DLMF:15.8.E25 (Q5082)

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DLMF:15.8.E25
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    Statements

    test
    F ⁡ ( a , b 1 2 ⁢ ( a + b + 1 ) ; z ) = Γ ⁡ ( 1 2 ⁢ ( a + b + 1 ) ) ⁢ Γ ⁡ ( 1 2 ) Γ ⁡ ( 1 2 ⁢ a + 1 2 ) ⁢ Γ ⁡ ( 1 2 ⁢ b + 1 2 ) ⁢ F ⁡ ( 1 2 ⁢ a , 1 2 ⁢ b 1 2 ; ( 1 - 2 ⁢ z ) 2 ) + ( 1 - 2 ⁢ z ) ⁢ Γ ⁡ ( 1 2 ⁢ ( a + b + 1 ) ) ⁢ Γ ⁡ ( - 1 2 ) Γ ⁡ ( 1 2 ⁢ a ) ⁢ Γ ⁡ ( 1 2 ⁢ b ) ⁢ F ⁡ ( 1 2 ⁢ a + 1 2 , 1 2 ⁢ b + 1 2 3 2 ; ( 1 - 2 ⁢ z ) 2 ) , Gauss-hypergeometric-F 𝑎 𝑏 1 2 𝑎 𝑏 1 𝑧 Euler-Gamma 1 2 𝑎 𝑏 1 Euler-Gamma 1 2 Euler-Gamma 1 2 𝑎 1 2 Euler-Gamma 1 2 𝑏 1 2 Gauss-hypergeometric-F 1 2 𝑎 1 2 𝑏 1 2 superscript 1 2 𝑧 2 1 2 𝑧 Euler-Gamma 1 2 𝑎 𝑏 1 Euler-Gamma 1 2 Euler-Gamma 1 2 𝑎 Euler-Gamma 1 2 𝑏 Gauss-hypergeometric-F 1 2 𝑎 1 2 1 2 𝑏 1 2 3 2 superscript 1 2 𝑧 2 {\displaystyle{\displaystyle F\left({a,b\atop\tfrac{1}{2}(a+b+1)};z\right)=% \frac{\Gamma\left(\tfrac{1}{2}(a+b+1)\right)\Gamma\left(\tfrac{1}{2}\right)}{% \Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{% 1}{2}\right)}F\left({\tfrac{1}{2}a,\tfrac{1}{2}b\atop\tfrac{1}{2}};(1-2z)^{2}% \right)+(1-2z)\frac{\Gamma\left(\tfrac{1}{2}(a+b+1)\right)\Gamma\left(-\tfrac{% 1}{2}\right)}{\Gamma\left(\tfrac{1}{2}a\right)\Gamma\left(\tfrac{1}{2}b\right)% }F\left({\tfrac{1}{2}a+\tfrac{1}{2},\tfrac{1}{2}b+\tfrac{1}{2}\atop\tfrac{3}{2% }};(1-2z)^{2}\right),}}
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    DLMF id
    DLMF:15.8.E25
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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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    Symbols used
    gamma function
    test
    Γ ⁡ ( z ) Euler-Gamma 𝑧 {\displaystyle{\displaystyle\Gamma\left(\NVar{z}\right)}}
    xml-id
    C5.S2.E1.m2aidec
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