Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

296

and thus

x Z 0

x Z 0

α

α

t 0 E 1

I =

2 , 1 ( − t

) dt =

E 2 ( − t

) dt .

Further, given that

∞ ∑ k = 0

∞ ∑ k = 0

∞ ∑ k = 0

( − t 2 ) k Γ ( 2 k + 1 )

( − 1 ) k t 2 k Γ ( 2 k + 1 )

( − 1 ) k ( t 2 ) k ( 2 k ) !

2 )=

E 2 ( − t

= cos t

=

=

the integral I is equal to

x Z 0

x Z 0

2 ) dt =

I =

E 2 ( − t

cos tdt = sin t .

Using the result from Example 208, we obtain x Z 0 E 2 ( − t α

2 )

) dt = xE 2 , 2 ( − x

which is the required result.

Problem 213 Let α > 0. Prove that

2 ) − xE

2 )

E α ( − x )= E 2 α ( x

2 α , α + 1 ( x

is valid for ML functions.

Solution Using the definition for ML functions with one and with two parameters, the right hand side becomes Ω ≡ E 2 α ( x 2 ) − xE 2 α , α + 1 ( x 2 )= ∞ ∑ k = 0 ( x 2 ) k Γ ( 2 α k + 1 ) − x ∞ ∑ k = 0 ( x 2 ) k Γ ( 2 α k α + 1 ) , or in the expanded form Ω = 1 + x 2 Γ ( 2 α + 1 ) + x 4 Γ ( 4 α + 1 ) + x 6 Γ ( 6 α + 1 ) + ··· − − x Γ ( α + 1 ) + x 3 Γ ( 3 α + 1 ) + x 5 Γ ( 5 α + 1 ) + ··· . Grouping the terms next to each power of − x , we obtain Ω = ( − x ) Γ ( α + 1 ) + ( − x ) 2 Γ ( 2 α + 1 ) + ( − x ) 3 Γ ( 3 α + 1 ) + ( − x ) 4 Γ ( 4 α + 1 ) + ( − x ) 5 Γ ( 5 α + 1 ) + ··· = = ∞ ∑ k = 0 ( − x ) k Γ ( k α + 1 ) = E α ( − x ) .