Mathematical Physics Vol 1

5.10 Examples

295

Problem 211 Let α > 0, β > 0, ρ > 0, x > 0 and a ∈ R . Prove that I ≡ x Z 0 t β − 1 E ρ α , β ( at α ) d t = x β E ρ α , β ( at α ) , where E ρ α , β ( at α ) is the ML function with three parameters.

Solution Using the definition of the ML function with three parameters, for the given integral we obtain

x Z 0

x Z 0

∞ ∑ k = 0

( at α ) k k !

( ρ k Γ ( α k + β )

t β − 1 E ρ

α

t β − 1

I =

( at

) d t =

d t =

α , β

x Z 0

x Z 0

∞ ∑ k = 0

∞ ∑ k = 0

a k k !

a k k !

( ρ ) k Γ ( α k + β )

( ρ ) k Γ ( α k + β )

t β − 1 · t α k d t

t α k + β − 1 d t

=

=

=

=    ∞ ∑ k = 0

   ·

∞ ∑ k = 0

x α k + β α k + β

( ax α ) k k !

a k k !

( ρ ) k Γ ( α k + β )

( ρ ) k ( α k + β ) Γ ( α k + β ) | {z } Γ ( α k + β + 1 ) ·

x β

=

.

The expression in brackets is the ML function with three parameters, and thus

β E ρ

α

I = x

( ax

) ,

α , β

which was to be proved.

Problem 212 Compute the value of the integral I = x R 0 parameter values: a = − 1, α = 2 i β = 1 = ρ .

t β − 1 E ρ

α , β ( at α ) d t (from Example 211), for

Solution Note, that in this special case, E 1

2 , 1 ( · )= E 2 ( · ) . Namely, as by definition

∞ ∑ k = 0

∞ ∑ k = 0

( · ) k k !

( · ) k k !

( ρ ) k Γ ( α k + β ) ·

Γ ( ρ + k ) Γ ( ρ )

1 Γ ( α k + β ) ( · ) k Γ ( α k + 1 )

E ρ

( · )=

⇒

=

α , β

Γ ( 1 + k ) Γ ( 1 )

∞ ∑ k = 0

∞ ∑ k = 0

( · ) k k !

1 Γ ( α k + 1 )

E 1

= E α ( · ) ,

α , 1 ( · )=

=

it follows that

E 1

2 , 1 ( · )= E 2 ( · ) ,