Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

246

Specially, for n = 1, from (5.123), we obtain

∞ Z 0

e − x d x = 1 .

Γ ( 1 )=

(5.124)

By partial integration, from (5.123) we obtain

∞ Z 0

: 0

− e − x x n − 1

∞ 0 +( n − 1 )

e − x x n − 2 d x | {z } Γ ( n − 1 ) ,

Γ ( n )=

(5.125)

and then for n > 1

Γ ( n )=( n − 1 ) · Γ ( n − 1 ) .

(5.126)

Substituting n by n + 1yields ( n = 1 , 2 ,... )

Γ ( n + 1 )= n · Γ ( n )= n · ( n − 1 ) · Γ ( n − 1 )= ··· = n !

(5.127)

If we substitute x by x 2 in (5.123), we obtain

∞ Z 0

∞ Z 0

e − x 2

e − x 2 x 2 n − 1 d x .

· x 2 n − 2 d ( x 2 )= 2

Γ ( n )=

(5.128)

From here, for n = 1 / 2, it follows

∞ Z 0

e − x 2 d x

Γ ( 1 / 2 )= 2

(5.129)

.

Let us now show that the integral

π / 2 Z 0

cos m τ · sin n τ d τ ,

(5.130)

can be expressed in terms of the Γ – function. Let us start from the integral

∞ Z 0

∞ Z 0

e − x 2 − y 2

· x 2 m − 1 · y 2 n − 1 d x d y .

u =

(5.131)

This double integral can be represented as a product of two single integrals

∞ Z 0

∞ Z 0

1 4

e − x 2

e − y 2

· x 2 m − 1 d x ·

· y 2 n − 1 d y =

Γ ( m ) · Γ ( n ) .

u =

(5.132)

On the other hand, by switching to polar coordinates

( x = r cos ϕ , y = r sin ϕ , d x d y = r d r d ϕ ) ,