Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

294

Problem 210 Let | x | < 1 and α > 0. Prove the validity of the following equation ∞ Z 0 e − t t β − 1 E α , β ( xt α ) d t = ∞ Z 0 e − t E α ( xt α ) d t = 1 1 − x

Solution According to the definition of the ML function with two parameters

∞ ∑ k = 0

z k Γ ( α k + β ) ,

E α , β ( z )=

that is, in our case

∞ ∑ k = 0

( xt α ) k Γ ( α k + β )

α

E α , β ( xt

)=

the required integral is

Z ∞ 0

∞ ∑ k = 0

( xt α ) k Γ ( α k + β )

e − t t β − 1

d t .

The same as in the case of derivation, we can alter here the sequence of the operations of integration and addition (convergent series!), and thus obtain

x k Γ ( α k + β ) Z

x k Γ ( α k + β ) Z x k Γ ( α k + β ) ·

∞ ∑ k = 0

∞ ∑ k = 0 ∞ ∑ k = 0

∞

∞

e − t t β − 1 t α k d t

0 e − t t α k + β − 1 d t | {z } Γ ( α k + β ) Γ ( α k + β )=

=

=

0

∞ ∑ k = 0

1 1 − x

x k =

=

,

(for | x | < 1), and it further follows that ∞ Z 0 e − t t β − 1 E

1 1 − x

α

α , β ( xt

) d t =

.

The right hand side of the equation does not depend on parameters α and β , and thus, if for the integral on the left hand side β = 1 is chosen, we obtain ∞ Z 0 e − t E α ( xt α ) d t = 1 1 − x , which was also to be proved.