SOLVED EXAMPLES ON BETA AND GAMMA FUNCTIONS
MFMCF-FUTA
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Recall: ⌈ = ∫ . Evaluate the following integrals as a function of gamma
1. ∫ (Let n – 1 = 2/3)
Thus, n = 2/3 + 1 = 5/3. ∫ ⌈( )
2. ∫ (Let n – 1 = -1/2)
Thus, n = -1/2 + 1 = 1/2. ∫ ⌈( )
3. ∫ (Let u = , solve the integration by substitution)
;
.
When When
(now we will substitute all the factors of u for x)
∫ ∫ =∫
∫ ∫ (Let y=3u, u=y/3)
,
When u=0, y=0 When u=∞, y=∞
∫ ∫ ∫
∫ (Now the expression can written as gamma, if n – 1=1)
Thus, n = 1+1 = 2. ∫ ∫ ⌈
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, 4. ∫ √ (√ )
⁄
∫ √ ∫ (Let n – 1= 1/2)
Thus, n = ½ + 1 = 3/2. ∫ √ ⌈
5. ∫ (Let u=x2 and x=√u: when x=0, u=0, when x=∞, u=∞)
√
(Substitute for x)
∫ ∫ ( ) ∫ ( ) ∫ (√ )
√ √
∫ √ ∫
Let n – 1 = 1/2. Thus, n = ½ + 1 = 3/2
∫ ∫ ⌈ (MFMCF-FUTA)
Evaluate the following and Recall: ⌈ ⌈ , ⌈( ) √
⌈
6. ⌈( ) Recall; ⌈
⌈( ) ⌈( ). ⌈( ) ⌈( )
⌈( ) ⌈( ). ⌈( ) ⌈( )
⌈( ) ⌈( ). ⌈( ) √
7. ∫ Let n – 1 = 3, therefore n = 4
∫ ⌈
08144345986 Youngscholar
MFMCF-FUTA
This document is created by Youngscholar
Recall: ⌈ = ∫ . Evaluate the following integrals as a function of gamma
1. ∫ (Let n – 1 = 2/3)
Thus, n = 2/3 + 1 = 5/3. ∫ ⌈( )
2. ∫ (Let n – 1 = -1/2)
Thus, n = -1/2 + 1 = 1/2. ∫ ⌈( )
3. ∫ (Let u = , solve the integration by substitution)
;
.
When When
(now we will substitute all the factors of u for x)
∫ ∫ =∫
∫ ∫ (Let y=3u, u=y/3)
,
When u=0, y=0 When u=∞, y=∞
∫ ∫ ∫
∫ (Now the expression can written as gamma, if n – 1=1)
Thus, n = 1+1 = 2. ∫ ∫ ⌈
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, 4. ∫ √ (√ )
⁄
∫ √ ∫ (Let n – 1= 1/2)
Thus, n = ½ + 1 = 3/2. ∫ √ ⌈
5. ∫ (Let u=x2 and x=√u: when x=0, u=0, when x=∞, u=∞)
√
(Substitute for x)
∫ ∫ ( ) ∫ ( ) ∫ (√ )
√ √
∫ √ ∫
Let n – 1 = 1/2. Thus, n = ½ + 1 = 3/2
∫ ∫ ⌈ (MFMCF-FUTA)
Evaluate the following and Recall: ⌈ ⌈ , ⌈( ) √
⌈
6. ⌈( ) Recall; ⌈
⌈( ) ⌈( ). ⌈( ) ⌈( )
⌈( ) ⌈( ). ⌈( ) ⌈( )
⌈( ) ⌈( ). ⌈( ) √
7. ∫ Let n – 1 = 3, therefore n = 4
∫ ⌈
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