Gamma and Beta Functions
Introduction
As introduced by the Swiss mathematician Leonhard Euler in18th century, gamma function is the extension
of factorial function to real numbers. Beta function (also known as Euler’s integral of the first kind) is closely
connected to gamma function; which itself is a generalization of the factorial function. Both Beta and Gamma
functions are very important in calculus as complex integrals can be moderated into simpler form using and
Beta and Gamma function.
I Gamma Function
∞
We define Gamma function as: Γ𝑛 = ∫0 𝑒 −𝑥 𝑥 𝑛−1 𝑑𝑥
Important results
1. 𝒊. 𝜞𝟏 = 𝟏
∞
Proof: 𝛤1 = ∫0 𝑒 −𝑥 𝑥 0 𝑑𝑥 = −[𝑒 −𝑥 ]∞
0 =1
𝟏
𝒊𝒊. 𝜞 = √𝝅
𝟐
1
1
∞
∞
2
Proof: 𝛤 = ∫0 𝑒 −𝑥 𝑥 −2 𝑑𝑥 = ∫0 𝑒 −𝑡 𝑡 −1 2𝑡 𝑑𝑡 , by putting 𝑥 = 𝑡 2
2
∞
2
= 2 ∫0 𝑒 −𝑡 𝑑𝑡 = √𝜋 ,
∞
2
⸪ ∫0 𝑒 −𝑥 𝑑𝑥 =
1
⸫ 𝛤 = 𝛤 (0.5) = √𝜋 = 1.772
2
2. Reduction formula for Γ𝒏 : Γ(𝒏 + 𝟏) = 𝒏𝚪𝒏
∞
We have Γ(𝑛 + 1) = ∫0 𝑒 −𝑥 𝑥 𝑛 𝑑𝑥
√𝜋
2
∞
𝑛−1 −𝑥
= −[𝑥 𝑛 𝑒 −𝑥 ]∞
𝑒 𝑑𝑥 = 0 + 𝑛Γ𝑛
0 + 𝑛 ∫0 𝑥
⸫ Γ(𝑛 + 1) = 𝑛Γ𝑛
∞
𝚪𝒏
3. ∫𝟎 𝒆−𝒌𝒙 𝒙𝒏−𝟏 𝒅𝒙 = 𝒏
𝒌
∞
Proof: We have Γ𝑛 = ∫0 𝑒 −𝑡 𝑡 𝑛−1 𝑑𝑡
Putting 𝑡 = 𝑘𝑥
⇒ 𝑑𝑡 = 𝑘𝑑𝑥
∞
∞
⸫ Γ𝑛 = ∫0 𝑒 −𝑘𝑥 (𝑘𝑥)𝑛−1 𝑘𝑑𝑥 = 𝑘 𝑛 ∫0 𝑒 −𝑘𝑥 𝑥 𝑛−1 𝑑𝑥
∞
⇒ ∫0 𝑒 −𝑘𝑥 𝑥 𝑛−1 𝑑𝑥 =
Γ𝑛
𝑘𝑛
Extension of Gamma function from factorial notation
Case 𝒊. When 𝒏 is a positive integer
We have Γ(𝑛 + 1) = 𝑛Γ𝑛
= 𝑛(𝑛 − 1)Γ(𝑛 − 1)
= 𝑛(𝑛 − 1)(𝑛 − 2)Γ(𝑛 − 2)
⋮
= 𝑛(𝑛 − 1)(𝑛 − 2) ⋯ 3.2.1Γ1 = 𝑛!
⸫ Γ2 = 1! , Γ3 = 2! , Γ4 = 3! etc.
case 𝒊𝒊. When 𝒏 is a positive rational number
Γ𝑛 = (𝑛 − 1)(𝑛 − 2) ⋯ upto a positive number in Γ function
7
5 3 1
1
15√𝜋
2
2 2 2
7 3
3
2
8
Illustration: Γ =
Also 𝛤
11
4
3
=
. . Γ =
. Γ
4 4
4
Now value of Γ can be obtained from table of gamma function.
4
case 𝒊𝒊𝒊. When 𝒏 is a negative rational number
Using Γ(𝑛 + 1) = 𝑛Γ𝑛
⇒ Γ𝑛 =
Γ(𝑛+1)
𝑛
=
=
=
(n+1)Γ(𝑛+1)
𝑛(𝑛+1)
Γ(𝑛+2)
𝑛(𝑛+1)
Γ(𝑛+3)
𝑛(𝑛+1)(𝑛+2)
⋮
Continuing in this manner, we get Γ𝑛 =
Γ(𝑛+𝑘+1)
𝑛(𝑛+1)…(𝑛+𝑘)
, where 𝑘 is the least positive integer such that
(𝑛 + 𝑘 + 1) > 0
Γ(−3.4+k+1)
Illustration: Γ(-3.4) = (−3.4)(−2.4)…(−3.4+𝑘) , (−3.4 + k + 1) > 0
⇒ 𝑘 > 2.4 ⇒ 𝑘 = 3
⸫ Γ(-3.4) =
Γ(−3.4+4)
(−3.4)(−2.4)(−1.4)(−0.4)
Γ0.6 can be found using tables.
=
Γ0.6
(−3.4)(−2.4)(−1.4)(−0.4)
Also, to evaluate Γ(-2.5),
Γ(−2.5+k+1)
Γ(-2.5) = (−2.5)(−1.5)…(−2.5+𝑘) , (−2.5 + k + 1) > 0
⇒ 𝑘 > 1.5 ⇒ 𝑘 = 2
⸫ Γ(-2.5) =
Γ(−2.5+3)
(−2.5)(−1.5)(−0.5)
=
Γ0.5
(−2.5)(−1.5)(−0.5)
=−
1.772
1.875
= −0.945
case 𝒊𝒗. 𝚪𝒏 is not defined when 𝒏 = 𝟎 or a negative integer
We know Γ𝑛 =
Γ(𝑛+𝑘+1)
𝑛(𝑛+1)…(𝑛+𝑘)
, 𝑛 = 0, −1, −2, …
For all 𝑛 = 0, −1, −2, …, we will have a zero in the denominator
For instance, Γ0 =
Γ(0+𝑘+1)
, Γ(−1) =
0(1)…(0+𝑘)
Γ(−1+𝑘+1)
,…
(−1)(0)…(−1+𝑘)
Hence, we can conclude that gamma function cannot be defined for zero or negative integers.
Example 1 If 𝑛 is a positive integer, show that
1
2𝑛 Γ(𝑛 + ) = 1.3.5 … (2𝑛 − 1)√𝜋
2
1
1
Solution: Γ(𝑛 + ) = Γ (𝑛 − + 1)
2
2
1
1
= (𝑛 − ) Γ (𝑛 − )
2
2
1
3
⸪ Γ(𝑛 + 1) = 𝑛Γ𝑛
3
= (𝑛 − ) (𝑛 − ) Γ (𝑛 − )
2
2
2
⋮
1
3
5
3 1
1
= (𝑛 − ) (𝑛 − ) (𝑛 − ) … . . Γ
2
2
2
2 2
2
2𝑛−1
=(
2
2𝑛−3
)(
2
2𝑛−5
)(
2
3 1
) … 2 . 2 √𝜋
1
⇒ 2𝑛 Γ(𝑛 + ) = 1.3.5 … (2𝑛 − 1)√𝜋
2
Example 2 Evaluate the following integrals
∞
2
𝑖. ∫0 𝑒 −𝑥 𝑥 2𝑛−1 𝑑𝑥 , 𝑛 > 1
1
∞
𝑖𝑖. ∫0 𝑒 −√𝑥 𝑥 4 𝑑𝑥
∞ 𝑥𝑎
𝑖𝑖𝑖. ∫0 𝑥
𝑎
𝑑𝑥
1
1 𝑛−1
𝑖𝑣. ∫0 (log )
𝑑𝑥
𝑥
,𝑛>0
∞
Solution: 𝑖. We have Γ𝑛 = ∫0 𝑒 −𝑡 𝑡 𝑛−1 𝑑𝑡, … ①
Putting 𝑡 = 𝑥 2 in ① , we get
∞
2
Γ𝑛 = ∫0 𝑒 −𝑥 𝑥 2𝑛−2 . 2𝑥𝑑𝑥
∞
2
⇒ ∫0 𝑒 −𝑥 𝑥 2𝑛−1 𝑑𝑥 =
Γ𝑛
2
𝑖𝑖. Putting 𝑡 = √𝑥 in ① , we get
𝑛 1
∞
1
1
1
∞
𝑛
Γ𝑛 = ∫0 𝑒 −√𝑥 𝑥 2 −2 . 𝑥 −2 𝑑𝑥 = ∫0 𝑒 −√𝑥 𝑥 2 −1 𝑑𝑥
2
2
𝑛
1
5
2
4
2
Substituting − 1 = , i.e. 𝑛 = , we get
5
1
1
∞
Γ( ) = ∫0 𝑒 −√𝑥 𝑥 4 𝑑𝑥
2
2
1
∞
5
3 1
1
⸫ ∫0 𝑒 −√𝑥 𝑥 4 𝑑𝑥 = 2Γ ( ) = 2. . Γ ( ) =
2
2 2
2
𝑖𝑖𝑖. Putting 𝑎 𝑥 = 𝑒 𝑡 or 𝑥 log 𝑎 = 𝑡 ⇒𝑑𝑥 =
⸫
∞ 𝑥𝑎
∫0 𝑎𝑥
𝑑𝑥 =
3√𝜋
2
𝑑𝑡
log 𝑎
𝑎 𝑑𝑡
∞ −𝑡
𝑡
∫0 𝑒 (log 𝑎) log 𝑎
= (log
∞
1
Γ(𝑎+1)
𝑒 −𝑡 𝑡 (𝑎+1)−1 𝑑𝑡 = (log 𝑎+1
∫
𝑎+1
0
𝑎)
𝑎)
∞
𝑖𝑣. We have Γ𝑛 = ∫0 𝑒 −𝑡 𝑡 𝑛−1 𝑑𝑡
1
Putting 𝑡 = log ⇒ −𝑡 = log 𝑥 ⇒ 𝑒 −𝑡 = 𝑥
𝑥
1
Also 𝑑𝑡 = − 𝑑𝑥
𝑥
as 𝑡 = 0 ⇒ 𝑥 = 1 , 𝑡 = ∞ ⇒ 𝑥 = 0
⸫ Γ𝑛 =
1
0
1 𝑛−1
1
(− 𝑥) 𝑑𝑥
∫1 𝑥 (log 𝑥)
1 𝑛−1
⇒ ∫0 (log )
𝑥
=
1
1 𝑛−1
𝑑𝑥
∫0 (log 𝑥)
𝑑𝑥 = Γ𝑛
II Beta Function
Beta function is defined as:
𝟏
𝜷(𝒎, 𝒏) = ∫𝟎 𝒙𝒎−𝟏 (𝟏 − 𝒙)𝒏−𝟏 𝒅𝒙 ,
𝒎, 𝒏 > 𝟎
Important Results
4. Beta function is symmetric i.e. 𝜷(𝒎, 𝒏) = 𝜷(𝒏, 𝒎)
1
Proof: 𝛽(𝑚, 𝑛) = ∫0 𝑥 𝑚−1 (1 − 𝑥 )𝑛−1 𝑑𝑥 ,
𝑚, 𝑛 > 0
1
= ∫0 (1 − 𝑥)𝑚−1 (1 − (1 − 𝑥 )𝑛−1 𝑑𝑥 , 𝑚, 𝑛 > 0
𝑎
𝑎
⸪ ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥
1
= ∫0 𝑥 𝑛−1 (1 − 𝑥 )𝑚−1 𝑑𝑥 ,
𝑛, 𝑚 > 0
= 𝛽 (𝑛, 𝑚)
5. Another definition of Beta function:
𝒙𝒎−𝟏
𝒅𝒙 ,
𝒎, 𝒏 >
(𝟏+𝒙)𝒎+𝐧
1
∫0 𝑦 𝑚−1 (1 − 𝑦)𝑛−1 𝑑𝑦 ,
1
1
∞
𝜷(𝒎, 𝒏) = ∫𝟎
Proof: 𝛽(𝑚, 𝑛) =
Putting 𝑦 =
⇒ 𝛽 (𝑚, 𝑛) =
1+𝑥
, 𝑑𝑦 = −
(1+𝑥)2
𝑚−1
0
1
− ∫∞ ( )
1+𝑥
∞
1 𝑚−1
= ∫0 ( )
1+𝑥
∞
= ∫0
∞
= ∫0
𝟎
𝑚, 𝑛 > 0
𝑑𝑥
1
𝑛−1
(1 − 1+𝑥)
𝑥
𝑛−1
(1+𝑥)
1
1
(1+𝑥)2
2
𝑑𝑥 , 𝑚, 𝑛 > 0
(1+𝑥) 𝑑𝑥
𝑥 𝑛−1
𝑑𝑥
(1+𝑥)𝑚+𝑛
𝑥 𝑚−1
(1+𝑥)𝑚+𝑛
𝑑𝑥
⸪𝛽 (𝑚, 𝑛) = 𝛽 (𝑛, 𝑚)
6. Another form of Beta function is given by:
𝝅
𝟐
𝜷(𝒎, 𝒏) = 𝟐 ∫𝟎 𝒔𝒊𝒏𝟐𝒎−𝟏 𝜽 𝒄𝒐𝒔𝟐𝒏−𝟏 𝜽𝒅𝜽
1
Proof: we have 𝛽 (𝑚, 𝑛) = ∫0 𝑥 𝑚−1 (1 − 𝑥 )𝑛−1 𝑑𝑥
Let 𝑥 = 𝑠𝑖𝑛2 𝜃 ⇒ 𝑑𝑥 = 2 sin 𝜃 cos 𝜃 𝑑𝜃
𝜋
2
⸫ 𝛽 (𝑚, 𝑛) = ∫0 (𝑠𝑖𝑛2 𝜃 )𝑚−1 (𝑐𝑜𝑠 2 𝜃 )𝑛−1 2 sin 𝜃 cos 𝜃
𝜋
2
= 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑛−1 𝜃𝑑𝜃
7. Relation between Beta Gamma functions:
𝜷(𝒎, 𝒏) =
𝚪𝒎𝚪𝒏
𝚪(𝐦+𝒏)
,
𝒎, 𝒏 > 𝟎
∞
Proof: Using result 3, ∫0 𝑒 −𝑘𝑥 𝑥 𝑚−1 𝑑𝑥 =
Replacing 𝑘 by 𝑦 , we get
Γ𝑚
𝑦𝑚
Γ𝑚
𝑘𝑚
…①
∞
= ∫0 𝑒 −𝑦𝑥 𝑥 𝑚−1 𝑑𝑥
∞
⇒ Γ𝑚 = ∫0 𝑒 −𝑦𝑥 𝑦 𝑚 𝑥 𝑚−1 𝑑𝑥
∞
⇒ 𝑒 −𝑦 𝑦 𝑛−1 Γ𝑚 = ∫0 𝑒 −𝑦(1+𝑥) 𝑦 𝑚+𝑛−1 𝑥 𝑚−1 𝑑𝑥
Integrating both sides with respect to 𝑦 within limits 0 to ∞
∞
∞
∞
Γ𝑚 ∫0 𝑒 −𝑦 𝑦 𝑛−1 𝑑𝑦 = ∫0 ∫0 𝑒 −𝑦(1+𝑥) 𝑦 𝑚+𝑛−1 𝑥 𝑚−1 𝑑𝑥 𝑑𝑦
∞
∞
⇒ Γ𝑚Γ𝑛 = ∫0 [∫0 𝑒 −(1+𝑥)𝑦 𝑦 (𝑚+𝑛−1) 𝑑𝑦] 𝑥 𝑚−1 𝑑𝑥
∞ Γ(𝑚+𝑛)
⇒ Γ𝑚Γ𝑛 = ∫0
⇒
⇒
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
=
(1+𝑥)𝑚+𝑛
∞ 𝑥 𝑚−1
∫0 (1+𝑥)𝑚+n
𝑥 𝑚−1 𝑑𝑥 , comparing with ①
𝑑𝑥
= 𝛽 (𝑚, 𝑛), using result 5
𝒑+𝟏
𝝅
𝟐
8. ∫𝟎 𝒔𝒊𝒏𝒑 𝜽 𝒄𝒐𝒔𝒒 𝜽𝒅𝜽 =
𝒒+𝟏
𝚪( 𝟐 )𝚪( 𝟐 )
𝒑+𝒒+𝟐
𝟐𝚪( 𝟐 )
𝜋
2
Proof: we have 𝛽 (𝑚, 𝑛) = 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑛−1 𝜃𝑑𝜃
⇒
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
𝜋
2
= 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑛−1 𝜃𝑑𝜃 ⸪𝛽(𝑚, 𝑛) =
Replacing 2𝑚 − 1 by 𝑝 and 2𝑛 − 1 by 𝑞
i.e 𝑚 =
𝜋
2
𝑝
𝑞
𝑝+1
2
⇒ ∫0 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃𝑑𝜃 =
and 𝑛 =
𝑞+1
2
𝑝+1
𝑞+1
)Γ( 2 )
2
𝑝+𝑞+2
2Γ( 2 )
Γ(
𝜋
2
…①
𝑝
Putting 𝑞 = 0 in ①, we get ∫0 𝑠𝑖𝑛 𝜃𝑑𝜃 =
𝜋
2
Putting 𝑝 = 0 in ①, we get ∫0 𝑠𝑖𝑛𝑝 𝜃𝑑𝜃 =
9. Duplication formula is given by:
𝑝+1
1
𝑞+1
1
Γ( 2 )Γ(2)
𝑝+2
2Γ( 2 )
Γ( 2 )Γ(2)
𝑞+2
2Γ( 2 )
Γ𝑚Γ𝑛
Γ(m+𝑛)
𝟏
𝚪𝒎𝚪 (𝒎 + ) =
𝟐
√𝝅 .𝚪(𝟐𝒎)
,
𝟐𝟐𝒎−𝟏
𝒎>𝟎
Γ𝑚Γ𝑛
Proof: We have 𝛽 (𝑚, 𝑛) =
Γ(𝑚+𝑛)
𝜋
2
= 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑛−1 𝜃𝑑𝜃
𝜋
2
⸫ 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑛−1 𝜃𝑑𝜃 =
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
…①
1
Putting 𝑛 = on both sides, we get
2
𝜋
2
2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃𝑑𝜃 =
Γ𝑚√𝜋
1
Γ(𝑚+2)
…②
Again Putting 𝑛 = 𝑚 in ①, we get
(Γ𝑚)2
Γ(2𝑚)
𝜋
2
= 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑚−1 𝜃𝑑𝜃
𝜋
2
2 sin 𝜃 cos 𝜃 2𝑚−1
= 2 ∫0 (
=
=
=
1
2
)
𝑑𝜃
𝜋
2
∫ 𝑠𝑖𝑛2𝑚−1 2𝜃𝑑𝜃
22𝑚−2 0
𝜋
1
𝑠𝑖𝑛2𝑚−1 𝑡𝑑𝑡 , Putting
∫
2𝑚−1
0
2
2
2𝜃 = 𝑡
𝜋
2
∫ 𝑠𝑖𝑛2𝑚−1 𝑡𝑑𝑡
22𝑚−1 0
2𝑎
𝑎
⸪∫0 𝑓(𝑥 )𝑑𝑥 = 2 ∫0 𝑓 (𝑥 )𝑑𝑥
, if 𝑓 (2𝑎 − 𝑥 ) = 𝑓(𝑥)
⇒
(Γ𝑚)2
Γ(2𝑚)
=
𝜋
2
2
∫
22𝑚−1 0
𝑠𝑖𝑛2𝑚−1 𝜃𝑑𝜃
𝜋
2
⇒2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃𝑑𝜃 =
22𝑚−1 (Γ𝑚)2
Comparing ② and ③, we get
1
⇒ Γ𝑚Γ (𝑚 + ) =
2
10. Γ𝒏 Γ(𝟏 − 𝒏) =
Γ𝑚√𝜋
1
2
Γ(𝑚+ )
=
22𝑚−1 (Γ𝑚)2
Γ(2𝑚)
√𝜋 Γ(2𝑚)
22𝑚−1
𝝅
𝐬𝐢𝐧 𝒏𝝅
Proof: we have 𝛽 (𝑚, 𝑛) =
⇒
…③
Γ(2𝑚)
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
=
, 𝟎<𝒏<𝟏
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
=
∞ 𝑥 𝑛−1
∫0 (1+𝑥)𝑚+n 𝑑𝑥
∞ 𝑥 𝑛−1
∫0 (1+𝑥)𝑚+n 𝑑𝑥
Putting 𝑚 = 1 − 𝑛 on both sides, we get
Γ𝑛 Γ(1 − 𝑛) =
∞ 𝑥 𝑛−1
∫0 1+𝑥 𝑑𝑥
𝑡
Putting 𝑥 = 𝑒 𝑡 , 𝑑𝑥 = 𝑒 𝑑𝑡
As 𝑥 → 0 , 𝑡 → −∞ and as 𝑥 → ∞ , 𝑡 → ∞
∞
⸫ Γ𝑛 Γ(1 − 𝑛) = ∫−∞
𝑒 𝑛𝑡
1+𝑒 𝑡
𝑑𝑡
Now by using complex integration, we have:
∞ 𝑒 𝑛𝑡
∫−∞ 1+𝑒 𝑡
𝑑𝑡 =
𝜋
sin 𝑛𝜋
, 0<𝑛<1
, 𝑚, 𝑛 > 0
⸫ Γ𝑛 Γ(1 − 𝑛) =
𝜋
sin 𝑛𝜋
, 0<𝑛<1
𝜋
2
𝜋
2
5
2
3
Example 3 Evaluate 𝑖. ∫0 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥𝑑𝑥
10
𝑖𝑖. ∫0 𝑠𝑖𝑛
2
𝜋
𝑖𝑣. ∫0 𝑥 𝑚−1 (2 − 𝑥 )𝑛−1 𝑑𝑥
5
2
3
Solution: 𝑖. ∫0 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥𝑑𝑥 =
𝜋
2
11
10
𝑖𝑖. ∫0 𝑠𝑖𝑛
𝑥𝑑𝑥 =
1
Γ( 2 )Γ(2)
2Γ6
=
1
Γ( 2 )Γ(2 2 )
7
=
5
3+ +2
2Γ( 22 )
1.Γ(4)
=
11 7 7
2. 4 .4Γ(4)
97531 1
1
. . . . Γ( )Γ(2)
22222 2
8
77
Γ2Γ( )
4
15
2Γ( )
4
⸪ Γ2 = 1! = 1, also Γ(𝑛 + 1) = 𝑛Γ𝑛
945𝜋
=
240
7680
⸪ Γ6 = 5! = 120, also Γ(𝑛 + 1) = 𝑛Γ𝑛
=
63𝜋
512
1
⸪Γ ( ) = √𝜋
2
𝜋
2
𝜋
2
1
2
𝑖𝑖𝑖. ∫0 √tan 𝜃 + √sec 𝜃 𝑑𝜃 = ∫0 (𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠
=
1
1
+1
− +1
2
Γ( 2 )Γ( 22 )
1 1
− +2
2Γ(2 22 )
1 𝑛−1
𝑣. ∫0 𝑠𝑖𝑛2 𝜃 (1 + cos 𝜃 )4 𝑑𝜃 𝑣𝑖. ∫0 𝑥 𝑚−1 (log )
𝑥
7
=
𝑖𝑖𝑖. ∫0 √tan 𝜃 + √sec 𝜃 𝑑𝜃
5
+1
3+1
𝜋
2
𝑥𝑑𝑥
𝜋
2
1
−2
+
𝜃 + 𝑐𝑜𝑠
1
−2
1
− +1
1
Γ(2)Γ( 22 )
1
− +2
2Γ( 22 )
𝜃)𝑑𝜃
𝑑𝑥
3
1
Γ( )Γ( )
4
4
=
2Γ1
1
1
+
1
1
Γ( )Γ( )
2
4
3
2Γ(4)
3
= Γ ( ) {Γ ( ) +
2
4
4
√𝜋
3
}
Γ(4)
2
𝑖𝑣. Let 𝐼 = ∫0 𝑥 𝑚−1 (2 − 𝑥 )𝑛−1 𝑑𝑥
Putting 𝑥 = 2𝑠𝑖𝑛2 𝜃,
𝜋
2
𝐼 = ∫0 2𝑚−1 𝑠𝑖𝑛2𝑚−2 𝜃. 2𝑛−1 𝑐𝑜𝑠 2𝑛−2 𝜃. 2 sin 𝜃 cos 𝜃 𝑑𝜃
⇒𝐼=2
𝑚+𝑛−2
𝜋
2
. 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑚−1 𝜃𝑑𝜃 = 2𝑚+𝑛−2 𝛽(𝑚, 𝑛)
𝜋
2
⸪ 2 ∫0 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠 2𝑚−1 𝜃𝑑𝜃 = 𝛽 (𝑚, 𝑛)
𝜋
𝑣. Let 𝐼 = ∫0 𝑠𝑖𝑛2 𝜃(1 + cos 𝜃 )4 𝑑𝜃
𝜋
𝜃 2
𝜃
𝜃 4
= ∫0 (2 sin cos ) (2𝑐𝑜𝑠 2 ) 𝑑𝜃
2
2
2
𝜋
𝜃
𝜃
2
2
= 64 ∫0 𝑠𝑖𝑛2 𝑐𝑜𝑠10 𝑑𝜃
Putting
𝜃
2
= 𝑥, 𝑑𝜃 = 2𝑥𝑑𝑥
𝜋
2
= 128 ∫0 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠10 𝑥 𝑑𝑥
3
=
11
64.Γ(2)Γ( 2 )
Γ7
1
=
1 97531
1
64. 2Γ(2).2.2.2.2.2Γ(2)
720
=
21𝜋
16
⸪ Γ7 = 6! = 720, also Γ(𝑛 + 1) = 𝑛Γ𝑛
𝑣𝑖.
1 𝑚−1
1 𝑛−1
Let 𝐼 = ∫0 𝑥
𝑑𝑥
(log 𝑥)
1
Putting log = 𝑡 or 𝑥 = 𝑒 −𝑡 ⇒𝑑𝑥
𝑥
0 −(𝑚−1)𝑡 𝑛−1 −𝑡
𝐼 = − ∫∞ 𝑒
𝑡
𝑒 𝑑𝑡
∞
= ∫0 𝑒 −𝑚𝑡 𝑡 𝑛−1 𝑑𝑡
= −𝑒 −𝑡 𝑑𝑡,
Putting 𝑚𝑡 = 𝑦
𝐼=
∞ −𝑦 𝑦 𝑛−1
𝑑𝑦
∫ 𝑒 (𝑚)
𝑚 0
1
=
1
∞
Γ𝑛
𝑒 −𝑦 𝑦 𝑛−1 𝑑𝑦 = 𝑛
∫
𝑛
0
𝑚
𝑚
Example 4 Prove that 𝑖. 𝛽 (𝑚, 𝑛) = 𝛽 (𝑚 + 1, 𝑛) + 𝛽 (𝑚, 𝑛 + 1)
𝑖𝑖.
𝛽(𝑚+1,𝑛)
𝛽(𝑚,𝑛)
1
=
𝑚
𝑚+𝑛
𝑖𝑖𝑖. 𝛽 (𝑚, ) = 22𝑚−1 𝛽(𝑚, 𝑚)
2
Solution: 𝑖. R.H.S. = 𝛽 (𝑚 + 1, 𝑛) + 𝛽 (𝑚, 𝑛 + 1)
=
=
=
=
Γ(𝑚+1)Γ𝑛
Γ(𝑚+𝑛+1)
+
Γ𝑚Γ(𝑛+1)
Γ(𝑚+𝑛+1)
𝑚Γ𝑚.Γ𝑛+Γ𝑚.𝑛Γ𝑛
Γ(𝑚+𝑛+1)
Γ𝑚.Γ𝑛(𝑚+𝑛)
(𝑚+𝑛)Γ(𝑚+𝑛)
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
= 𝛽 (𝑚, 𝑛) = L.H.S.
𝑖𝑖. L.H.S.=
=
𝛽(𝑚+1,𝑛)
𝛽(𝑚,𝑛)
=
Γ(𝑚+1)Γ𝑛
Γ(𝑚+𝑛+1)
𝑚Γ𝑚Γ𝑛
(𝑚+𝑛)Γ(𝑚+𝑛)
1
.
.
Γ(𝑚+𝑛)
Γ𝑚Γ𝑛
Γ(𝑚+𝑛)
Γ𝑚Γ𝑛
=
𝑚
𝑚+𝑛
= R.H.S.
1
𝑖𝑖𝑖. We have 𝛽 (𝑚, ) =
2
ΓmΓ(2)
…①
1
Γ(𝑚+ )
2
1
Again, by Duplication formula Γ𝑚Γ (𝑚 + ) =
2
1
⸫ Γ (𝑚 + ) =
2
√𝜋 Γ(2𝑚)
22𝑚−1 Γ𝑚
√𝜋 Γ(2𝑚)
22𝑚−1
…②
1
1
Using ② in ①, we get 𝛽 (𝑚, ) =
2
ΓmΓ( )22𝑚−1 Γ𝑚
2
√𝜋 Γ(2𝑚)
= 22𝑚−1
ΓmΓ𝑚
Γ(2𝑚)
1
⸪ Γ ( ) = √𝜋
2
= 22𝑚−1 𝛽(𝑚, 𝑚)
Example 5 Express the following integrals in terms of Beta function
1
1
𝑖. ∫0 𝑥 𝑚 (1 − 𝑥 2 )𝑛 𝑑𝑥 𝑖𝑖. ∫0
1
Solution: 𝑖. Let 𝐼 = ∫0 𝑥 𝑚 (1 −
Putting 𝑥 2 = 𝑦
⇒ 𝐼=
𝑑𝑥
√1−𝑥 5
1 1
𝑥 2 )𝑛 𝑑𝑥 = ∫0 𝑥 𝑚−1 (1
2
⇒2𝑥𝑑𝑥 = 𝑑𝑦
1 𝑚−1
∫ 𝑦 2 (1
2 0
1
𝑥2
− 𝑦)𝑛 𝑑𝑦
− 𝑥 2 )𝑛 2𝑥𝑑𝑥
1
1
= ∫0 𝑦
2
𝑚+1
−1
2
1
𝑚+1
2
2
𝑥2
= 𝛽(
1
𝑖𝑖. Let 𝐼 = ∫0
1
, 𝑛 + 1)
√1−𝑥
Putting 𝑥 5 = 𝑦
(1 − 𝑦)(𝑛+1)−1 𝑑𝑦
⇒5𝑥 4 𝑑𝑥 = 𝑑𝑦
2
1
1
1
1
−2 (
5 )−2
𝑑𝑥
=
𝑥
1
−
𝑥
5𝑥 4 𝑑𝑥
∫
5
5 0
1
⇒ 𝐼 = ∫0 𝑦 −5 (1 − 𝑦)−2 𝑑𝑦
5
=
1 3−1
∫ 𝑦 5 (1
5 0
3
1
− 𝑦)
1
−1
2
1
3 1
𝑑𝑦 = 𝛽 ( , )
5
5 2
3
1
Example 6 Prove that 𝑖. Γ ( − 𝑥) Γ ( + 𝑥) = ( − 𝑥 2 ) 𝜋. sec 𝜋𝑥
2
2
4
𝑏
𝑖𝑖. ∫𝑎 (𝑥 − 𝑎)𝑚 (𝑏 − 𝑥 )𝑛 𝑑𝑥 = (𝑏 − 𝑎)𝑚+𝑛+1 𝛽(𝑚 + 1, 𝑛 + 1)
3
3
2
2
Solution 𝑖. L.H.S.= Γ ( − 𝑥) Γ ( + 𝑥)
1
1
1
1
2
2
2
2
= ( − 𝑥) Γ ( − 𝑥) . ( + 𝑥) Γ ( + 𝑥)
⸪ Γ(𝑛 + 1) = 𝑛Γ𝑛
1
1
1
= ( − 𝑥 2 ) Γ ( + 𝑥) Γ (1 − ( + 𝑥))
4
2
2
1
1
2
2
⸪ Γ ( − 𝑥) = Γ (1 − ( + 𝑥))
1
= ( − 𝑥2)
4
𝜋
1
sin(2+𝑥)𝜋
1
,0< +𝑥 <1
2
⸪ Γ𝑛 Γ(1 − 𝑛) =
1
𝜋
sin 𝑛𝜋
,0<𝑛<1
𝜋
= ( − 𝑥2)
4
cos 𝜋𝑥
1
1
1
= ( − 𝑥 2 ) 𝜋. sec 𝜋𝑥 , − < 𝑥 <
4
2
2
𝑏
𝑖𝑖. Let 𝐼 = ∫𝑎 (𝑥 − 𝑎)𝑚 (𝑏 − 𝑥 )𝑛 𝑑𝑥
𝑏−𝑎
= ∫0
𝑦 𝑚 (𝑏 − 𝑎 − 𝑦)𝑛 𝑑𝑦
By putting 𝑥 − 𝑎 = 𝑦
1
= ∫0 (𝑏 − 𝑎)𝑚 𝑡 𝑚 (𝑏 − 𝑎 − (𝑏 − 𝑎)𝑡)𝑛 (𝑏 − 𝑎)𝑑𝑡
By putting 𝑦 = (𝑏 − 𝑎)𝑡
1
1
𝐼 = ∫0 𝑥 𝑚 (1 − 𝑥 𝑛 )𝑝 𝑑𝑥 = ∫0 (𝑏 − 𝑎)𝑚 𝑡 𝑚 (𝑏 − 𝑎)𝑛 (1 − 𝑡)𝑛 (𝑏 − 𝑎)𝑑𝑡
1
= (𝑏 − 𝑎)𝑚+𝑛−1 ∫0 𝑡 𝑚 (1 − 𝑡)𝑛 𝑑𝑡
1
= (𝑏 − 𝑎)𝑚+𝑛−1 ∫0 𝑡 (𝑚+1)−1 (1 − 𝑡)(𝑛+1)−1 𝑑𝑡
= (𝑏 − 𝑎)𝑚+𝑛+1 𝛽(𝑚 + 1, 𝑛 + 1)
1
Example 7 Express the integral ∫0 𝑥 𝑚 (1 − 𝑥 𝑛 )𝑝 𝑑𝑥 in terms of gamma function and hence evaluate
1 3
∫0 𝑥 2 (1
1
1 2
2
− 𝑥 ) 𝑑𝑥
1
Solution: Let 𝐼 = ∫0 𝑥 𝑚 (1 − 𝑥 𝑛 )𝑝 𝑑𝑥
Putting 𝑥 𝑛 =t, so that 𝑛𝑥 𝑛−1 𝑑𝑥 = 𝑑𝑡, we get
1
1 𝑚
𝐼 = ∫0 𝑡 𝑛 (1 − 𝑡)𝑝 𝑡 −
𝑛
⸫𝐼 =
1
∫0 𝑥 𝑚
𝑛−1
𝑛
1
𝑚+1
𝑛
𝑛
𝑛 )𝑝
𝑑𝑥 = 𝛽 (
3
1
1
2
2
2
(1 − 𝑥
1 𝑚+1−1
1
𝑑𝑡 = ∫0 𝑡
𝑛
𝑛
(1 − 𝑡)𝑝+1−1 𝑑𝑡
𝑚+1
1 Γ( 𝑛 )Γ(𝑝+1)
𝑛 Γ(𝑚+1+𝑝+1)
𝑛
, 𝑝 + 1) =
…①
Putting 𝑚 = , 𝑛 = , 𝑝 = in ①, we get
1 3
∫0 𝑥 2 (1
1
1 2
2
3
+1
2
1
2
3
2Γ(5)Γ(2)
3
Γ(5+ )
2
− 𝑥 ) 𝑑𝑥 = 2𝛽 (
=
1
3
, + 1) = 2𝛽 (5, )
2
2
3
=
2.4! Γ(2)
13
Γ( )
2
3
48Γ(2)
= 11 9 7 5 3
3
. . . . .Γ( )
2 2222
2
∞
=
512
3465
∞
Example 8 Evaluate 𝑖. ∫0 𝑥 𝑛−1 cos 𝑎𝑥 𝑑𝑥 𝑖𝑖. ∫0 𝑥 𝑛−1 sin 𝑎𝑥 𝑑𝑥
∞
∞
Solution: let 𝐼 = ∫0 𝑥 𝑛−1 𝑒 −𝑖𝑎𝑥 𝑑𝑥 = ∫0 𝑥 𝑛−1 (cos 𝑎𝑥 − 𝑖sin 𝑎𝑥) 𝑑𝑥
Putting 𝑖𝑎𝑥 = 𝑡 , 𝑑𝑥 =
⸫𝐼 =
=
1
𝑖𝑎
𝑛−1
𝑡
∞
∫ 𝑒 −𝑡 (𝑖𝑎)
𝑖𝑎 0
Γ𝑛
𝑎𝑛
(−𝑖
𝑑𝑡
)𝑛
=
=
Γ𝑛
𝑎𝑛
Γ𝑛
𝑑𝑡 =
1
∞
Γ𝑛
∫ 𝑒 −𝑡 𝑡 𝑛−1 𝑑𝑡 = 𝑖 𝑛 𝑎𝑛
𝑖 𝑛 𝑎𝑛 0
𝜋 𝑛
𝜋
(cos 2 − 𝑖 sin 2 )
(cos
𝑎𝑛
𝑛𝜋
2
− 𝑖 sin
∞
⸫∫0 𝑥 𝑛−1 (cos 𝑎𝑥 − 𝑖sin 𝑎𝑥) 𝑑𝑥 =
Comparing real and imaginary parts we get,
𝑛𝜋
2
Γ𝑛
𝑎𝑛
)
(cos
𝑛𝜋
2
− 𝑖 sin
𝑛𝜋
2
)
∞
𝑖. ∫0 𝑥 𝑛−1 cos 𝑎𝑥 𝑑𝑥 =
∞
𝑖𝑖. ∫0 𝑥 𝑛−1 sin 𝑎𝑥 𝑑𝑥 =
Γ𝑛
𝑛𝜋
𝑎
Γ𝑛
2
𝑛𝜋
𝑛 cos
𝑎𝑛
sin
2
Exercise
𝜋
2
𝜋
1. Show that ∫0 𝑡𝑎𝑛𝑛 𝑥 𝑑𝑥 =
𝑛𝜋
sec ( ) , 0 < 𝑛 < 1
2
2
8
5
2. Given that Γ ( ) = 0.8935 , find the values of Γ (− ) and Γ (−
5
2
3 1
3. Find 𝑖. 𝛽 ( , )
2 2
1
4 5
𝑖𝑖. 𝛽 ( , )
3 3
4. Evaluate ∫0 𝑥 2 (1 − 𝑥 2 )4 𝑑𝑥
∞
2
5. Prove that Γ𝑛 = 2 ∫0 𝑒 −𝑥 𝑥 2𝑛−1 𝑑𝑥
6. Prove that for 𝑎, 𝑏 > 0
∞
Γ𝑛
𝑖. ∫0 𝑥 𝑛−1 𝑒 −𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 = 2 2
∞
𝑖𝑖. ∫0 𝑥 𝑛−1 𝑒 −𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 =
7. Show that
𝛽(𝑚+1,𝑛)
𝜋
2
8. Show that ∫0
9. Show that
𝑚
1
=
𝑏
𝑛
) ⁄2
(𝑎 +𝑏
Γ𝑛
𝑛
(𝑎2 +𝑏2 ) ⁄2
𝛽(𝑚,𝑛)
𝑚+𝑛+1
𝑑𝑥.
𝜋
2
√sin 𝑥 𝑑𝑥 = 𝜋
∫0
√sin 𝑥
1 𝑥 𝑚−1 +𝑥 𝑛−1
∫0 (1+𝑥)𝑚+𝑛 𝑑𝑥
= 𝛽 (𝑚, 𝑛)
cos (𝑛 tan−1 )
𝑎
𝑏
sin (𝑛 tan−1 )
𝑎
12
5
)
1
10. Prove that 𝛽 (𝑚, ) = 22𝑚−1 . 𝛽 (𝑚, 𝑚)
2
Answers
2. −
3. 𝑖.
4.
8
15
𝜋
2
128
3465
√𝜋 , −1.108
𝑖𝑖.
2𝜋
9√3