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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