Gamma Function and Beta Function Beta Function: It is denoted by
(m, n) and is defined by 1
( m, n ) =
0
x m 1 (1 x) n 1 d x , m, n
0.
It is also known as First Eulerian Integral. Gamma Function: It is denoted by (n) and is defined by
x n 1 e x dx ,
( n)
n
0.
0
It is also known as Second Eulerian Integral.
Prove that (1) = 1 Proof: From the definition of gamma function
x n 1 e x dx , n 0 … … … (1)
( n) 0
Setting n = 1 in Equation (1) (1)
x 0 e x dx
= 0
e x dx
= 0
=
[e x ]0
=
[e
e0 ]
= 1
If n is a positive integer then show that
(n 1)
n!
Proof: From the definition of gamma function
x n 1 e x dx , n 0 … … … (1)
( n) 0
Writing n 1 for n in (1) ( n 1) = x n e
x
dx
0
=
[ x ne x ] 0
n x n 1 e x dx 0
x n 1e x dx
= n 0
= n
( n)
… … … … … … … (2)
Putting n 1 , n 2 , … … 3, 2, 1 for n successively in (2) (n ) n 1 n 1 n 1 n 2 n 2 ………………… …………………… ( 4) 3 3 (3) 2 2 ( 2) 1 1 n 1
n (n 1) (n 2)
3 2 1 3 2 1
= n ( n 1) (n 2) = n!
(1)
[Proved]
Show that (m, n)
( n, m ) .
Proof: From the definition of Beta function ( m, n )
1
=
0 0
=
1 1
=
0 1
=
0
=
xm 1 1 x
n 1
dx
1 y
m 1
1
1 y
m 1
y n 1dy
xn 1 1 x
m 1
1 y
n 1
Let, x 1 y dx dy
dy
When x 1, y 0 x 0, y 1
dx
( n, m)
Relation between Gamma function and Beta function Answer: From the definition of Gamma function
x n 1 e x dx , n 0
( n) 0
=
0
zy
n 1
e
zy
z dy
Let, x
z y where z is constant dx
When, x = zn
(n) = zn
0
0
yn 1 e
yn 1 e
zy
zy
dy
0, y
0 x
… … … (1)
dy
Multiplying both sides of (1) by z m 1 e z we get
… … … (2)
,y
zdy ,
(n ) z m 1 e
= zn zm 1 e
z
z
yn 1 e
0
yz
dy
Integrating both sides with respect to z from 0 to (n) 0 z m 1 e z dz (n) m
0
{ 0 zm
0
(m) ( n) ( m n)
Or,
0 n 1
zm e
n 1
e
z (1 y )
z
yn
1
e
dz } y n
1
dy
0
(m n ) n 1 y dy (1 y ) m n
0
yz
dy dz
[Comparing with equation (1)]
yn 1 dy (1 y ) m n
… … … (3)
From the definition of Beta Function 1
(m, n) Let x
1 x
1 1 y
0
xm 1 1 x
When,
(m, n)
1 1 y
1 1 x 1 y 1 y 1 y 1 y
x 1, y 0 x 0, y 0
dx , m , n 0
dy 2 1 y
then dx
1 y Or y
1 x 1
n 1
1
1 m 1
1 y
y
n 1
1 y =
=
=
0
0
yn 1 m 1 1 y yn 1 m 1 y (m ) (n ) ( m n)
n
dy 1 y
n 1 2
2
dy
dy
[using (3)]
1 2
Show that
Proof: From the definition of Beta function 1
m,n Let m
0
xm 1 1 x
dx , m, n
0
1 then 2
n 1 1 , 2 2
1
x 0
1 2
1 2 1 2
1 2
1 x
1 2
1
x 0
1 2
1 2
dx
1 x
1 2
dx
2
1 2 1
dx
1 0
1 2
x 1 x
Let, x sin 2 When x
n 1
dx
0,
2 sin cos d
0 and x 1,
2
1 2
1 2
2 0
2
2 sin cos d sin cos
2 0 2d = 2
2 0
=2
2
=
1 2
Prove that
2 0
sin m x cos n x dx
m 1 n 1 2 2 . m n 2 2 2
Proof: Let, I = =
2
sin m x cos n x dx
2
sin m 1 x cos n 1 x sin x cos x dx
0
0
= =
2
sin 2 x
m 1 2
cos 2 x
2
sin 2 x
m 1 2
1 sin 2 x
0
0
Let sin 2 x
I
i.e.
#
#
2
0 and x
1 2
=
1 2
=
1 2
1
z 0 1 0
z
m 1 2
1 z
m 1 1 2
1 z
2
n 1 2
dz
,z 1
n 1 1 2
dz
m 1 n 1 2 2 m n 2 2 2
sin m xdx =
2
cos n xdx =
Answer:
sin x cos x dx
dz
sin m x cos n x dx
Evaluate
n 1 2
m 1 n 1 , 2 2
2 0
0
=
=
0
sin x cos x dx
z then 2 sin x cos xdx
When, x 0, z
I
n 1 2
2 0 2 0
m 1 n 1 2 2 m n 2 2 2
m 1 0 1 2 2 m 0 2 2 2 0 1 n 1 2 2 0 n 2 2 2
sin 7 x cos 4 xdx
sin 7 x cos 4 xdx
7 1 4 1 2 2 = 7 4 2 2 2 5 4 2 = 13 2 2 5 3! 2 = 11 9 7 5 5 2 2 2 2 2 2 3 2 2 2 2 = 11 9 7 5 48 = 3465 Use the definition of Beta function to evaluate
2 0
sin 7 x cos 4 x dx
Answer: 2 0
= = =
sin 7 x cos 4 x dx 2
0
6 2
3
2
sin 2 x
2
sin 2 x 1 sin 2 x 2 sin x cos x dx
0
0
sin 6 x cos 3 x sin x cos x dx cos 2 x 2 sin x cos x dx 3
3
3 2
dz 2
1 1 41 z 1 z 2 0 1 5 = 4, 2 2 5 4 2 = 5 2 4 2 5 4 2 = 13 2 2 48 = 3465
5 1 2
= =
1 0
3
z 1 z
dz
sin 2 x z 2 sin x cos xdx
x 0, z dz
x
2
0
,z 1
