
𝑟𝑒−(𝑟2)𝑑𝑟𝑑𝜃
Set q=r2,
dq=2rdr
Γ
1
2
2
=
4
2 0
𝜋
2
0
∞
𝑒−𝑞𝑑𝑞𝑑𝜃 =2 0
𝜋
2
1 𝑑𝜃
=π
𝜞
𝟏
𝟐
= 𝝅
Properties (5)

8. Evaluate 𝜞
𝟑
𝟐
= 𝛤
1
2
+ 1
Γ 𝑛 + 1 = n Γ 𝑛
Question
= 𝛤
1
2
+ 1
=
1
2
𝛤
1
2
=
1
2
𝜋
Evaluate 𝜞
𝟓
𝟐
= 𝛤
3
2
+ 1 =
3
4
𝜋
Answer
Answer

9. Evaluate 𝜞
𝟗
𝟐
Γ 𝑛 + 1 = n Γ 𝑛
Question
= (
7
2
)(
5
2
)(
3
2
)(
1
2
) 𝜋