DLMF:15.8.E28 (Q5085)

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DLMF:15.8.E28
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    Statements

    test
    2 ⁢ z ⁢ Γ ⁡ ( - 1 2 ) ⁢ Γ ⁡ ( a + b - 1 2 ) Γ ⁡ ( a - 1 2 ) ⁢ Γ ⁡ ( b - 1 2 ) ⁢ F ⁡ ( a , b ; 3 2 ; z ) = F ⁡ ( 2 ⁢ a - 1 , 2 ⁢ b - 1 ; a + b - 1 2 ; 1 2 - 1 2 ⁢ z ) - F ⁡ ( 2 ⁢ a - 1 , 2 ⁢ b - 1 ; a + b - 1 2 ; 1 2 + 1 2 ⁢ z ) , 2 𝑧 Euler-Gamma 1 2 Euler-Gamma 𝑎 𝑏 1 2 Euler-Gamma 𝑎 1 2 Euler-Gamma 𝑏 1 2 Gauss-hypergeometric-F 𝑎 𝑏 3 2 𝑧 Gauss-hypergeometric-F 2 𝑎 1 2 𝑏 1 𝑎 𝑏 1 2 1 2 1 2 𝑧 Gauss-hypergeometric-F 2 𝑎 1 2 𝑏 1 𝑎 𝑏 1 2 1 2 1 2 𝑧 {\displaystyle{\displaystyle\frac{2\sqrt{z}\Gamma\left(-\tfrac{1}{2}\right)% \Gamma\left(a+b-\tfrac{1}{2}\right)}{\Gamma\left(a-\tfrac{1}{2}\right)\Gamma% \left(b-\tfrac{1}{2}\right)}F\left(a,b;\tfrac{3}{2};z\right)=F\left(2a-1,2b-1;% a+b-\tfrac{1}{2};\tfrac{1}{2}-\tfrac{1}{2}\sqrt{z}\right)-F\left(2a-1,2b-1;a+b% -\tfrac{1}{2};\tfrac{1}{2}+\tfrac{1}{2}\sqrt{z}\right),}}
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    DLMF id
    DLMF:15.8.E28
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    constraint
    | ph ⁡ z | < π phase 𝑧 𝜋 {\displaystyle{\displaystyle|\operatorname{ph}z|<\pi}}
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    | ph ⁡ ( 1 - z ) | < π phase 1 𝑧 𝜋 {\displaystyle{\displaystyle|\operatorname{ph}\left(1-z\right)|<\pi}}
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