Mathematical Physics Vol 1

5.6 Special functions that are not a result of the Frobenius method

247

the integral (5.131) becomes

π / 2 Z 0

∞ Z 0

e − r 2

( r cos ϕ ) 2 m − 1 · ( r sin ϕ ) 2 n − 1 r d r d ϕ =

u =

π / 2 Z 0 ( cos ϕ ) 2 m − 1 · ( sin ϕ ) 2 n − 1 d ϕ =

∞ Z 0

e − r 2 r 2 ( m + n ) − 1 d r

(5.133)

=

π / 2 Z 0 ( cos ϕ ) 2 m − 1 · ( sin ϕ ) 2 n − 1 d ϕ

1 2

Γ ( m + n )

=

From (5.132) and (5.133) we obtain

π / 2 Z 0 ( cos ϕ ) 2 m − 1 · ( sin ϕ ) 2 n − 1 d ϕ ,

1 4

1 2

Γ ( m ) Γ ( n )=

Γ ( m + n )

(5.134)

u =

that is

π / 2 Z 0 ( cos ϕ ) 2 m − 1 · ( sin ϕ ) 2 n − 1 d ϕ =

Γ ( m ) Γ ( n ) Γ ( m + n ) .

1 2

(5.135)

If we now introduce the following substitutions

m ′ + 1 2

2 m − 1 = m ′ ⇒ m = 2 n − 1 = n ′ ⇒ n =

,

(5.136)

n ′ + 1 2

,

the integral (5.135) becomes

Γ

2 ·

Γ

2

m ′ + 1

n ′ + 1

π / 2 Z 0

( cos ϕ ) m ′ ( sin ϕ ) n ′ d ϕ = 1 2

(5.137)

Γ

2

.

m ′ + n ′ + 2

Note that m ′ > − 1 and n ′ > − 1, which follows from the condition that m > 0 and n > 0. In the special case, when m ′ = n ′ = 0, we obtain π / 2 Z 0 d ϕ = 1 2 [ Γ ( 1 / 2 )] 2 Γ ( 1 ) ⇒ π 2 = [ Γ ( 1 / 2 )] 2 2 ⇒ Γ ( 1 / 2 )= √ π .

(5.138)

Further, from (5.126), we obtain:

1 2

1 2 √ π , Γ ( 5 / 2 )=

1 · 3 2 2

1 · 3 · 5 2 3

√ π , Γ ( 7 / 2 )=

√ π ,

Γ ( 3 / 2 )=

Γ ( 1 / 2 )=

(5.139)

etc.

From (5.127) it follows that

Γ ( n + 1 ) n

Γ ( n )=

(5.140)

,

and then Γ ( n ) → ∞ , when n → + 0.