1
Pertemuan VII
Fungsi Gamma dan Fungsi Beta
Fungsi Gamma
Euler's gamma function is defined by the integral
Some further values of the Gamma function for small arguments are:
1,0
1,0000
1,1
0,9514
1,2
0,9182
1,3
0,9875
1,4
0,8873
1,5
0,8862
1,6
0,8935
1,7
0,9086
1,8
0,9314
1,9
0,9618
2,0
1,0000
(1/5)=4.5909 ,
(1/4)=3.6256
(1/3)=2.6789 ,
(2/5)=2.2182
(3/5)=1.4892 ,
(2/3)=1.3541
(3/4)=1.2254 ,
(4/5)=1.1642 .
2
If x is an integer n = 1, 2, 3, ..., then
Γ(n) =

x
n 1  x
e dx
0
Γ(n+1) = n Γ(n)
3

Γ (p) Γ (1-p) =

x p 1
0 1  x dx =
0 < p < 1
sin p

0 < p < 1
sin p
,
,

Contoh
a.
x e
3 -x
.
dx  ( 4 )  3!  3 x 2 x 1  6
0

x e
b.
3 2x
0

1
3
(4) 
16
8


Contoh. 
x e 3
x
3
 2x 
dx     e  2 x d ( 2 x / 2)
0 2 
dx ,
misalkan
y 3 x
0

4
(3) 
27
x = 1/9 y2

2/27
 y. e
2 y
dy 
2/27
0
dx =
2/9
y dy

Contoh. 
3
x e- x dx
misalkan
3
x
 y
0
x = y3
dx = 3 y2 dy

y
3/ 2
e- y 3y 2dy
0

 3  y 7/2e  y dy  3 (9/2) 
0
 3x
7 5 3 1
2 2 2 2


315 
16
4


Contoh.
x e-
4
x
dx
x y
mis
0
x = y2
dx = 2y dy

(y )
2 1/ 4
e- y 2y dy
0


2 y1 / 2 e- y y dy
=
0
5
2 y 3 / 2 e- y dy  2  
2
0
= 2.
1
 
2 2 2
3 1

3
2



Contoh
mis x 3  y
3
e- x dx
0
x = y1/3
dx = 1/3 y-2/3 dy

1/3
y
2 / 3
1 1
 
3 3

e- y dy
0
atau Γ(4/3)

Contoh.
2 - 4x
 x e dx
2
.
misalkan
4x 2  y
0
x = ½ y1/2
dx = ¼ y-1/2 dy
1
4

-y
ye
0
1
y-1/2 dy
=
4


y
16
1/ 2
0
=
Contoh .
1
=
3
 
16  2 
1
1

32

 4x 2
- 4x 2 ln 2
2
dx

e
dx


0
e - y dy
0
misalkan 4 x2 ln 2 = y
5
1
2
-
 y
x= 

 4 ln 2 


2
- 4x
y 2
dx 
dy
4 ln 2

2
1

dx 
0
0
e
-y
y
-
1
2
4 ln 2
1
 
2
  
4 ln 2
dy


4 ln 2
1

Contoh .
3
misalkan ln x  - y
ln( 1/ x) dx .
0
x = e-y
1


1
0
3
 ln x dx .  -  y 3 e  y dy


0
atau dx = - e-y dy
1
-y
 y 3 e dy
0
4
1
1
( )  ( )
3
3
3
Fungsi Beta.
Fungsi Beta dinyatakan sebagai berikut :
1
a.
B(m , n) =
x
m 1
(1  x) n 1 dx
convergen untuk n dan m  0
0
b.
B(m , n) = B(n , m)
π/2
c.
 sin
2n-1
 cos2m-1  d
0

d.
e.

1
B(m, n)
2
x p1
π
0 1  x dx  Γ (p) ( 1 - p)  sin pπ
Γ( m ) . Γ ( n )
B(m,n ) 
Γ(mn)
(0<p<1)
1
Contoh
: Hitung
x
8
( 1 - x ) 4 dx
 B( 9,5 )
0

( 9 )  ( 5 )
 ( 14 )

8! 4!
1

13 !
6435
6
1
Contoh
x
: Hitung
4
( 1 - x )5 dx
x  y
misal
0
x = y2
dx = 2ydy
1
y
1
8
2  y9 (1  y)5 dy

( 1 - y )5 2ydy
0
0
2  ( 10 )  ( 6 )
2 B ( 10 , 6 ) =
(16 )
1
=
15015
1
Contoh
 (1- x
: Hitung
3
misalkan x 3  y
)-1/2 dx
0
x = y1/3
dx =
1
 (1- y )
-
1
2
2
3
y
dy
3
0
1
3 0
1

1
-
2
3
1 - 2/3
y
dy
3
y (1- y )
-
1
2
dy
1
( ) ( )
1
1 1
1
3
2
B( , ) 
2
5
3
3
3
( )
6
Contoh
: Hitung
2
x2

2-x
0
misalkan x  2y
dx
dx = 2 dy
1

(2y) 2 2 dy
2 - 2y
0
8

1
B(3 , ) 
2
2
8
2
8
2
1
y
2
(1- y )
-
1
2
dy
0
(3) (
(
1
)
2
7
)
2

64 2
15
7
Contoh
: Hitung
3

0
3
1
3x - x

dx 
2
0
dx
x 3- x
misalkan x  3y
dx = 3 dy
1
 (3 y)
1

2
( 3 - 3y )
1
2
1
3 dy  B (
2
0
,
1
)
2
1
1
  ( ) ( )  
2 2
2
Contoh
 (4-x
: Hitung
2
misalkan x 2  4y
)3/2 dx
0
x = 2y1/2
dx = y-1/2 dy
1
 ( 4 - 4y )
3/2
y-1/2 dy
0
1
43/2  ( 1 - y )3/2 y-1/2 dy
0
(1 / 2) ( 5/2 )
= 8 B ( ½ , 5/2 ) = 8

(3)
8 
3 1

2 2
2
7

Contoh
4
misalkan x  4y  3
( x - 3 ) ( 7 - x ) dx
3
dx = 4 dy
1
 ( 4y  3 - 3 )
1/4
( 7 - 4y - 3 )1/4 4 dy
0
1
4  ( 4y )1/4 ( 4 - 4y )1/4 dy  8 B (
0
8
1 1
4 4
( 1/4 )(1 / 4)
(5 / 2)

2
3 
1

0
5
,
4
)
4
((1/4)) 2
1
( 1 - y ) 4 3 y 2 dy
5
 3  y2 ( 1 - y ) 4 dy
0
 3
8
3 B ( 3 , 5)  3
(3) ( 5 )
( 8 )
1
35

10
Contoh

: Hitung
misalkan x 2  100y
100 - x 2 dx
0
1
x = 10 y 2
dx = 5 y
1
 10( 1 - y )
1
2
1
2
1
2
dy
1
2
 50  y ( 1 - y ) dy
0
0
3 1
50
50 B ( , ) 
2 2
2
 /2
Contoh
1
2
1
5 y dy


Hitung
1
1
( ) ( )
2
2
( 2 )
 25 
ctg  d
0
 /2
=

( cos  )1/2 (sin  ) -1/2 d
1
3 1
B( , )
2
4 4

0
2m  1 
1
2
2n  1  
m
1
2
n 
3
4
1
4
1
3 1
B( , )
2
4 4

3
1

2
1
( ) ( )
4
4
( 1 )


2
Soal - soal
1
1.
Hitung
dx
 (1- x
0
n 1/ n
)
misalkan x n  sin 2
2
2.
x
0
3
8 - x 3 dx
misalkan x 3  8y
9
 /2
3.
/2
d
cos 

0
 /2

4.
 cos

0
/2
tg  d

0
 sin
 cos-1/2 d
1/2
0

5.
 d
-1/2
dx
 1 x
Hitung
misalkan
4
x4  y
0
1
6.
Buktikan
 (1 x )
p -1
( 1 - x )q -1 dx  2p  q 1
1

7.
Carilah

x e -x dx
0
( p ) ( q )
.
( p  q )
3
Jwb .  
2

8.
Carilah
m x
 e dx
2 2
Jwb .
0

9.
Carilah
x
2
2
e- 2x dx
Jwb .
0

10.
Carilah
x
 1  x dx
0
Jwb .

2m
2
16

3 3
TERIMA KASIH