Theorem: Kummer's Hypergeometric Theorem

:$\map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }$

Proof:
From Euler's Integral Representation of Hypergeometric Function, we have:
:$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$
Where $a, b, c \in \C$.
and $\size x < 1$
and $\map \Re c > \map \Re b > 0$.
Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to -1$.
By symmetry, we have:
:$\ds \map F {n, -x; x + n + 1; -1} = \ds \map F {-x, n; x + n + 1; -1}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map F {-x, n; x + n + 1; -1}
      | r = \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + n + 1 - n } } \int_0^1 t^{n - 1} \paren {1 - t}^{x + n + 1 - n - 1} \paren {1 - \paren {-1} t}^{- \paren {-x} } \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t}^x \paren {1 + t}^x \rd t
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t^2}^x \rd t
      | c = simplifying further: $\paren {1 - t^2} = \paren {1 - t}\paren {1 + t}$
}}
{{end-eqn}}
We now apply a u-substitution:
Let $u = t^2$
{{begin-eqn}}
{{eqn | l = u 
      | r = t^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = t
      | r = \sqrt u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d t}
      | r = 2 t
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \d t
      | r = \frac {\d u} {2 t}
      | c = 
}}
{{eqn | ll= \leadsto
      | r = \frac {\d u} {2 \sqrt u}
      | c = 
}}
{{end-eqn}}
Substituting back into our equation, we have:
{{begin-eqn}}
{{eqn | l = \map F {-x, n; x + n + 1; -1}
      | r = \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 \paren {\sqrt u}^{n - 1} \paren {1 - u}^x \frac {\d u} {2 \sqrt u}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 u^{\frac n 2 - 1} \paren {1 - u}^x \d u
      | c = 
}}
{{eqn | ll= \leadsto
      | r = \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \dfrac {\map \Gamma {\dfrac n 2 } \map \Gamma {x + 1 } } {\map \Gamma {\dfrac n 2 + x + 1 } }
      | c = {{Defof|Beta Function}}
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\dfrac 1 2 \map \Gamma {x + n + 1} } {\map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }
      | c = simplifying and canceling $\map \Gamma {x + 1 }$
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\dfrac n 2 \map \Gamma {x + n + 1} } {n \map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }
      | c = multiplying numerator and denominator by $n$
}}
{{eqn | ll= \leadsto
      | r = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }
      | c = {{Defof|Gamma Function}}
}}
{{end-eqn}}
{{qed}}