Integral Representation Of Confluent Hypergeometric Function

Integral Representation Of Confluent Hypergeometric Function:


 Theorem :


            If γ>α>0 , then the function F(α;γ;x) can be expressed as
                               1
   Γ(γ)/Γ(γ)Γ(γ-α)  ∫ eˣᵗ t^(α-1)(1-t)^(γ-α-1) dt.
                               0

Proof :


          We know that 

  B(α+n,γ-α)/B(α,γ-α) = 

       Γ(α+n)Γ(γ-α)/Γ(γ+n) / Γ(α)Γ(γ-α)/Γ(γ)

     = Γ(α+n)/Γ(α) / Γ(γ+n)/Γ(γ)

But Γ(α+n)/Γ(α) = (α)ₙ and Γ(γ+n)/Γ(γ) = (γ)ₙ

Therefore,

(α)ₙ/(γ)ₙ  = B(α+n,γ-α)/Β(α,γ-α) 
                                 1
  = Γ(γ)/Γ(α)Γ(γ-α) ∫ t^(α+n-1) (1-t)^(γ-α-1) dt
                                0
                            ∞
Now  F(α;γ;x) = Σ (α)ₙxⁿ/n!(γ)ₙ
                           n=0
                                         
  = Γ(γ)/Γ(α)Γ(γ-α)
                  ∞           1
                  Σ xⁿ/n! ∫ t^(α-1) tⁿ(1-t)^(γ-α-1) dt
                n=0        0

Interchanging the order of summation and integration we have 

F(α;γ;x) = Γ(γ)/Γ(α)Γ(γ-α)×
           1                                       ∞
          ∫ t^(α-1) (1-t)^(γ-α-1) dt Σ (xt)ⁿ/n!
          0                                      n=0

                       
                               1
 = Γ(γ)/Γ(α)Γ(γ-α) ∫ eˣᵗ t^(α-1) (1-t)^(γ-α-1) dt
                               0                     ............(1)

 We can use this integral representation to deduce an important relation satisfied by the function F(α;γ;x).

Putting t= 1-s in the above result we get 


F(α;γ;x) = Γ(γ)/Γ(α)Γ(γ-α)×
       0
       ∫ e⁻ˣˢ s^(γ-α-1) (1-s)^(α-1) (-ds)
      1


             = Γ(γ)eˣ/Γ(α)Γ(γ-α) ×
    1
    ∫ s^(γ-α-1) (1-s)^(α-1)e⁻ˣˢ ds 
   0

Let γ-α = α₁; then γ-α₁ = α.

Hence 

F(α;γ;x) = Γ(γ)eˣ/Γ(γ-α₁)Γ(α₁)×
             1
             ∫ s^(α₁-1) (1-s)^(α-1) e⁻ˣˢ ds 
            0

            = eˣ F(α₁;γ;-x)

           = eˣ F(γ-α;γ;-x),    .........(2)

Which is known as Kummer's result.

Whittaker Functions :


     If we substitute y= x^-γ/2 e^x/2 W(x) in the equation 

    xy" + (γ-x)y' -αy = 0,

 We find that the function W(x) satisfies the differential equation 

d²W/dx² + [-1/4 + k/x + (1/4 - m²) /x²]W =0,

            where k stands for γ/2 -α and m for 
1/2  (1-γ). Solution of this equation are known as Whittaker's confluent hypergeometric functions .


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