Gamma beta functions-1
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The document defines the gamma and beta functions and provides examples of using them to evaluate integrals. The gamma function Γ(n) generalizes the factorial function to real and complex numbers. It satisfies properties like Γ(n+1)=nΓ(n). The beta function B(m,n) defines integrals over the interval [0,1]. It relates to the gamma function as B(m,n)=Γ(m)Γ(n)/Γ(m+n). Several integrals are evaluated using these functions, including changing variables to match their definitions. Proofs are also given for relationships between beta function integrals over [0,1] and [0,π/2].
![Examples:
Example(1)
Evaluate 0∫1
x4
(1 – x ) 3
dx
Solution
0∫1
x 4
(1 – x ) 3
dx = x 5-1
(1 – x ) 4-1
dx
= B(5,4) = Γ(5) Γ(4) / Γ(9) = 4! . 3! / 8! = 3!/(8.7.6.5) = 1/ (8.7.5) = 1/280
Exercise
Evaluate 0∫1
x2
(1 – x ) 6
dx
Example(2)
Evaluate I = 0∫1
[ 1 / 3
√[x2
(1 – x )] ] dx
Solution
I = 0∫1
x -2/3
(1 – x ) -1/3
dx = 0∫1
x 1/3 - 1
(1 – x ) 2/3 - 1
dx
= B(1/3,2/3) = Γ(1/3) Γ(2/3) / Γ(1)
Γ(1/3) Γ(2/3) = Γ(1/3) Γ(1- 1/3) = π /sin(π/3) = π / ( √3/2) = 2π / √3
6](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-6-320.jpg)
![Exercise
Evaluate I = 0∫1
[ 1 / 4
√[x3
(1 – x )] ] dx
Example(3)
Evaluate I = 0∫1
√x . (1 – x ) dx
Solution
I = 0∫1
x 1/2
(1 – x ) dx = 0∫1
x 3/2 - 1
(1 – x ) 2 - 1
dx
= B(3/2 , 2) = Γ(3/2) Γ(2) / Γ(7/2)
Γ(3/2) = ½ √π
Γ(5/2) = Γ(3/2+ 1) = (3/2) Γ(3/2 ) = (3/2) . ½ √π = 3√π / 4
Γ(7/2) = Γ(5/2+ 1) = (5/2) Γ(5/2 ) = (5/2) . (3√π / 4) = 15 √π / 8
Thus,
I = (½ √π ) . 1! / (15 √π / 8) = 4/15
Exercise
Evaluate I = 0∫1
√x5
. (1 – x ) dx
7](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-7-320.jpg)
![Example(3)
Evaluate: I = 0∫∞
xm
e – k x^n
dx
Solution:
Letting y = k xn
, we get
I = [ 1 / ( n . k (m+1)/n
) ] 0∫∞
y [(m+1)/n – 1]
e -y
dy = [ 1 / ( n . k (m+1)/n
) ] Γ[(m+1)/n ]
9](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-9-320.jpg)
![Using Formula (3)
Example(3)
Evaluate: I = 0∫∞
dx / ( 1+x4
)
Solution:
Letting x4
= y , we get
I = (1 / 4) 0∫∞
y -3/4
dy / (1 + y ) = (1 / 4) . Γ (1/4) . Γ (1 - 1/4 )
= (1/4) . [ π / sin ( ¼ . π ) ] = π √2 / 4
Using Formula (2)
Example(4)
a. Evaluate: I = 0∫π/2
sin 3
. cos 2
x dx
b. Evaluate: I = 0∫π/2
sin 4
. cos 5
x dx
Solution:
12](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-12-320.jpg)
![Details
I.
Example(1)
Evaluate: I = 0∫∞
x 6
e -2x
dx
x = y/2
x 6
= y 6
/64
dx = (1/2)dy
x 6
e -2x
dx = y 6
/64 e –y
. (1/2)dy
Example(2)
I = 0∫∞
√x e –x^3
dx x=y1/3
√x= y1/6
dx=(1/3)y-2/3
dy
√x e –x^3
dx = y1/6
e –y
. (1/3)y-2/3
dy
Example(3)
Evaluate: I = 0∫∞
xm
e – k x^n
dx
y = k xn
x = y1/n
/ k1/n
xm
= ym/n
/ km/n
dx = (1/n) y(1/n-1)
/ k1/n
dy
xm
e – k x^n
dx = ( ym/n
/ km/n
) . e – y
. (1/n) y(1/n-1)
/ k1/n
dy
m/n + 1/n – 1 = (m+1)/n - 1
-m/n – 1/n = - (m+1)/n
I = [ 1 / ( n . k (m+1)/n
) ] 0∫∞
y [(m+1)/n – 1]
e -y
dy
14](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-14-320.jpg)
![Formula (3)
We have,
I = 0∫∞
x q-1
/ (1+x) dx
Let
y = x / (1+x)
Hence, x = y / 1-y
, 1 + x = 1 + (y / 1-y) = 1/(1-y)
& dx = - [ (1- y) – y(-1)] / (1-y)2
. dy= 1 / (1-y)2
. dy
whn x = 0 , we have y = 0
when x→∞ , we have y = lim x→∞ x / (1+x) = 1
Thus,
I = 0∫∞
[ x q-1
/ (1+x) ] dx = 0∫∞
[ ( y / 1-y ) q-1
/ (1/(1-y)) ] . 1 / (1-y)2
. dy
= 0∫1
[ y q-1
/ (1-y) -q
] dy
= B(q , 1-q) = Γ(q) Γ(1-q)
18](https://image.slidesharecdn.com/gamma-betafunctions-1-140502014136-phpapp01/85/Gamma-beta-functions-1-18-320.jpg)
Gamma beta functions-1
- 1. Gamma & Beta Functions I. Gamma Function Definition Γ(n) = 0∫∞ x n-1 e -x dx ; n > 0 & Γ(n) = Γ(n+1) / n ; n Є R - Z≤0 Results: (1) Γ(n+1) = n Γ(n) ; n > 0 , where Γ(1) = 1 (2) Γ(n+1) = n! ; n Є N ( convention: 0! = 1) (3) Γ(n) Γ(1- n) = π /sin(nπ) ; 0 < n < 1 In Particular; Γ(1/2) = √π 1
- 2. Examples: Example(1) Evaluate 0∫∞ x 4 e -x dx Solution 0∫∞ x 4 e -x dx = 0∫∞ x 5-1 e -x dx = Γ(5) Γ(5) = Γ(4+1) = 4! = 4(3)(2)(1) = 24 Exercise Evaluate 0∫∞ x 5 e -x dx Example(2) Evaluate 0∫∞ x 1/2 e -x dx 0∫∞ x 1/2 e -x dx = 0∫∞ x 3/2-1 e -x dx = Γ(3/2) 3/2 = ½ + 1 Γ(3/2) = Γ(½+ 1) = ½ Γ(½ ) = ½ √π Exercise Evaluate 0∫∞ x 3/2 e -x dx 2
- 3. Example(3) Evaluate 0∫∞ x 3/2 e -x dx 0∫∞ x 3/2 e -x dx = 0∫∞ x 5/2-1 e -x dx = Γ(5/2) 5/2 = 3/2 + 1 Γ(5/2) = Γ(3/2+ 1) = 3/2 Γ(3/2 ) = 3/2 . ½ Γ(½ ) = 3/2 . ½ . √π = ¾ √π Exercise Evaluate 0∫∞ x 5/2 e -x dx Example(4) Find Γ(-½) (-½) + 1 = ½ Γ(-1/2) = Γ(-½ + 1) / (-½) = - 2 Γ(1/2 ) = - 2 √π 3
- 4. Example(5) Find Γ(-3/2) (-3/2) + 1 = - ½ Γ(-3/2) = Γ(-3/2 + 1) / (-3/2) = Γ(-1/2 ) / (-2/3) = ( - 2 √π ) / (-2/3) = 4 √π /3 Exercise Evaluate Γ(-5/2) 4
- 5. II. Beta Function Definition B(m,n) = 0∫1 x m-1 (1 – x ) n-1 dx ; m > 0 & n > 0 Results: (1) B(m,n) = Γ(m) Γ(n) / Γ(m+ n) (2) B(m,n) = B(n,m) (3) 0∫π/2 sin 2m-1 x . cos 2n-1 x dx = Γ(m) Γ(n) / 2 Γ(m+ n) ; m>0 & n>0 (4) 0∫∞ x q-1 / (1+x) . dx = Γq) Γ(1-q) = Π / sin(qπ) ; 0<q<1 5
- 6. Examples: Example(1) Evaluate 0∫1 x4 (1 – x ) 3 dx Solution 0∫1 x 4 (1 – x ) 3 dx = x 5-1 (1 – x ) 4-1 dx = B(5,4) = Γ(5) Γ(4) / Γ(9) = 4! . 3! / 8! = 3!/(8.7.6.5) = 1/ (8.7.5) = 1/280 Exercise Evaluate 0∫1 x2 (1 – x ) 6 dx Example(2) Evaluate I = 0∫1 [ 1 / 3 √[x2 (1 – x )] ] dx Solution I = 0∫1 x -2/3 (1 – x ) -1/3 dx = 0∫1 x 1/3 - 1 (1 – x ) 2/3 - 1 dx = B(1/3,2/3) = Γ(1/3) Γ(2/3) / Γ(1) Γ(1/3) Γ(2/3) = Γ(1/3) Γ(1- 1/3) = π /sin(π/3) = π / ( √3/2) = 2π / √3 6
- 7. Exercise Evaluate I = 0∫1 [ 1 / 4 √[x3 (1 – x )] ] dx Example(3) Evaluate I = 0∫1 √x . (1 – x ) dx Solution I = 0∫1 x 1/2 (1 – x ) dx = 0∫1 x 3/2 - 1 (1 – x ) 2 - 1 dx = B(3/2 , 2) = Γ(3/2) Γ(2) / Γ(7/2) Γ(3/2) = ½ √π Γ(5/2) = Γ(3/2+ 1) = (3/2) Γ(3/2 ) = (3/2) . ½ √π = 3√π / 4 Γ(7/2) = Γ(5/2+ 1) = (5/2) Γ(5/2 ) = (5/2) . (3√π / 4) = 15 √π / 8 Thus, I = (½ √π ) . 1! / (15 √π / 8) = 4/15 Exercise Evaluate I = 0∫1 √x5 . (1 – x ) dx 7
- 8. II. Using Gamma Function to Evaluate Integrals Example(1) Evaluate: I = 0∫∞ x 6 e -2x dx Solution: Letting y = 2x, we get I = (1/128) 0∫∞ y 6 e -y dy = (1/128) Γ(7) = (1/128) 6! = 45/8 Example(2) Evaluate: I = 0∫∞ √x e –x^3 dx Solution: Letting y = x3 , we get I = (1/3) 0∫∞ y -1/2 e -y dy = (1/3) Γ(1/2) = √π / 3 8
- 9. Example(3) Evaluate: I = 0∫∞ xm e – k x^n dx Solution: Letting y = k xn , we get I = [ 1 / ( n . k (m+1)/n ) ] 0∫∞ y [(m+1)/n – 1] e -y dy = [ 1 / ( n . k (m+1)/n ) ] Γ[(m+1)/n ] 9
- 10. II. Using Beta Function to Evaluate Integrals Formulas (1) 0∫1 x m-1 (1 – x ) n-1 dx = B(m,n) = Γ(m) Γ(n) / 2 Γ(m+ n) ; m > 0 & n > 0 (3) 0∫π/2 sin 2m-1 x . cos 2n-1 x dx = (1/2) B(m,n) ; m>0 & n>0 (4) 0∫∞ x q-1 / (1+x) . dx = Γ(q) Γ(1-q) = Π / sin(qπ) ; 0 < q < 1 Using Formula (1) Example(1) Evaluate: I = 0∫2 x2 / √(2 – x ) . dx Solution: Letting x = 2y, we get I = (8/√2) 0∫1 y 2 (1 – y ) -1/2 dy = (8/√2) . B(3 , 1/2 ) = 64√2 /15 10
- 11. Example(2) Evaluate: I = 0∫a x4 √ (a2 – x2 ) . dx Solution: Letting x2 = a2 y , we get I = (a6 / 2) 0∫1 y 3/2 (1 – y )1/2 dy = (a6 / 2) . B(5/2 , 3/2 ) = a6 /3 2 Exercise Evaluate: I = 0∫2 x √ (8 – x3 ) . dx Hint Lett x3 = 8y Answer I = (8/3) 0∫1 y-1/3 (1 – y ) 1/3 . dy = (8/3) B(2/3 , 4/3 ) = 16 π / ( 9 √3 ) 11
- 12. Using Formula (3) Example(3) Evaluate: I = 0∫∞ dx / ( 1+x4 ) Solution: Letting x4 = y , we get I = (1 / 4) 0∫∞ y -3/4 dy / (1 + y ) = (1 / 4) . Γ (1/4) . Γ (1 - 1/4 ) = (1/4) . [ π / sin ( ¼ . π ) ] = π √2 / 4 Using Formula (2) Example(4) a. Evaluate: I = 0∫π/2 sin 3 . cos 2 x dx b. Evaluate: I = 0∫π/2 sin 4 . cos 5 x dx Solution: 12
- 13. a. Notice that: 2m - 1 = 3 → m = 2 & 2n - 1 = 2 → m = 3/ 2 I = (1 / 2) B( 2 , 3/2 ) = 8/15 b. I = (1 / 2) B( 5/2 , 3 ) = 8 /315 Example(5) a. Evaluate: I = 0∫π/2 sin6 dx b. Evaluate: I = 0∫π/2 cos6 x dx Solution: a. Notice that: 2m - 1 = 6 → m = 7/2 & 2n - 1 = 0 → m = 1/ 2 I = (1 / 2) B( 7/2 , 1/2 ) = 5π /32 b. I = (1 / 2) B( 1/2 , 7/2 ) = 5π /32 Example(6) a. Evaluate: I = 0∫π cos4 x dx b. Evaluate: I = 0∫2π sin8 dx Solution: a. I = 0∫π cos4 x = 2 0∫π/2 cos4 x = 2 (1/2) B (1/2 , 5/2 ) = 3π / 8 b. I = I = 0∫π sin8 x = 4 0∫π/2 sin8 x = 4 (1/2) B (9/2 , 1/2 ) = 35π / 64 13
- 14. Details I. Example(1) Evaluate: I = 0∫∞ x 6 e -2x dx x = y/2 x 6 = y 6 /64 dx = (1/2)dy x 6 e -2x dx = y 6 /64 e –y . (1/2)dy Example(2) I = 0∫∞ √x e –x^3 dx x=y1/3 √x= y1/6 dx=(1/3)y-2/3 dy √x e –x^3 dx = y1/6 e –y . (1/3)y-2/3 dy Example(3) Evaluate: I = 0∫∞ xm e – k x^n dx y = k xn x = y1/n / k1/n xm = ym/n / km/n dx = (1/n) y(1/n-1) / k1/n dy xm e – k x^n dx = ( ym/n / km/n ) . e – y . (1/n) y(1/n-1) / k1/n dy m/n + 1/n – 1 = (m+1)/n - 1 -m/n – 1/n = - (m+1)/n I = [ 1 / ( n . k (m+1)/n ) ] 0∫∞ y [(m+1)/n – 1] e -y dy 14
- 15. II. Example(1) Example(1) I = 0∫2 x2 / √(2 – x ) . dx x = 2y dx=2dy x2 = 4 y2 √(2 – x ) = √(2 – 2y ) =√2 √(1 – y ) x2 / √(2 – x ) . dx = 4 y2 / √2 √(1 – y ) 2dy y=0 when x=0 y=1 when x=2 Example(2) Evaluate: I = 0∫a x4 √ (a2 – x2 ) . dx x2 = a2 y , we get x4 = a4 y2 x= a y1/2 15
- 16. dx= (1/2)a y-1/2 dy √ (a2 – x2 ) = √ (a2 – a2 y ) = a (1 – y )1/2 x4 √ (a2 – x2 ) . dx = a4 y2 a (1 – y )1/2 (1/2)a y-1/2 dy y=0 when x=0 y=1 when x=a Example(3) I = 0∫∞ dx / ( 1+x4 ) x4 = y x=y1/4 dy= (1/4) y-3/4 dy dx / ( 1+x4 ) = (1 / 4) y -3/4 dy / (1 + y ) 16
- 17. Proofs of formulas (2) & (3) Formula (2) We have, B(m,n) = 0∫1 x m-1 (1 – x ) n-1 dx Let x = sin2 y Then dy = 2 sinx cox dx & x m-1 (1 – x ) n-1 dx = (sin2 y)m-1 ( cos2 y )n-1 ( dy / 2 sinx cox ) = 2 sin 2m-1 y . cos 2n-1 y dy When x=0 , we have y = 0 When x=1, we hae y = π/2 Thus, I = 2 0∫π/2 sin 2m-1 y . cos 2n-1 y dy I = 0∫π/2 sin 2m-1 y . cos 2n-1 y dy = B(m,n) / 2 17
- 18. Formula (3) We have, I = 0∫∞ x q-1 / (1+x) dx Let y = x / (1+x) Hence, x = y / 1-y , 1 + x = 1 + (y / 1-y) = 1/(1-y) & dx = - [ (1- y) – y(-1)] / (1-y)2 . dy= 1 / (1-y)2 . dy whn x = 0 , we have y = 0 when x→∞ , we have y = lim x→∞ x / (1+x) = 1 Thus, I = 0∫∞ [ x q-1 / (1+x) ] dx = 0∫∞ [ ( y / 1-y ) q-1 / (1/(1-y)) ] . 1 / (1-y)2 . dy = 0∫1 [ y q-1 / (1-y) -q ] dy = B(q , 1-q) = Γ(q) Γ(1-q) 18
- 19. Proving that Γ(1/2 ) = √π Γ(1/2) = 0∫∞ x 1/2-1 e -x dx = 0∫∞ x -1/2 e -x dx Let y = x ½ x = y2 dx = 2y dy Γ(1/2) = 0∫∞ y -1 e –y^2 2y dy = 2 0∫∞ e –y^2 dy =2 (√π / 2 ) = √π 19