Btech_II_ engineering mathematics_unit2
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1. The document discusses the gamma and beta functions, which are defined in terms of improper definite integrals and belong to the category of special transcendental functions. 2. Several properties and examples involving the gamma and beta functions are provided, including their relationship via the equation β(m,n)= Γ(m)Γ(n)/Γ(m+n). 3. Dirichlet's integral and its extension to calculating areas and volumes are covered. Four examples demonstrating the application of gamma and beta functions are worked out.
![Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 7
Γ𝑛 . Γ(𝑛 +
1
2
) =
√𝜋 Γ(2𝑛)
22𝑛−1
4.2 EXAMPLES:
Example 1: Evaluate ∫ 𝒙 𝟒
(𝟏− √ 𝒙)
𝟓
𝒅𝒙
𝟏
𝟎
Solution: Let √ 𝑥 = 𝑡 ⟹ 𝑥 = 𝑡2
so that 𝑑𝑥 = 2𝑡 𝑑𝑡
∫ 𝑥4
(1− √ 𝑥)
5
𝑑𝑥 =
1
0
∫( 𝑡2)4 (1 − 𝑡)5
(2𝑡 𝑑𝑡)
1
0
= 2 ∫ 𝑡9
(1 − 𝑡)5
𝑑𝑡
1
0
= 2 𝐵(10,6)
= 2
Γ10 Γ6
Γ16
= 2 ×
9!5!
15!
=
2×1×2×3×4×5
15×14×13×12×11×10
=
1
11×13×7×15
=
1
15015
∴ ∫ 𝑥4
(1 − √ 𝑥)
5
𝑑𝑥 =
1
0
1
15015
_________ Ans.
Example 2: Find the value of 𝚪 (
𝟏
𝟐
).
Solution: We know that,
∫ 𝑠𝑖𝑛 𝑝
𝜃 𝑐𝑜𝑠 𝑝
𝜃 𝑑𝜃 =
Γ(
𝑝+1
2
).(
𝑞+1
2
)
2Γ(
𝑝+𝑞+2
2
)
𝜋
2⁄
0
Putting 𝑝 = 𝑞 = 0, we get ∫ 𝑑𝜃 =
𝚪(
𝟏
𝟐
) 𝚪(
𝟏
𝟐
)
2 𝚪𝟏
𝜋
2
0
[ 𝜃]0
𝜋 2⁄
=
1
2
(Γ
1
2
)
2](https://image.slidesharecdn.com/b-150317055747-conversion-gate01/85/Btech_II_-engineering-mathematics_unit2-7-320.jpg)
![Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 9
= 2 𝑝+𝑞
∫ 𝑠𝑖𝑛2𝑞−1
𝜃 𝑐𝑜𝑠2𝑝−1
𝜃
∞
0
𝑑𝜃
= 2 𝑝+𝑞
Γ(
2𝑞
2
) Γ(
2𝑝
2
)
2Γ(
2𝑝+2𝑞
2
)
= 2 𝑝+𝑞−1 Γ( 𝑝) Γ( 𝑞)
Γ( 𝑝+𝑞)
__________Ans.
Example 5: Show that 𝚪( 𝒏) 𝚪( 𝟏− 𝒏) =
𝝅
𝒔𝒊𝒏 𝒏𝝅
(𝟎 < 𝑛 < 1)
Solution: We know that
𝛽( 𝑚, 𝑛) = ∫
𝑥 𝑛−1
(1 + 𝑥) 𝑚+𝑛
𝑑𝑥
∞
0
Γ𝑚 Γ𝑛
Γ(𝑚+𝑛)
= ∫
𝑥 𝑛−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
∞
0
Putting 𝑚 + 𝑛 = 1 𝑜𝑟 𝑚 = 1 − 𝑛, we get
Γ(1−𝑛) Γ𝑛
Γ1
= ∫
𝑥 𝑛−1
(1+𝑥)1
𝑑𝑥
∞
0
Γ(1 − 𝑛)Γ𝑛 = ∫
𝑥 𝑛−1
1+𝑥
𝑑𝑥
∞
0
[∵ ∫
𝑥 𝑛−1
1+𝑥
𝑑𝑥
∞
0
=
𝜋
𝑠𝑖𝑛 𝑛𝜋
]
∴ Γ( 𝑛)Γ(1− 𝑛) =
𝜋
𝑠𝑖𝑛 𝑛𝜋
______proved.
4.3 EXERCISE:
1) Evaluate ∫ (1 − 𝑥3)−1 2⁄
𝑑𝑥
1
0
2) Evaluate ∫
𝑥 𝑚−1+𝑥 𝑛−1
(1+𝑥) 𝑚+𝑛
𝑑𝑥
1
0
3) Evaluate ∫ (
𝑥3
1−𝑥3
)
1
2
𝑑𝑥
1
0
4) Prove that Γ (
1
4
) Γ (
3
4
) = 𝜋√2
5) Show that 𝛽( 𝑝, 𝑞) = 𝛽( 𝑝 + 1, 𝑞) + (𝑝, 𝑞 + 1)
5.1 DIRICHLET’S INTEGRAL:
If 𝑙, 𝑚, 𝑛 are all positive, then the triple integral
∭ 𝑥 𝑙−1
𝑦 𝑚−1
𝑧 𝑛−1
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑉
=
Γ(l)Γ(m)Γ(n)
Γ(𝑙 + 𝑚 + 𝑛 + 1)](https://image.slidesharecdn.com/b-150317055747-conversion-gate01/85/Btech_II_-engineering-mathematics_unit2-9-320.jpg)
![Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 11
= ∫ 𝑡9
(1− 𝑡)4
𝑑𝑡
1
0
= 𝛽(10,5) ________(3)
∴ 𝐼 = 𝐼1 + 𝐼2
= 𝛽(5,10) + 𝛽(10,5) [Using(2) and (3)]
= 𝛽(5,10) + 𝛽(5,10) [𝛽( 𝑚, 𝑛) = 𝛽(𝑛, 𝑚)]
= 2𝛽(5,10)
=
2Γ5Γ10
Γ15
=
2×4!×9!
14!
=
2×4×3×2×1×9!
14×13×12×11×10×9!
=
1
5005
_______ Proved.
5.3 EXERCISE:
1) Find the value of ∫
𝑥3−2𝑥4+𝑥5
(1+𝑥)7
𝑑𝑥
1
0
2) Show that ∫
𝑥 𝑚−1(1−𝑥) 𝑛−1
(𝑎+𝑥) 𝑚+𝑛
𝑑𝑥 =
𝛽(𝑚,𝑛)
𝑎 𝑛(1+𝑎) 𝑚
1
0
3) 𝛽( 𝑚 + 1, 𝑛) =
𝑚
𝑚+𝑛
𝛽(𝑚, 𝑛)
6.1 Application to Area & Volume:
Liouville’s extension of dirichlet theorem:](https://image.slidesharecdn.com/b-150317055747-conversion-gate01/85/Btech_II_-engineering-mathematics_unit2-11-320.jpg)
![Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 12
∭ 𝑓(𝑥 + 𝑦 + 𝑧)𝑥 𝑙−1
𝑦 𝑚−1
𝑧 𝑛−1
𝑑𝑥 𝑑𝑦 𝑑𝑧
=
Γ(l)Γ(m)Γ(n)
Γ(l + m + n)
∫ 𝑓( 𝑢) 𝑢𝑙+𝑚+𝑛−1
𝑑𝑢
ℎ2
ℎ1
Example1: Show that ∭
𝒅𝒙 𝒅𝒚 𝒅𝒛
(𝒙+𝒚+𝒛+𝟏) 𝟑
=
𝟏
𝟐
𝒍𝒐𝒈𝟐−
𝟓
𝟏𝟔
, the integral being
takenthroughout the volume bounded by
𝒙 = 𝟎, 𝒚 = 𝟎, 𝒛 = 𝟎, 𝒙 + 𝒚 + 𝒛 = 𝟏.
Solution: By Liouville’s theorem when 0 < 𝑥 + 𝑦 + 𝑧 < 1
∭
𝑑𝑥 𝑑𝑦 𝑑𝑧
(𝑥+𝑦+𝑧+1)3
= ∭
𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧
(𝑥+𝑦+𝑧+1)3
(0 ≤ 𝑥 + 𝑦 + 𝑧 ≤ 1)
=
Γ1Γ1Γ1
Γ(l+m+n)
∫
1
(u+1)3
u3−1
du
1
0
=
1
2
∫
𝑢2
(𝑢+1)3
𝑑𝑢
1
0
= ∫ [
1
𝑢+1
−
2
(𝑢+1)2
+
1
(𝑢+1)3
] 𝑑𝑢
1
0
(Partial fractions)
=
1
2
[log( 𝑢 + 1) +
2
𝑢+1
−
1
2(𝑢+1)2
]
0
1
=
1
2
[𝑙𝑜𝑔2 + 2(
1
2
− 1) − (
1
8
−
1
2
)]
=
1
2
𝑙𝑜𝑔2 −
5
16
∴ ∭
𝒅𝒙 𝒅𝒚 𝒅𝒛
(𝒙+𝒚+𝒛+𝟏) 𝟑
=
𝟏
𝟐
𝒍𝒐𝒈𝟐−
𝟓
𝟏𝟔
_______Proved.
Example 2: Find the mass of an octant of the ellipsoid
𝒙 𝟐
𝒂 𝟐
+
𝒚 𝟐
𝒃 𝟐
+
𝒛 𝟐
𝒄 𝟐
= 𝟏,
the density at any point being 𝝆 = 𝒌 𝒙 𝒚 𝒛.
Solution: Mass = ∭ 𝜌 𝑑𝑣
= ∭( 𝑘 𝑥 𝑦 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧](https://image.slidesharecdn.com/b-150317055747-conversion-gate01/85/Btech_II_-engineering-mathematics_unit2-12-320.jpg)
Btech_II_ engineering mathematics_unit2
- 1. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 1 Course: B.Tech- II Subject: Engineering Mathematics II Unit-2 RAI UNIVERSITY, AHMEDABAD
- 2. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 2 Unit-II: GAMMA, BETA FUNCTION Sr. No. Name of the Topic Page No. 1 Definition of Gamma function 2 2 Examples Based on Gamma Function 3 3 Beta function 5 4 Relation between Beta and Gamma Functions 5 5 Dirichlet’s Integral 9 6 Application to Area & Volume: Liouville’s extension of dirichlet theorem 11 7 Reference Book 13
- 3. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 3 GAMMA, BETA FUNCTION The Gamma function and Beta functions belong to the category of the special transcendental functions and are defined in terms of improper definite integrals. 1.1 Definition of Gamma function : The gamma function is denoted and defined by the integral Γ𝑚 = ∫ 𝑒−𝑥 𝑥 𝑚−1 𝑑𝑥 (𝑚 > 0) ∞ 0 1.2 Properties of Gamma function : 1) Γ( 𝑚 + 1) = 𝑚Γ𝑚 2) Γ( 𝑚 + 1) = 𝑚! When m is a positive integer. 3) Γ( 𝑚 + 𝑎) = ( 𝑚 + 𝑎 − 1)( 𝑚 + 𝑎 − 2)……… 𝑎Γ𝑎, when n is a positive integer. 4) Γ𝑚 = 2 ∫ 𝑒−𝑥2 𝑥2𝑚−1 𝑑𝑥 ( 𝑚 > 0) ∞ 0 5) Γ𝑚 𝑡 𝑚 = ∫ 𝑒−𝑡𝑥 𝑥 𝑚−1 𝑑𝑥 ( 𝑚 > 0) ∞ 0 6) Γ 1 2 = √ 𝜋 7) ∫ 𝑒−𝑥2 𝑑𝑥 = √𝜋 2 ∞ 0 8) ∫ 𝑥 𝑛 (𝑙𝑜𝑔 𝑥) 𝑚 𝑑𝑥 = (−1) 𝑚 ( 𝑛+1) 𝑚+1 Γ(𝑚 + 1) 1 0 2.1 Examples Based on Gamma Function: Example 1: Evaluate 𝚪(− 𝟏 𝟐 ).
- 4. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 4 Solution: We know that Γ( 𝑚 + 1) = 𝑚Γ𝑚 Γ (− 1 2 + 1) = − 1 2 Γ (− 1 2 ) Γ ( 1 2 ) = − 1 2 Γ(− 1 2 ) √ 𝜋 = − 1 2 Γ (− 1 2 ) ∴ 𝚪(− 𝟏 𝟐 ) = −𝟐√ 𝝅. __________Ans. Example 2: Evaluate ∫ √ 𝒙𝟒 𝒆−√𝒙 𝒅𝒙 ∞ 𝟎 Solution: Let 𝐼 = ∫ 𝑥 1 4 𝑒−√𝑥 𝑑𝑥 ∞ 0 __________(i) Putting √ 𝑥 = 𝑡 ⟹ 𝑥 = 𝑡2 so that 𝑑𝑥 = 2𝑡 in (i), we get 𝐼 = ∫ 𝑡1 2⁄ 𝑒−𝑡 2𝑡 𝑑𝑡 ∞ 0 = 2∫ 𝑡3 2⁄ 𝑒−𝑡 𝑑𝑡 ∞ 0 = 2∫ 𝑡 5 2 −1 𝑒−𝑡 𝑑𝑡 ∞ 0 = 2Γ( 5 2 ) = (2 × 3 2 )Γ ( 3 2 ) = (2 × 3 2 × 1 2 )Γ ( 1 2 ) = 3 2 √ 𝜋 ∴ ∫ √ 𝒙𝟒 𝒆−√𝒙 𝒅𝒙 ∞ 𝟎 = 𝟑 𝟐 √ 𝝅 ________Ans. Example 3: Evaluate ∫ 𝒙 𝒂 𝒂 𝒙 𝒅𝒙 ∞ 𝟎 . Solution: Let 𝐼 = ∫ 𝑥 𝑎 𝑎 𝑥 𝑑𝑥 ∞ 0 _______ (i)
- 5. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 5 Putting 𝑎 𝑥 = 𝑒 𝑡 ⟹ 𝑥 log 𝑎 = 𝑡 ⟹ 𝑥 = 1 log 𝑎 ⟹ 𝑑𝑥 = 𝑑𝑡 log 𝑎 in (i), we have 𝐼 = ∫ ( 𝑡 log 𝑎 ) 𝑎 𝑒−𝑡∞ 0 𝑑𝑡 log 𝑎 = 1 (log 𝑎) 𝑎+1 ∫ 𝑒−𝑡 𝑡 𝑎 𝑑𝑡 ∞ 0 = 1 (log 𝑎) 𝑎+1 ∫ 𝑡( 𝑎+1)−1 𝑒−𝑡 𝑑𝑡 ∞ 0 = 1 (log 𝑎) 𝑎+1 Γ(𝑎 + 1) ∴ ∫ 𝒙 𝒂 𝒂 𝒙 𝒅𝒙 ∞ 𝟎 = 𝟏 ( 𝐥𝐨𝐠 𝒂) 𝒂+𝟏 𝚪(𝒂 + 𝟏) ________ Ans. Example 4: Prove that ∫ ( 𝒙 𝒍𝒐𝒈𝒙) 𝟒 𝒅𝒙 = 𝟒! 𝟓 𝟓 𝟏 𝟎 Solution: We know that ∫ 𝑥 𝑛 (𝑙𝑜𝑔 𝑥) 𝑚 𝑑𝑥 = (−1) 𝑚 ( 𝑛+1) 𝑚+1 Γ(𝑚 + 1) 1 0 _______(i) Now, ∫ ( 𝑥 𝑙𝑜𝑔𝑥)4 𝑑𝑥 = 1 0 ∫ 𝑥41 0 ( 𝑙𝑜𝑔𝑥)4 𝑑𝑥 Putting 𝑛 = 𝑚 = 4 in (i), we get ∫ 𝑥4 1 0 ( 𝑙𝑜𝑔𝑥)4 𝑑𝑥 = (−1)4 (4 + 1)4+1 Γ(4 + 1) = Γ5 55 = 4! 55 __________ proved. 2.2 EXERCISE:
- 6. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 6 1) Evaluate: (a) Γ(− 3 2 ) (b) Γ ( 7 2 ) (c)Γ(0) 2) ∫ 𝑒−ℎ2 𝑥2 𝑑𝑥 ∞ 0 3) ∫ 𝑑𝑥 √−𝑙𝑜𝑔𝑥 1 0 4) ∫ ( 𝑥 𝑙𝑜𝑔𝑥)3 𝑑𝑥 1 0 3.1 BETA FUNCTION: Definition: The Beta function denoted by 𝛽( 𝑚, 𝑛) or 𝐵(𝑚, 𝑛) is defined as 𝐵( 𝑚, 𝑛) = ∫ 𝑥 𝑚−1 (1 − 𝑥) 𝑛−1 𝑑𝑥, (𝑚 > 0, 𝑛 > 0) 1 0 3.2 Properties of Beta function: 1) B(m,n)= B(n,m) 2) 𝐵( 𝑚, 𝑛) = 2 ∫ 𝑠𝑖𝑛2𝑚−1 𝜃 𝑐𝑜𝑠2𝑛−1 𝜃 𝑑𝜃 𝜋 2⁄ 0 3) 𝐵( 𝑚, 𝑛) = ∫ 𝑥 𝑚−1 (1+𝑥) 𝑚+𝑛 𝑑𝑥 ∞ 0 4) 𝐵( 𝑚, 𝑛) = ∫ 𝑥 𝑚−1+𝑥 𝑛−1 (1+𝑥) 𝑚+𝑛 𝑑𝑥 1 0 4.1 RelationbetweenBeta and Gamma Functions: Relation between Beta and gamma functions is 𝛽( 𝑚, 𝑛) = Γm .Γn Γ(m+n) Using above relation we can derive following results: ∫ 𝑠𝑖𝑛 𝑝 𝜃 𝑐𝑜𝑠 𝑝 𝜃 𝑑𝜃 = 1 2 𝛽 ( 𝑝+1 2 , 𝑞+1 2 ) = Γ( 𝑝+1 2 ).( 𝑞+1 2 ) 2Γ( 𝑝+𝑞+2 2 ) 𝜋 2⁄ 0 Γ ( 1 2 ) = √ 𝜋 Euler’s formula: Γ𝑛 . Γ(1 − 𝑛) = 𝜋 sin 𝑛𝜋 Duplication formula:
- 7. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 7 Γ𝑛 . Γ(𝑛 + 1 2 ) = √𝜋 Γ(2𝑛) 22𝑛−1 4.2 EXAMPLES: Example 1: Evaluate ∫ 𝒙 𝟒 (𝟏− √ 𝒙) 𝟓 𝒅𝒙 𝟏 𝟎 Solution: Let √ 𝑥 = 𝑡 ⟹ 𝑥 = 𝑡2 so that 𝑑𝑥 = 2𝑡 𝑑𝑡 ∫ 𝑥4 (1− √ 𝑥) 5 𝑑𝑥 = 1 0 ∫( 𝑡2)4 (1 − 𝑡)5 (2𝑡 𝑑𝑡) 1 0 = 2 ∫ 𝑡9 (1 − 𝑡)5 𝑑𝑡 1 0 = 2 𝐵(10,6) = 2 Γ10 Γ6 Γ16 = 2 × 9!5! 15! = 2×1×2×3×4×5 15×14×13×12×11×10 = 1 11×13×7×15 = 1 15015 ∴ ∫ 𝑥4 (1 − √ 𝑥) 5 𝑑𝑥 = 1 0 1 15015 _________ Ans. Example 2: Find the value of 𝚪 ( 𝟏 𝟐 ). Solution: We know that, ∫ 𝑠𝑖𝑛 𝑝 𝜃 𝑐𝑜𝑠 𝑝 𝜃 𝑑𝜃 = Γ( 𝑝+1 2 ).( 𝑞+1 2 ) 2Γ( 𝑝+𝑞+2 2 ) 𝜋 2⁄ 0 Putting 𝑝 = 𝑞 = 0, we get ∫ 𝑑𝜃 = 𝚪( 𝟏 𝟐 ) 𝚪( 𝟏 𝟐 ) 2 𝚪𝟏 𝜋 2 0 [ 𝜃]0 𝜋 2⁄ = 1 2 (Γ 1 2 ) 2
- 8. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 8 𝜋 2 = 1 2 (Γ 1 2 ) 2 (Γ 1 2 ) 2 = 𝜋 Γ ( 1 2 ) = √ 𝜋 _______Ans. Example 3: show that ∫ √ 𝒄𝒐𝒕𝜽 𝒅𝜽 = 𝟏 𝟐 𝚪( 𝟏 𝟒 ) 𝚪 ( 𝟑 𝟒 ) 𝝅 𝟐 𝟎 Solution: We know that, ∫ 𝑠𝑖𝑛 𝑝 𝜃 𝑐𝑜𝑠 𝑝 𝜃 𝑑𝜃 = Γ( 𝑝+1 2 ).( 𝑞+1 2 ) 2Γ( 𝑝+𝑞+2 2 ) 𝜋 2⁄ 0 ∫ √ 𝑐𝑜𝑡𝜃𝑑𝜃 = ∫ 𝑐𝑜𝑠1 2⁄ 𝜃 𝑠𝑖𝑛1 2⁄ 𝜃 𝑑𝜃 𝜋 2 0 𝜋 2 0 = ∫ 𝑠𝑖𝑛−1 2⁄ 𝜃 𝜋 2 0 𝑐𝑜𝑠1 2⁄ 𝜃 𝑑𝜃 On applying formula (1), we have ∫ √ 𝑐𝑜𝑡𝜃𝑑𝜃 = Γ( − 1 2 +1 2 ) Γ( 1 2 +1 2 ) 2Γ( − 1 2 + 1 2 +2 2 ) 𝜋 2 0 = Γ( 1 4 ) Γ( 3 4 ) 2 Γ(1) = 1 2 Γ ( 1 4 )Γ ( 3 4 ) ∴ ∫ √ 𝑐𝑜𝑡𝜃 𝑑𝜃 = 1 2 Γ ( 1 4 )Γ ( 3 4 ) 𝜋 2 0 __________Ans. Example 4: Evaluate ∫ ( 𝟏 + 𝒙) 𝒑−𝟏 ( 𝟏 − 𝒙) 𝒒−𝟏 𝒅𝒙 +𝟏 −𝟏 Solution: Put 𝑥 = 2cos 2𝜃, then 𝑑𝑥 = −2sin 2𝜃 𝑑𝜃 in ∫ (1 + 𝑥) 𝑝−1 (1 − 𝑥) 𝑞−1 𝑑𝑥 +1 −1 = ∫ (1 + 𝑐𝑜𝑠2𝜃) 𝑝−1(1 − 𝑐𝑜𝑠2𝜃) 𝑞−1(−2𝑠𝑖𝑛2𝜃) 0 𝜋 2 = ∫ (1 + 2𝑐𝑜𝑠2 𝜃 − 1) 𝑝−1(1 − 1 + 2𝑠𝑖𝑛2 𝜃) 𝑞−1(−4𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑑𝜃) 0 𝜋 2 = 4∫ 2 𝑝−1 𝑐𝑜𝑠2𝑝−2 𝜃 . 2 𝑞−1 𝑠𝑖𝑛2𝑞−2 𝜃 . 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑑𝜃 𝜋 2 0
- 9. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 9 = 2 𝑝+𝑞 ∫ 𝑠𝑖𝑛2𝑞−1 𝜃 𝑐𝑜𝑠2𝑝−1 𝜃 ∞ 0 𝑑𝜃 = 2 𝑝+𝑞 Γ( 2𝑞 2 ) Γ( 2𝑝 2 ) 2Γ( 2𝑝+2𝑞 2 ) = 2 𝑝+𝑞−1 Γ( 𝑝) Γ( 𝑞) Γ( 𝑝+𝑞) __________Ans. Example 5: Show that 𝚪( 𝒏) 𝚪( 𝟏− 𝒏) = 𝝅 𝒔𝒊𝒏 𝒏𝝅 (𝟎 < 𝑛 < 1) Solution: We know that 𝛽( 𝑚, 𝑛) = ∫ 𝑥 𝑛−1 (1 + 𝑥) 𝑚+𝑛 𝑑𝑥 ∞ 0 Γ𝑚 Γ𝑛 Γ(𝑚+𝑛) = ∫ 𝑥 𝑛−1 (1+𝑥) 𝑚+𝑛 𝑑𝑥 ∞ 0 Putting 𝑚 + 𝑛 = 1 𝑜𝑟 𝑚 = 1 − 𝑛, we get Γ(1−𝑛) Γ𝑛 Γ1 = ∫ 𝑥 𝑛−1 (1+𝑥)1 𝑑𝑥 ∞ 0 Γ(1 − 𝑛)Γ𝑛 = ∫ 𝑥 𝑛−1 1+𝑥 𝑑𝑥 ∞ 0 [∵ ∫ 𝑥 𝑛−1 1+𝑥 𝑑𝑥 ∞ 0 = 𝜋 𝑠𝑖𝑛 𝑛𝜋 ] ∴ Γ( 𝑛)Γ(1− 𝑛) = 𝜋 𝑠𝑖𝑛 𝑛𝜋 ______proved. 4.3 EXERCISE: 1) Evaluate ∫ (1 − 𝑥3)−1 2⁄ 𝑑𝑥 1 0 2) Evaluate ∫ 𝑥 𝑚−1+𝑥 𝑛−1 (1+𝑥) 𝑚+𝑛 𝑑𝑥 1 0 3) Evaluate ∫ ( 𝑥3 1−𝑥3 ) 1 2 𝑑𝑥 1 0 4) Prove that Γ ( 1 4 ) Γ ( 3 4 ) = 𝜋√2 5) Show that 𝛽( 𝑝, 𝑞) = 𝛽( 𝑝 + 1, 𝑞) + (𝑝, 𝑞 + 1) 5.1 DIRICHLET’S INTEGRAL: If 𝑙, 𝑚, 𝑛 are all positive, then the triple integral ∭ 𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑉 = Γ(l)Γ(m)Γ(n) Γ(𝑙 + 𝑚 + 𝑛 + 1)
- 10. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 10 Where V is the region 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 and 𝑥 + 𝑦 + 𝑧 ≤ 1. Note: ∭ 𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑉 = Γ(l)Γ(m)Γ(n) Γ(𝑙+𝑚+𝑛+1) ℎ𝑙+𝑚+𝑛 Where V is the domain, 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 and 𝑥 + 𝑦 + 𝑧 ≤ ℎ 5.2 Corollary: Dirichlet’s theorem for n variables, the theorem status that ∭…∫ 𝑥1 𝑙1−1 𝑥2 𝑙2−1 … 𝑥 𝑛 𝑙 𝑛−1 𝑑𝑥1 𝑑𝑥2 𝑑𝑥3 … 𝑑𝑥 𝑛 = Γ𝑙1Γ𝑙2Γ𝑙3 … Γ𝑙 𝑛 Γ(1 + 𝑙1 + 𝑙2 + ⋯+ 𝑙 𝑛) ℎ𝑙1+𝑙2+⋯+𝑙 𝑛 Example 1: Prove that ∫ 𝒙 𝟒(𝟏+𝒙 𝟓) (𝟏+𝒙 𝟏𝟓) 𝒅𝒙 = 𝟏 𝟓𝟎𝟎𝟓 ∞ 𝟎 Solution: Let 𝐼 = ∫ 𝒙 𝟒(𝟏+𝒙 𝟓) (𝟏+𝒙) 𝟏𝟓 𝒅𝒙 ∞ 𝟎 𝐼 = ∫ 𝑥4 (1+𝑥)15 𝑑𝑥 ∞ 0 + ∫ 𝑥9 (1+𝑥)15 𝑑𝑥 ∞ 0 𝐼 = 𝐼1 + 𝐼2 __________ (i) Now, put 𝑥 = 𝑡 1+𝑡 , when 𝑥 = 0, 𝑡 = 0; when 𝑥 = ∞, 𝑡 = 1 1 + 𝑥 = 1 + 𝑡 1−𝑡 = 1 1−𝑡 𝑑𝑥 = 𝑑𝑡 (1−𝑡)2 ∴ 𝐼1 = ∫ ( 𝑡 1−𝑡 ) 4 . (1 − 𝑡)15 . 1 (1−𝑡)2 𝑑𝑡 1 0 = ∫ 𝑡4 (1− 𝑡)9 𝑑𝑡 1 0 = 𝛽(5,10) _______(2) And 𝐼2 = ∫ ( 𝑡 1−𝑡 ) 9 . (1 − 𝑡)15 . 1 (1−𝑡)2 𝑑𝑡 1 0
- 11. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 11 = ∫ 𝑡9 (1− 𝑡)4 𝑑𝑡 1 0 = 𝛽(10,5) ________(3) ∴ 𝐼 = 𝐼1 + 𝐼2 = 𝛽(5,10) + 𝛽(10,5) [Using(2) and (3)] = 𝛽(5,10) + 𝛽(5,10) [𝛽( 𝑚, 𝑛) = 𝛽(𝑛, 𝑚)] = 2𝛽(5,10) = 2Γ5Γ10 Γ15 = 2×4!×9! 14! = 2×4×3×2×1×9! 14×13×12×11×10×9! = 1 5005 _______ Proved. 5.3 EXERCISE: 1) Find the value of ∫ 𝑥3−2𝑥4+𝑥5 (1+𝑥)7 𝑑𝑥 1 0 2) Show that ∫ 𝑥 𝑚−1(1−𝑥) 𝑛−1 (𝑎+𝑥) 𝑚+𝑛 𝑑𝑥 = 𝛽(𝑚,𝑛) 𝑎 𝑛(1+𝑎) 𝑚 1 0 3) 𝛽( 𝑚 + 1, 𝑛) = 𝑚 𝑚+𝑛 𝛽(𝑚, 𝑛) 6.1 Application to Area & Volume: Liouville’s extension of dirichlet theorem:
- 12. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 12 ∭ 𝑓(𝑥 + 𝑦 + 𝑧)𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧 = Γ(l)Γ(m)Γ(n) Γ(l + m + n) ∫ 𝑓( 𝑢) 𝑢𝑙+𝑚+𝑛−1 𝑑𝑢 ℎ2 ℎ1 Example1: Show that ∭ 𝒅𝒙 𝒅𝒚 𝒅𝒛 (𝒙+𝒚+𝒛+𝟏) 𝟑 = 𝟏 𝟐 𝒍𝒐𝒈𝟐− 𝟓 𝟏𝟔 , the integral being takenthroughout the volume bounded by 𝒙 = 𝟎, 𝒚 = 𝟎, 𝒛 = 𝟎, 𝒙 + 𝒚 + 𝒛 = 𝟏. Solution: By Liouville’s theorem when 0 < 𝑥 + 𝑦 + 𝑧 < 1 ∭ 𝑑𝑥 𝑑𝑦 𝑑𝑧 (𝑥+𝑦+𝑧+1)3 = ∭ 𝑥 𝑙−1 𝑦 𝑚−1 𝑧 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑𝑧 (𝑥+𝑦+𝑧+1)3 (0 ≤ 𝑥 + 𝑦 + 𝑧 ≤ 1) = Γ1Γ1Γ1 Γ(l+m+n) ∫ 1 (u+1)3 u3−1 du 1 0 = 1 2 ∫ 𝑢2 (𝑢+1)3 𝑑𝑢 1 0 = ∫ [ 1 𝑢+1 − 2 (𝑢+1)2 + 1 (𝑢+1)3 ] 𝑑𝑢 1 0 (Partial fractions) = 1 2 [log( 𝑢 + 1) + 2 𝑢+1 − 1 2(𝑢+1)2 ] 0 1 = 1 2 [𝑙𝑜𝑔2 + 2( 1 2 − 1) − ( 1 8 − 1 2 )] = 1 2 𝑙𝑜𝑔2 − 5 16 ∴ ∭ 𝒅𝒙 𝒅𝒚 𝒅𝒛 (𝒙+𝒚+𝒛+𝟏) 𝟑 = 𝟏 𝟐 𝒍𝒐𝒈𝟐− 𝟓 𝟏𝟔 _______Proved. Example 2: Find the mass of an octant of the ellipsoid 𝒙 𝟐 𝒂 𝟐 + 𝒚 𝟐 𝒃 𝟐 + 𝒛 𝟐 𝒄 𝟐 = 𝟏, the density at any point being 𝝆 = 𝒌 𝒙 𝒚 𝒛. Solution: Mass = ∭ 𝜌 𝑑𝑣 = ∭( 𝑘 𝑥 𝑦 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧
- 13. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 13 = 𝑘 ∭( 𝑥 𝑑𝑥)( 𝑦 𝑑𝑥)(𝑧 𝑑𝑧) _______(1) Putting 𝑥2 𝑎2 = 𝑢, 𝑦2 𝑏2 = 𝑣, 𝑧2 𝑐2 = 𝑤 and 𝑢 + 𝑣 + 𝑤 = 1 So that 2𝑥 𝑑𝑥 𝑎2 = 𝑑𝑢, 2𝑦 𝑑𝑦 𝑏2 = 𝑑𝑣, 2𝑧 𝑑𝑧 𝑐2 = 𝑑𝑤 Mass= 𝑘∭ ( 𝑎2 𝑑𝑢 2 )( 𝑏2 𝑑𝑣 2 )( 𝑐2 𝑑𝑤 2 ) = 𝑘 𝑎2 𝑏2 𝑐2 8 ∭ 𝑑𝑢 𝑑𝑣 𝑑𝑤, Where 𝑢 + 𝑣 + 𝑤 ≤ 1 = 𝑘 𝑎2 𝑏2 𝑐2 8 ∭ 𝑢𝑙−1 𝑣 𝑙−1 𝑤 𝑙−1 𝑑𝑢 𝑑𝑣 𝑑𝑤 = 𝑘 𝑎2 𝑏2 𝑐2 8 Γ1Γ1Γ1 Γ3+1 = 𝑘 𝑎2 𝑏2 𝑐2 8×6 = 𝑘 𝑎2 𝑏2 𝑐2 48 ∴ 𝑴𝒂𝒔𝒔 = 𝒌 𝒂 𝟐 𝒃 𝟐 𝒄 𝟐 𝟒𝟖 Ans.
- 14. Unit-2 GAMMA, BETA FUNCTION RAI UNIVERSITY, AHMEDABAD 14 6.2 EXERCISE: 1) Find the value of ∭ 𝑙𝑜𝑔( 𝑥 + 𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 the integral extending over all positive and zero values of 𝑥, 𝑦, 𝑧 subject to the condition 𝑥 + 𝑦 + 𝑧 < 1. 2) Evaluate ∭ √1−𝑥2−𝑦2−𝑧2 1+𝑥2+𝑦2+𝑧2 𝑑𝑥 𝑑𝑦 𝑑𝑧, integral being taken over all positive values of 𝑥, 𝑦, 𝑧 such that 𝑥2 + 𝑦2 + 𝑧2 ≤ 1. 3) Find the area and the mass contained m the first quadrant enclosed by the curve ( 𝑥 𝑎 ) 𝛼 + ( 𝑦 𝑏 ) 𝛽 = 1 𝑤ℎ𝑒𝑟𝑒 𝛼 > 0, 𝛽 > 0 given that density at any point 𝑝(𝑥𝑦) is 𝑘 √ 𝑥𝑦. 7.1 REFERENCEBOOK: 1) Introduction to Engineering Mathematics By H. K. DASS. & Dr. RAMA VERMA 2) Higher Engineering Mathematics By B.V.RAMANA 3) A text bookof Engineering Mathematics By N.P.BALI 4) www1.gantep.edu.tr/~olgar/C6.SP.pdf