DLMF:15.8.E25 (Q5082): Difference between revisions

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DLMF:15.8.E25
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    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:15.8.E25
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