1. The document discusses the gamma and beta functions, which are defined in terms of improper definite integrals involving exponential and power functions. 2. Examples are provided to demonstrate properties and applications of the gamma function, including evaluating integrals involving the gamma function. 3. The beta function is defined in terms of an integral from 0 to 1, and its relationship to the gamma function is described.

1. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 1
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

2. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 2
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
).
Solution: We know that Γ( 𝑚 + 1) = 𝑚Γ𝑚

3. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 3
 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)
Putting 𝑎 𝑥
= 𝑒 𝑡

4. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 4
⟹ 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:

5. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 5
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:

6. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 6
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

7. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 7

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

8. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 8
= 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)

9. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 9
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

10. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 10
= ∫ 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:

11. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 11
∭ 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 octantof 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

12. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 12
= 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.

13. Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 13
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